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  An exercise in Geometric Topology for physicists

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Compute the first, second and third Hirzebruch L-polynomials L1(p1), L2(p1,p2) and L3(p1,p2,p3)  using the fact that CP2, CP4, CP6, CP2×CP2, CP2×CP2×CP2 and CP4×CP2  have signature 1 and the Hirzebruch signature formula.·

asked Sep 19, 2015 in Mathematics by juancho (1,130 points) [ no revision ]

1 Answer

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We look for Hirzebruch L-polynomials with the form

L1(p1)=ap1

L2(p1,p2)=b1p21+b2p2

L3(p1,p2,p3)=c1p31+c2p1p2+c3p3

where the coefficients a,b1,b2,c1,c2.c3 must be computed.

The Hirzebruch signature formula says that for any closed smooth oriented 4-manifold M4 we have

τ(M4)=M4L1(p1(M4))

The Hirzebruch signature formula says that for any closed smooth oriented 8-manifold M8 we have

τ(M8)=M8L2(p1(M8),p2(M8))

The Hirzebruch signature formula says that for any closed smooth oriented 12-manifold M12 we have

τ(M12)=M12L3(p1(M12),p2(M12),p3(M12))

The total Chern classes for the considered complex projective spaces are

c(CP2)=1+3f+3f2

c(CP4)=1+5g+10g2+10g3+5g4

c(CP6)=1+7h+21h2+35h3+35h4+21h5+7h6

c(CP2×CP2)=1+3f2+3f1+3f22+9f1f2+3f12+9f1f22+9f12f2+9f12f22

c(CP2×CP2×CP2)=1+(3f3+3f2+3f1)+(3f32+9f3f2+9f3f1+3f22+9f1f2+3f12)+(9f32f2+9f32f1+9f3f22+27f3f1f2+9f3f12+9f1f22+9f12f2)+(9f32f22+27f32f1f2+9f32f12+27f3f1f22+27f3f12f2+9f12f22)+(27f32f1f22+27f32f12f2+27f12f22f3)+27f12f22f32

c(CP2×CP4)=1+(3f+5g)+(3f2+15gf+10g2)+(15gf2+30g2f+10g3)+(30g2f2+30g3f+5g4)+(30g3f2+15g4f)+15g4f2

where the cohomological generators are normalized according to

CP2f2=1

CP4g4=1

CP6h6=1

CP2×CP2f21f22=CP2f21CP2f22=(1)(1)=1

CP2×CP2×CP2f21f22f23=CP2f21CP2f22CP2f23=(1)(1)(1)=1

CP2×CP4f2g4=CP2f2CP4g4=(1)(1)=1

Now using the following expressions for the Pontryagin classes in terms of the Chern classes

p1=2c2+c12

p2=2c1c3+c22+2c4

p3=2c2c4+c322c6+2c1c5

we obtain that

p1(CP2)=3f2

p1(CP4)=5g2

p1(CP6)=7h2

p1(CP2×CP2)=3f22+3f12

p1(CP2×CP2×CP2)=3f32+3f22+3f12

p1(CP2×CP4)=3f2+5g2

p2(CP4)=10g4

p2(CP6)=21h4

p2(CP2×CP2)=9f12f22+9f24+9f14=9f12f22+9(0)+9(0)=9f12f22

p2(CP2×CP2×CP2)=9f12f22+9f24+9f14+9f32f22+9f32f12+9f34=9f12f22+9(0)+9(0)+9f32f22+9f32f12+9(0)=9f12f22+9f32f22+9f32f12

p2(CP2×CP4)=15g2f2+10g4+9f4=15g2f2+10g4

p3(CP6)=35h6

p3(CP2×CP2×CP2)=27f12f22f32+27f14f22+27f12f24+27f32f14+27f32f24+27f34f12+27f34f22=27f12f22f32

p3(CP2×CP4)=30g4f2+45g2f4=30g4f2+45g2(0)=30g4f2

Using these results we have that

L1(p1(CP2))=ap1(CP2)=a(3f2)=3af2

L2(p1(CP4),p2(CP4))=b1p1(CP4)2+b2p2(CP4)=b1(5g2)2+b2(10g4)=25b1g4+10b2g4=(25b1+10b2)g4

L2(p1(CP2×CP2),p2(CP2×CP2))=b1p1(CP2×CP2)2+b2p2(CP2×CP2)=b1(3f21+3f22)2+b2(9f21f22)=b1(9f41+18f21f22+9f42)+9b2f21f22=18b1f21f22+9b2f21f22=(18b1+9b2)f21f22

L3(p1(CP6),p2(CP6),p3(CP6))=c1p1(CP6)3+c2p1(CP6)p2(CP6)+c3p3(CP6)=c1(7h2)3+c2(7h2)(21h4)+c3(35h6)=343c1h6+147c2h6+35c3h6=(343c1+147c2+35c3)h6

L3(p1(CP2×CP2×CP2),p2(CP2×CP2×CP2),p3(CP2×CP2×CP2))=c1p1(CP2×CP2×CP2)3+c2p1(CP2×CP2×CP2)p2(CP2×CP2×CP2)+c3p3(CP2×CP2×CP2)=c1(3f12+3f22+3f32)3+c2(3f12+3f22+3f32)(9f12f22+9f32f22+9f32f12)+27c3f12f22f32=27f12f32f22(6c1+3c2+c3)

L3(p1(CP2×CP4),p2(CP2×CP4),p3(CP2×CP4))=c1p1(CP2×CP4)3+c2p1(CP2×CP4)p2(CP2×CP4)+c3p3(CP2×CP4)=c1(3f2+5g2)3+c2(3f2+5g2)(15g2f2+10g4)+c3(30g4f2)=15f2g4(15c1+7c2+2c3)

Now, the Hirzebruch signature formula says that for CP2 we have

τ(CP2)=CP2L1(p1(CP2))=1

CP23af2=1

3aCP2f2=1

3a=1

a=13

then we obtain

L1(p1)=13p1

The Hirzebruch signature formula says that for CP4 we have

τ(CP4)=CP4L2(p1(CP4),p2(CP4))=1

CP4(25b1+10b2)g4=1

(25b1+10b2)CP4g4=1

25b1+10b2=1

The Hirzebruch signature formula says that for CP2×CP2 we have

τ(CP2×CP2)=CP2×CP2L2(p1(CP2×CP2),p2(CP2×CP2))=1

CP2×CP2(18b1+9b2)f21f22=1

(18b1+9b2)CP2×CP2f21f22=1

18b1+9b2=1

Solving the equations for b1 and b2 we obtain

b1=145

b2=745

then we have that

L2(p1,p2)=145p12+745p2

The Hirzebruch signature formula says that for CP6 we have

τ(CP6)=CP6L3(p1(CP6),p2(CP6),p3(CP6))=1

CP6(343c1+147c2+35c3)h6=1

(343c1+147c2+35c3)CP6h6=1

343c1+147c2+35c3=1

The Hirzebruch signature formula says that for CP2×CP2×CP2 we have

τ(CP2×CP2×CP2)=CP2×CP2×CP2L3(p1(CP2×CP2×CP2),p2(CP2×CP2×CP2),p3(CP2×CP2×CP2))=1

CP2×CP2×CP227f12f32f22=1

27(6c1+3c2+c3)CP2×CP2×CP2f12f32f22=1

27(6c1+3c2+c3)=1

The Hirzebruch signature formula says that for CP2×CP4 we have

τ(CP2×CP4)=CP2×CP4L3(p1(CP2×CP4),p2(CP2×CP4),p3(CP2×CP4))=1

CP2×CP415f2g4(15c1+7c2+2c3)=1

15(15c1+7c2+2c3)CP2×CP4f2g4=1

15(15c1+7c2+2c3)=1

Solving the equations for c1, c2 and c3 we obtain

c1=2945

c2=13945

c3=62945

then we have that

L3=2945p1313945p1p2+62945p3

answered Sep 20, 2015 by juancho (1,130 points) [ revision history ]
edited Sep 29, 2015 by juancho

But I've still never understood why the L-genus looks like it does. Is there some differential equation I expect it to satisfy?

Your question is very interesting and the answer is the theory of the "Elliptic genus".  I am planning to make a post with some explanations about how the Hirzebruch genus arises from the elliptic genus.  For the time is being please look at https://en.wikipedia.org/wiki/Genus_of_a_multiplicative_sequence#L_genus_and_the_Hirzebruch_signature_theorem

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