MA 620 – Langlands Program  13 January 1997

We’ll be talking about nonabelian reciprocity

Maybe a (-1) step would be quadratic reciprocity, at least over ℚ.

Abelian reciprocity laws might be enumerated as follows.

1. Artin reciprocity. Given a global field F, there’s a reciprocity law map rec_{F} : 𝔸_{F}^{×}/F^{×} → Gal(F^{ab}, F), so that after passing to connected components on the left, we get an isomorphism

𝔸_{F}^{×}/F^{×}  \overline{^{rec_{F}}} Gal(F^{ab}, F)
[figure]
π_{0}(𝔸_{F}^{×}/F^{×})

2. There’s a way of formulating a result like this in terms of analysis; see last semester, if you were there, on L-functions. To χ : Gal(F^{ab}, F) → ℂ^{×} corresponds some L(χ, s). Now suppose given a finite order character ρ : 𝔸_{F}^{×}/F^{×} → ℂ^{×}. We define L(ρ, s) = L(ρ · ω_{s}), where ω_{s} is the absolute value character. The statement in terms of L-functions of the reciprocity law is that these two L-functions are the same; you get the same L-functions. This relates Galois-theoretic information to data from harmonic analysis.

So much for abelian L-functions. How shall we generalize this to nonabelian stuff? Let F be global. We don’t know what Gal(\overline{F} , F) is, and there aren’t even any conjectural, simple descriptions.

Want to understand representations on finite-dimensional complex vector spaces V, Gal(\overline{F} , F) →^{ρ} GL(V), possibly in terms of some objects from [harmonic] analysis. In this context, for instance, there’s a famous conjecture.

Well, for such a ρ there’s an Artin L-function L(ρ, s).

Conjecture [Artin] If (ρ|triv) = 0, i.e., ρ does not contain the trivial representation of its image, then L(ρ, s) is an entire function of s ∈ ℂ.

Anyways, its believed that the L-functions which come up L(ρ, s) should be a [product of] automorphic L-functions.

Further generalizations: Let X/F be an algebraic variety over a global field F. We can define something called the zeta function of X, Z_{X}, or even Z_{X,S}, where S is a finite set of primes of F and X has good reduction outside of S. For us for now everthing in sight has an Euler product expansion. Define

Z_{X,S} =^{def} exp(∑_{𝔭∉S} ∑_{m≥1} \frac{#X_{𝔭}(𝔽_{q^{m}})}{m}q^{−ms}).

If you’re having troubles remembering the exact definition, remember that log(1 − λq^{−s})^{−1} = ∑_{m≥1} \frac{λ^{m}q^{−ms}}{m}.

Anyways, Z_{X,S} should be a quotient of products of automorphic L-functions. There’s not a lot of evidence either way, but we believe it anyway.

Another generalization: There’s nothing especially sacred about finite-dimensional complex representations. So let V be a finite-dimensional vector space over something like E_{λ}, where λ is a finite place of a number field E. Then [in contrast to the complex situation] the image can be infinite. So suppose that Gal(\overline{F} , F) →^{ρ} GL(V) is reasonable, e.g., absolutely irreducible, det ρ = 1, stuff like that. Then we still expect to get automorphic L-functions.

Now try letting F be a local field, i.e., locally compact local field. It’s a finite extension of one of ℚ_{p}, ℝ, or 𝔽_{q}[[t]]. Now, local abelian reciprocity’s usual formulation is a reciprocity law

F^{×} ^{_{⊂} \begin{matrix}rec & F \\  \end{matrix}}Gal(F^{ab}, F)

with dense image. In particular, finite quotients on right and left sides are the same.

What about the nonabelian generalization? From what we’ve been talking about, it’s easy to imagine that we’ll be dealing with representations of one of these groups. Before we try this, we’ll make a slight modification. Consider the profinite group Gal(\overline{F} , F). For some reason we’ll defer for now, it’s better to work with W_{F} ⊆ Gal(\overline{F} , F), where the Weil group is – if F is nonarchimedean –

=^{def} \{σ ∈ Gal(\overline{F} , F)|σ|_{F^{unram}} = Fr^{n} some n\}.

It’s not hard to see that Gal(\overline{F} , F) → Gal(F^{unram}, F) ≅ ℤ̂ [the isomorphism is canonical up to choice of Frobenius, i.e., up to sign]; and under this identificaiton, W_{F} →^{∼} ℤ.

Now, if F is archimedean, then F is either ℂ or ℝ. Define W_{ℂ} = F^{×}. If F = ℝ, then W_{F} is an extension of the Galois group of ℝ by ℂ^{×};

[figure: 1; \overline{F}^{×}; W_{F}; ℤ/2ℤ; 0]

Set theoretically,

W_{F} = \overline{F}^{×} ⊔ \overline{F}^{×}j

where j^{2} = −1. And the action of Ad(j) on ℂ is the complex conjugation.

In all cases, there’s map W_{F} → Gal(\overline{F} , F). And experience shows that it’s easier to work with representations of W_{F}. Consider W_{F} →^{ρ} GL(V), V a finite-dimensional vector space over some field, probably either ℚ_{l} or ℂ. Morally, such ρ correspond to representations GL_{n}(F) →^{π} Aut(U), where U is a complex vector space, probably infinite-dimensional. And GL_{n}(F) is either a real Lie group or a p-adic group. These representations are called admissible irreducible representations. And here, n = dim V. We hope to use representations of the Weil group to parametrize representations of these Lie groups. This correspondence need not be 1-1. The idea is that ρ should somehow parametrize π; want a nice functorial correspondence.

Now, compose ρ with a representation GL(V) → GL_{m}. If you believe in all this then you get a representation W_{F} →^{ψ∘ρ} GL_{m}. The effect is to send π to ψ(π) = GL_{m}(F) → U′. This is called the functoriality principle.

And the same should be true for automorphic representations.

Situation: F a locally compact non-archimedean local field, G = GL_{2}. We’ll consider G(F) = GL_{2}(F) and study its representations.

G contains a maximal compact subgroup K = GL_{2}(O_{F}), which we’ll think of as somehow a good maximal compact subgroup.

Philosophy: Think of these as studying symmetries. If you’re familiar with Hecke algebras, what we’re doing is sort of Hecke algebras with only one prime. Our goal is to essentially classify all [irreducible] representations of our group, at least the interesting ones.

Fix a nontrivial additive character (F, +) →^{ψ_{F}} ℂ_{1}^{×}, nontrivial. [Image has to be finite, which is why it sits inside ℂ_{1}^{×}.]

We’re trying to represent a p-adic group on a complex vector space, which is a little weird.

Definition An admissible representation of G(F) is a [continuous] representation of G(F) on a ℂ-vector space V, G(F) →^{π} GL(V), so that

i. π is smooth, i.e., stabilizers are open.

ii. π|_{K} = ⊕_{ω∈K̂}V (ω) is an algebraic direct sum, and dim_{ℂ} V (ω) < ∞.

Let d^{×}y be a Haar measure [and with the usual coordinates on GL_{2} there’s a fairly obvious one]. Let H_{F} = H_{G(F)} be the space of locally constant [complex-valued] functions with compact support on G(F). That’ll be our Hecke algebra for G(F). This has a multiplication – convolution – but it depends on the choice of Haar measure. So, more canonically, one could describe it as the space of compactly supported densities, or measures.

Given an irreducible representation V, we want to define its contragradient or dual representation V̌. The naïve dual is too big; we’ll use the algebraic dual

V̌ = ⊕_{ω∈K̂}V (ω)^{∨}.

This has the usual duality properties, e.g., canonical pairing V ×V̌; andifWisasubrepresentation, thenW^{⊥} ⊆ V̌. Moreover, V →^{∼} (V̌)^{∨}, W →^{∼} (W^{⊥})^{⊥}.

Exercise An irreducible finite-dimensional admissible representation of G(F) has the form

G(F) →^{det} F^{×} → ℂ^{×}.

[Basically, look at the kernel of the representation; it’s an open normal subgroup of GL_{2}(F), and there are precious few such.]

Theorem [Existence of Kirillov model] Let (π, V) be an infinite dimensional admissible irreducible repesentation of G(F). Then there exists a unique^{1} subspace V′^{2} of the space of locally constant functions on F^{×} with action π′ by G(F) such that

1up to choice of additive character ψ_{F}

2Which we’ll later call K(π).

i. There exists an action on V′ such that (π′, V′) ≅ (π, V).

ii. There’s a functional equation; let ξ ∈ V′. Then

(π′(( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ))ξ)(x) = ψ_{F} (bx)ξ(ax).

iii. For every ξ ∈ V′, if |x| ≫ 0 then ξ(x) = 0.

iv. V′ ⊇ (F^{×}) the Schwarz functions, and dim(V′/S(f^{×})) < ∞.

Exercise S(F^{×}) is irreducible under \{( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ) ∈ G(F)\}.

Exercise ψ_{F} (bx)ξ(x) − ξ(x) ∈ S(F^{×}).

Relation with Whittaker models Given ξ ∈ K(π), define a function W_{ξ} : G(F) → ℂ by g ↦ W_{ξ}(g) = (π(g)ξ)(1). We see that W_{π(g)ξ}(x) = (π(x)π(g)ξ)(1) = W_{ξ}(xg). In other words, on the space of all such W_{ξ}, G(F) acts by right-translation.

Also, think about W_{ξ}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g). It’s gotta be

W_{ξ}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = (π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))π(g)ξ)(1)
= ψ_{F} (x)(π(g)ξ)(1)
= ψ_{F} (x)W_{ξ}(g)

So the unipotent group acts via the additive character ψ_{F}.

Moreover, W_{ξ} determines ξ; ξ(x) = W_{ξ} ( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ).

So these Whittaker models give you pretty much the same thing. We’ve got W(π) = {W_{ξ}} with action by G(F); which determines K(π), and vice-versa.

Theorem [Existence of Whittaker models] Let π be an admissible irreducible infinite dimensional representation of G(F). Then there exists a unique subrepresentation (W(π), π′) of the

space of locally constant functions W on G(F) [equipped with right-translation] with W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) =

ψ_{F} (x)W (g) equivalent to π.

Definition (π, V) as usual is supercuspidal if K(π) = S(F^{×}). What this is trying to get at is that (π, V) isn’t induced from a representation of a subgroup.

Last time we were talking about Kirillov models. Not having done the proof, we missed the following

Fact K(π) ⊇ S(F^{×}); the Kirillov model contains the Schwarz functions. Moreover, let w be the

Weyl element, ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )^{3} Then

K(π) = S(F^{×}) + wS(F^{×}).

Definition (π, V) an infinite-dimensional admissible irreducible representation is supercuspidal if K(π) = S(F^{×}).

There are a couple different ways of saying this. The following are equivalent:

i. (π, V) is supercuspidal.

ii. ∀ξ, ∫_{𝔭^{−n}} π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ξdx = 0 if n ≫ 0.

iii. ∀ξ ∈ V and η ∈ V̆, g ↦ π(g)ξ, η has compact support modulo the center Z(F) = Z(G(F)) =

\{( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) |a ∈ F^{×}\}.^{4}

Note that (ii) and (iii) are intrinsic conditions, which don’t depend on realizing anything in the Kirillov model.

Why? Well, assume supercuspidal. (1 ⇒ 2) We know that (π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ξ)(y) = ψ_{F} (xy)ξ(y). So, to

say that this integral is zero is to say, well,

(∫_{𝔭^{−n}} π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ξ)(y) = (∫_{𝔭^{−n}} ψ_{F} (xy)dx))ξ(y)

Well, when is this integral zero? Suppose that ξ ∈ S(F^{×}), that is, has compact support. Then supp(ξ) ⊆ {q^{−n} ≤ |y| ≤ q^{n}}. Go back to the integrand on the right. dx is an additive Haar measure. When can we possibly get something nonzero? What can this value be? Well, remember that if you integrate an irreducible representation over a compact group you get zero. Thus, (∫_{𝔭^{−n}} ψ_{F} (xy)dx))ξ(y) is zero if x ↦ ψ(xy) is not trivial on 𝔭^{−n}. So choose ψ an additive character appropriately and you’re done.

3Unless we use the inverse!

4Exercise. The point is that this is an intrinsic definition; explore what happens when you realize V in its Kirillov model. Use the Cartan decomposition, G(F) = KAK; K is the maximal compact subgroup GL_{2}(O_{F}), A = A(F) = \{( \begin{matrix}a & 0 \\ 0 & d \end{matrix} ) |a, d ∈ F^{×}\}. So look at the matrix coefficients, restricting g to elements in A.

We’re only integrating over supp(ξ) ∩ 𝔭^{−n}. When n is quite big, this becomes nontrivial, and so integrates to zero.

Claim, to be precise, that n ≫ 0 ⇒ x ↦ ψ(xy) is nontrivial on 𝔭^{−n}. Now, ψ = ψ_{ℚ_{p}} ∘ tr_{F,ℚ}. And ψ is trivial on some 𝔭^{−n_{0}}, but not on 𝔭^{−n_{0}−1}. That number n_{0} is essentially the conductor.

In the other direction (2 ⇒ 1) suppose that for all ξ and suffciently large n, ∫_{𝔭^{−n}} π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ξdx =

0; deduce supercuspidality. Let ξ ∈ K(π). We need to show ξ(y) = 0 for y in some neighborhood of 0, i.e., |y| ≤ q^{−m} for some m.

Well, if not, then no m works; unable to find a value of n so that x ↦ ψ(xy) is nontrivial for all y ∈ supp(ξ).◇

BTW, (π, V) admissible irreducible, then there’s a central character, i.e., look at Z(F); it operates

by a character; ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) acts as ω_{π}(a). This is a sort of infinite-dimensional Schur’s lemma. Use

admissibility to get down to finite dimension.

So what? SO this. We said last time that supercuspidal representations spring to life when you pass from 𝔾_{m} to GL_{2}. 𝔾_{m} is totally classical, and is covered in any algebriaic number theory text. Now, for GL_{2}, most representations come from GL_{1}; they’re induced. But life isn’t totally simple, and we get extra representations showing up.

Induced representations ^{5} We’ll be using the Borel subgroup B(F). For GL_{2}, this is just B(F) =

\{( \begin{matrix}a & b \\ 0 & d \end{matrix} ) ∈ G(F)\}. The Borel subgroup has lots of characters, e.g., (μ_{1}, μ_{2}), ( \begin{matrix}a & b \\ 0 & d \end{matrix} ) ↦

μ_{1}(a)μ_{2}(d), μ_{i} [quasi?]characters F^{×} → ℂ^{×}. So that’s a bunch of representations right there, and there’s standard way of inducing to G.

Define Ind_{B(F)}^{G(F)}(μ_{1}, μ_{2}) = \{f : G(F) → ℂ locally constant so that f(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) x) = μ_{1}(a)μ_{2}(d)|\frac{a}{d} |^{\frac{1}{2}} f(x)\}.

What’s that extra factor |\frac{a}{d} |^{\frac{1}{2}} doing?

The support of such an f is canonically supp(f) ⊆ B(F)\G(F) ≅ ℙ^{1}(F), and so the support is compact.

Anyway, about the extra factor. Naïvely, with this factor we normalize things for unitary representations. More precisely, if μ_{i} unitary, then this induced representation is also unitary, i.e., on V there’s a nondegenerate Hermitian form, positive definite. Or something.

5Assume we know what’s going on for finite groups....

And for the real reason....? To answer this, we’ll have to delve into Haar measures.

Ind_{B(F)}^{G(F)}(μ_{1}, μ_{2}) should be the space of sections of

ℂ ×^{B(F)} G(F)

a ℂ-bundle over B(F)\G(F). Look at sections of this vector bundle; it’s the space of functions we mentioned before without the annoying factor. Then |\frac{a}{d} |^{\frac{1}{2}} means twist by the half-density bundle. And that’s the canonical meaning. [We want our sections to be densities so that we can actually integrate them.] Quick recap of last time elided. Recall that Ind_{B(F)}^{G(F)}(μ_{1}, μ_{2}) is the space of locally constant G(F) →^{f} ℂ transforming well. There was a factor in there which looked a little noncanonical, but we tried to explain it away.

In general, one an define a duality of sorts, a perfect pairing

Ind(μ_{1}, μ_{2}) × Ind(μ_{1}^{−1}, μ_{2}^{−1})  \overline{}ℂ

If we didn’t have the normalization factor built in we’d need to fudge.

Haar measures For a little while, let G be just any locally compact topological group. With a mild abuse of notation, we’ll let d_{l}g be a left-invariant Haar measure. In other words, ∫ f(g)d_{l}g = ∫ f(xg)d_{l}g.

Translation on the right will in general be a little more complicated. Consider d_{l}(gg_{0}^{−1}), by which we mean ∫ f(g)d_{l}gg_{0}^{−1}) = ∫ f(gg_{0})d_{l}g. This is another left-invariant Haar measure, as left- and right-translation commute. Therefore, d_{l}(gg_{0}^{−1}) = m_{G}(g_{0})d_{l}(g), for some number m_{G}(g_{0}); and m_{G} : G → ℝ_{+}^{×} is a continuous homomorphism called the modular function. Observe that d + l(gg_{0}^{−1}) = d_{l}(g_{0}gg_{0}^{−1}); this notation might show up in other books.

Then m_{G}(g)d_{l}(g) is invariant under right-translation, and we’ll denote it by d_{r}(g). By its very definition,

d_{r}(g_{0}g) = m_{G}(g_{0})d_{r}(g)

Exercise d_{l}(g^{−1}) is clearly a right-invariant measure; d_{l}(g^{−1}) =?d_{r}(g). The exercise is to show that ? is 1.^{6}

Back to our case, I think. Consider L(G, B) =^{def} \{G(F) →^{f} ℂ|f(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) · g) = |\frac{a}{d} |f(g)∀g ∈

G(F), ( \begin{matrix}a & b \\ 0 & d \end{matrix} ) ∈ B(F)\}. There’s a right-invariant integral

L(G, B) \overline{\begin{matrix}∫_{B\G} dġ \\  \end{matrix}}ℂ

We won’t do it here but this is the sort of thing you can read in books.

In general, let P ⊆ G be a closed subgroup [arbitrary G]. C_{c}(G) is the set of continuous functions with compact support. Fix a left-invariant Haar measure d_{l}p On P. Can talk about

6Don’t be too weirded out by this. It’s because we took care of normalization at the beginning.

[figure: C_{c}(G); L_{c}(G, P); f; ϕ_{f} : g ↦ ∫_{P} f(pg)d_{l}(p)]

Then ϕ_{f}(p_{1}g) = ∫_{P} f(pp_{1}g)d_{l}p = m_{P} (p_{1})ϕ_{f}(g). Here, L_{c}(G, P) means the functions ϕ on such that ϕ(p_{1}g)m_{P} (_{1})ϕ(g) with compact support in P \G. Think of these as sections of some line bundle if you like.

The usual proof goes like this.

[figure: C_{c}(G); L_{c}(G, P)]

ℂ^{^{∫G}}

Sorry, I was busy listening.

Assume G is unimodular, meaning that the modular function is 1. ^{7} Then

∫_{P \G} dġ ∫_{P} f(pg)d_{l}(p) = ∫_{G} f(g)dg

And that’s where our normalization for ∫_{B\G} comes from.

Properties of ∫_{P \G}(·)d^{•}g. If there’s a closed subgroup M ⊆ G so that P ∩ M is compact and P M = G up to a set of measure zero, then [up to ℝ_{+}^{×}]

∫_{P \G} ϕ(g)dġ = ∫_{M} ϕ(mx)d_{r}(m)

for ϕ ∈ L_{c}(G, P), all x ∈ G.

7Certainly, semisimple groups are unimodular – everything factors through the commutator, the diagonals – and even reductive groups are unimodular. But a Borel usually isn’t unimodular. For now, just remember that GL_{2}, SL_{2}, N =^{def} \{( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} )\} are unimodular, but B isn’t.

For us, G = G(F), P = B(F), M = K = GL_{2}(O); G(F) = B(F)K = A(F)N(F)K is the Iwasawa

decomposition. Pick ϕ so that ϕ(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) g) = |\frac{a}{d} |ϕ(g). Then the assertion is that

∫_{B(F)\G(F)} ϕ(g)dġ = ∫_{K} ϕ(x)dx.

The function x ↦ ϕ(w^{−1}x)^{8} determines ϕ. Why? Well, consider B(F)w^{−1}N(F), equivalent to B(F)w^{−1}N(F)w which is dense in G(F), even though it doesn’t give you everything.^{9} Our assertion is

∫_{B(F)\G(F)} ϕ(g)dg = ∫_{F} ϕ(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))dx
= ∫_{GL_{2}(O_{F})} ϕ(xg)d^{×}x.

Exercise Show the duality pairing is nondegenerate: Ind(μ_{1}, μ_{2})^{∨} = Ind(μ_{1}^{−1}, μ_{2}^{−1}).

8Here, we’re using x ↭ ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ).

9This is the big cell in the Bruhat decomposition.

Proposition If (π, V) an irreducible infinite-dimensional admissible representation is not supercupsidal, then it is isomorphic to a subquotient of an induced representation.

Proof We have K(π) ⊋ S(F^{×}), and K(π)/S(F^{×}) is finite dimensional. Note that S(F^{×}) is stable under the action of B(F), almost by definition. Actually,

Exercise S(F^{×}) is irreducible under \{( \begin{matrix}a & b \\ 0 & 1 \end{matrix} )\}.

Moreover, N(F) = \{( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} )\} operates trivially on X as, for all ξ ∈ K(π), x ↦ ξ(x)−ψ(bx)ξ(x) ∈

S(F^{×}).

Then B(F) acts via B(F) → B(F)/N(F) ←^{≅} A(F),^{10} where A(F) is commutative. So there’s a B(F)-equivariant K(π) → X ⇀ ℂ. We want (π, V) → ind_{B(F)}^{G(F)}(μ_{1}, μ_{2}). But to give such a G(F)- equivariant map is to give [Frobenius reciprocity] a map on the subgroups.^{11} So we need some B(F)-equivariant π|_{B(F)} → ℂ_{μ_{1},μ_{2}}.

Main result on induced representations

• ρ(μ_{1}, μ_{2}) is irreducible unless μ =^{def} \frac{μ_{1}}{μ_{2}} = |·|^{±1}.

• If μ = |·|, then the space ρ(μ_{1}, μ_{2}) contains an irreducible subrepresentation of codimen-

sion one,^{12} consisting of \{f ∈ B(μ_{1}, μ_{2})| ∫_{B(F)\G(F)} f(g)μ^{−1}(det g)·|det g|^{\frac{1}{2}} dġ = 0 = ∫_{F} f(w^{−1}(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )))dx\}.

• Dually, if μ = |·|^{−1}, then ρ(μ_{1}, μ_{2}) contains a 1-dimensional subrepresentation, generated by g ↦ μ_{1}(det g) · |det g|^{\frac{1}{2}} , and the quotient is irreducible.

Define

B(μ_{1}, μ_{2})  \overline{}Functions on F
f  ξ_{f} : x ↦ f(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))

10Is that really B(F)/N(F)?

11In general, given H ⊆ G, Hom_{G}((π, V), Ind_{H}^{G}(ρ, W)) = Hom_{H}((π, V)|_{H}, (ρ, W)), canonically.

12You can ask how the center ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) acts on ρ(μ_{1}, μ_{2}); in our situation it’s gotta be via μ_{1} · μ_{2}.

MA 620 – Langlands Program  22 January 1997

13

Now, one can verify easily that [assume c ≠ 0]

( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}c^{−1} det g & a \\ 0 & c \end{matrix} ) w^{−1} ( \begin{matrix}1 & \frac{d}{c} \\ 0 & 1 \end{matrix} )
f(( \begin{matrix}a & b \\ c & d \end{matrix} )) = μ_{1}(c^{−1} det g)μ_{2}(c)|\frac{det g}{c^{2}} |^{\frac{1}{2}} ξ_{f}(\frac{d}{c})
= μ(c)^{−1}μ_{1}(detg)|det g|^{\frac{1}{2}} |c|^{−1}ξ_{f}(\frac{d}{c})

The next thing to do is pin down the asyptotic behavior of ξ_{f}. Before we do that, observe that

w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & −x^{−1} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}x^{−1} & 0 \\ 0 & x \end{matrix} ) ( \begin{matrix}1 & 0 \\ x^{−1} & 1 \end{matrix} )
f(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )) = μ^{−1}(x)|x|^{−1}f(( \begin{matrix}1 & 0 \\ x^{−1} & 1 \end{matrix} ))
= μ^{−1}|x|^{−1}f(e)

Then ξ_{f}(x)|x|μ(x) is constant for |x| ≫ 0.

Exercise This condition characterizes ξ_{f}. Umm, henceforth ξ = Φ? Φ_{f}(x)|x|μ(x) is constant for |x| ≫ 0. Now take the Fourier transform; ξ_{f} = μ_{2}(x)|x|^{\frac{1}{2}} ∫_{F} f(w^{−1} ( \begin{matrix}1 & y \\ 0 & 0 \end{matrix} ))\overline{ψ_{F} (xy)}dy, where f(w^{−1} ( \begin{matrix}1 & y \\ 0 & 0 \end{matrix} )) = Φ_{f}(y).

Jeff Achter  14  Ching-Li Chai

13Recall that B(μ_{1}, μ_{2}) is the space of all locally constant functions G →^{f} ℂ with f(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) g) = μ_{1}(a)μ_{2}(d)| \frac{a}{d} |^{\frac{1}{2}} f(g).

We’ve got the space of the induced representation B(μ_{1}, μ_{2}), a space of functions on the group. We’ll have f ∈ B(μ_{1}, μ_{2}) correspond to Φ_{f}, where

Φ_{f}(x) = f(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )).

Moreover,

μ|·|Φ is constant for x ≫ 0

This property actually characterizes the Φ_{f}’s.

Another formula:

f(( \begin{matrix}a & b \\ c & d \end{matrix} )) = μ_{1}(det g)|det g|^{\frac{1}{2}} μ^{−1}(c)|c|^{−1}Φ(\frac{d}{c})

That should be easy; and if it’s not, then you’re in trouble.

Moreover, these Φ_{f} correspond, by a twisted Fourier transform, to

ξ_{f} = μ_{2}(x)|x|^{\frac{1}{2}} ∫_{F} Φ_{f}(y)\overline{ψ_{F} (xy)}dy
= μ_{2}(x)|x|^{\frac{1}{2}} ∫_{F} f(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))\overline{ψ_{F} (xy)}dy

The spirit is that if you hit f with ( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ), the transform sort of behaves like the Kirillov model.

For instance, adjust by ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ); recover the factor ψ_{F} (xb). Or, adjust by ( \begin{matrix}a & 0 \\ 0 & 1 \end{matrix} ), and get a

fudge factor μ_{2}(a)^{−1}|a|^{− \frac{1}{2}} .

A formal consequence of all this [exercise] is

ξ_{ρ(μ_{1},μ_{2})( \begin{matrix}a & b \\ 0 & 1 \end{matrix} )f}(x) = ψ_{F} (bx)ξ_{f}(ax)

So what? So this. The thing to remember is that there’s something we really want to prove; we want to study whether an induced representation is irreducible. If not, we want to decompose them. The thing is, suppose all these formulas work. Then it becomes some sort of Kirillov model. And usually, a Kirillov model contains the space of Schwarz functions with finite codimension. Also, we know that the space of such functions is irreducible under the action

ρ(μ_{1}, μ_{2}) ( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ). Then the action on our ξ_{f}, or whatever, can’t be that far from irreducible.

what the hell is going on?

Lemma Let μ be a character of F^{×}, and F_{μ} = { locally constant Φ on F such that μ(x)|x|Φ(x) = const for |x| ≫ 0}. For Φ ∈ F_{μ}, consider the sum

Φ̂(x) =^{def} ∑_{n∈ℤ}∫_{v(y)=n} Φ(y)\overline{ψ(xY)}dy

for x ∈ F^{×}. Then

i. The sum conveges uniformly on compact sets of F^{×}.

ii. Φ ↦ Φ̂ is injective unless μ = |·|^{−1}.

iii. If μ = |·|^{−1} then ker Φ ↦ Φ̂ consists of constant functions.

iv. im(Φ ↦ Φ̂) is the space of locally constant functions Ψ on F^{×} such that (a) Ψ(x) vanishes for |x| ≫ 0.

(b) The behavior of Ψ(x) near 0 is given by

ψ(x) = \{ \begin{matrix}aμ(x) + b & μ ≠ triv, |·|^{−1} \\ av_{F} (x) + b & μ = triv \\ b & μ = |·|^{−1} \end{matrix} .

Exercise 1. Show that if μ = |·|^{−1}, then the Fourier transform of a constant function is zero.

2. Show that the Fourier transform of a Schwarz function is a Schwarz function. Now, F_{μ} isn’t much bigger than S(F); there’s a one-dimensional family of constants that μ(x)|x|Φ(x) can take for large x. So F_{μ} = S(F) ⊕ ℂΦ_{μ}, where

Φ_{μ}(x) = \{ \begin{matrix}μ(x)^{−1}|x|^{−1} & x ≥ 1 \\ 0 & |x| < 1 \end{matrix} .

So to verify (i), we need only check it for Φ_{μ}. And there are two cases, as usual. (A) Suppose μ is ramified. We’re thinking about ∑_{n≤0} ∫_{v(y)=n} μ(x)^{−1}|x|^{−1}\overline{ψ_{F} (xy)}dy. Apparently if I remember anything from Tate’s thesis I’d recognize local factors ϵ(χ) materializing. We’re looking at

∑_{n≤0} ∫_{v(y)=n} μ(y)^{−1}\overline{ψ_{F} (xy)}d^{×}y.

The ϵ factor is, essentially, ∫_{F^{×}} χ^{−1}(y)ψ_{F} (y)d^{×}y. The claim is that at most one of these integrals is nonzero. Specifically, it’s zero unless v(x) + n = −d − f where d = cond(μ) > 0 and f = cond(ψ); the conductors of μ and f. By definition, f is such that ψ_{F} |_{𝔭^{−f} O_{F}} is trivial but ψ_{F} |_{𝔭^{−f−1}O_{F}} is nontrivial. We’ll leave out this calculation; it works, but the mechanics of it aren’t terribly enlightening. ... we explicitly check kernel and image. Elementary properties of Fourier transforms tell us that S(F) is preserved; checking injectivity comes down to looking at Φ_{μ}(x). So at this point we can verify all of (i) through (iv) directly. (B) μ = |·|^{s} for some s ∈ ℂ. Then we can get a fairly explicit formula for Φ̂(x). Specifically,

∫_{v(y)=n} μ^{−1}(y)\overline{ψ_{F} (xy)}d^{×}y = q^{−ns}(∫_{O} \overline{ψ_{F}} (x^{n}z)dz − ∫_{𝔭} \overline{ψ_{F}} (π^{n}xz)dz)

So

Φ̂_{m}u(x) = F_{s}(x) − |π|F_{s}(−x)

where F_{s} = ∑_{−d−v(x)≤n≤0} q^{−ns}, a finite geometric series. And that, pretty much, is that.◇ We’ve got the map

B(μ_{1}, μ_{2})

TFT_{μ_{1},μ_{2}}

Twisted Fourier Transform

[figure]
K_{μ_{1},μ_{2}}
f  ξ_{f} : x ↦ μ_{2}(x)|x|^{\frac{1}{2}} ∫_{F} f(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))\overline{ψ_{F} (xy)}dy

Here, K_{μ_{1},μ_{2}} is just the name of the image. K_{μ_{1},μ_{2}} ⊇ S(F^{×}); and the image is really the set of all functions on F^{×} vanishing for |x| ≫ 0; and for x in a neighborhood of zero, the behavior near zero is given by one of these three possibilities:

• |x|^{\frac{1}{2}} (aμ_{1}(x) + bμ_{2}(x)) if μ ≠ triv, |·|^{−1}.

• |x|^{\frac{1}{2}} (aμ_{1}(x)v_{F} (x) + bμ_{2}(x)) if μ = triv.

• b|x|^{\frac{1}{2}} μ_{2}(x) if μ = |·|^{−1}.

And ker(TFT_{μ_{1},μ_{2}}) is

• 0 if μ ≠ |·|^{−1}.

• ℂμ_{1}(det g)|det(g)|^{\frac{1}{2}} if μ = |·|^{−1}.

where μ =^{def} μ_{1}μ_{2}^{−1}.

We also know that TFT_{μ_{1},μ_{2}}(ρ(μ_{1}, μ_{2}) ( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ) f) = ψ_{F} (bx) TFT_{μ_{1},μ_{2}}(f)(ax).

Remark S(F^{×}) is irreducible under \{( \begin{matrix}a & b \\ 0 & 1 \end{matrix} )\}.

A. Suppose μ ≠ |·|^{−1}. Then we know that TFT_{μ_{1},μ_{2}} is an isomorphism, by its very definition and the stated injection. Given a subrepresentation (0) ≠ V ⊆ B(μ_{1}, μ_{2}), then V ⊇ {f −

ρ(μ_{1}, μ_{2}) ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) f|f ∈ V \}. In other words^{14} V^{⊥} is fixed under all ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ), b ∈ F.^{15}

14Maybe we should have done this a little earlier. It’s a fact that ρ(μ_{1}, μ_{2})^{∨} = ρ(μ_{1}^{−1}, μ_{2}^{−1}). And that’s cause we know how to write down things like ∫_{B\G} fdġ.

15We’re trying to prove there are very few subrepresentations.

Well, V^{⊥} is a subspace of B(μ_{1}^{−1}, μ_{2}^{−1}): and it’s still fixed by ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ). We want to know

what sorts of things can happen.

A function f ∈ B(μ_{1}^{−1}, μ_{2}^{−1}) fixed under all ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ), b ∈ F has the property that f(( \begin{matrix}t_{1} & ∗ \\ 0 & t_{2} \end{matrix} ) w^{−1} ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} )) =

f(( \begin{matrix}t_{1} & ∗ \\ 0 & t_{2} \end{matrix} ) w^{−1}) = μ_{1}(t_{1})^{−1}μ_{2}(t_{2})^{−1}|\frac{t_{1}}{t_{2}} |^{\frac{1}{2}} f(w^{−1}). Thus, dim(V^{⊥}) ≤ 1; the only undeter-

mined number is f(w^{−1}).

On the other hand, we have w^{−1} ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & −b \\ 0 & 1 \end{matrix} )^{−1} ( \begin{matrix}b^{−1} & 0 \\ 0 & b \end{matrix} ) ( \begin{matrix}1 & 0 \\ b^{−1} & 1 \end{matrix} ), b ≠ 0,

whence f(w^{−1} ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} )) = μ_{1}(b)μ_{2}^{−1}(b)|b|^{−1}f(( \begin{matrix}1 & 0 \\ b^{−1} & 0 \end{matrix} )) = μ(b)|b|^{−1}f(( \begin{matrix}1 & 0 \\ b^{−1} & 0 \end{matrix} )), for

any b ≠ 0.

So take t_{1} = t_{2} = 1. Then it’s hard to make these two formulae compatible. In short, if μ ≠ |·|, then V^{⊥} = (0).

B. We might blow these off. Use the same formula and earlier work to describe the irreducibility. So, suppose μ = |·|. Follow the same route to analyze it. Then figure out that f can be just one fixed function, etc.

C. μ = |·|^{−1}. Then use duality.

Recall that ρ(μ_{1}, μ_{2}) is

• irreducible if μ =^{def} μ_{1}μ_{2}^{−1} ≠ |·|^{±1}.

• if μ = |·|, then B(μ_{1}, μ_{2}) contains an irreducible subrepresentation of codimension one, namely,

\{f ∈ B(μ_{1}, μ_{2})| ∫_{F} f(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))dx = 0\}.

Of course, up to a nonzero constant, this integral is the same as ∫_{B(F)\G(F)} f(g)μ^{−1}(det g)|det g|^{\frac{1}{2}} dġ.

• dually, if μ = |·|^{−1}, then [use ker TFT] there is a one-dimensional subrepresentation given by g ↦ μ_{1}(det g)|det g|^{\frac{1}{2}} and the quotient map is irreducible.

Nomenclature If μ ≠ |·|^{±1}, then π(μ_{1}, μ_{2}) = ρ(μ_{1}, μ_{2}); this is called a principal series representation

If μ = |·|^{±1}, then π(μ_{1}, μ_{2}) is the one-dimensional subquotient; and σ(μ_{1}, μ_{2}) is the infinitedimensional subquotient, sometimes called a special representation.

F’rinstance, σ(|·|^{\frac{1}{2}} , |·|^{− \frac{1}{2}} ) is a special representation. And because of the duality, we know that π(μ_{1}, μ_{2}) = π(μ_{2}, μ_{1}). Well, we don’t really know it yet, but we’ll prove it later by understanding π^{∨}.

It’s straight forward from the definition that ρ(μ_{1}χ, μ_{2}χ) = ρ(μ_{1}, μ_{2}) · (χ ∘ det). So once you have your hands on one representation, you can twist it by a one-dimensional representation, that is, a [quasi]character.

Concretely, we’ve seen that if π is an irreducible infinite dimensional representation, we’ve seen before that the Kirillov model contains the Schwarz functions with finite index;

dim_{ℂ}(K(π)/S(F^{×})) = \{ \begin{matrix}0 & π supercuspidal \\ 1 & π special \\ 2 & π principal \end{matrix} .

MA 620 – Langlands Program  3 February 1997

We correct a slight oversight.

How to express concretely the contragradient representation of an irreducible admissible representation

Theorem Let (π, V) be an admisible repesentation, ω_{π} its central character. Then

1. π̌ is isomorphic to (π′, V′) where V′ = V, and π′(g) : v ↦ ω_{π}(det g)^{−1}π(g)(v).^{16}

2. K(π̌) = {x ↦ ω_{π}(x)^{−1}ξ(x)|ξ ∈ K(π)}.

3.

[figure: K(π); Φ; K(π̌); η; η′ = (x ↦ ω_{π}(x)^{−1}η(x))]

makes the following diagram commute.

[figure: K(π); Φ; K(π̌); K(π); ω_{π}(g)^{−1}ω_{π}; Φ; π̌(g); K(π̌)]

But this isn’t anything new.

4. The pairing is given by

K(π) × K(π̌)  \overline{}ℂ
ξ × η  ⟨ξ, η⟩ =^{def} ∫_{F^{×}} ξ_{1}(x)η(−x)d^{×}x + ∫_{F^{×}} ξ_{2}(x)(π̌(w^{−1})η)(−x)d^{×}x

ξ = ξ_{1} + π(w)ξ_{2}, ξ_{i} ∈ S(F^{×}) Equivalently, the following bilinear form on K(π)

16( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) acts as ω_{π}(a). By the very definition of contragredient, it’s now gotta operate by ω_{π}(a)^{−1}.

ξ = ξ_{1} + π(w)ξ_{2}
ξ_{i} ∈ S(F^{×})
⟨ξ, η⟩′ = ∫_{F^{×}} ξ_{1}(x)η(−x)ω_{π}(x)^{−1}d^{×}x + ∫_{F^{×}} ξ_{2}(x)(π(w)^{−1}η)(−x)ω_{π}(x)^{−1}d^{×}x

We have to check that this is well-defined, and ⟨π(g)ξ, η⟩′ = ⟨ξ, π′(g^{−1})η⟩′ for all ξ, η ∈ K(π), where π′(g) = ω_{π}(det g)^{−1}π(g).

Checking this bilinearity isn’t bad; we know how ( \begin{matrix}a & b \\ 0 & 1 \end{matrix} ) and ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) act on the Kirillov

model. That gives us the Borel. Now go check for w, which is chosen in such a way that it works out.

The Ultimate Truth [Local Langlands correspondence] Let F be a nonarchimedean local field, G = GL_{2}. We should have

admissible irred reps of G(F) ↭ 2-dim reps of Gal(F^{sep}, F) Gal(F^{sep}, F) → GL_{2}(ℂ) .

This can’t quite be true, so we make some minor modifications. On the right-hand side, there aren’t enough representations of the Galois group. You see, the Galois group is a profinite group. And with a profinite group, any representation has to have finite image, a very severe restrcition; there just aren’t enough . This only gives us reps which factor through a finite-dimensional Galois extension.

So, replace the right-hand side by 2-dimensional representations of the Weil group W_{F}. Recall that the Weil group fits into

[figure: Gal(F^{sep}, F); Gal(κ_{F}^{sep}, κ_{F})]

ℤ̂^{≅}

[figure: W_{F}; ∪; ℤ; ∪]

We need a topology on W_{F}.

[figure: 0; I_{F}; Gal(F^{sep}, F); Gal(κ_{F}^{sep}, κ_{F})]

[figure: 0; I_{F}; =; open; W_{F}; ∪; ℤ; ∪]

But now the right-hand side isn’t big enough to get the special representations. Two possible fixes; replace

W_{F} by \{ \begin{matrix}W_{F} × SL_{2}(ℂ) \\ W_{F} ×|𝔾_{a} \end{matrix} .

Here, this 𝔾_{a} is \{( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} )\}. By semidirect product, we mean that, wxw^{−1} = ∥w∥^{±1} x.^{17}

Given a representation W_{F} × SL_{w}(ℂ) →^{ρ} GL_{2}(ℂ), we want a representation of W_{F} ×|𝔾_{a}. And the

sort of obvious action should be (w · x) ↦ ρ(w)ρ(( \begin{matrix}∥w∥^{\frac{1}{2}} & 0 \\ 0 & ∥w∥^{− \frac{1}{2}} \end{matrix} ))ρ(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} )). Check; does

this really factor through the semidirect product? Yep.

Ad nowadays, it’s more common to use W_{F} × SL_{2}(ℂ). Now supercuspdial representations will correspond to irreducible representations of W_{F}, so that SL_{2}(ℂ) acts trivially. And principal se-

ries π(μ_{1}, μ_{2}) correspond to ( \begin{matrix}μ_{1} & 0 \\ 0 & μ_{2} \end{matrix} ); use W_{F} ⇀ F^{×} from class-field theory. Finally[?] a

special representation corresponds to a nontrivial action of SL_{2}(ℂ). And Schur’s lemma says that then W_{F} must operate by a 1-dimensional character.

This all seems a little ad hoc, but hopefull it’ll get us into L-series.

17We’re not yet sure which normalization we’ll use. It sort of depends on whether we use arithmetic or geometric, Frobenius. w ↦ Fr^{n}, the geometric Frobenius; ∥w∥ = q^{\frac{n}{2}} .

Langlands tells us what to expect to be true. Irreducible admissible representations of GL_{2}(F) should match up with two-dimensional complex representations of W_{F} × SL_{2}(ℂ).

Grant this. What should the correspondence be?

• supercuspidal ↭ irreducible representations of W_{F}.

• principal series π(μ_{1}, μ_{2}) ↭ ( \begin{matrix}μ_{1} & 0 \\ 0 & μ_{2} \end{matrix} ).

• special representations ↭ a 1-dimensional representation of W_{F} and the standard representation of SL_{2}(ℂ).

Now let ρ be a representation of the Weil group. The local L-factor is

L(ρ, s) = det(1 − ρ(Fr)q^{−s}|V^{I_{F}})^{−1}

where I_{F} is the inertia group and Fr is the geometric Frobenius.

Now, the corespondence π ↭ ρ is so nice that it respects twisting. So let χ be a character; composing with det gives another representation. We can twist π by χ, getting π · χ ∘ det; and then π · χ ∘ det ↭ ρ · χ. Moreover, there’s a central character ω_{π} of a representation, which is a character of F^{×}. By classfield theory, you canthink of this as a character of the Galois group, or even of the Weil group; and π is to ω_{π} as ρ is to det ρ.

(π ⇝ ω_{π}) ↭ (ρ ⇝ det ρ)
supercuspidal ↭ irreducible representation of W_{F}
π(μ_{1}, μ_{2}) ↭ ( \begin{matrix}μ_{1} & 0 \\ 0 & μ_{2} \end{matrix} )
unitary ↭ unitary central character

Functional Equation Let π be admissible, irreducible, infinite-dimensional representation of GL_{2}. We have Whittaker and Kirillov models W(π) and K(π). We’d like to define local zetafunctions just as in the 1-dimensional, or abelian, case.

For every W ∈ W(π) we define

ζ(g, χ, W, s) = ∫_{F^{×}} W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x)|x|^{s− \frac{1}{2}} d^{×}x

where g ∈ G(F), χ a character of F^{×} [we’re in effect twisting by χ]. The ^{s− \frac{1}{2}} is to get a formula later on to work out.

Theorem

1. ζ(g, χ, W, s) converges for ℜ(s) ≫ 0.

2. There’s an Euler factor^{18} L(π, χ, s) = L(π ⊗ χ, s) such that

\frac{ζ(g, χ, W, s)}{L(π, χ, s)}

is an entire function in s for every χ and W; and such that there’s a W^{0} with

ζ(1, χ, W^{0}, s) = L(π, χ, s).

3. ζ(g, χ, W, s) has a meromorphic continuation to the s-plane ℂ = ℂ_{s}, satisfying the functional equation

\frac{ζ(g, χ, W, s)}{L(π, χ, s)}ϵ(π, χ, s) = \frac{ζ(wg, χ^{−1}ω_{π}^{−1}, W, 1 − s)}{L(π, ω_{π}^{−1} · χ^{−1}, 1 − s)}

where w is the Weyl element w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ).

Of course, we can do all this for the Kirillov model. If W ↭ ξ, then W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g) = π(g)ξ(x).

For the moment, let’s assume all this. What is it really saying? Let’s try to figure out what the local L-factors should be.

Work first with a supercuspidal representation π. Then all the functions ξ are really Schwarz functions. As such, we know that the local zeta function

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)χ(s)|x|^{s− \frac{1}{2}} d^{×}x

is always an entire function; we don’t have to worry about convergence. Moreover, if we take

ξ_{0}(x) = \{ \begin{matrix}χ(x) & x ∈ O_{F}^{×} \\ 0 & x ∉ O_{F}^{×} \end{matrix}

then ζ_{ξ_{0},χ}(s) is a nonzero constant, the the local L-factor is L(π, χ, s) = 1.

18Broadly speaking, for an Euler factor, take a finite poly in q^{−s} whose constant term is one. Then an Euler factor is the reciprocal of such a thing.

A little more interesting will be the principal series. Take π(μ_{1}, μ_{2}), μ_{i} ≠ triv, μ_{1}μ_{2}^{−1} ≠ |·|^{±1}. Then K(π(μ_{1}, μ_{2})) = \{ξ : F^{×} → ℂ|ξ(x) = 0 for |x| ≫ 0, ξ(x) = |x|^{\frac{1}{2}} (aμ_{1}(x) + bμ_{2}(x)) for x close to 0\}. A typical element is ξ(x) = |x|^{\frac{1}{2}} μ_{1}(x)f_{1}(x) + |x|^{\frac{1}{2}} μ_{2}(x)f_{2}(x), f_{i}(x) ∈ S(F). Recall that, classically, ζ(f, χ, s) = ∫_{F^{×}} f(x)χ(x)|x|sd^{×}x. So for us, take the integral and get

ζ_{ξ,χ}(s) = ζ(f_{1}, μ_{1}χ, s) + ζ(f_{2}, μ_{2}χ, s);

that’s what the^{s− \frac{1}{2}} is for. Then L(π(μ_{1}, μ_{2}), χ, s) = L(μ_{1}, χ, s) · L(μ_{2}, χ, s).

Today we’ll talk about “structural constants” of admissible irreducible representations of G(F).

We define formal Mellin transforms in the following way. Let F^{×} →^{ϕ} ℂ be locally constant and vanishing for |x| large. Fix a uniformizer ϖ in 𝔭_{F}. For a character ν : O_{F}^{×} → ℂ^{×} and n ∈ ℤ, we define

ϕ̂_{n}(ν) = ∫_{O_{F}^{×}} ϕ(ϵϖ^{n})ν(ϵ)d^{×}ϵ

where d^{×}ϵ is a normalized Haar measure

Using t for some formal variable, write

ϕ̂(ν, t) = ∑_{n∈ℤ}ϕ_{n}(ν)t^{n}

a formal Laurent series.

We get Gauss sums: for ν : O_{F}^{×} → ℂ a character and x ∈ F^{×}, define a number

η(μ, x) = ∫_{O_{F}^{×}} μ(ϵ)ψ(ϵx)d^{×}ϵ.

It’s a well-known fact that this thing η(μ, 0) is zero unless ν(x) = −cond(μ) − cond(ψ), minus the sum of the conductors.

Finally, we can use all this to define structural constants. Again, O_{F}^{×} →^{ν} ℂ^{×} is a character of our compact group, and n ∈ ℤ. First of all, define a function

ϕ_{ν} =^{def} \{ \begin{matrix}(ω_{π} · ν)(x) & x ∈ O_{F}^{×} \\ 0 & x ∉ O_{F}^{×} \end{matrix} .

Then ϕ_{ν}(x) ∈ K(π). We can hit it with the Weyl element, getting another function π(w)ϕ_{ν}, albeit not necessarily one with compact support. Now do a little Fourier analysis.

c_{n}(ν) =^{def} { \atop (π(w)ϕ_{ν})}_{n}(ν)
c(ν, t) =^{def} ∑_{n} c_{n}(ν)t^{n}

Knowing these number c(ν, t) determines ϕ, I think. In general, we’re thinking about ∫_{O_{F}^{×}} π(w)ϕ_{ν}(ϵ· ϖ^{n})μ(ϵ)d^{×}ϵ. Now think about how π(w)ϕ_{ν}(ϵϖ^{n}) transforms under local units. Bearing in mind

that ( \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) = ( \begin{matrix}0 & ϵ \\ −1 & 0 \end{matrix} ) = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & ϵ \end{matrix} ) = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}ϵ^{−1} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}ϵ & 0 \\ 0 & ϵ \end{matrix} ).

Use this identity to explain how ϵ acts; then π(( \begin{matrix}ϵ & 0 \\ 0 & 1 \end{matrix} ))π(w)ϕ_{ν} = ω_{π}(ϵ)ω_{π}(ϵ)^{−1}ν(ϵ)^{−1}π(w)ϕ_{n}u = ν(ϵ)^{−1}π(w)ϕ_{ν}. So if we translate our integrand by ϵ, all we’re doing is multiplying by ν(ϵ)^{−1}. Thus,

∫_{O_{F}^{×}} π(w)ϕ_{ν}(ϵ · ϖ^{n})μ(ϵ)d^{×}ϵ = 0 unless μ = ν.

Proposition

1. For δ ∈ O_{F}^{×}, l ∈ ℤ,

{ \atop π(( \begin{matrix}δϖ^{l} & 0 \\ 0 & 1 \end{matrix} ))ϕ}(ν, t) = t^{−l}ν(δ)^{−1}ϕ̂(ν, t).

2. For x ∈ F,

{ \atop π(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))ϕ}(ν, t) = ∑_{n} t^{n}( ∑_{O_{F}^{×}→^{μ}ℂ^{×}} ν(μ^{−1}ν, ϖ^{n}x) · ϕ̂_{n}(μ)).

3.

{ \atop π(w)ϕ}(ν, t) = c(ν, t)ϕ̂(ν^{−1}ω_{0}^{−1}, t^{−1}ω_{π}(ϖ)^{−1}).

where ω_{0} = ω_{π}|_{O_{F}^{×}}. Well, let’s look at the first.

∫ ϕ(ϵδϖ^{n+l})ν(ϵ)d^{×}ϵ = ν(δ)^{−1} ∫_{O_{F}^{×}} ϕ(ϵ′ϖ^{n+l})ν(ϵ′)d^{×}ϵ′
= ν(δ)vϕ_{n+l}(ν)

On to the second. Initially, recall the Fourier expansion

f = ∑_{μ}(∫_{G} fμ^{−1}dg)
ϕ(ϵϖ^{n}) = ∑_{μ} ϕ̂_{n}(μ)μ^{−1}(ϵ)

∫_{O_{F}^{×}} ψ(xϵϖ^{n})ϕ(ϵϖ^{n})ν(ϵ)d^{×}ϵ = ∑_{μ} ϕ̂_{n}(μ) ∫_{O_{F}^{×}} ψ(xϵϖ^{n})(νμ^{−1})(ϵ)d^{×}ϵ
= ∑_{μ} ϕ̂_{n}(μ)η(νμ^{−1}, xϖ^{n})

For (3) it suffices to check for ϕ_{ν}’s and their translates by ϖ^{l}. We only need that, if it’s true for ϕ, then it’s true for π(( \begin{matrix}ϖ^{l} & 0 \\ 0 & 1 \end{matrix} ))ϕ. For

( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}ϖ^{l} & 0 \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & ϖ^{l} \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )
= ( \begin{matrix}ϖ^{l} & 0 \\ 0 & ϖ^{l} \end{matrix} ) ( \begin{matrix}ϖ^{−l} & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )
{ \atop π(w)π(( \begin{matrix}ϖ^{l} & 0 \\ 0 & 1 \end{matrix} ))}(ν, t) = multiply by t^{l}ω_{π}(ϖ^{l})

ummm,

ϕ̂(ν, t) t^{−l}ϕ̂(ν, t)

so right-hand side got multiplied by t^{l}ω_{π}(ϖ^{l}), cool.

Last time we had, where δ ∈ O_{F}^{×},

{ \atop π ( \begin{matrix}δϖ^{l} & 0 \\ 0 & 1 \end{matrix} ) · ϕ}(ν, t) = t^{−l}ν^{−1}(δ)ϕ̂(ν, t)
{ \atop π ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) ϕ}(ν, t) = ∑_{n} t^{n}(∑_{μ} η(μ^{−1}ν, ϖ^{n}x), ϕ̂_{n}(μ))
{ \atop π(w)ϕ}(ν, t) = C(ν, t)ϕ̂(ν^{−1}ω_{0}^{−1}, t^{−1}ω_{π}(ϖ)^{−1})

and recall that the η(μ^{−1}ν, ϖ^{n}x) is a Gauss sum. We now want to relate these structural constants to the functional equations.

Local zeta function Let ξ ∈ K(π),

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)|x|^{s−?}d^{×}x.

Recall that ϕ_{n}(ν) = ∫_{O_{F}^{×}} ϕ(ϵϖ^{n})ν(ϵ)d^{×}ϵ. So

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)|x|^{s−?}d^{×}x
= ∑_{n} ∫_{O_{F}^{×}} ξ(ϵϖ^{n})χ(ϵϖ^{n})q^{−n(s− \frac{1}{2} )}
= ∑_{n} ξ_{n}(χ) · (χ(ϖ) · q^{−(s− \frac{1}{2} )})^{n}
= ξ̂(χ, χ(ϖ)q^{−(s− \frac{1}{2} )})

Assume that d^{×} is normalized appropriately. Usually, we mean d^{×}x = \frac{dx}{|x|}. And obviously the functional equation depends on the choice of Haar measure. But for now, let’s normalize so that ∫_{O_{F}^{×}} d^{×}x = 1. Recall that we changed ξ by hitting it with the Weyl element; and this gives

ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s) = { \atop π(w)ξ}(χ^{−1}ω_{π}^{−1}, χ^{−1}(ϖ) · ω_{π}^{−1}(ϖ) · q^{s− \frac{1}{2}}

So now it makes sense to investigate

\frac{ζ_{ξ,χ}(s)}{L(π ⊗ χ, s)}ϵ(π ⊗ χ, s) = \frac{ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s)}{L(π ⊗ ω_{π}^{−1} ⊗ χ^{−1}, 1 − s)}
\frac{ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s)}{ζ_{ξ,χ}(s)} = \frac{L(π ⊗ ω_{π}^{−1} ⊗ chi^{−1}, 1 − s)}{L(π ⊗ χ, s)}ϵ(π ⊗ χ, s)
= C(χ^{−1}ω_{π}|_{O_{F}^{×}}, (χ^{−1}ω_{π}^{−1})(ϖ)q^{s− \frac{1}{2}} )

Let that quotient of L-factors be γ(π ⊗ χ, s).

Corollary [Assume the existence of the functional equation.] If for all χ L(π⊗χ, s) = L(π′⊗χ, s) and ϵ(π ⊗ χ, s) = ϵ(π′ ⊗ χ, s) then π ≅ π′. We also get

Corollary γ(π ⊗ χ, s) exists.

Corollary ϵ(π ⊗ χ, s) exists. In principal, we’ve proved the functional equation. Just to clean up for special representations....

L(σ(μ_{1}, μ_{2}), s) = \{ \begin{matrix}L(μ_{1}χ, s) & μ_{1}μ_{2}^{−1} = |·| \\ L(μ_{2}χ, s) & μ_{1}μ_{2}^{−1} = |·|^{−1} \end{matrix} .

The standard case to think about is L(σ(|·|^{\frac{1}{2}} , |·|^{− \frac{1}{2}} ), s) = L(|·|^{\frac{1}{2}} ) = _{\frac{1}{1−q^{−s− \frac{1}{2}} )}}. [The foregoing is an exercise; use the same method as in the principal series case.] BTW, if you believe in this sort of stuff, then you should believe that these L-factors are really just one factor; for it corresponds to a map W_{F} × SL_{2}(ℂ) → GL_{2}(ℂ) which is an abelian representation of W_{F}; and that’s the representation from which the L-factor comes. We know that

ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s) = γ(π ⊗ χ, s) · ζ_{ξ,χ}(s)
ξ = π(w)π(w^{−1}ξ)
= ω_{π}(−1)π(w)π(w)ξ

= ω_{π}^{−1}π(w)ξ_{1}
ξ_{1} = π(w)ξ
χ = ω_{π}^{−1} · (χ^{−1}ω_{π}^{−1})^{−1}
ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s) = ω_{π}(−1)γ(π ⊗ χ, s)γ(π ⊗ χ^{−1}ω_{π}^{−1}, 1 − s) · ζ_{π(w)ξ,χ^{−1}ω_{π}^{−1}}(1 − s)

γ(π ⊗ χ, s)γ(π ⊗ ω_{π}^{−1} ⊗ χ^{−1}, 1 − s) = ω_{π}(−1)

Compare this to the classical case: We’ve got a local zeta function ζ, f a Schwarz function on F. Then

\frac{ζ(f, χ, s)}{L(χ, s)}ϵ(χ, s) = \frac{ζ(f̂, χ^{−1}, 1 − s)}{L(χ^{−1}, 1 − s)}
\frac{ζ(f̂, χ^{−1}, 1 − s)}{ζ(f, χ, s)} = \frac{L(χ^{−1}, 1 − s)}{L(χ, s)}ϵ(χ, s)

actually, our epsilon factors are ϵ(χ, s, ψ, dx). Now write the right-hand side of the last equation as γ(χ, s); then try to play a similar game for this gamma factor;

γ(χ, s)γ(χ^{−1}, 1 − s) = ?

Exercise Figure out ?; it’s probably χ(−1).

We’re talking about ϵ-factors. We have

\frac{ζ_{ξ,χ}(s)}{L(π ⊗ χ, s)}ϵ(π, χ, s) = \frac{ζ_{π(w)ξ,ω_{π}^{−1}χ^{−1}}(1 − s)}{L(π ⊗ ω_{π}^{−1} ⊗ χ^{−1}, 1 − s)}.

Remark Clasisically [in the abelian case] ϵ depends on an additive character and a Haar measure, ϵ(χ, s, ψ_{F} , dx). But in our situation... ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)χ(x)|x|^{s− \frac{1}{2}} d^{×}x. So there’s an apparent dependency on the Haar measure, but as we always use the same it doens’t matter. But ψ_{F} really is an important choice; remember, the Kirillov model is canonical only up to ψ_{F}.

When π is supercuspidal, the L-factors are all trivial, since the zeta functions are entire functions. So L(π⊗χ, s) = 1 and ϵ(π⊗χ, s) is a finite Laurent series in q^{−s}. If we sort of iterate the functional equation, we get

ϵ(π ⊗ χ, s) · ϵ(π ⊗ ω_{π}^{−1} ⊗ χ^{−1}, 1 − s) = ω_{π}(−1) = ±1.

Thus, ϵ(π ⊗ χ, s) has the form A · q^{−ns}, some n ∈ ℤ.

Computation of ϵ-factors for π(μ_{1}, μ_{2}) What’s the answer?

ϵ(π(μ_{1}, μ_{2}) ⊗ χ, s) = ϵμ_{1}χ, s) · ϵ(μ_{2}χ, s)

Which looks good, but what does it mean? ϵ(μ_{1}χ, s) = ϵ(μ_{1}χ, s, ψ_{F} , dx) where dx is the Haar measure on F self-dual with respect to ψ_{F}.

[and the same sort of thing should work for special representations]

Recall the functional equation for the abelian local zeta function. Let f ∈ S(F). Then

ζ(f, χ, s) = ∫_{F^{×}} f(x)χ(x)|x|^{s}d^{×}x
\frac{ζ(f, χ, s)}{L(χ, s)}ϵ(χ, s) = \frac{ζ(f̂_{ψ}, χ^{−1}, 1 − s)}{L(χ^{−1}, 1 − s)}

Pick ξ ∈ S(F^{×}) ∩ π(w)S(F^{×}); think of ξ as being realized in its Kirillov model; it correpsonds to some ϕ ∈ B(μ_{1}, μ_{2}) the representation space; and

ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} )) = ∫_{F} μ_{2}^{−1}(x)|x|^{− \frac{1}{2}} ξ(x)ψ_{F} (xy)dx
ξ(x) = μ_{2}(x)|x|^{\frac{1}{2}} ∫_{F} ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))\overline{ψ_{F} (xy)}dy

are dual to each other under this twisted Fourier transform. Moreover,

ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) w) = ∫ μ_{2}^{−1}(x)|x|^{− \frac{1}{2}} (π(w)ξ)(x)ψ_{F} (xy)dx

=

ϕ(( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}\frac{1}{y} & 0 \\ 0 & y \end{matrix} ) w ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} )) = μ_{1}(y^{−1})μ_{2}()|y|^{−1}ϕ(w ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ))

Hmmm. Make some temporary definitions.

f(x) = μ_{2}^{−1}(x)|x|^{− \frac{1}{2}} ξ(x)
= ∫_{F} ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))\overline{ψ_{F} (xy)}dy

Then

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)χ(x)|x|^{s− \frac{1}{2}}
= ∫_{F^{×}} f(x)(χ · μ_{2})(x)|x|^{\frac{1}{2}} d^{×}x
= ζ(f, χ · μ_{2}, s)
= γ(χ^{−1} · μ_{2}^{−1}, s)^{−1}ζ(f̂, χ^{−1} · μ_{2}^{−1}, 1 − s)

here, we’ve used

\frac{ζ(f̂, χ^{−1}, 1 − s)}{ζ(f, χ, s)} = \frac{L(χ^{−1}, 1 − s)}{L(χ, s)}ϵ(χ, s) = γ(χ, s).

We’ve got a functional equation

\frac{ζ(F, χ, s)}{L(χ, s)}ϵ(χ, s) = \frac{ζ(F̂, χ^{−1}, 1 − s)}{L(χ^{−1}1 − s)}

or

ζ(F̂, χ^{−1}, 1 − s) = γ(χ, s)ζ(F, χ, s).

For ξ ∈ K(μ_{1}, μ_{2}) the Kirillov model, we have

f(x) =^{def} μ_{2}^{−1}(x) · |x|^{− \frac{1}{2}} ξ(x) = ∫_{F} ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))\overline{ψ(xy)}dy.

Well, actually, ξ(x) ∈ S(F^{×}) ∩ pi(w)S(F^{×}). As usual, we know

w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) w = ( \begin{matrix}1 & 0 \\ −y & 1 \end{matrix} )
= ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}\frac{1}{y} & 0 \\ 0 & y \end{matrix} ) w ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} )
ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) w) = ω_{π}(−1)μ_{1}(y^{−1})μ_{2}(y)|y|^{−1}ϕ(w^{−1} ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ))
= π(w)ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))

Now

η(x) =^{def} μ_{2}^{−1}(x)|x|^{− \frac{1}{2}} π(w)ξ(x) = ω_{π}(−1) ∫_{F} μ(y^{−1})|y|^{− \frac{1}{2}} ϕ(w^{−1} ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 0 \end{matrix} ))\overline{ψ(xy)}dy

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)χ(s)|x|^{− \frac{1}{2}} d^{×}x
= ∫_{F^{×}} f(x)μ_{2}(x)χ(s)|x|^{s}d^{×}x
= ζ(f, μ_{2}χ, s)
= γ(μ_{2}^{−1}χ^{−1}, 1 − s) ∫_{F^{×}} ϕ(w^{−1} ( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ))μ_{2}(y)^{−1}χ(y)^{−1}d^{×}|y|^{1−s}d^{×}y

Remembering that ω_{π} = μ_{1}μ_{2} is the central character....

ζ_{π(w)ξ,ω_{π}^{−1}χ^{−1}}(1 − s) = ∫_{F^{×}}(π(w)ξ)(x)(μ_{1}^{−1}μ_{2}^{−1}χ^{−1})(x)|x|^{\frac{1}{2} −s}d^{×}x
= ∫_{F^{×}} η(x)μ_{2}(x)μ_{1}^{−1}(x)μ_{2}^{−1}(x)χ^{−1}(x)|x|^{1−s}d^{×}x
= ζ(η, μ_{1}^{−1}χ^{−1}, 1 − s)

But η itself is the Fourier transform; and then we want to use the abelian local functional equation.

ζ(η, μ_{1}^{−1}χ^{−1}, 1 − s) = γ(μ_{1}χ, s) · ω_{π}(−1) ∫_{F^{×}} μ(y)^{−1}|y|^{−1}ϕ(w^{−1} ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ))μ_{1}(y)χ(y)|y|^{s}d^{×}y
(∗) = γ(μ_{1}χ, s) · μ_{1}(−1)μ_{2}(−1) ∫_{F^{×}} ϕ(w^{−1} ( \begin{matrix}1 & −\frac{1}{y} \\ 0 & 1 \end{matrix} ))μ_{2}(y)χ(y)|y|^{s−1}d^{×}y
z =^{def} −\frac{1}{y}
(∗) = γ(μ_{1}χ, s)μ_{1}(−1)\overline{μ_{2}(−1)} ∫_{F^{×}} ϕ(w^{−1} ( \begin{matrix}1 & z \\ 0 & 1 \end{matrix} ))μ_{2}(z^{−1})χ(z)^{−1}|z|^{1−s}d^{×}z

[Remember that μ = μ_{1}μ_{2}^{−1}.] Which is the same as the integral we’d got before for ζ_{ξ,χ}(s), cool. And

γ(π(μ_{1}, μ_{2}) ⊗ χ, s) = \frac{ζ_{π(w)ξ,ω_{π}^{−1}χ^{−1}}(1 − s)}{ζ_{ξ,χ}(s)}
= \frac{γ(μ_{1}χ, s)}{γ(μ_{2}^{−1}χ^{−1}, 1 − s)}μ_{1}(−1)χ(−1)

But we’d actually expected a factor of μ_{2}(−1), instead of μ_{1}(−1), bummer. For we know that γ(μ_{2}^{−1}χ^{−1}, 1 − s)γ(μ_{2}χ, s) = μ_{2}(−1)χ(−1). And we expect

γ(π(μ_{1}, μ_{2}) ⊗ χ, s) = γ(μ_{1}χ, s) · γ(μ_{2}χ, s)

so we’ve blown a sign somewhere. We want

ϵ(π(μ_{1}, μ_{2}) ⊗ χ, s) = ϵ(μ_{1}χ, s)ϵ(μ_{2}χ, s).

It’s the only way that the Langlands parametrization is going to work out.

We suspect that the introduction of ω_{π}(−1) above is exactly what’s messing it up.

Unfortunately, I missed class last time.

Unitary irreducible admissible representations of G(F).

0 Unitary characters of F^{×}.

1 Supercuspidal with unitary central character.

2 Principal series π(μ_{1}, μ_{2}) such that μ_{1} and μ_{2} are unitary.

3 Special representation with unitary central character.

4 [Complementary series] π(μ_{1}, μ_{2}) such that μ_{1}(x)\overline{μ_{2}(x)} = 1 for all x ∈ F^{×} and (μ_{1}·μ_{2}^{−1})(x) = |x|^{σ} with 0 < σ < 1.

There’s an invariant hermitian pairing on (π, V) ⟺ there’s a G-equivariant map \overline{π} → π̌.

Consider a principal series and its complex conjugate \overline{π(μ_{1}, μ_{2})} = π(\overline{μ}_{1}, \overline{μ}_{2}). Question: When is this the same as π(μ_{1}, μ_{2})ˇ = π(μ_{1}^{−1}, μ_{2}^{−1})? Obviously, this happens when μ_{1}\overline{μ}_{1} = μ_{2}\overline{μ}_{2} = 1 or μ_{1}\overline{μ}_{2}μ_{2}\overline{μ}_{1} = 1. In the former case, this just means each μ_{i} is a unitary character; we now consider the latter, which for us is still unfamiliar territory.

Assume that μ(x) = μ_{1}(x) · μ_{2}(x)^{−1} = |μ_{1}(x)|^{2} = |x|^{σ}, σ > 0. [For if σ > 0, then μ_{1} = μ_{2} unitary.] We can write down an intertwining operator; consider the intertwining operator

[figure: B(μ_{1}, μ_{2}); A; B(μ_{2}, μ_{1}) = B(\overline{μ_{1}}^{−1}, \overline{μ_{2}}^{−1}); ϕ; Aϕ : g ↦ ∫_{F} ϕ(w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)dx]

Exercise This integral really converges, as ϕ(w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) ≫ |x|^{−σ−1} for |x| ≫ 0. And

this last is from a simple computation, w ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & −x^{−1} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}−x^{−1} & 0 \\ 0 & −x \end{matrix} ) ( \begin{matrix}1 & 0 \\ x^{−1} & 1 \end{matrix} ).

Exercise Aϕ ∈ B(\overline{μ}_{1}^{−1}, \overline{μ}_{2}^{−1}). Now, write down a hermitian pairing.

⟨ϕ_{1}|ϕ_{2}⟩ = c ∫_{B(F)\G(F)}(Aϕ_{1})(g)\overline{ϕ_{2}(g)}dg

= c ∫_{F} A(ϕ_{1})(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))\overline{ϕ(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))}dx

Here, c is some uniquely correct scaling factor in ℂ^{×}. [The c is what makes it G-equivariant.] Exercise You can work with the integrals and change it to something like

⟨ϕ_{1}|ϕ_{2}⟩ = c ∫ ∫ ϕ(w^{−1} ( \begin{matrix}1 & x + y \\ 0 & 1 \end{matrix} ))\overline{ϕ_{2}(w^{−1} ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))}|y|^{σ}dxd^{×}y.

We started with a principal series, assume it’s not induced from unitary characters. We want to know whether it’s unitary. We wrote down an obvious condition, which we henceforth assumed. Now we assume there’s a hermitian pairing; the representation π(μ_{1}, μ_{2}) is unitary if and only if we can find some c ∈ ℂ^{×} so that the pairing is positive definite.

We’ll need a famous harmonic analysis theorem, Bochner[?]’s Theorem, which will get us out of the hole.

Classically, a function f is positive definite if ∑ a_{n}\overline{a}_{m}f(x_{n} − x_{m}) ≥ 0 for any finite set of points x_{i} and complex numbers a_{i}.

Theorem [Bochner] A continuous positive definite function is the Fourier transform of a positive semidefinite measure on the dual group.

In our context, we’re looking at something like an integral ∫ f_{1}(x + y)\overline{f_{2}(x)}dx. But this is clearly a convolution, f_{1} ∗ f̌_{2}, where f̌_{2}(x) = \overline{f_{2}(−x)}.

Anyways, π(μ_{1}, μ_{2}) is unitary ⟺ there’s a c ∈ ℂ^{×} so that this pairing is positive definite ⟺ FT(|y|^{σ}d^{×}y) is a positive definite measure.

So what’s the Fourier transform of |y|^{σ}d^{×}y? Now, classically for the zeta function, we were able to relate ∫ f(x)|x|^{s}d^{×}x and ∫ f̂(x)|x|^{1−s}d^{×}x. With σ fixed, this functional equation tells us that the Fourier transform of this is, up to a constant, that second integrand.

Specifically, the functional equation for abelian local zeta functions says FT(|y|^{σ}) = (const)|x|^{1−σ}d^{×}x. And 1 − σ > 0 is exactly the same as saying |x|^{1−σ}d^{×}x extends to a measure on F.

And that, essentially, is the proof.

We are, at least temporarily, done with the non-archimedean case. We now briefly treat the archimedean situation.

Now we’re dealing with two possible groups; GL_{2}(ℝ) and GL_{2}(ℂ). We’ll be talking about the representation theory of these two groups. These are nice, easy, reductive Lie groups; as such, they admit a variety of approaches. A variety of textbooks, even.

Prefatory remark: In the archimedean case, there’s nothing like a supercuspidal representation. So you should never expect something like matrix coefficients with compact support, ’cause a real analytic function with compact support is dead. A corollary of which is that you can build these models very concretely, especially if you just want the characters.

We reiterate: There are no supercuspidal representations.

In general, for a real semisimple Lie group, there’s a discrete [modulo center] family of representations; these are the discrete series representations. Essentially, the matrix coefficients are L^{2} modulo the center.

Actually, you get a discrete series on a connected real semisimple Lie group exactly when, up to conjugation, there’s a unique maximal compact subgroup.

So GL_{2}(ℝ) has discrete series, but GL_{2}(ℂ) doesn’t; this makes the latter easier than the former.

Admissible representations Let G be a reductive real Lie group. For us, for now, we’ll define this as G(ℝ) where G/ℝ is reductive as an algebraic group. Such a G has a maximal compact subgroup K. [If we’re being really careful, we’ll work with G(ℝ)^{0}, the connected component of identity.] An admissible representation of G on a Banach space V is one such that each irreducible representation of K occurs with finite multiplicity.

i. If we look at V_{K} ⊂ V the K-finite vectors [the sum of all the finite-dimensional representations of K inside V], then every V_{K} ⊆ V^{∞} the smooth vectors.

Consequently, V_{K} is a (𝔤, K)-module.

ii. Theres a correspondence between G-invariant closed subspaces of V and (𝔤, K)-invariant subsapces of K. To go ⇝, take K-finite vectors; to go ←, take closure.

We’ve got a bunch of facts.

• Any irreducible unitary representation of G is automatically admissible.

• If (π_{i}, V_{i}) is a pair of irreducible unitary representations of G, such that (π_{1}, 1)_{K} ≅ (π_{2}, V_{2})_{K} [they’re isomorphic as (𝔤, K)-modules] then π_{1} ≅ π_{2} as unitary representations.

Theorem Let X be a (𝔤, K)-module. Among other things, this means X is locally K-finite. Then

i. X is admissible.

ii. X is isomorphic to the Harish-Chandra module of an irreducible representation of G on a Hilbert space, such that elements of K operate via unitary operators.

Now we’ll consider GL_{2}(ℝ) and GL_{2}(ℂ); let F be ℝ or ℂ. Then the Hecke algebra is H_{F}, distribu-

tions supported at \{ \begin{matrix}( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) & F = ℂ \\ ( \begin{matrix}±1 & 0 \\ 0 & 1 \end{matrix} ) & F = ℝ \end{matrix} .

In the first case, this is just the universal enveloping algebra U(𝔤); for F = ℝ, this is two copies of it. You can think of this as invariant differential operators on each component. Somehow, if you use only differentiation, then you can only hope to control that connected component; it never gives you information outside the component. Lie theory tells you that if you know how the algebra acts, then you can recover the representation.

We still have induced representations. They are B(μ_{1}, μ_{2}) = \{G(F) →^{f} ℂ|f(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) g) =

μ_{1}(a)μ_{2}(d)|\frac{a}{d} |F^{\frac{1}{2}} f(g)\}, where all the f are assumed smooth, K-finite. The associated representation is ρ(μ_{1}, μ_{2}).

Theorem Every irreducible admissible representation of H_{F} is isomorphic to a subrepresentation of a ρ(μ_{1}, μ_{2}).

We’ll just leave that as a statement for now.

Theorem Suppose F = ℝ. Then ρ(μ_{1}, μ_{2}) is irreducible, except when μ =^{def} μ_{1}μ_{2}^{−1} has the form μ(t) = t^{n} sgn(t). If μ(t) = t^{n} sgn(t), then:

• If n > 0, then B(μ_{1}, μ_{2}) contains exactly one non-trivial irreducible subrepresentation. This representation is infinite dimensional. The quotient is finite-dimensional and irreducible, of dimension n.

• If n < 0, then ρ(μ_{1}, μ_{2}) contains exactly one non-trivial irreducible subrepresentation; it is finite dimensional, and its dimension is |n|. In fact, this subrepresentation consists of

\{G(F) →^{ϕ} ℂ|ϕ(( \begin{matrix}a & b \\ c & d \end{matrix} )) = μ_{1}∘det(g)·|det g|^{\frac{1}{2}} f(c, d) where f is a homogeneous polynomial

of degree |n| − 1}. The quotient is infinite-dimensional.

Use π(μ_{1}, μ_{2}) to denote these irreducible representations. To be compatible with Langlands stuff, we’ll think of this as π(μ_{1} ⊕ μ_{2}); the direct sum of two representations of the Weil group. Then Langlands predicts a representation os GL_{2}(F).

So, over ℝ, take n > 0 and look at π(ω_{\frac{1}{2}} t^{n}, ω_{− \frac{1}{2}} ) = Symm^{n}(std). This isn’t surprising; the representation of dimension n of SL_{2} is usually the n^{th} symmetric power of the standard one.

Remarks on B(μ_{1}, μ_{2}). Recall that B(μ_{1}, μ_{2}) is \{GL_{2}(ℝ) →^{f} ℂ|f is smooth, K-finite, f(( \begin{matrix}a & b \\ 0 & d \end{matrix} ) g) =

μ_{1}(a)μ_{2}(d)|\frac{a}{d} |ℝ^{\frac{1}{2}}f(g)\}.

What’s a typical element? Well, B(ℝ) is the upper-triangular matrices; K^{+} is SO_{2}(ℝ); and B(ℝ) · SO_{2}(ℝ) = G(ℝ). Note that B(ℝ) isn’t connected [determinant]. Anyways, this is the Iwasawa decomposition of G. Actually, K should be O_{2}, instead of SO_{2}. So, f is determined by f|_{K^{+}},

and K^{+} is a 1-dimensional torus group. Consider f_{n} : ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) ↦ e^{inθ}. Then every

f ∈ B(μ_{1}, μ_{2}) is a linear combination of these f_{n}.

Fact F = ℂ.

“Theorem”

• ρ(μ_{1}, μ_{2}) is irreducible, except when μ =^{def} μ_{1}μ_{2}^{−1} = t^{q} · \overline{t}^{q} with p · q > 0.

0 Every infinite-dimensional irreducible admissible representation of GL_{2}(ℂ) is a ρ(μ_{1}, μ_{2}).

1 μ = t^{p}\overline{t}^{q}, p, q < 0, then B(μ_{1}, μ_{2}) contains an irreducible finite-dimensional subrepresenta-

tion, namely \{G(F) →^{ϕ} ℂ|ϕ(( \begin{matrix}a & b \\ c & d \end{matrix} )) = μ_{1}(det g) · |det g|_{ℂ}^{− \frac{1}{2}}Φ_{1}(c, d)\overline{Φ_{2}(c, d)}\} where Φ_{1} is a

homogeneous polynomial of degree |p| − 1, and Φ_{2} is homogeneous of degree |q| − 1; and the quotient is irreducible.

2 If p, q > 0, then ρ(μ_{1}, μ_{2}) contains an irreducible infinite-dimensional subrepresentation; the quotient is finite-dimensional and irreducible.

For case 1 and 2, we’d like to write these infinite-dimensional subquotients σ(μ_{1}, μ_{2}) π(ν_{1}, ν_{2}) for some ν_{i}. It turns out that the right thing to do is

σ(μ_{1}, μ_{2}) = π(ν_{1}, ν_{2})
with ν_{1}(t) = \overline{t}^{−q}μ_{1}(t)
ν_{2}(t) = \overline{t}^{q}μ_{2}(t)

For then

(ν_{1}ν_{2}^{−1})(t) = t^{q} · \overline{t}^{−q}.

As an aide memoire we offer the following.

Symm^{⊗n}(std) ⊗ Symm^{⊗m}(\overline{std}) = π(|·|ℂ^{\frac{1}{2}}z^{n}\overline{z}^{m}, |·|_{ℂ}^{− \frac{1}{2}})

What are the equivalence relations? If F = ℝ, there aren’t many. π(μ_{1}, μ_{2}) ≅ π(μ_{2}, μ_{1}), and that’s all. For the discrete series reprsentation, σ(μ_{1}, μ_{2}) ≅ σ(μ_{2}, μ_{1}). Additionally, σ(μ_{1}, μ_{2}) ≅ σ(μ_{1} · sgn, μ_{2} · sgn). And these are all.

If F = ℂ, we don’t have the sign character; and so π(μ_{1}, μ_{2}) ≅ π(μ_{2}, μ_{1}), and that’s it.

So now we have a complete list of possible irreducible admissible representations, including the finite-dimensional ones.

Remarks on discrete series and their parameters for GL_{2}(ℝ). Now, there’s an obvious decomposition

GL_{2}(ℝ) = SL_{2}(ℝ)^{±} × ℝ_{+}^{×}.

Now we write down a representation D_{l}^{+} of SL_{2}(ℝ). Here, l ∈ ℕ is some parameter. The space is,

roughly speaking, {𝔥 →^{f} ℂ| holomorphic with }. Actually, we take the f with ∥f∥^{2} < ∞. The norm is the Petersson inner product, -ish, ∥f∥^{2} =^{def} ∫ ∫_{𝔥} |f(z)||y|^{l+1} \frac{dxdy}{y^{2}} . This is a good inner product to use on modular forms of weight l + 1, which thus form a Hardy space.

Possibly at odds with the usual yoga of modular forms, we’ll have G act on the left. The action

will be as follows. If ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℝ), the action is

f  z ↦ (bz + d)^{−l−1}f(\frac{az + c}{bz + d})

Then set

D_{l} =^{def} Ind_{SL_{2}(ℝ)}^{SL_{2}^{±}(ℝ)} D_{l}^{+}.

Then we can look at D_{l} ⊗ ω_{t}, ω_{t} = |det(·)|^{t}; this is an irreducible discrete series representation. Moreover, every discrete series representation is of this form.

What’s its central character? It’s obviously controlled by the ω_{t} part. So an element of the center gets sent, up to absolute value, to |·|^{2t}. We just need to figure out the sign. We claim that the central characer is actually sgn^{l+1} |·|^{2t}.

Think about SL_{2}(ℝ) ↪ SL_{2}^{±}(ℝ).

Notes gratefully copied from Roger Lin....

More on discrete series Given D_{l} ⊗ ω_{t}, we want to relate this with σ(μ_{1}, μ_{2}). It’s known that D_{l} ⊗ ω_{t} ≅ σ(μ_{1}, μ_{2}) for some μ_{i}, at least if l > 0.

Look at the central character. On the left-hand side we get (sgn)^{l+1}|·|^{2t}; on the right-hand side we get μ_{1}, μ_{2}, with μ_{1}μ_{2}^{−1}t^{n} sgn for n > 0¿ Thus, n = l, i.e.,

σ(μ_{1}, μ_{2}3) = σ(|·|^{t+ \frac{l}{2}} sgn^{l+1}, |·|^{t− \frac{l}{2}} ).

Philosophy/Langlands correspondence At least in the 2-dimensional case, we get

[figure: (W_{F} → GL_{2}(ℂ)); (representations of GL_{2}(F))]

In the more general situation we get

[figure: (W_{F} → ^{L}G admissible ); (packets of representations of GL_{2}(F))]

Here, of course, W_{F} is the Weil group and^{L}G is the Langlands d ual.

F’rinstance, if G = ℍ^{×} the Hamiltonian quaternions, then^{L}G is essentially GL_{2}(ℂ) with a twist.

The upshot is that given an irreducible representation of ℍ^{×}¡ we have an admissible representation W_{F} → ^{L}G, and thus we get W_{F} → GL_{2}(ℂ); we attach an admissible irreducible representation of GL_{2}(F).

ℍ^{×}/center is a compact symplectic group, or a complexified SL_{2}(ℂ). Topologically, it’s ≅ S^{3}. But anyways, we understand all representations of SL_{2}; end of story.

Another parameter For every representaiton σ of ℍ^{×} we get a representation π(σ) of GL_{2}(ℝ) in the discrete series. Logical question: what is π(σ)?

The representation σ has a central character ω_{σ}(t), t ∈ ℝ^{×}. If π(σ) ≅ pi(μ_{1}, μ_{2}) then they must have the same central character, i.e., ω_{σ} = μ_{1}μ_{2}, μ = μ_{1}μ_{2}^{−1} = t^{n} sgn.

σ is parametrized up to a central character by 0, 1, 2, 3 ⋯ with associated deg σ = 1, 2, 3, 4...; and both parameters determine D_{l} ⊗ ω_{t}.

Weil group We’ll define the Weil groups of the various locally compact fields. If F = ℂ, then W_{F} = F^{×}.

For F = ℝ, there’s a non split extension

[figure: 1; ℂ^{×}; W_{ℝ}; {−1, 1}; 1]

Set theoretically, W_{ℝ} = ℂ^{×} ⊔ jℂ^{×}, where j ∈ W_{ℝ} is such that j^{2} = −1 ∈ ℂ^{×} and jzj^{−1} = \overline{z} for z ∈ ℂ^{×}. Note the isomorphism

[figure: Wℝ^{ab}; ≅; ℝ^{×}; [j]; −1; [\sqrt{a}]_{a>0}; a]

Langlands parametrization

• F ≅ ℂ. Then every 2-dimensional representation of W_{F} = F^{×} is iomprhic to μ_{1} ⊕ μ_{2}, where μ_{i} are characters of W_{F} = F^{×}. Under the Langlands correspondence, this attaches to π(μ_{1}, μ_{2}) [the irreducible one, or the finite dimensional subquotient if ρ(μ_{1}, μ_{2}) is reducible].

• F ≅ ℝ. Then W_{ℝ} has a large normal subgroup ℂ^{×}, and Mackey’s theorem applies. SUppose σ is a 2-dimensional representation W_{ℝ} →^{σ} GL_{2}(V), dim_{ℂ} V = 2. Then the only possibilities are

1. σ ≅ μ_{1} ⊕ μ_{2}; then local Artin reciprocity gives

[figure: W_{ℝ}; μ_{i}; Wℝ^{ab}; ≅; ℝ^{×}; ℂ^{×}]

2. σ|_{\overline{F}}_{×} ≅ θ ⊕ θ^{τ}. Here, θ = |·|_{ℂ}^{r}z^{m}\overline{z}^{n}, where z : \overline{F} →^{∼} ℂ, r ∈ ℂ, m, n ∈ ℤ_{≥0}, min(m, n) = c,

max(m, n) > 0; and θ^{τ} is conjugation by j. Then σ = IndW_{F} ; \overline{F} (θ) ≅ IndW_{F} ; \overline{F} (θ^{τ}), and π(σ) is a discrete series representaiton.

MA 620 – Langlands Program  28 February 1997

Let F be an Archimedean local field. Whittaker models Theorem Let π be an admissible irreducible infinite dimensional representation [i.e., an admissible irreducible (𝔤, K)-module] of GL_{2}(F), and let ψ : F → ℂ^{×} be a nontrivial additive character of F. Then there’s a unique space W_{ψ}(F) of functions on GL_{2}(F) so that

1. For all W ∈ W_{ψ}(π), W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x)W (g) for all g ∈ GL_{2}(F) and all x ∈ F.

2. Every W ∈ W_{ψ}(π) is smooth, K-finite, and W (( \begin{matrix}t & 0 \\ 0 & 1 \end{matrix} )) = O(|t|_{F}^{N}) as |t|_{F} → ∞. [N may depend on W.]

3. 𝔤 Operates on W_{ψ}(π) via right translation, making W_{ψ}(π) a (𝔤, K)-module isomorphic to π. Of course, this also tells us the existence of Kirillov models. Functional equations The general form is: Well, fix, F^{×} →^{χ} ℂ^{×}; set

ζ(g, χ, W, s) = ∫_{F^{×}} W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x)|x|_{F}^{s− \frac{1}{2}} d^{×}x.

Then

ϵ(π ⊗ χ, ψ, s)\frac{ζ(g, χ, s)}{L(π ⊗ χ, s)} = \frac{L(wg, χ^{−1}, W, 1 − s)}{L(ω_{π}^{−1} ⊗ χ^{−1} ⊗ π, 1 − s)}.

In terms of ξ in the Kirillov model, let ξ(x) = W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} )) ∈ K_{ψ}(π). Then

ζ_{ξ,χ}(s) = ∫_{F^{×}} ξ(x)χ(x)|x|_{F}^{s− \frac{1}{2}} d^{×}x,

so that the functional equation is

ϵ(π ⊗ χ, ψ, s)\frac{ζ_{ξ,χ}(s)}{L(π ⊗ χ, s)} = \frac{ζ_{π(w)ξ,χ^{−1}}(1 − s)}{L(π ⊗ ω_{π}^{−1} ⊗ χ^{−1}, 1 − s)}.

Jeff Achter  46  Ching-Li Chai

[Of course, π ⊗ ω_{π}^{−1} = π̌.] Next, we’d like to try for a more explicit form for the ϵ and L-factors. First, we recall the abelian case. Let F = ℝ. Then set ψ : F → ℂ^{×}, x ↦ e^{2π\sqrt{−1}ux}. There are Γ factors

Γ_{ℝ}(s) = π^{\frac{s}{2}} Γ(\frac{s}{2})
Γ_{ℂ}(s) = 2(2π)^{−s}Γ(s)
= Γ_{ℝ}(s)Γ_{ℝ}(s + 1)

Consider

[figure: ℝ^{×}; μ; ℂ^{×}; t; |t|_{ℝ}^{r} sgn(t)^{m}, m ∈ {0, 1}]

Then

L(μ, s) = Γ_{ℝ}(s + r + m).

For the complex case, define μ by

[figure: ℂ^{×}; μ; ℂ^{×}; z; |z|_{ℂ}^{r}z^{m}\overline{z}^{n}, m, n ≥ 0, m · n = 0]

Here, z is an equivalence from F to ℂ. Then

L(μ, s) = Γ_{ℂ}(s + r + m + n).

Anyways, for the real case, we have ϵ(μ, ψ, s) = (\sqrt{−1} sgn u)^{m}|u|_{ℝ}^{s+r− \frac{1}{2}}, where the left-hand side is defined with respect to the same self-dual measure; and {−1} is the same choice as in ψ(x) = e^{2π\sqrt{−1}ux}. Back in the complex case, ϵ(μ, ψ ∘ tr_{F,ℝ}, s)μ(u)|u|_{ℂ}^{s− \frac{1}{2}} \sqrt{−1}^{m+n}.

In both situations,

L(π(μ_{1}, μ_{2}), s) = L(μ_{1}, s) · L(μ_{2}, s)
ϵ(π(μ_{1}, μ_{2}), ψ, s) = ϵμ_{1}, ψ, s)ϵ(μ_{2}, ψ, s)
L(π(Ind_{W\overline{F}}^{W_{F}} (θ)), s) = L(θ, s)
ϵ(π(Ind_{W\overline{F}}^{W_{F}} (θ)), ψ, s) = (\sqrt{−1} sgn u)ϵ(θ, ψ ∘ tr_{\overline{F} ,F} , s)

where the last two only make sense, of course, if F ≅ ℝ. Move on to the general theory of automorphic representations. Set up notation as follows:

• F a number field

• 𝔸 = 𝔸_{F} = ∏′_{v} F_{v}.

• (G𝔸) = GL_{2}(𝔸_{F}) ≅ ∏′_{v} GL_{2}(F_{v}), where the standard compact subgroup is GL_{2}(O_{v}).

An automorphic representation is some sort of representation of GL_{2}(𝔸_{F}). We’re supposed to study spectral decomposition of functions living in L^{2}(G(F)\G(𝔸_{F})). Actually,

L^{2}(G(F)\G(𝔸_{F})) = ∫_{ω} L^{2}(G(F)\G(𝔸_{F}), ω)

where Z(𝔸_{F}) operates via 𝔸_{F}^{×}/F^{×} →^{ω} ℂ_{1}^{×}.

MA 620 – Langlands Program  3 March 1997

Spectral decomposition: F is a global field, usually a number field. Let ω be a unitary character ω ∈ (F^{×}\𝔸_{F}^{×})^{∧}; then

L^{2}(G(F)\G(𝔸_{F})) = ∫_{ω} L^{2}(G(F)\G(𝔸_{F}), ω)dω.

Here, the group action of G(𝔸_{F}) is via right translation. Standard representation theory tells us that there’s an important subspace L_{0}^{2}(G(F)\G(𝔸_{F}), ω) ⊂ L^{2}(G(F)\G(𝔸_{F}), ω), the cuspidal

functions or cuspidal forms. It’s the set \{f : G(F)\G(𝔸_{F}) →?| ∫_{F \𝔸_{F}} T (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ))fdx = 0\},

where T is the right-translation action. In other words, ∫_{F \𝔸_{F}} f(xg)dx = 0 almost everywhere.

General Facts

• For every unitary idèle class character ω and every irreducible representation θ of K,

L_{0}^{2}(G(F)\G(𝔸_{F}), ω)(θ)

is finite-dimensional. Especially, every irreducible subrepresentation of L_{0}^{2}(G(F)\G(𝔸 − F), ω) is admissible.

The proof uses classical reduction theory. [Which deals with fundamental domains and that sort of thing.]

• L_{0}^{2}(G(F)\G(𝔸_{F}), ω) is a discrete direct sum of irreducible subrepresentations.

The proof relies on the fact that T (f), where f is a Schwartz function on G(𝔸_{F}), is compact.

• F a global function field, then every cuspidal automorphic form^{19} [element of L_{0}^{2}(G(F)\G(𝔸_{F}), ω)] has compact support mod G(F)Z(𝔸_{F}).

• If F is a number field, then every cuspidal automorphic form in L^{2}(G(F)\G(𝔸_{F}), ω) decreases rapidly.

Let 𝔖 be a Siegel set; then ∪_{g∈G(F)}g𝔖 = G(𝔸_{F}); and {g ∈ G(F)|g𝔖 ∩ 𝔖 ≠ ∅} is finite. We’ll take

it to be of the following form: \{( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & d \end{matrix} ) k||b| ≤ c_{0}, |\frac{a}{d} | ≥ c_{1}, k ∈ K\}. The condition on

|\frac{a}{d} | essentially means we go to infinity in only one way.

Anyways, with this setup, as |\frac{a}{d} | → ∞, for any g_{0} ∈ G(F), f(g_{0} ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & d \end{matrix} ) k) = O(|\frac{a}{d} |^{−N}).

So the blowup isn’t any worse than polynomial.

19Terminology: An L^{2} automorphic form is a L^{2} function on G(F)\G(𝔸_{F}) which is K-finite and Z(𝔸_{F})-finite.

Sometimes, people also insist that, for each Archimedean component v, it’s also finite under Z(U(𝔤_{v})) the center of the universal enveloping algebra.

Theorem ^{20} Every irreducible admissible representation π of G(𝔸_{F}) is a restricted tensor product

π = ⊗′_{v}π_{v}

of admissible irreducible representations.

What does ⊗′ mean, anyway? For this to make sense we need to know that, for almost all v, π_{v} is spherical.^{21}

Now suppose that S is a finite set of primes, such that π_{v} is spherical for all v ∉ S. For each v ∈ S, choose a vector w_{v} ∈ π_{v} fixed under K_{v} = G(O_{v}). [For unitary representations, choose it with ∥w_{v}∥ = 1.] The the space ⊗′π_{v} is generated by ⊗_{v}u_{v}, where u_{v} ∈ π_{v}¡ u_{v} = w_{v} for almost all v; differently put, it’s

∪_{T ⊃S,|T |<∞}{v ⊗ ⊗_{v∉S}w_{v}|v ∈ ⊗_{v∈T} }.

Fourier expansions Let f ∈ L^{2}(G(F)\G(𝔸_{F}), ω). Consider the function x ↦ f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g), g

some fixed element in the adelic group. Anyways, this is a function on F \𝔸_{F}, a compact group. So we can take its Fourier transform, whose Fourier expansion is

f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ∑_{ξ∈F} ϕ_{ξ}(g)ψ(ξx).

where ψ is a fixed nontrivial ψ : F \𝔸_{F} → ℂ_{1}^{×}. The Fourier coefficients are

ϕ_{ξ}(g) = ∫_{F \𝔸_{F}} f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)\overline{ψ(ξx)}dx.

Let W = W_{f} = ϕ_{1}: W (g) = ∫_{F \𝔸_{F}} f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)\overline{ψ(x)}dx. By its very construction, you see that

W (( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) g) = ψ(y)W (g). And that’s the functional equation satisfied by functions in the

Whittaker model.

Moreover, ϕ_{0}(g) = ∫_{F \𝔸_{F}} f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)dx vanishes if f is a cusp form.

20Or fact, anyway

21If v is nonarchimedean, this means the maximal compact subgroup has a fixed vector, i.e., the representation is an unramified principal series. So π_{v} = π(μ_{1v}, μ_{2v}), μ_{iv} unramified, μ_{1}μ_{2}^{−1} ≠ |·|_{v}^{±1}.

MA 620 – Langlands Program  3 March 1997

Exercise For ξ ∈ F^{×}, ϕ_{ξ}(g) = W (( \begin{matrix}ξ & 0 \\ 0 & 1 \end{matrix} ) g). Consequently, f(g) = ϕ_{0}(g) + ∑_{ξ∈F^{×}} W_{f}(( \begin{matrix}ξ & 0 \\ 0 & 1 \end{matrix} ) g.

Remarks about Whittaker models

• Let F be archimedean. We have the following estimate. For all W ∈ W(π),

W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} )) = O(|x|^{−N})∀N

This is an archimedean analogue of the corresponding fact that, in non-archimedean Kirillov models, the functions vanish as the absolute value gets very big.

• Spherical representations. Let F be non-archimedean. We have an unramified principal series representation π(μ_{1}, μ_{2}), μ_{i} : F^{×} → ℂ^{×} unramified, μ_{1}μ_{2}^{−1} ≠ |·|_{F}^{±}. Our addiitve character is ψ, with some conductor d = d(ψ); so 𝔭^{−d} is the largest ideal on which ψ is trivial.

Then W(π) contains a unique function right invariant under K = G(O_{F}), such that the

value at ( \begin{matrix}ϖ^{−d} & 0 \\ 0 & 1 \end{matrix} ) is 1. Specifically, define a function by

W^{0}(( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} )) = \{ \begin{matrix}0 & x ∉ 𝔭^{−d} \\ |x · ϖ^{d}|^{\frac{1}{2}} ∑_{i,j≥0,i+j=v(x)+d} μ_{1}(ϖ^{i})μ_{2}(ϖ^{j}) & x ∈ 𝔭^{−d} \end{matrix} .

Global Whittaker models Let π = ⊗′_{v}π_{v} be an admissible representation of G(𝔸_{f}). Then W_{ψ}(π) = ⊗′_{v}W_{ψ_{v}}(π_{v}) with respect to the W_{π_{v}}^{0}’s.

The following comes pretty easily from the existence and uniqueness of local Whittaker models.

Proposition W_{ψ}(π) is the unique model of π in the space of functions on G(𝔸_{F}) which are real analytic [have moderate growth], K-finite, and satisfying the right functional equation

W (( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g) = ψ(x)W (g)

for all x ∈ 𝔸_{F}, g ∈ G(𝔸_{F}). And

W (( \begin{matrix}x & 0 \\ 0 & 0 \end{matrix} ) g) = ω_{π}(x)W (g).

In fact,

W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g) = O(|x|^{−N})∀N, W ∈ W(π).

Recall:

f(g) = ϕ_{0}(g) + ∑_{ξ∈F^{×}} W_{f} ( \begin{matrix}ξ & 0 \\ 0 & 1 \end{matrix} ) g)
where ϕ_{0}(g) = ∫_{F \𝔸_{F}} f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)dx
W_{f}(g) = ∫_{F \𝔸_{F}} f(( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) g)\overline{ψ(x)}dx

Moreover, W satisfies

W_{f}(( \begin{matrix}1 & y \\ 0 & 1 \end{matrix} ) g) = ψ(y)W_{f}(g)

We have the fairly surprising

Theorem [Multiplicity one] Every irreducible cuspidal automorphic representation occurs with multiplicity one in L_{0}^{2}(G(F)\G(𝔸_{F}), ω). This follows from our [somewhat blackboxed] theory of Whittaker models. [Multiplicity = Plancherel measure? Whatever.]

Euler products and L-functions Let π = ⊗′_{v}π_{v} be a unitary admissible representation. We know immediately that each π_{v} is unitary. For almost all v, π_{v} = π(μ_{1v}, μ_{2v}), where π_{iv} are un-

ramified; |μ_{1,v}(x)| = |x|v^{\frac{σv}{2}} , |μ_{2,v}(x)| = |x|_{v}^{− \frac{σv}{2}} , 0 ≤ σ_{v} ≤ 1.

Now set

L(π ⊗ χ, s) =^{def} ∏_{v} L(π_{v} ⊗ χ_{v}, s).

This product converges for Re(s) ≫ 0.

Theorem Let π = ⊗′π_{v} be a cuspidal automorphic representation of G(𝔸_{F}). Then for every Hecke character χ : F^{×}\𝔸_{F}^{×} → ℂ^{×} the L-function L(π ⊗ χ, s) extends to an entire function, bounded on every vertical strip, and satisfies the functional equation

L(π ⊗ χ, s) = ϵ(π ⊗ χ, s)L(π̌ ⊗ χ^{−1}, s)

where ϵ(π ⊗χ, s) is actually a finite product ∏ ϵ(π_{v} ⊗χ_{v}, s) [we only have to worry about v’s where ψ or χ ramify].

Proof Let f ∈ π be a cusp form. Then f ↭ W; f(g) = ∑_{ξ∈F^{×}} W (\binom{0}{0 1} g). Consider the

zeta integral

ζ(g, χ, W, s) = ζ_{f}(g, χ, s) =^{def} ∫_{F^{×}\𝔸_{F}^{×}} f(( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x)|x|^{s− \frac{1}{2}} d^{×}x
= ∫_{𝔸_{F}^{×}} W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x0|x|s − \frac{1}{2}d^{×}x

From this form, it’s clearly an entire function; f decreases rapidly at infinity. Similarly, as only Re(s) seems to matter, it’s bounded on vertical strips.

And we see that the same is true for L(π ⊗ χ, s), as the L-function is controlled by the [local] zeta functions.

It remains to show that

ζ_{f}(g, χ, s) = ζ_{f}(wg, χ^{−1} ⊗ ω_{π}^{−1}, 1 − s).

Well,

ζ_{f}(wg, χ^{−1} ⊗ ω_{π}^{−1}, 1 − s) = ∫_{F^{×}\𝔸_{F}^{×}} f(( \begin{matrix}x & 0 \\ 0 & w \end{matrix} ) g)χ(x)^{−1}ω_{π}(x)^{−1}|x|^{\frac{1}{2} −s}d^{×}x
= ∫_{F^{×}\𝔸_{F}^{×}} f(w ( \begin{matrix}x & 0 \\ 0 & w \end{matrix} ) g)χ(x)^{−1}ω_{π}(x)^{−1}|x|^{\frac{1}{2} −s}d^{×}x
= ∫_{F^{×}\𝔸_{F}^{×}} ω_{π}(x)f(( \begin{matrix}x^{−1} & 0 \\ 0 & 1 \end{matrix} ) g)ω_{π}(x)^{−1}χ(x)^{−1}|x|^{\frac{1}{2} −s}d^{×}x
= ∫_{F^{×}𝔸_{F}^{×}} f(( \begin{matrix}y & 0 \\ 0 & 1 \end{matrix} ) y)χ(y)|y|^{s− \frac{1}{2}} d^{×}y
= ζ_{f}(g, χ, s)

◇ Last time, we had: Let π = ⊗′_{v}π_{v} be an ireducible cuspidal automorphic representaiton of G(𝔸_{F}). Then L(π, s) is entire and bounded on vertical strips; so is L(π ⊗ χ, s) for all Hecke characters χ, and satisfies

L(π, s) = ϵ(π, s) · L(π̌, 1 − s).

A natural question might be the following. Given an admissible irreducible representation π = ⊗′_{v}π_{v}, when is it automorphic?

We already know a necessary condition for π cuspidal. Specifically, the L-function should be entire, bounded on every vertical strip, and it should satisfy a functional equation L(π, s) = ϵ(π, s)L(π̌, 1 − s) as usual. Depending on how you think of the last term, it could be the dual representation; it’s also π tensored with the inveres of its central character [although this last interpretation depends on GL_{2}].

The converse theorem for GL_{2}, due to Weil, says that something like this is always true. Specifically, if π = ⊗′_{v}π_{v} is an admissible irreducible representation of G(𝔸_{F}) such that for all Hecke characters π L(π ⊗ χ, s) is entire, is bounded in every vertical strip [this also means the same is true for π̌] and satisfies a functional equation of the right sort,

L(π ⊗ χ, s = ϵ(π ⊗ χ, s)L(π̌ ⊗ χ^{−1}, 1 − s),

then π is a cuspidal automorphic representation.

Example ^{22} Given K_{v} a quadratic extension of F_{v} and a character θ_{v} : K_{v}^{×} → ℂ^{×}.... We’ll think of K_{v}^{×} as res_{K_{v},F_{v}} 𝔾_{m}(F_{v}), a two-dimensional torus over F_{v} we’ll call T_{v} = T^{K_{v}/F_{v}}(F_{v}).

Like any torus, T_{v} has a [Langlands] dual group

[figure: 1; ^{L}T^{0}; ^{L}T_{v}; W_{F}; 1]

Here,^{L}T^{0}, the connected dual group, is the complex torus whose cocharacter group is the character group of T. And the thing in the middle is a semidirect product with the canonical action of W_{F}.

Now to θ_{v} corresponds a homomorphism W_{F} → ^{L}T_{v}.

Well, we have

22We’ll look at this in more detail later.

W_{F}  \overline{}^{L}T_{v}

composition

GL_{2}(ℂ) × W_{F}

and our homomorphism is one over W_{F}.

If you believe in the Langlands correspondence, this should give you a representation of GL_{2}(F_{v}) and a good understanding of its local L- and ϵ-factors.

Similarly, globally, if θ : 𝔸_{K}^{×}/K^{×} → ℂ^{×}, we can get an admissible irreducible representation π(θ) of G(𝔸_{F}). [K/F quadratic still.] To check π(θ) is cuspidal automorphic, just use the converse theorem.

Proof [of converse theorem] 1 Let W = W(π) be the global Whittaker model of π.^{23}^{24} For W ∈

W, let f_{W} (g) = ∑_{ξ∈F^{×}} W (( \begin{matrix}ξ & 0 \\ 0 & 1 \end{matrix} ) g); this is our candidate for the corresponding automorphic

form. At this point, we’ll state some properties as exercises.

Exercise

i. This formal sum can be replaced by a sum ∑_{ξ∈F^{×}∩𝔞}, where 𝔞 is a fractional ideal, the largest ideal on which ψ is nontrivial, more or less.

ii. Convergence.

2 Consider ζ(g, χ, W, s) = ∫_{𝔸_{F}^{×}} W (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x)|x|^{s− \frac{1}{2}} d^{×}x = ∫_{F^{×}\𝔸_{F}^{×}} f_{W} (( \begin{matrix}x & 0 \\ 0 & 1 \end{matrix} ) g)χ(x)|x|^{s− \frac{1}{2}} d^{×}x.

Assume W is a product of local terms. This zeta function diffs from L(π ⊗ χ, s) by at most finitely many factors [since most of the terms will be W_{0}]. The conclusion is that ζ(g, χ, W, s) is entire, bounded on vertical strips, and satisfies the functional equation

ζ(g, χ, W, s) = ζ(wg, χ^{−1}ω_{π}^{−1}, W, 1 − s).

3 We derive some condition; then the inverse Fourier transform tells us f_{W} (wg) = f(g), and consequently [do some simple analysis] the zero’th Fourier coefficient vanishes; f ∈ L_{0}^{2}(G(F)\G(𝔸_{K}), ω_{π}).

23To say that π is automorphic is to say you can realize it in the space of functions of some special form; so to start, you’d damned well better put it in a space of functions. So we might as well start with the Whittaker model, functions on the adelic group. And then we’ll prove these functions are of the right sort.

24Remember from Fourier stuff that the functions which come up in the Whittaker model end not to be automorphic themselves, although the first Fourier coefficient does determine the automorphic function.

Last time we were at the converse theorem; if you have an admissible irreducible representation of an adelic group, then it’s an automorphic cuspidal representation^{25} if and only if the L-functions and their twists satisfy the right functional equations. We did this by looking at the Whittaker model; then the issue is whether elements in the Whittaker model, and the sums thereof which look like they might be cusp forms, really are cusp forms. In other words, they’re invariant under the rational points of the group; as it’s obvious for the Borel, we need only verify for the Weyl element.

The Shale-Weil representation Before we get into the thick of things – the main reference is

Weil, 1964[?], Sur certaines groupes des operateurs unitaires, first paper in volume three of his complete works.

There’s a group called the Heisenberg group. Let V be a free module over k of finite rank, where k for us will be a local field or the ring of adeles of a global field. [Weil actually took V a locally compact abelian group.]

For the moment, write V^{∗} for the Pontryagin dual Hom_{cont}(V, ℂ_{1}^{×}). The Heisenberg group is a certain extension

[figure: 1; ℂ_{1}^{×}; H(V); V × V^{∗}; 1]

As a set,

H(V) = {(u, u^{∗}, t)|u ∈ V, u^{∗} ∈ V^{∗}, t ∈ ℂ_{1}^{×}}.

The model to think about is this. Consider L^{2}(V). For Φ ∈ L^{2}(V) there are a couple operations. First, we can translate by an element, as in (u, 0, 1) : Φ ↦ (x ↦ Φ(x + u)). Secondly, we can multiply by a character; (0, u^{∗}, 1) : Φ ↦ (x ↦ ⟨x, u^{∗}⟩Φ(x)). And in general, we’ll define

Φ \begin{matrix}(u, u^{∗}, t) \\  \end{matrix}x ↦ t⟨x, u^{∗}⟩Φ(x + u)

Call this action ρ. We can try to figure out the group law from this:

(u_{1}, u_{1}^{∗}, t_{1})(u_{2}, u_{2}^{∗}, t_{2})Φ = x ↦ t_{1}⟨x, u_{1}^{∗}⟩((u_{2}, u_{2}^{∗}, t_{2})Φ)(x + u_{1}))
= x ↦ t_{1}t_{2}⟨x, u_{1}^{∗}⟩⟨x + u_{1}, u_{2}^{∗}⟩Φ(x + u_{1} + u_{2})
(u_{1}, u_{1}^{∗}, t_{1})(u_{2}, u_{2}^{∗}, t_{2}) = (u_{1} + u_{2}, u_{1} + u_{2}, t_{1}t_{2}⟨u_{1}, u_{2}^{∗}⟩)

25the model may be realized in the space of cusp forms

Which will hopefully motivate the general definition for the group law on H(V);

(u_{1}, u_{1}^{∗}, t_{1}) · (u_{2}, u_{2}^{∗}, t_{2}) = (u_{1} + u_{2}, u_{1}^{∗} + u_{2}^{∗}, ⟨u_{1}, u_{2}^{∗}⟩t_{1}t_{2})

So H(V) is a central extension of V × V^{∗} by ℂ_{1}^{×}. It’s almost commutative. Call (u_{i}, u_{i}^{∗}) w_{i}, and ⟨u_{1}, u_{2}^{∗}⟩ F (w_{1}, w_{2}). Then the commutator is

[(u_{1}, u_{1}^{∗}, t_{1}), (u_{2}, u_{2}^{∗}, t_{2})] = ⟨u_{1}, u_{2}^{∗}⟩⟨u_{2}, u_{1}^{∗}⟩^{−1} = F (w_{1}, w_{2}) · F (w_{2}, w_{1})^{−1}.

On V ×V^{∗} there’s a canonical symplectic pairing (w_{1}, w_{2}) ↦ F (w_{1}, w_{2})F (w_{2}, w_{1})^{−1}. If we were using additive notation instead of the multiplicative, it’s the usual symplectic pairing. This makes sense; V × V^{∗} is the cotangent bundle of V, and our pairing is the canonical 2-form.

Theorem [Stone-von Neumann etc.] H(V) has exactly one irreducible unitary representation

so that ℂ_{1}^{×} operates via ℂ_{1}^{×} ↪^{id} ℂ^{×}.

Actually, our ρ above does the trick.

There’s a little group-theoretic fact about the Heisenberg group. Let B(V) = Aut(H(V)). Sitting inside here is B_{0}(V) ⊆ B(V), which is {α ∈ B(V)|α|_{ℂ_{1}^{×}} = id}. Remember, ℂ_{1}^{×} is the center of the group. Clearly, an element α ∈ B_{0}(V) gives an automorphism of the quotient V × V^{∗}; and it’s actually a symplectic automorphism of V × V^{∗}, as the commutator and thus the symplectic form are already encoded in the group structure.

Conversely, every element of Sp(V) = Sp(V × V^{∗}) lifts to an element of B_{0}(V). This isn’t hard, but it’s maybe not immediate, either.

Grant the theorem. For every α ∈ B_{0}(V) consider the representation ρ [as above] and twist; ρ^{α}. In other words, ρ^{α}(x) = ρ(x^{α}). This is clearly another representation as in the Stone-von Neumann theorem. Thus, there’s a unitary operator on L^{2}(V) U so that Uρ(x)U^{−1} = ρ(x^{α}). Moreover, this U is unique up to ℂ_{1}^{×}

This gives a projective unitary representation of Sp(V) on L^{2}(V). This gives rise to the metaplectic, or Shale-Weil oscillator, representation.

And that’s the opening move of the game.

The metaplectic group is a double covering Mp(V) → Sp(V). It’s still a double cover if we change to the adelic points 𝔸_{k}, whence

Mp(V_{𝔸_{k}}) \overline{\begin{matrix}2 − 1 \\  \end{matrix}}Sp(V_{𝔸_{k}})

[figure: Mp(V_{k_{v}}); 2 − 1; Sp(V_{k_{v}})]

So where’s the arithmetic already? Well, inside these groups are the rational points Sp(V)(k) ↪ Sp(V_{𝔸_{k}}). And it turns out that we get a canonical splitting

Sp(V)(k)

\frac{}{r}

[figure: Mp(V_{𝔸_{k}}); 2 − 1; Sp(V_{𝔸_{k}})]

[figure: Mp(V_{k_{v}}); 2 − 1; Sp(V_{k_{v}})]

and there’s a linear functional

V (ρ) = L^{2}(V_{𝔸_{k}}) \overline{Θ} ℂ

invariant under r(Sp(V)(k)).

This lets us cook up lots of nice functions; f’rinstance, fix some v, and consider f_{v} : g ↦ Θ(gv). Then f_{v} is invariant under the image of the global elements r(Sp(V)(k)); these should correspond to automorphic forms on the metaplectic group.

We start out wondering why the Heisenberg representation is irreducible. Let V = ℂ_{1}^{×}; V^{∗} = ℤ. Then any f ∈ L^{2}(V) may be written as f = ∑ a_{n}(f)e^{2π\sqrt{−1}nz}. We want to know that, starting from one, we can generate everything else. We’ve got two actions. One is translation; take u ∈ V; what’s the Fourier expansion of T_{u}(f) : z ↦ f(zu)? Well,

a_{n}(T_{u}f) = ∫ f(zu)z^{−n}d^{×}z
= f(z)(zu^{−1})^{−n}d^{×}z
= u^{n}a_{n}(f)

Thus, a′_{n}(f) =^{def} a_{n}(T_{u}f) = u^{n}a_{n}(f). Now try averaging the translation operator; want to cook up a g(u) so that ∫_{u} g(u)T_{u} picks off just one coefficient and kills the others. Well, just use u^{−n_{0}}. Then

the operator (∫_{u} u^{−n_{0}}T_{u})(f) is some f′ so that a_{n}(f′) = \{ \begin{matrix}a_{n_{0}}(f) & n = n_{0} \\ 0 & n ≠ n_{0} \end{matrix} .

Cool.

Now, we get all the functions; start with one function; pick off each coefficient; then take linear combinations. In this way, we see that the space L^{2}(V) is irreducible.

And an argument like this should work for a general V.

BTW, references are a little hard to come up with beyond Weil’s original paper. For real groups, one might profitably consult Lion-Vergne in the Birkhauser series. For the metaplectic group Mp(2), Gelbart has a good book.

Also, although we defined our Heisenberg group as 1 → ℂ_{1}^{×} → H(V) → V × V^{∗} → 1; but some

people use a sort of additive form, 1 → k → H′(V) → V × V^{∗} → 1. A character k →^{ψ} ℂ_{1}^{×} pushes out to our group.

Gory details Unfortunately, initially this will look a little unmotivated.

Definition A character of the second degree on a group locally compact abelian group G is a continuous function G →^{f} ℂ^{×} such that (x, y) ↦ f(x + y)f(x)^{−1}f(y)^{−1} is bimultiplicative on G × G. [Call this map q; then q(x_{1} + x_{2}, y) = q(x_{1}, y)q(x_{2}, y).]

So what? So this. We had B_{0}(V), automorphisms of the Heisenberg group inducing the identity on the center. We expect [Aut(H(V)) : B_{0}(V)] = 2. Anyway, we want a way to parametrize elements of B_{0}(V). Denote a typical element by s. From now on, we’ll adopt the convention that these s operate on H(V) on the right, so that the symplectic group operates on the right of V × V^{∗}. Let (w, t) ∈ H(V), where w = (x, x^{∗}) ∈ V × V^{∗}. Write (w, t)s = (wσ, f(w)t), where σ ∈ Sp(V) is a symplectic transformation on V × V^{∗}. [If dim V = 1, this is what people usually write as Sp(2).] Describe this situation as “s = (σ, f)”.

The condition for (σ, f) to come from some s ∈ B_{0}(V) is that

(w_{1}, t_{1})  (w_{1}σ, f(w_{1})t_{1})
(w_{2}, t_{2})  (w_{2}σ, f(w_{2})t_{2})
(w_{1}, t_{1})(w_{2}, t_{2})
= (w_{1} + w_{2}, F (w_{1}, w_{2})t_{1}t_{2}) ((w_{1} + w_{2})σ, F (w_{1}, w_{2})f(w_{1} + w_{2})t_{1}t_{2})

The condition is thus that

F (w_{1}σ, w_{2}σ) · f(w_{1})f(w_{2}) = F (w_{1}, w_{2})f(w_{1} + w_{2}),

i.e.,

f(w_{1} + w_{2})f(w_{1})^{−1}f(w_{2})^{−1} = F (w_{1}, w_{2})^{−1}F (w_{1}σ, w_{2}σ)

[Recall that F (w_{1}, w_{2}) = ⟨x_{1}, x_{2}^{∗}⟩.] As a start, observe that this means f is a character of second order; f is quadratic.

Now B_{0}(V) shows up as an extension in a natural way;

\begin{aligned}1 & \overline{} & V^{∗} × V & & B_{0}(V) & \overline{} & Sp(V) & \overline{} & 1 \\ & Hom(V × V^{∗}, ℂ_{1}^{×}) \end{aligned}

Let’s go compute the group law on B_{0}(V). Well,

[figure: (w, t); (σ_{1}, f_{1}); (wσ_{1}, f_{1}(w)t); (σ_{2}, f_{2}); (wσ_{1}σ_{2}, f_{1}(w)f_{2}(wσ_{1})t)]

Thus – remember the conventions about acting on the right –

(σ_{1}, f_{1})(σ_{2}, f_{2}) = (σ_{1}σ_{2}, f_{1}(w)f_{2}(wσ_{1})).

Sometimes it’s convenient to use a matricial notation. This might be impossible to T_{E}X, we’ll see. For σ ∈ Sp(V),

(x, x^{∗}) \overline{\begin{matrix}σ \\  \end{matrix}}(x, x^{∗}) ( \begin{matrix}α & β \\ γ & δ \end{matrix} )

where

[figure: V; α; V; V; β; V^{∗}; V^{∗}; γ; V; V^{∗}; δ; V^{∗}]

With a certain amount of risk, we return to using, symbolically, w = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) the Weyl element. Say what? We actually mean

[figure: V × V^{∗}; w; V^{∗} × V; (x, x^{∗}); w; (−x^{∗}, x)σ]

define another element σ^{∗} by σ^{∗} = ( \begin{matrix}α∗ & γ^{∗} \\ β^{∗} & δ^{∗} \end{matrix} );

V^{∗} × V \overline{\begin{matrix}σ^{∗} = ( \begin{matrix}α∗ & γ^{∗} \\ β^{∗} & δ^{∗} \end{matrix} ) \\  \end{matrix}}V^{∗} × V

Finally, we define an involution by σ^{I} = wσ^{∗}w^{−1} = ( \begin{matrix}δ^{∗} & −β^{∗} \\ −γ^{∗} & α^{∗} \end{matrix} );

V × V^{∗} \overline{\begin{matrix}σ^{I} \\  \end{matrix}}V × V^{∗}

Note that, for a symplectic matrix, ( \begin{matrix}A & B \\ C & D \end{matrix} ) ( \begin{matrix}D^{t} & −B^{t} \\ −C^{t} & A^{t} \end{matrix} ) ( \begin{matrix}I_{n} & 0 \\ 0 & I_{n} \end{matrix} );

equivalently, ( \begin{matrix}D^{t} & −B^{t} \\ −C^{t} & A^{t} \end{matrix} ) ( \begin{matrix}A & B \\ C & D \end{matrix} ) = ( \begin{matrix}I_{n} & 0 \\ 0 & I_{n} \end{matrix} ).

Anyways, σ ↦ σ^{I} is an involution; and σσ^{I} = id ⟺ σ ∈ Sp(V) ⟺ σ^{I}σ = id.

Warning: The next little bit will be computational and somewhat unenlightening, but it’ll eventually get better. Trust us.

We’ve got Sp(V), the group of symplectic transformations on V × V^{∗}. A typical element Sp(V) ∋

σ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) acts on the right on V ×V^{∗}. We want to write down an explicit condition for (σ, f)

to be in B_{0}(V). We already know that a condition is

f(w_{1} + w_{2})f(w_{1})^{−1}f(w_{2})^{−1} = F (w_{1}σ, w_{2}σ)F (w_{1}, w_{2})^{−1}. (∗)

Now we’d like to explain this in terms of α, β, γ, δ. Set w_{i} = (u_{i}, u_{i}^{∗}), and let f′(u, u^{∗}) = f(u, u^{∗})⟨u^{∗}γ, −uβ⟩. Then (*) becomes

f′(u_{1} + u_{2}, u_{1}^{∗} + u_{2}^{∗})f′(u_{1}, u_{1}^{∗})^{−1}f′(u_{2}, u_{2}^{∗})^{−1} ·
⟨(u_{1}^{∗} + u_{2}^{∗})γ, (u_{1} + u_{2})β⟩⟨u_{1}^{∗}γ, −u_{1}β⟩⟨u_{2}^{∗}γ, −u_{2}β⟩ = ⟨u_{1}α + u_{1}^{∗}γ, u_{2}β + u_{2}^{∗}δ⟩⟨u_{1}, u_{2}^{∗}⟩^{−1}

We did this just by looking at it. The final answer we’re going for is

f′(w_{1} + w_{2})f′(w_{1})^{−1}f′(w_{2})^{−1} = ⟨u_{1}, u_{2}αβ^{∗}⟩⟨u_{1}^{∗}γδ^{∗}, u_{2}^{∗}⟩

We’ll cheat a little bit and try to verify, while using the symplectic identities

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ( \begin{matrix}δ^{∗} & −β^{∗} \\ −γ^{∗} & α^{∗} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )
( \begin{matrix}δ^{∗} & −β^{∗} \\ −γ^{∗} & α^{∗} \end{matrix} ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

that f′(u_{1} + u_{2}, u_{1}^{∗} + u_{2}^{∗})f′(u_{1}, u_{1}^{∗})^{−1}f′(u_{2}, u_{2}^{∗})^{−1}

= ⟨(u_{1}^{∗} + u_{2}^{∗})γ, −(u_{1} + u_{2})β⟩⟨u_{1}^{∗}γ, u_{1}β⟩⟨u_{2}^{∗}γ, u_{2}β⟩⟨u_{1}α + u_{1}^{∗}γ, u_{2}β + u_{2}^{∗}δ⟩⟨u_{1}, u_{2}^{∗}⟩^{−1}
= ⟨u_{1}^{∗}γ, −u_{1}β⟩⟨u_{1}^{∗}γ, −u_{2}β⟩⟨u_{2}^{∗}γ, −u_{1}β⟩⟨u_{2}^{∗}γ, −u_{2}β⟩⟨u_{1}^{∗}γ, u_{1}β⟩⟨u_{2}^{∗}γ, u_{2}β⟩⟨u_{1}αu_{2}β⟩⟨u_{1}α, u_{2}^{∗}δ⟩⟨u_{1}^{∗}γ, u_{2}β⟩⟨u_{1}^{∗}γ, u_{2}^{∗}δ⟩⟨u_{1}, u_{2}^{∗}⟩^{−1}
= ⟨u_{2}^{∗}γ, −u_{1}β⟩⟨u_{2}^{∗}γ, −u_{2}β⟩⟨u_{1}α, u_{2}β⟩⟨u_{1}α, u_{2}^{∗}δ⟩⟨u_{1}^{∗}γ, u_{2}β⟩⟨u_{1}^{∗}γ, u_{2}^{∗}δ⟩⟨u_{1}, u_{2}^{∗}⟩^{−1}
= to hell with it
= ⟨u_{1}, u_{2}αβ^{∗}⟩⟨u_{1}^{∗}γδ^{∗}, u_{2}^{∗}⟩

Now, write g(u) = f′(u, 0), h(u^{∗}) = f′(0, u^{∗}). Then g and h are characters of second degree, on V and V^{∗}, respectively. So

g(u_{1} + u_{2})g(u_{1})^{−1}g(u_{2})^{−1} = ⟨u_{1}, u_{2}αβ^{∗}⟩
h(u_{1}^{∗} + u_{2}^{∗})h(u_{1}^{∗})^{−1}h(u_{2}^{∗})^{−1} = ⟨u_{1}^{∗}γδ^{∗}, u_{2}^{∗}⟩
f′(u, u^{∗}) = g(u)h(u^{∗})

Putting all this together gives us

f(u, u^{∗}) = g(u)h(u^{∗})⟨u^{∗}γ, uβ⟩

where g(u), a symmetric homomorphism V → V^{∗}, is associated with αβ^{∗} and h(u^{∗}), a symmetric homomorphism V^{∗} → V, with γδ^{∗}.

And that, pretty much, is our answer.

Typically, 2 : V → V is invertible, in which case we can take f(u, u^{∗}) = ⟨u, 2^{−1}uαβ^{∗}⟩⟨u^{∗}, 2^{−1}γδ^{∗}⟩⟨u^{∗}γ, uβ⟩. The right-hand side is a formula which gives a [set-theoretic] lifting of B_{0}(V) → Sp(V).

It’s a fact, although not a stunningly deep one, that this lifting is actually a group homomorphism.

Which tells us that, as long as 2 is invertible, B_{0}(V) is a semidirect product B_{0}(V) = Sp(V)×|(V × V^{∗}).

This might be explained by the vanishing of a cohomology group like H^{1}(Sp(V), V × V^{∗}); but we’re not promising anything.

Various special liftings (1) Let σ = ( \begin{matrix}α & 0 \\ 0 & α^{∗}^{−1} \end{matrix} ). Then take f(u, u^{∗}) = 1, trivial. We, following

Weil, thus get a lifting

[figure: GL(V); d_{0}; B_{0}(V); α; (( \begin{matrix}α & 0 \\ 0 & α^{∗}^{−1} \end{matrix} ) , 1)]

(2) Similarly, given an isomorphism γ : V^{∗} →^{∼} V, set σ = ( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ). Then the associated f

is f(u, u^{∗}) = ⟨uu^{∗}γ, −uγ^{∗}^{−1}⟩ = ⟨−u, u^{∗}⟩, whence a lifting

[figure: Iso(V, V^{∗}); d′_{0}; B_{0}(V); γ; (( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩)]

We start by recalling some special liftings. Remember that in general, σ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ↦ (σ, f)

where f(u, u^{∗}) = ⟨u, 2^{−1}uαβ^{∗}⟩⟨2^{−1}u^{∗}γδ^{∗}, u^{∗}⟩⟨u^{∗}γ, uβ⟩.

1. Let σ = ( \begin{matrix}α & 0 \\ 0 & α^{∗−1} \end{matrix} ); then set d_{0}(α) = (( \begin{matrix}α & 0 \\ 0 & α^{∗−1} \end{matrix} ) , 1).

2. σ = ( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ), d′_{0}(γ) = (( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩).

3. Take σ = ( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ), β : V → V^{∗}, β^{∗} = β; then take f to be ⟨u, 2^{−1}uβ⟩ [at least assuming [2] :

V →^{∼} V]. More generally, if β is attached to V →^{f} ℂ_{1}^{×} – that is, f(x + y, x + y)f(x)^{−1}f(y)^{−1} =

⟨x, yβ⟩ – there’s a lifting t_{0}(f) = (( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) , f) depending naturally on f.

4. Similarly, if γ : V^{∗} → V, γ^{∗} = γ is attached to f′ : V^{∗} → ℂ_{1}^{×} a character of second order on

V^{∗}, then we may lift it as t′_{0}(f′) = (( \begin{matrix}1 & 0 \\ γ & 1 \end{matrix} ) , f′).

We’re supposed to think of the matrix in 2 as a family of Weyl elements, especially if γ is symmetric.

We want to lift B_{0}(V) to the normalizer of ρ(H(V)) in U(L^{2}(V)), the unitary automorphisms of L^{2}(V).

There’s a certain convention regarding normalizations. For any s ∈ B_{0}(V) there exists some s̃ ∈ U(L^{2}(V)) such that, when w = (x, x^{∗}, 1), s^{−1}ρ(w, t)s̃ = ρ((w, t)s). Of course, if s = (σ, f), then ρ((w, t)s) = ρ(wσ, tf(w)).

Let h = (w, t); then

ρ(hs) = s^{−1}ρ(h)s̃
ρ(hs_{1}s_{2}) = s_{2}^{−1}ρ(hs_{1})s̃_{2}
= s_{2}^{−1}s_{1}^{−1}ρ(h)s̃_{1}s̃_{2}
s̃_{1}s_{2} = λs̃_{1}s̃_{2}

where λ = λ(s_{1}, s_{2}) ∈ ℂ_{1}^{×} is a cocycle. In other words, we get a projective unitary representation on L^{2}(V).

Explicit liftings We’ve got an exact sequence

[figure: !; V^{∗} × V; B_{0}(V); Sp(V); 1]

If s = (1, f) then f is an honest to god character of V × V^{∗}, and ρ((x, x^{∗}, t)s) = f(x, x^{∗})ρ(x, x^{∗}, t). We thus assert that there’s some s̃ ∈ U(L^{2}(V)) so that s^{−1}ρ(x, x^{∗}, t)s̃ = f(x, x^{∗})ρ(x, x^{∗}, t).

So let’s figure out what f can be. There are v^{∗} ∈ V and v ∈ V so that f(x, x^{∗}) = ⟨x, v^{∗}⟩⟨v, x^{∗}⟩ [by duality].

The assertion is true ⟺ the commutator [s^{−1}, ρ(x, x^{∗})] = f(x, x^{∗}) = ⟨x, v^{∗}⟩⟨v, x^{∗}⟩.

Recall the Heisenberg commutation relation; [(y, y^{∗}, 1), (x, x^{∗}, 1)] = ⟨y, x^{∗}⟩⟨x, y^{∗}⟩^{−1}. So take

s̃ = ρ(−v, v^{∗}).

We’ll now try to get explicit liftings for our special ones.

1. s = (( \begin{matrix}α & 0 \\ 0 & α^{∗}^{−1} \end{matrix} ) , 1). Let, for Φ ∈ L^{2}(V), d̃_{0}(α)Φ = (x ↦ |α|^{\frac{1}{2}} Φ(xα)).

We have to check that ρ((u, u^{∗}) ( \begin{matrix}( & α \\ 0 & 0 \end{matrix} ) α^{∗}^{−1}, 1))Φ = d̃_{0}(α)^{−1}ρ(u, u^{∗}, 1)d̃_{0}(α)Φ. In other

words, is it true that d_{0}(α)ρ(uα, u^{∗}α^{∗}^{−1}, 1)Φ = ρ(u, u^{∗}, 1)d̃_{0}(α)Φ? Well, start computing by evaluating at xα

|α|^{\frac{1}{2}} (ρ(uα, u^{∗}alpha^{∗}^{−1}, 1)Φ)(xα) = |α|^{\frac{1}{2}} Φ(xα + uα)⟨xα, u^{∗}α^{∗}^{−1}⟩
ρ(u, u^{∗}, 1)d̃_{0}(α)Φ = (d̃_{0}(α)Φ)(x + u)⟨x, u^{∗}⟩
= |α|^{\frac{1}{2}} Φ(x + u)α⟨x, u^{∗}⟩

So that’s okay.

2. s = (( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩): then d̃′_{0}(s)Φ = (x ↦ |γ|^{− \frac{1}{2}} Φ̂(−xγ^{∗}^{−1}) where Φ̂ is the Fourier

transform of Φ.

3. t̃_{0}(f)Φ : x ↦ f(x)Φ(x)

Last time we wrote down a few special liftings to U(L^{2}(V));

1. s = (( \begin{matrix}α & 0 \\ 0 & α^{∗}^{−1} \end{matrix} ) , 1) lifts to s̃ = d̃_{0}(α) as s̃(Φ) : x ↦ |α|^{\frac{1}{2}} Φ(xα).

2. s = (( \begin{matrix}0 & γ \\ γ^{∗}^{−1} & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩) lifts to (d̃′_{0}(s)Φ)(x) = |γ|^{− \frac{1}{2}} Φ̂(−xγ^{∗}^{−1}).

3. Let f be a character of second degree on V. Then (t̃_{0}(f)Φ)(x) = f(x)Φ(x).

Remark When [2] : V →^{∼} V we had a formula for lifting B_{0}(V) ⇀ Sp(V) which is a group homomorphism. To check that this lifting r really is a homomorphism when V is a vector space over k, char k ≠ 2, V →^{∼} V^{∗}, is pretty easy. Just observe that for any element σ ∈ Sp(V), r(σ) = (σ, f) with f of the form χ ∘ f̃with f̃: V → k a quadratic form and k →^{χ} ℂ_{1}^{×} some fixed character.

But, for general V, we still don’t know a fast, pure-thought proof.

Anyways, the general setup is this. We’ve got

[figure: 1; V^{∗} × V; B_{0}(V); Sp(V); 1]

r

Ω_{0}(V) =^{def} \{( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ Sp(V)|γ : V^{∗} →^{∼} V \}^{^{∪} }

Here, Ω_{0}(V) is the “big cell”. We’ve got a formula for explicit lifting of elements in Ω_{0}(V). But it’s pretty unreasonable to assume there’s a formula for r for every element in the group; the best you can lift is to a double cover.

We have other models for ρ, e.g., Λ ⊆ V a closed subgroup, and we realize ρ in a space of functions on V × V^{∗} satisfying transformation laws for Λ × Λ^{⊥}. Then we’ll be able to lift a subgroup Sp(V, Λ) by an explicit closed formula.

[The situation to keep in mind is where V = V_{0} ⊗ 𝔸_{K}, Λ = V_{0} ⊂ V.]

Let’s check that (3), above, really is a good lifting. We need to check that t̃_{0}(f)^{−1}U((u, u^{∗}, 1))t_{0}(f) =

U((u, u^{∗}, 1)t_{0}(f)). Now, t_{0}(f)(( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) , f). Recall that β : V →^{V}^{∗} is attached to f, so that

f(x + y)f(x)^{−1}f(y)^{−1} = ⟨x, yβ⟩. So compute already.

U((u, u^{∗}, 1)t_{0}(f = U(u, u^{∗}, 1)(( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) , f))
= U((u, uβ + u^{∗}, f(u)))
U((u, u^{∗}, 1)t_{0}(f)Φ(x) = U((u, uβ + u^{∗}, f(u)))Φ(x)
= ⟨x, uβ + u^{∗}⟩Φ(x + u)f(u)

On the other hand,

(t̃_{0}(f)^{−1}U((u, u^{∗}, 1))t̃_{0}(f)Φ)(x) = f(x)^{−1}⟨x, u^{∗}⟩(t̃_{0}(f)Φ)(x + u)
= f(x)^{−1}⟨x, u^{∗}⟩f(x + u)Φ(x + u)

So then we wonder,

f(x)^{−1}f(x + u) = ⟨x, uβ⟩f(u)?

And the answer, happily, is yes. Now we check (2). We need to know

d̃′_{0}(γ)^{−1}U((u, u^{∗},))d̃′_{0}(γ)Φ(x)) = U((u, u^{∗}, 1)(( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩).

Again,

U((u, u^{∗}, 1)(( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) , ⟨u, −u^{∗}⟩)Φ(x) = U((u^{∗}γ, −uγ^{∗}^{−1}, ⟨u, −u^{∗}⟩)Φ)(x)
= ⟨u, −u^{∗}⟩⟨x, −uγ^{∗}^{−1}⟩Φ(x + u^{∗}γ)

And we now need to know if

(U((u, u^{∗}, 1))d̃′_{0}(γ)Φ(x)) =^{?} (d′_{0}(γ)U((u^{∗}γ, 0uγ^{∗}^{−1}, ⟨u, −u^{∗}⟩)Φ)(x)
(d̃′_{0}(γ)Φ)(x + u)⟨x, u^{∗}⟩ =^{?} |γ|^{− \frac{1}{2}} ∫_{V} ⟨u, −u^{∗}⟩⟨y, −uγ^{∗}^{−1}⟩Φ(y + u^{∗}γ)⟨y, −xγ^{∗}^{−1}⟩dy
|γ|^{− \frac{1}{2}} ∫_{V} Φ(z)⟨z, −(x + u)γ^{∗}^{−1}⟩dz =^{?} |γ|^{− \frac{1}{2}} ∫_{V} ⟨u, −u^{∗}⟩⟨y, −uγ^{∗}^{−1}⟩Φ(y + u^{∗}γ)⟨y, −xγ^{∗}^{−1}⟩dy

So set z = y + u^{∗}γ, and pray.◇ One should also verify that these formulae preserve the L^{2} norm. Incidentally, the point of all this is that, if k is global and V/k, well, there’s an adelized version V_{𝔸_{k}}. We claim there’s a lifting of Sp(V) and a linear functional Θ : L^{2}(V_{𝔸_{k}}) → ℂ so that

Θ(r(s)Φ) = Θ(Φ)

for every s ∈ Sp(V). This should be a big generalization of the Poisson summation formula.

As before, let Ω_{0}(V) = \{(( \begin{matrix}α & β \\ γ & δ \end{matrix} ) , f) ∈ B_{0}(V)|γ : V^{∗} →^{∼} V \}. We have some formulas.

The first is a classical formula.

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}1 & αγ^{−1} \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & −γ^{∗} \\ γ & 0 \end{matrix} ) ( \begin{matrix}1 & γ^{−1}δ \\ 0 & 1 \end{matrix} ) .

Then any s = (( \begin{matrix}α & β \\ γ & δ \end{matrix} ) , f) ∈ Ω_{0}(V) may be written in a canonical way as a sum of standard

elements:

s = t_{0}(f_{1})d′_{0}(γ)t_{0}(f_{2})
f_{1}(u) = f(u, −uαγ^{−1}) a character of second order
f_{2}(u) = f(0, uγ^{−1})

Now, (σ′, f′)(σ′′, f′′) = (σ′σ′′, f′(w)f′′(wσ′)); armed with this and symplectic identities, one can verify the foregoing.

The following is an easy but significant consequence of this. Change notation a little bit; call β ρ : V →^{∼} V^{∗}, and now assume it’s symmetric and nondegenerate. Assume it’s attached to f : V → ℂ_{1}^{×} a character of second order on V. Then one can define f′ : V^{∗} → ℂ_{1}^{×} as f′(u^{∗}) = f(−u^{∗}ρ^{−1}). What’s the symmetric homomorphism V^{∗} → V attached to f′ is, surprise, ρ^{−1}. We’ll write down

t′_{0}(f′) = (( \begin{matrix}1 & 0 \\ ρ^{−1} & 1 \end{matrix} ) , f′) in two ways.

1. By a direct calculation,

t′_{0}(f′) = d′_{0}(ρ^{−1})t_{0}(f^{−1})d′_{0}(−ρ^{−1})

Project down on to its first component, get

( \begin{matrix}1 & 0 \\ ρ^{−1} & 1 \end{matrix} ) = ( \begin{matrix}0 & −ρ^{∗} \\ ρ^{−1} & 0 \end{matrix} ) ( \begin{matrix}1 & −ρ \\ 0 & 1 \end{matrix} ) ( \begin{matrix}0 & ρ^{∗} \\ −ρ^{−1} & 0 \end{matrix} ) .

2. From the general formula [“there’s a canonical way of writing s as...”] gives

t′_{0}(f′) = t_{0}(f_{1})d′_{0}(ρ^{−1})t_{0}(f_{2})
f_{1}(u) = f(u)
f_{2}(u) = f(−u)

Notationally, f^{−}(x) = f(−x).

Consequently,

d′_{0}(ρ^{−1})t_{0}(f^{−1})d′_{0}(−ρ^{−1}) = t_{0}(f)d′_{0}(ρ^{−1})t_{0}(f^{−})

t_{0}(f^{−1})d′_{0}(−ρ^{−1}) = d′_{0}(−ρ^{−1})t_{0}(f)d′_{0}(ρ^{−1})t_{0}(f^{−})

In sum, we’ve written down a bunch of generators for the symplectic group; and things like what we just boxed give relations among the generators. That, pretty much, is the significance of the work thus far.

Now, there’s a γ(f) ∈ ℂ_{1}^{×} with

γ(f)t̃_{0}(f^{−1})d̃′0(−ρ^{−1}) = d̃′_{0}(−ρ^{−1})t̃_{0}(f)d̃′_{0}(ρ^{−1})t̃_{0}(f^{−}).

Apply both sides to some Φ ∈ l^{2}(V), or even Φ ∈ S(V). Well, the left-hand side is^{26}

LHS = γ(f)t̃_{0}(f^{−1})d̃′_{0}(−ρ^{−1})Φ(x)
= γ(f)f(x)^{−1}(d̃′_{0}(ρ^{−1}Φ)(x)
= γ(f)f(x)^{−1}|ρ|^{\frac{1}{2}} Φ^{∧}(xρ)
RHS = |ρ|^{\frac{1}{2}} (t̃_{0}(f)d′_{0}(ρ^{−1})t̃_{0}(f^{−})Φ)^{∧}(xρ)

And that Fourier transform is

y ↦ f(y)|ρ|^{\frac{1}{2}} (t̃_{0}(f^{−})Φ)^{∧}(−yρ)
= f(y)|ρ|^{\frac{1}{2}} ∫_{V} f(−z)Φ(z)⟨z, −y⟩dy

so

RHS = |ρ| ∫_{V} f(y)⟨y, xρ⟩dy ∫_{V} f(−z)Φ(z)⟨−z, −yρ⟩dz
= |ρ| ∫_{V} ⟨y, xρ⟩dy ∫_{V} f(y − z)Φ(z)dz
∫_{V} f(y − z)Φ(z)dz = Φ ∗ f = f ∗ Φ

26For now, anyway, Φ^{∧} is the Fourier transform of Φ.

MA 620 – Langlands Program  28 March 1997

Let’s actually assume Φ ∈ S(V) so that the convolution Φ ∗ f makes sense, and

RHS = |ρ|(Φ ∗ f)^{∧}(xρ)

In conclusion, Theorem There’s some γ(f) ∈ ℂ_{1}^{×} so that

(Φ ∗ f)^{∧} = γ(f)|ρ|^{\frac{1}{2}} gΦ^{∧}

where g(x^{∗}) = f(x^{∗}ρ^{−1})^{−1}. Shorthand for this is that F(f) = γ(f)|ρ|^{\frac{1}{2}} f(x^{∗}ρ^{−1})^{−1}. This, essentially, is Weil’s Theorem 2.

Last time we started with V →^{f} ℂ a character of second order on V f(x+y)f(x)^{−1}f(y)^{−1} = ⟨x, yρ⟩ such that ρ : V →^{∼} V^{∗}. From this, we deduce an equality of elements in B_{0}(V)

d′_{0}(−ρ^{−1})t_{0}(f)d′_{0}(ρ^{−1})t_{0}(f^{−}) = t_{0}(f^{−1})d′_{0}(−ρ^{−1}).

Thus,

d̃′_{0}(−ρ^{−1})t̃_{0}(f)d̃′_{0}(ρ^{−1})t̃_{0}(f^{−}) = γ(f)t̃_{0}(f^{−1})d̃′_{0}(−ρ^{−1})

We can interpret this as a sort of cocycle condition.

Remember: s = (σ, f) = (( \begin{matrix}α & β \\ γ & δ \end{matrix} ) , f), γ : V^{∗} →^{∼} V. Then s = t_{0}(f_{1})d′_{0}(γ)t_{0}(f_{2}) with f_{1}(u) =

f(u, −uαγ^{−1}) and f_{2}(u) = f(−, uγ^{−1}). Define s̃ = t_{0}(f_{1})d̃′_{0}(γ)t̃_{0}(f_{2})

Exercise (σ_{1}, f_{1})(σ_{2}, f_{2}) = (σ_{3}, f_{3}) with γ_{i} : V^{∗} →^{∼} V. Then s̃_{1}s̃_{2} = λ(s_{1}, s_{2})s̃_{3}, where λ(s_{1}, s_{2}) is a

2-cocycle. Identify this cocycle in terms of the γ(f)’s above.

Umm,

t̃_{0}(f_{1,1}d̃′_{0}(γ_{1})t̃_{0}(f_{1,2})t̃_{0}(f_{2,1})d̃′_{0}(γ_{2})t̃_{0}(f_{2,2}) = λ(s_{1}, s_{2})t̃_{0}(f_{3,1})d̃′_{0}(γ_{3})t̃_{0}(f_{3,2})

So what’s this all for, anyway? Let V be a locally compact abelian group, probably a vector space. Let Γ ⊂ V be closed; take Γ^{⊥} ⊂ V^{∗}, γ^{⊥} = {ξ^{∗} ∈ V^{∗}|⟨ξ, ξ^{∗}⟩ = 1∀ξ ∈ Γ}, the Pontrjagin perp space.

We’re moving closer to a general formalism for theta functions. It works like this. For Φ ∈ L^{2}(V), we’ll produce a function Θ(x, x^{∗}) = ∫_{Γ} Φ(x + ξ)⟨ξ, x^{∗}⟩dξ. Think of this as a partial Fourier transform of Φ_{x}, a family of functions: Φ_{x} : ξ ↦ Φ(x + ξ). Thus, for ξ ∈ Γ, ξ^{∗} ∈ Γ^{⊥},

θ(x, x^{∗} + ξ^{∗}) = θ(x, x^{∗})
θ(x + ξ, x^{∗}) = ∫_{Γ} Φ(x + ξ + η)⟨η, x^{∗}⟩dη
= ∫_{Γ} Φ(x + η)⟨η − ξ, x^{∗}⟩dη
= ⟨ξ, −x^{∗}⟩θ(x, x^{∗})

Remix: Write λ = (ξ, xi^{∗}), w = (x, x^{∗}) in Γ × Γ^{⊥} and V × V^{∗}, respectively. Then

(∗) θ(w + λ) = F (λ, w)^{−1}θ(w)

In short, we get a map

\begin{matrix}L^{2}(V) & \overline{\begin{matrix}Z \\ \end{matrix}} & H(V, Γ) \\ Φ & ↦ & θ \end{matrix}

Conversely, given θ ∈ H(V, Γ), hopefully applying an inverse Fourier transform will give us back some Φ; just remember that Γ^{∗} = V^{∗}/Γ^{⊥}. In short, Φ(x) =^{def} ∫_{V^{∗}/Γ^{⊥}} θ(x, x^{∗})d\overline{x}^{∗}; evaluate the Fourier transform at zero.

We’ll sometimes think of H(V, Γ) as the set of all θ satisfying (∗) so that ∫_{V ×V^{∗}/Γ×Γ^{⊥}} |θ|^{2}d\overline{x}d\overline{x}^{∗} < ∞.

Then Z : L^{2}(V) → H(V, Γ) is actually an isomorphism. So far all we’ve got is another model. But, if you’ve seen theta functions before, you’ll feel good about it.

Let’s try to put this back in the context of the Heisenberg representation. Our goals are twofold:

1. Get a formula for the Heisenberg representation ρ using the model H(V, Γ).

2. For a nice subgroup of B_{0}(V), lift by easy formulae. Lift the Shale-Weil representation over this nice subgroup as an honest representation, and get a homomorphism L^{2}(V) → ℂ which is invariant under the lift of this homomorphism.

This last bit means

[figure: 1; ℂ_{1}^{×}; Mp(V); B_{0}(V); 1]

Sp(V, Γ)^{^{∪} }

Classical example: V/k over a number field. We’ve got V_{𝔸} = V_{k} ⊗_{k} 𝔸, 𝔸 = 𝔸_{k}. There’s V_{k} ⊂ V_{𝔸}. Then given a function Φ on V_{𝔸}, we send it to, essentially, ∑_{ξ∈V_{k}} Φ(x + ξ)⟨ξ, x^{∗}⟩. And this, essentially, is a theta series. And the functional L^{2}(V) → ℂ is just Φ ↦ θ ↦ θ(0, 0).

To be totally classical, V = k = ℚ. Then Φ is an L^{2} function, Φ = ∏_{v} Φ_{v}. At v finite, let’s try Φ_{v} the characteristic function of ℤ_{v}. At v = ∞, use, say, e^{−πx^{2}}. So what’s θ? Take an extreme example x = (x_{∞}, 0). Then we’re looking at

((x_{∞}, 0), x^{∗}) ↦ ∑_{n∈ℤ}e^{−π(x+n)^{2}}⟨ξ, x^{∗}⟩.

And if that’s not a theta series, nothing is...

MA 620 – Langlands Program  2 April 1997

Last time we had Γ ⊆ V a closed subset of a locally compact abelian group. We’re looking for some explanation of the Poisson summation formula. There’s an isomorphism

L^{2}(V)  \overline{}H(V, Γ)
Φ  Θ : V × V^{∗} → ℂ
x × x^{∗} ↦ ∫_{Γ} Φ(x + ξ)⟨ξ, x^{∗}⟩dξ

Conversely,

Φ(x) = ∫_{V^{∗}/Γ^{⊥}} θ(x, x^{∗})d\overline{x}^{∗}.

We can look for a Heisenberg representation in the model H(V, Γ). Let (u, u^{∗}, t) ∈ H(V) be an element of the Heisenberg group. How should it act on a function in H(V, Γ)?

Recall that

H(V, Γ) = \{V × V^{∗} →^{θ} ℂ|θ(x + x^{∗} + ξ^{∗}) = ⟨−ξ, x^{∗}⟩θ(x, x^{∗}), ∫_{V ×V^{∗}/Γ×Γ^{⊥}} d\overline{x}d\overline{x}^{∗} < ∞\}.

Then

(ρ(u, u^{∗}, t)θ)(x, x^{∗}) = ∫_{Γ}(ρ(u, u^{∗}, t)Φ)(x + ξ)⟨ξ, x^{∗}⟩dξ
= ∫_{Γ} Φ(x + ξ + u)t⟨x + ξ, u^{∗}⟩⟨ξ, x^{∗}⟩dξ
= t⟨x, u^{∗}⟩θ(x + u, x^{∗} + u^{∗})

From which we conclude

(ρ(u, u^{∗}, t)θ)(x, x^{∗}) = tF ((x, x^{∗}), (u, u^{∗}))θ(x + u, x^{∗} + u^{∗})

Essentially, the first factor is necessary to turn the naïve translate back into a theta function.

The real significance is that we can define a subgroup

B_{0}(V, Γ) =^{def} {(σ, f) ∈ B_{0}(V)|f(ξ, ξ^{∗}) = 1∀ξ ∈ Γ, ξ^{∗} ∈ Γ^{⊥}
σ preserves Γ × Γ^{⊥}}

We still have a projective representation of B_{0}(V) on H(V, Γ), and we can restrict to get a projective representation of B_{0}(V, Γ). What’ll happen, nicely enough, is that we can lift to an honest [non-projective] representation of B_{0}(V, Γ). Moreover, this lifting is given by a nice formula.

As promised, there’s a formula for lifting B_{0}(V, Γ) to a unitary representation of B_{0}(V, Γ) on H(V, Γ). Write s = (σ, f) ∈ B_{0}(V, Γ). Then set

(r_{Γ}(s)θ)(x, x^{∗}) = f(x, x^{∗})θ((x, x^{∗})σ)

A bold assertion. Some things need checking.

1. The right-hand side really is a theta function, that is, an element of H(V, Γ).

2. r_{Γ}(s)^{−1}ρ((u, u^{∗}, t))r_{Γ}(s) = ρ((u, u^{∗})σ, tf(u, u^{∗})).

3. For all s_{1}, s_{2} ∈ B_{0}(V, Γ), r_{Γ}(s_{1}, s_{2}) = r_{Γ}(s_{1})r_{Γ}(s_{2}).

So let’s try it.

(1) We need to check whether f(x + ξ, x^{∗} + ξ^{∗})θ((x + u, x^{∗} + u^{∗})σ) = ⟨ξ, −x^{∗}⟩f(x, x^{∗})θ(x, x^{∗})σ). Well,

LHS : f(x + ξ, x^{∗} + ξ^{∗}) = f(x, x^{∗})f(ξ, xi^{∗})F ((x, x^{∗})σ, (ξ, ξ^{∗})σ)F ((x, x^{∗}), (ξ, ξ^{∗}))^{−1}
= f(x, x^{∗})f(ξ, ξ^{∗})F ((ξ, ξ^{∗})σ, (x, x^{∗})σ)F ((ξ, ξ^{∗}), (x, x^{∗}))^{−1}
LHS = f(x, x^{∗})f(ξ, ξ^{∗})F ((ξ, ξ^{∗})σ, (x, x^{∗})σ)F ((ξ, ξ^{∗}), (x, x^{∗}))^{−1} ·
F ((ξ, ξ^{∗})σ, (x, x^{∗})σ)^{−1}θ(x, x^{∗})σ)

But f(ξ, ξ^{∗}) = 1, so LHS = RHS, and we’re done.

2. This one is to show that our r_{Γ}(s) really is a lift, that is, winds up in the class we want. So, we want to show that ρ(u, u^{∗}, t)r_{Γ}(s) = r_{Γ}(s)ρ((u, u^{∗})σ, f(u, u^{∗})t). Well, apply both sides to θ and evaluate on (x, x^{∗}). Set t = 1.

(ρ(u, u^{∗}, 1)r_{Γ}(s)θ)(x, x^{∗}) = F ((x, x^{∗}), (u, u^{∗}))(r_{Γ}(s)θ)(x + u, x^{∗} + u^{∗})
= F ((x, x^{∗}), (u, u^{∗}))f(x + u, x^{∗} + u^{∗})θ((x + u, x^{∗} + u^{∗})σ)
(r_{Γ}(s)ρ((u, u^{∗})σ, f(u, u^{∗}))θ)(x, x^{∗}) = f(x, x^{∗})(ρ((u, u^{∗})σf(u, uθ)((x, x^{∗})σ)
= f(x, x^{∗})f(u, u^{∗})F ((x, x^{∗})σ, (u, u^{∗})σ)θ((x + u, x^{∗} + u^{∗})σ)

Using f(w_{1} + w_{2})f(w_{1})^{−1}f(w_{2})^{−1} = F (w_{1}σ, w_{2}σ)F (w_{1}, w_{2})^{−1}, we’re done.

3. We’re making sure this really is a group homomorphism.

(r_{Γ}(s_{1})r_{Γ}(s_{2})θ)(x, x^{∗}) = (r_{Γ}(s_{2})θ)((x, x^{∗})σ_{1})f1(x, x^{∗})
= θ(x, x^{∗})σ_{1}σ_{2})f_{2}((x, x^{∗})σ)f_{1}(x, x^{∗})
= θ(x, x^{∗})σ_{1}σ_{2})f_{3}(x, x^{∗})

where (σ_{1}, f_{1})(σ_{2}, f_{2}) = (σ_{3}, f_{3}). We can write down a linear functional, essentially theta-null.

S(V)  \overline{}ℂ
Φ  ∫_{Γ} Φ(ξ)dξ
= θ(0, 0)

where Φ ⇝ θ. To be honest, this functional is really only a distribution; evaluating an L^{2} function at a point doesn’t make a lot of sense. So what? So this. For s ∈ B_{0}(V, Γ), ∫_{Γ}(r_{Γ}(s)Φ)(ξ)dξ = ∫_{Γ} Φ(ξ)dξ. By the formula, this is obvious; for Φ corresponds to some θ, then.... So the distribution is B_{0}(V, Γ)-invariant. Weil makes the cryptic remark that this is a generalization of the Poisson summation formula. Whyizzat? Take a Weyl element, in coordinates something like ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ). Invariantly, this is

some ( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ) whose square is −1. F’rinstance, start with V/k, k a number field, V finite dimensional. Apply all this to S(V_{𝔸}). Then V_{𝔸}^{∗} →^{∼} V_{𝔸} after picking a nontrivial additive character. Then this identification gives a Weyl element, etc. The Poisson summation formula says

∑_{ξ∈V_{k}} Φ(ξ) = ∑_{ξ∈V_{k}} Φ^{∧}(ξ)

[up to some normalization]. For taking s to be the Weyl element and acting with r_{Γ}(s) essentially takes the Fourier transform.

We’re in a pretty good situation, which we summarize now. We’ve got

\begin{matrix}1 & \overline{} & ℂ_{1}^{×} & \overline{} & H(V) & \overline{} & V × V^{∗} & \overline{} & 1 \\ 1 & \overline{} & ℂ_{1}^{×} & \overline{} & Mp(V) & \overline{} & B_{0}(V) & 1 \end{matrix}

Inside B_{0}(V) lives the big cell Ω_{0}(V), a dense set but not a subgroup. Over Ω_{0}(V) we have a lifting Ω_{0}(V) \underrightarrow{r̃_{0}} Mp(V).^{27}

Anyways, inside B_{0}(V) also lies B_{0}(V, Γ). This also has a lifting, and is a genuine group.

If we take Λ ⊆ V closed, what’s B(V, (0)), i.e., where Λ = (0); or, dually, Λ = V and look at B(V, V). Well, B_{0}(V, V) = {s = (σ, f)|σ : V × (0) →^{∼} V × (0) and f is 1 on V }. Now, f(x, x^{∗}) = f(x, 0)f(0, x^{∗})F ((x, 0)σ, (0, x^{∗})σ)F ((x, 0), (0, x^{∗}))^{−1}. Now, our condition means that [since the action is on the right]^{28} the matrix is lower triangular. Among other things, this include the d_{0}(α)’s.

Exercise So, using this Γ,

r_{Γ=V} (d_{0}(α)) = d̃′_{0}(α).

Proposition Let Ω_{0}(V, Γ) = {(σ, f) ∈ Ω_{0}(V) ∩ B_{0}(V, Γ)|γ : V^{∗} →^{∼} V, γ : Γ^{⊥} →^{∼} Γ}. Then r̃_{0}(s) = r_{Γ}(s); the liftings agree for s ∈ Ω_{0}(V, Γ). Moreover, the modulus of the maps V^{∗}/Γ^{⊥} → V/Γ, Γ^{⊥} → Γ are equal to the modulus of V^{∗} → V, namely, |γ|^{\frac{1}{2}} .

Proof Let s = (( \begin{matrix}α & β \\ γ & δ \end{matrix} ) , f); then r̃_{0}(s) = t̃_{0}(f_{1})d̃′_{0}(γ)t̃_{0}(f_{2}). And the t_{0}(f_{i}) are both in B_{0}(V, Γ)

[by the formula]. Also, trivially, t̃_{0}(f_{i}) = r_{Γ}(t_{0}(f_{i})); it’s the same as using the theta lift. The key is to show that

r_{Γ}(d′_{0}(γ)) = d̃′_{0}(γ)

28(x 0) ( \begin{matrix}α & 0 \\ γ & δ \end{matrix} )

27Aside on Bruhat decomposition. For GL_{2}, the Bruhat decomposition is \overline{N}AN, where N = ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ), \overline{N} = ( \begin{matrix}1 & 0 \\ ∗ & 1 \end{matrix} ). In general, given a semisimple Lie group, well, take A ⊆ G a maximal subtorus. You get the roots Φ = Φ_{+} ⊔ Φ_{−}. For each root α there’s a corresponding unipotent group U_{α}. Then [more or less] U_{+} = ∏_{α∈U_{+}} U_{α}, and similarly for U_{−}. This decomposition is stable under conjugation by A; and U_{−}AU_{+} is the big cell.

Not surprisingly, the strategy is just write down the formula and pray. Two items of business:

• (d̃′_{0}(γ)Φ)(x)|γ|^{− \frac{1}{2}} Φ^{∧}(−xγ^{∗}^{−1}).

•

(r_{Γ}(d′_{0}(γ))Φ)(x) = ∫_{V^{∗}/Γ}(r_{Γ}(d′_{0}(γ))θ)(x, x^{∗})d\overline{x}^{∗}
= ∫_{V^{∗}/Γ^{⊥}} ⟨−x, x^{∗}⟩θ((x, x^{∗}) ( \begin{matrix}0 & −γ^{∗}^{−1} \\ γ & 0 \end{matrix} ))d\overline{x}^{∗}
= ∫_{V^{∗}/Γ^{⊥}} ⟨−x, x^{∗}⟩θ(x^{∗}γ, −xγ^{∗}^{−1})d\overline{x}^{∗}
Use u = x^{∗}γ
= absγ_{V^{∗}/Γ^{⊥}}_{v} ∫_{V/Γ} ⟨x, −uγ^{−1}⟩θ(u, −xγ^{∗}^{−1})d\overline{u}
= |γ_{V^{∗}/Γ^{⊥}} | ∫_{V/Γ} ⟨x, −uγ^{−1}⟩d\overline{u} ∫_{ξ∈Γ} Φ(u + ξ)⟨ξ, −xγ^{∗}^{−1}⟩dξ
= |γ_{V^{∗}/Γ^{⊥}} |^{−1} ∫_{V} Φ(y)⟨y, −xγ^{∗} ∈⟩dy
= |γ_{V^{∗}/Γ^{⊥}} |^{−1}Φ^{∧}(−xγ^{∗}^{−1})

Which gets us close, since we’re at worst off by a scalar factor. But γ and γ_{V^{∗}/Γ^{⊥}} are both unitary operators, and we’re done.◇

”Theorem” Let f be a character of second order such that i. f|_{Γ} = 1. ii. The symmetric homomorphism V → V^{∗} atached to f is an isomorphism.

Then γ(f) = 1. This follows from the following, which involves a fairly lengthy calculation.

Theorem (σ, f)(σ′, f′) = (σ′′, f′′). If σ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ), σ′ = ( \begin{matrix}α′ & β′ \\ γ′ & δ′ \end{matrix} ), etc., and γ, γ′, γ′′ : V^{∗} →^{∼}

V. Let f_{0}(u) = f(0, uγ^{−1})f′(u, −uα′γ′^{−1}). Then f_{0} is nondegenerate, and gives rise to V → V^{∗} γ^{−1}γ′γ′^{−1}. Moreover, we have the cocycle relation

r̃_{0}(s)r̃_{0}(s′) = r(f0(r̃_{0}(s′′)

Last time we were trying to interpret the γ_{f}’s. We have (σ, f)(σ′, f′) = (σ′′, f′′), ss′ = s′′. Moreover, s = t_{0}(f_{1})d′_{0}(γ) – we’re assuming all the γ’s involved are isomorphisms. So we have the standard liftings

s = t_{0}(f_{1})d′_{0}(γ)t_{0}(f_{2})
s′ = t_{0}(f′_{1})(d′_{0}(γ′)t_{0}(f′_{2})
s′′ = t_{0}(f′′_{1})d′_{0}(γ′′)t_{0}(f′′_{2})

We use our preferred choice of liftings to the metaplectic group

t̃_{0}(f_{1})d̃′_{0}(γ)t̃_{0}(f_{2})t̃_{0}(f′_{1})d̃′_{0}(γ′)t̃′_{0}(f′_{2}) = λ(s_{1}, s_{2})t̃_{0}(f′′_{1})d̃′_{0}(γ′′)t̃_{0}(f′′_{2})
d̃′_{0}(γ)t̃_{0}(f_{0})d̃′_{0}(γ′) = λ(s_{1}, s_{2})t̃_{0}(f_{3})d̃′_{0}(γ′′)t̃_{0}(f_{4})

where

f_{0} = f_{2}f′_{1}
f_{0}(u) = F (0, uγ^{−1})f′(u, −uα′γ′^{−1})
f_{3} = f_{1}^{−1}f′′_{1}
f_{4} = f′′_{2}f′_{2}^{−1}

Theorem

λ(s_{1}, s_{2}) = γ(f_{0})

To prove this, we’ll applying everything to some Φ ∈ S(V). On the right-hand side we get

RHS = (λ(s_{1}, s_{2})t̃_{0}(f_{3})d̃′_{0}(γ′′)t̃_{0}(f_{4})Φ)(x)
= λ(s_{1}, s_{2})f_{3}(x)(d̃′_{0}(γ′′)t̃_{0}(f_{4})Φ)(x)
= λ(s_{1}, s_{2})f_{3}(x)|γ′′|^{− \frac{1}{2}} (f_{4}Φ)^{∧}(−xγ^{′′∗}^{−1})
= λ(s_{1}, s_{2})|γ′′|^{− \frac{1}{2}} ∫_{V} f_{3}(x)f_{4}(u)Φ(u)⟨u, −xγ′′∗^{−1}⟩du

whereas on the left-hand side we get

LHS = (d̃′_{0}(γ)t̃_{0}(f_{0})d̃′_{0}(γ′)Φ)(x)
= |γ|^{− \frac{1}{2}} F(f_{0}d̃′_{0}(γ′)Φ)(−xγ^{∗}^{−1})
= |γ|^{− \frac{1}{2}} (F(f_{0}) ∗ F(d̃′_{0}(γ′)Φ))(−xγ^{∗}^{−1})
F(f_{0}(x^{∗}) = γ(f_{0})|ρ_{0}|^{− \frac{1}{2}} f_{0}(x^{∗}ρ_{0}^{−1})^{−1}
(d̃′_{0}(γ′)Φ)(y) = |γ|^{− \frac{1}{2}} Φ^{∧}(−yγ^{′∗}^{−1})
∫_{V} ψ(−yγ^{′∗}^{−1})⟨y, z^{∗}⟩dy = ∫_{V^{∗}} ψ(y^{∗})⟨−y^{∗}γ^{′∗}, z⟩|γ′|dy^{∗}
= ∫_{V} ψ(y^{∗})⟨y^{∗}, −zγ^{′∗}^{−1}⟩|γ′|dy^{∗}
= ψ^{∧}(−zγ^{′∗})
LHS = |γ|^{− \frac{1}{2}} |ρ_{0}|^{− \frac{1}{2}} γ(f_{0})|γ′|^{\frac{1}{2}} Φ(z^{∗}γ′)f_{0}(z^{∗}ρ_{0}^{−1})^{−1}
= γ(f_{0})|γ|^{\frac{1}{2}} |ρ_{0}|^{− \frac{1}{2}} |γ′|^{\frac{1}{2}} ∫_{V^{∗}} f_{0}(−xγ^{∗}^{−1} − z^{∗})Φ(z^{∗}γ′)dz^{∗}
= γ(f_{0})|γ|^{\frac{1}{2}} |ρ_{0}|^{− \frac{1}{2}} |γ′|^{\frac{1}{2}} |γ′|^{− \frac{1}{2}} ∫_{V} f_{0}(−xγ^{∗}^{−1} − uγ′^{−1})Φ(u)du

[where ρ_{0} is the symmetric V → V^{∗} attached to f_{0}]

Now, hopefully, we’re done. Now, L^{2}(V × V) = L^{2}(V) ⊗^{∧} L^{2}(V). We’re trying to show the two kernels of integration are the same, or something very much like that. Some structural constants show up; λ(s_{1}, s_{2}) and γ(f_{0}) are their angles, and actually identical.◇

Corollary Let f be a character of second order on V such that

i. f|_{Γ} = 1.

ii. The symmetric homomorphism ρ : V → V^{∗} attached to f induces an isomorphism (V, Γ) →^{∼} (V^{∗}, Γ^{∗}).

Then γ(f) = 1.

Why a corollaryUse the theorem and the fact that the theta series really give us a represent. For in this situation, yoiu can easily write down a pair s and s′′ so that they all lie in B_{0}(V, Γ), and f_{0} is the given f. All you need to do is take f_{2} = 1, f′_{1} to be f. Take a situation to make that true. Then you win, since on the one hand, from the theta series lifting you know you get a genuine representation, and λ is 1. On the other hand, the theorem shows this is γ(f).

This will, somehow, turn out to be quadratic reciprocity.

Properties of the symbol γ(f) Here, we restrict to a slightly different situation, namely: Let k be a local [locally compact] field; fix ψ : k → ℂ_{1}^{×} a nontrivial additive character. Then there’s another version of our Heisenberg groups, as follows. Define an extension which pushes out to the original one.

[figure: 1; k; H′(V); V × V^{∗}; 1; 1; ℂ_{1}^{×}; ψ; H(V); V × V^{∗}; =; 1]

[Now, the symbol ⟨x, x^{∗}⟩ is a pairing not with the Pontryagin dual but with the ilnear dual.]

Then B′_{0}(V) = {(σ, f)|f a quadratic form, all else as before }. In other words, f is symmetric wrt multiplication by -1. Then we can define everything as before, and think fo B′_{0}(V) as Sp(V). We now get a symbol γ(f) for any nondegenerate quadratic form.

• f_{1} ∼ f_{2} [as quadratic forms] then γ(f_{1}) = γ(f_{2}).

• γ(f_{1} ⊕ f_{2}) = γ(f_{1})γ(f_{2}).

• If f is hyperbolic, then γ(f) = 1.

In conclusion, γ depends only on the image of f in the Witt group W (k).