{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 42 " Our first example is \+ G_2/U(2) = G_2(R^7)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 65 "Here the metric depends on 3 parameters, with no trivia l summand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "The intermediate subalgebras are h+p_3 = SU(3) and , h+p_1 \+ = SO(4) " }}{PARA 0 "" 0 "" {TEXT -1 41 "The graph has two vertices n ot connected." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 65 "The scalar curvature function, accord ing to Megan , is equal to:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "S:=(a,b,c)->8*(1/a+2/b+2/c)- (4/3)*(a/b^2+2/a)-2*(a/(b*c)+b/(a*c)+c/(a*b));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "Now lets try to elim inate a from vol = 1 to get a=1/(bc)^2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "S(1/(b*c)^2,b,c); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "T:=(b,c)->16/3*b^2*c^ 2+16/b+16/c-4/3*1/(b^4*c^2)-2/(b^3*c^3)-2*b^3*c-2*c^3*b;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "We comput e the hessian and index and coindex of critical points:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(l inalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Hess:=matrix(2,2 ,(i,j)->D[i,j](T));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "There are 3 solutions to the Einstein equation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "The \+ first one is the symmetric space one." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "bE1:=evalf((1/18)^(1/5) );cE1:=evalf(3*(1/18)^(1/5));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "bE2:=( (.59713)^(-2)*(1.22554)^(-2) )^(1/5)*.59713 ; cE2:=( (. 59713)^(-2)*(1.22554)^(-2) )^(1/5)*1.22554;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "bE3:=( (5.35063)^(-2)*(5.25153)^(-2) )^(1/5)*5. 35063 ; cE3:=( (5.35063)^(-2)*(5.25153)^(-2) )^(1/5)*5.25153;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "He re are the approximate critical values:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "evalf(T(bE1,cE1));" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "evalf(T(bE2,cE2));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "evalf(T(bE3,cE3));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "No w we compute the index of the critical points:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf(Hess( bE1,cE1));eigenvals(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " evalf(Hess(bE2,cE2));eigenvals(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf(Hess(bE3,cE3));eigenvals(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "The symmetric s pace is a local max, and the other two are saddle!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "And finally a picture tha t shows the behaviour of the critical points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "The symmetric space metr ic is hardly visible, especialy on a global scale. It forces the exist ence of the saddle point nearby." }}{PARA 0 "" 0 "" {TEXT -1 58 "But o nly the third saddle point exists for global reasons!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "plot3d(T (b,c),b=0.45..0.8,c=1.2..2,view=29.6..29.8,axes=boxed,style=patchconto ur);" }}{PARA 13 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "plot3d(T(b,c),b=0.1..3,c=0.1..3,view=20..40,axes=boxe d,style=patchcontour);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "p lot3d(T(exp(b),exp(c)),b=-5..5,c=-5..5,view=0..100,axes=boxed,style=pa tchcontour);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 256 "" 0 "" {TEXT 257 45 "Scalar curvature of SU(9)/S( U(2)U(3 )U(4) )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "S:=(a,b,c)->6/a+8/b+12/c-(12/9)*(a/(b*c)+b/(a*c)+c/(a*b));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "The volume is a^6b^8c^12 , so we e liminate c :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "S(a,b,1/(sqrt(a)*b^(2/3)));" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "The following is S as a function of a,b alone :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "T:=(a,b)->6*1/a+8/b+12*sqrt (a)*b^(2/3)-4/3*a^(3/2)/(b^(1/3))-4/3*b^(5/3)/(sqrt(a))-4/3*1/(a^(3/2) *b^(5/3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Tb:=D[2](T);T a:=D[1](T);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 34 "There are 4 real critical points:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "The first 3 are the Kaehl er Einstein metrics, which are not isometric to each other." }}{PARA 0 "" 0 "" {TEXT -1 39 "The fourth one is a non-Keahler metric." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "Here are \+ the 4 critical points and their approximate coordinates." }}{PARA 0 " " 0 "" {TEXT -1 75 "We scale them to have volume 1 ( find t that does) and check that vol = 1 ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "t:=evalf(5^(-3/13)*12^(-4/13)*7^(-6 /13));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "aE1:=5*t;bE1:=12* t;cE1:=7*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "aE1^3*bE1^4* cE1^6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "t:=evalf(5^(-3/13 )*6^(-4/13)*11^(-6/13));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "aE2:=5*t;bE2:=6*t;cE2:=11*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "aE2^3*bE2^4*cE2^6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "t:=evalf(13^(-3/13)*6^(-4/13)*7^(-6/13));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 28 "aE3:=13*t;bE3:=6*t;cE3:=7*t;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "aE3^3*bE3^4*cE3^6;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 40 "t:=evalf(5^(-3/13)*6^(-4/13)*7^(-6/13));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "aE4:=5*t;bE4:=6*t;cE4:=7*t; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "aE4^3*bE4^4*cE4^6;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "He re are the critical values and their approximations:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "T(aE1,bE1 );evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "T(aE2,bE2);e valf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "T(aE3,bE3);eval f(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "T(aE4,bE4);evalf(% );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 151 " The critical values are extremely close, The smallest one is \+ the non-Kaehler metric and the third Kaehler metrics gets picked out \+ to be the best one." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "And now we comp ute the eigenvalues of the Hessian:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Hess:=matrix(2,2,(i,j)->D[i,j](T)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Hess(aE1,bE1);eigenvals (%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Hess(aE2,bE2);eigen vals(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Hess(aE3,bE3);e igenvals(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Hess(aE4,bE 4);eigenvals(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 78 "So the 3 Kaehler metrics are saddle points and the non-Kaehler is a local min." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "Now 4 pictures with varying features. Rotate t hem and also \"look from above\" for the contour lines." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "In the first two \+ the range of S is extremely small to see the features of the 4 critica l points, one in x_i coordinates and the second in exponential coordin ates." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 " The last two are global pictures (in both coordinates) where the criti cal points are hard to see." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "plot3d(T(exp(a),exp(b)),a=-1 ..1,b=-1..1,view=21.8..22.5,axes=boxed,style=patchcontour,contours=30) ;" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 259 33 "Scalar curvature of S^3 xS^3/S^1" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 146 "This Example examines the unique critica l point of S^3xS^3/S^1 where S^1=(exp(ipt)exp(,iqt)) and the manifol d is parametrized by 0< r = q/p <1" }}{PARA 0 "" 0 "" {TEXT -1 78 "r =0 is the product metric on S^3xS^2 and r=1 the Stiefel manifold SO(4) /SO(2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "The scalar curvature function is equal to:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "S:=(a,b,c)->8/b+8/c-(2*r^2) /(1+r^2)*(a/b^2)-(2/(1+r^2))*(a/c^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "S(1/(b^2*c^2),b,c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "F:=(b,c)->8*1/b+8*1/c-2*r^2/((1+r^2)*b^4*c^2)-2*1/((1 +r^2)*b^2*c^4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(lin alg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Hess:=matrix(2,2,( i,j)->D[i,j](F));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 108 "Here are the solutions that Rodionov describes, s caled so that vol = 1 : ( E stands for Einstein solution)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "s ol:=solve(8*y^3-8*r*y^2+4*y-r=0,y):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "The first solution is real, th e second and third (which I am hiding, try to replace : by ; to se e it) are complex" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 10 "y:=sol[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "aE:=((r^2*(2*y^2+1)^4)/(4*y^2*(r^2+1)^4))^(-1/5);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "bE:=aE*(r*(2*y^2+1)/(2*y*(r ^2+1)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "cE:=aE*(2*y^2+1 )/(r^2+1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "As you can see fairly complicated, but the pictures are n ice:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plot([aE,bE,cE],r=0..1,color=[red,green,blue]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "W e can try to compute the critical value, but as you can see (if you re place : ) this is already too complicated:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Scal:=F(bE,cE):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "But we can draw it:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(Scal,r=0..1,12..15);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "The Hessi an at the critical point and its eigenvalues is too much for Maple." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "So now \+ we specify the value of r and compute the index of the critical poi nt and draw pictures of the graph of S ( or better F )" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "When you are done with one value of r, you can go back and change it ( start with r:= 'r'; to erase the old value ) :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "r:='r';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "r:=1/100;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "We now look at things numerically: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "Firs t the a,b,c value, then Scal , and then we check again it is acritic al point and compute the eigenvalues of the Hessian:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(aE) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(bE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(cE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalf(Scal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "map(evalf,Hess(bE,cE));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eigenvals(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 34 "The critical point has coindex 1 !" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Next a p icture of the graph. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Now lets use our \"geodesic\" coordinates which actually show a much better pi cture of the saddle point:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 72 "A:=plot3d(F(exp(b),exp(c)),b=-10..10,c=-10..10 ,view=-14..40,axes=boxed):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "B:=plot3d(0,b=-10..10,c=-10..10,view=-14..40,color=blue,style=patc hnogrid,axes=boxed):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "dis play(\{A,B\},view=-14..40);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 175 "In this picture you can see the advanta ge of geodesic coordinates ! 0 -PS sequences are visible, but now th e region where it goes to + infty and -infty are very pronounced !" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Now go b ack and try a different value of r , e.g. r=1/100 is somewhat differen t, but not qualitatively." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 258 55 "Here \+ is the Sakane example with no Einstein metrics." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "S:=(a,b,c)->15/(14*a)+12/(7 *b)+9/c-15*a/(28*c^2)-9*b/(14*c^2)-5*a/(28*b^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "S(a,b,1/(a^(5/18)*b^(1/3)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "The follo wing is S as a function of a,b alone since the volume is a^5b^6c^18=1 \+ ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "T:=(a,b)->15/14*1/a+12/7/b+9*a^(5/18)*b^(1/3)-15/28* a^(14/9)*b^(2/3)-9/14*b^(5/3)*a^(5/9)-5/28*a/(b^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "Now a picture:. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "plot3d(T(exp(a),exp(b)),a=-3 ..4,b=-3..4,view=-10..30,axes=boxed,style=patchcontour,contours=30);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 98 "One can see that there could easily be a critical point , but there is no reason for its existence." }}}}{MARK "1 0 0" 8 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }