An exercise in Geometric Topology for physicists

Question

Compute the first, second and third Hirzebruch $L$-polynomials $L_1(p_1)$, $L_2(p_1,p_2)$ and $L_3(p_1,p_2,p_3)$  using the fact that $\mathbb{C}P^2$, $\mathbb{C}P^4$, $\mathbb{C}P^6$, $\mathbb{C}P^2 \times\mathbb{C}P^2$, $\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2$ and $\mathbb{C}P^4 \times\mathbb{C}P^2$  have signature 1 and the Hirzebruch signature formula.·

Answer 1

We look for Hirzebruch $L$-polynomials with the form

$$L_1(p_1)=ap_1$$

$$L_2(p_1,p_2)=b_1p_1^2+b_2p_2$$

$$L_3(p_1,p_2,p_3)=c_1p_1^3+c_2p_1p_2+c_3p_3$$

where the coefficients $a,b_1, b_2,c_1,c_2.c_3$ must be computed.

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

$$\tau \left( {M}^{4} \right) =\int_{M^4}L_{{1}} \left( p_{{1}} \left( {M}^
{4}\right)  \right) $$

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

$$\tau \left( {M}^{8} \right) =\int_{M^8}L_{{2}} \left( p_{{1}} \left( {M}^
{8} \right) ,p_{{2}} \left( {M}^{8} \right)  \right) $$

The Hirzebruch signature formula says that for any closed smooth oriented 12-manifold $M^{12}$ we have

$$\tau \left( {M}^{12} \right) =\int_{M^{12}}L_{{3}} \left( p_{{1}} \left( {M}
^{12} \right) ,p_{{2}} \left( {M}^{12} \right) ,p_{{3}} \left( {M}^{12
} \right)  \right)$$

The total Chern classes for the considered complex projective spaces are

$$c (\mathbb{C}P^{2})=1+3\,f+3\,{f}^{2}$$

$$c (\mathbb{C}P^{4})= 1+5\,g+10\,{g}^{2}+10\,{g}^{3}+5\,{g}^4$$

$$c (\mathbb{C}P^{6})=1+7\,h+21\,{h}^{2}+35\,{h}^{3}+35\,{h}^{4}+21\,{h}^{5}+7\,{h}^{6}$$

$c(\mathbb{C}P^2 \times\mathbb{C}P^2)=\\1+3\,f_{{2}}+3\,f_{{1}}+3\,{f_{{2}}}^{2}+9\,f_{{1}}f_{{2}}+3\,{f_{{1}}
}^{2}+9\,f_{{1}}{f_{{2}}}^{2}+9\,{f_{{1}}}^{2}f_{{2}}+9\,{f_{{1}}}^{2}{f_{{2}}}^{2}$

$c(\mathbb{C}P^2 \times\mathbb{C}P^2\times\mathbb{C}P^2)=\\1+ \left( 3\,f_{{3}}+3\,f_{{2}}+3\,f_{{1}} \right) +\\ \left( 3\,{f_{{3}}}^{2}+9\,f_{{3}}f_{{2}}+9\,f_{{3}}f_{{1}}+3\,{f_{{2}}}^{2}+9\,f_{{1
}}f_{{2}}+3\,{f_{{1}}}^{2} \right) + \\\left( 9\,{f_{{3}}}^{2}f_{
{2}}+9\,{f_{{3}}}^{2}f_{{1}}+9\,f_{{3}}{f_{{2}}}^{2}+27\,f_{{3}}f_{{1}
}f_{{2}}+9\,f_{{3}}{f_{{1}}}^{2}+9\,f_{{1}}{f_{{2}}}^{2}+9\,{f_{{1}}}^
{2}f_{{2}} \right) +\\ \left( 9\,{f_{{3}}}^{2}{f_{{2}}}^{2}+27\,{
f_{{3}}}^{2}f_{{1}}f_{{2}}+9\,{f_{{3}}}^{2}{f_{{1}}}^{2}+27\,f_{{3}}f_
{{1}}{f_{{2}}}^{2}+27\,f_{{3}}{f_{{1}}}^{2}f_{{2}}+9\,{f_{{1}}}^{2}{f_
{{2}}}^{2} \right) +\\ \left( 27\,{f_{{3}}}^{2}f_{{1}}{f_{{2}}}^{
2}+27\,{f_{{3}}}^{2}{f_{{1}}}^{2}f_{{2}}+27\,{f_{{1}}}^{2}{f_{{2}}}^{2
}f_{{3}} \right) +27\,{f_{{1}}}^{2}{f_{{2}}}^{2}{f_{{3}}}^{2}$

$c(\mathbb{C}P^2 \times\mathbb{C}P^4)=\\1+ \left( 3\,f+5\,g \right) + \left( 3\,{f}^{2}+15\,gf+10\,{g}^{2} \right) + \\\left( 15\,g{f}^{2}+30\,{g}^{2}f+10\,{g}^{3}
 \right) + \left( 30\,{g}^{2}{f}^{2}+30\,{g}^{3}f+5\,{g}^{4}
 \right) +\\ \left( 30\,{g}^{3}{f}^{2}+15\,{g}^{4}f \right) +15\,{g}^{4}{f}^{2}$

where the cohomological generators are normalized according to

$$\int_{\mathbb{C}P^2}f^2 = 1$$

$$\int_{\mathbb{C}P^4}g^4 = 1$$

$$\int_{\mathbb{C}P^6}h^6 = 1$$

$$\int_{\mathbb{C}P^2 \times\mathbb{C}P^2}f_1^2f_2^2 =\int_{\mathbb{C}P^2}f_1^2 \int_{\mathbb{C}P^2}f_2^2 = (1)(1)=  1$$

$$\int_{\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2}f_1^2f_2^2f_3^2 =\int_{\mathbb{C}P^2}f_1^2 \int_{\mathbb{C}P^2}f_2^2 \int_{\mathbb{C}P^2}f_3^2 = (1)(1)(1)=  1$$

$$\int_{\mathbb{C}P^2 \times\mathbb{C}P^4}f^2g^4=\int_{\mathbb{C}P^2}f^2 \int_{\mathbb{C}P^4}g^4 = (1)(1)=  1$$

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

$$p_1= -2\,c_{{2}}+{c_{{1}}}^{2}$$

$$p_{{2}}=-2\,c_{{1}}c_{{3}}+{c_{{2}}}^{2}+2\,c_{{4}}$$

$$p_{{3}}=-2\,c_{{2}}c_{{4}}+{c_{{3}}}^{2}-2\,c_{{6}}+2\,c_{{1}}c_{{5}}$$

we obtain that

$$p_{{1}}(\mathbb{C}P^2) =3\,{f}^{2}$$

$$p_{{1}}(\mathbb{C}P^4) =5\,{g}^{2}$$

$$p_{{1}}(\mathbb{C}P^6) =7\,{h}^{2}$$

$$ p_1(\mathbb{C}P^2 \times\mathbb{C}P^2) =3\,{f_{{2}}}^{2}+3\,{f_{{1}}}^{2}$$

$$ p_1(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2 )= 3\,{f_{{3}}}^{2}+3\,{f_{{2}}}^{2}+3\,{f_{{1}}}^{2}$$

$$ p_1(\mathbb{C}P^2 \times\mathbb{C}P^4) =3\,{f}^{2}+5\,{g}^{2}  $$

$$p_{{2}}(\mathbb{C}P^4) =10\,{g}^{4}$$

$$p_{{2}}(\mathbb{C}P^6) =21\,{h}^{4}$$

$ p_2(\mathbb{C}P^2 \times\mathbb{C}P^2) =9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9\,{f_{{2}}}^{4}+9\,{f_{{1}}}^{4}=9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9(0)+9(0)=9\,{f_{{1}}}^{2}{f_{{2}}}^{2}$

$ p_2(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)=\\9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9\,{f_{{2}}}^{4}+9\,{f_{{1}}}^{4}+9\,{f_{{3}}}^{2}{f_{{2}}}^{2}+9\,{f_{
{3}}}^{2}{f_{{1}}}^{2}+9\,{f_{{3}}}^{4}=\\ \\9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9(0)+9(0)+9\,{f_{{3}}}^{2}{f_{{2}}}^{2}+9\,{f_{
{3}}}^{2}{f_{{1}}}^{2}+9(0)=\\9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9\,{f_{{3}}}^{2}{f_{{2}}}^{2}+9\,{f_{
{3}}}^{2}{f_{{1}}}^{2}$

$$ p_2(\mathbb{C}P^2 \times\mathbb{C}P^4) =15\,{g}^{2}{f}^{2}+10\,{g}^{4}+9\,{f}^{4}=15\,{g}^{2}{f}^{2}+10\,{g}^{4}$$

$$p_{{3}}(\mathbb{C}P^6) = 35\,{h}^{6}$$

$p_3(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)=\\ 27\,{f_{{1}}}^{2}{f_{{2}}}^{2}{f_{{3}}}^{2}+27\,{f_{{1}}}^{4}{f_{{2}}}^{2}+27\,{f_{{1}}}^{2}{f_{{2}}}
^{4}+27\,{f_{{3}}}^{2}{f_{{1}}}^{4}+27\,{f_{{3}}}^{2}{f_{{2}}}^{4}+27
\,{f_{{3}}}^{4}{f_{{1}}}^{2}+\\27\,{f_{{3}}}^{4}{f_{{2}}}^{2} =27\,{f_{{1}}}^{2}{f_{{2}}}^{2}{f_{{3}}}^{2}$

$$ p_3(\mathbb{C}P^2 \times\mathbb{C}P^4) =30\,{g}^{4}{f}^{2}+45\,{g}^{2}{f}^{4}= 30\,{g}^{4}{f}^{2}+45\,{g}^{2}(0)=30\,{g}^{4}{f}^{2}$$

Using these results we have that

$$L_1(p_1(\mathbb{C}P^2))=ap_1(\mathbb{C}P^2)=a(3\,{f}^{2})= 3af^2$$

$L_2(p_1(\mathbb{C}P^4),p_2(\mathbb{C}P^4))=b_1p_1(\mathbb{C}P^4)^2+b_2p_2(\mathbb{C}P^4)=b_1(5\,{g}^{2})^2+b_2(10\,{g}^{4})=\\ 25b_1g^4+10b_2g^4= (25b_1+10b_2)g^4$

$L_2(p_1(\mathbb{C}P^2 \times\mathbb{C}P^2),p_2(\mathbb{C}P^2 \times\mathbb{C}P^2))=\\b_1p_1(\mathbb{C}P^2 \times\mathbb{C}P^2)^2+b_2p_2(\mathbb{C}P^2 \times\mathbb{C}P^2)= b_1(3f_1^2+3f_2^2)^2+b_2(9f_1^2 f_2^2)=\\b_1(9f_1^4+18f_1^2f_2^2+9f_2^4)+9b_2f_1^2f_2^2=18b_1f_1^2f_2^2+9b_2f_1^2f_2^2=(18b_1+9b_2)f_1^2f_2^2$

$L_3(p_1(\mathbb{C}P^6),p_2(\mathbb{C}P^6),p_3(\mathbb{C}P^6))=\\c_1p_1(\mathbb{C}P^6)^3+c_2p_1(\mathbb{C}P^6)p_2(\mathbb{C}P^6)+c_3p_3(\mathbb{C}P^6)=\\ c_1(7h^2)^3+c_2(7h^2)(21h^4)+c_3(35h^6)=343c_1h^6+147c_2h^6+35c_3h^6   =\\ (343c_1+147c_2+35c_3)h^6$

$L_3(p_1(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2),p_2(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2),p_3(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2))=\\c_1p_1(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)^3+\\c_2p_1(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)p_2(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)+c_3p_3(\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2)=\\c_{{1}} \left( 3\,{f_{{1}}}^{2}+3{f_{{2}}}^{2}+3\,{f_{{3}}}^{2}
 \right) ^{3}+\\c_{{2}} \left( 3\,{f_{{1}}}^{2}+3{f_{{2}}}^{2}+3\,{f_{{3}
}}^{2} \right)  \left( 9\,{f_{{1}}}^{2}{f_{{2}}}^{2}+9\,{f_{{3}}}^{2}{
f_{{2}}}^{2}+9\,{f_{{3}}}^{2}{f_{{1}}}^{2} \right) +27\,c_{{3}}{f_{{1}
}}^{2}{f_{{2}}}^{2}{f_{{3}}}^{2}=\\27\,{f_{{1}}}^{2}{f_{{3}}}^{2}{f_{{2}}}^{2} \left( 6\,c_{{1}}+3\,c_{{2
}}+c_{{3}} \right)$

$L_3(p_1(\mathbb{C}P^2 \times\mathbb{C}P^4),p_2(\mathbb{C}P^2 \times\mathbb{C}P^4),p_3(\mathbb{C}P^2 \times\mathbb{C}P^4))=\\c_1p_1(\mathbb{C}P^2 \times\mathbb{C}P^4)^3+c_2p_1(\mathbb{C}P^2 \times\mathbb{C}P^4)p_2(\mathbb{C}P^2 \times\mathbb{C}P^4)+c_3p_3(\mathbb{C}P^2 \times\mathbb{C}P^4)=\\c_1(3f^2+5g^2)^3+c_2(3f^2+5g^2)(15g^2f^2+10g^4)+c_3(30g^4f^2)=\\15\,{f}^{2}{g}^{4} \left( 15\,c_{{1}}+7\,c_{{2}}+2\,c_{{3}} \right) $

Now, the Hirzebruch signature formula says that for $\mathbb{C}P^2$ we have

$$\tau \left( \mathbb{C}P^2 \right) =\int_{\mathbb{C}P^2}L_{{1}} \left( p_{{1}} \left(\mathbb{C}P^2\right)  \right)=1 $$

$$\int_{\mathbb{C}P^2}3af^2=1 $$

$$3a\int_{\mathbb{C}P^2}f^2=1 $$

$$3a=1$$

$$a=\frac{1}{3}$$

then we obtain

$$L_1(p_1)=\frac{1}{3}p_1$$

The Hirzebruch signature formula says that for $\mathbb{C}P^4$ we have

$$\tau \left( \mathbb{C}P^4 \right) =\int_{\mathbb{C}P^4}L_{{2}} \left( p_{{1}} \left( \mathbb{C}P^4 \right) ,p_{{2}} \left( \mathbb{C}P^4 \right)  \right) =1$$

$$\int_{\mathbb{C}P^4}(25b_1+10b_2)g^4=1$$

$$(25b_1+10b_2)\int_{\mathbb{C}P^4}g^4=1$$

$$25b_1+10b_2=1$$

The Hirzebruch signature formula says that for $\mathbb{C}P^2 \times\mathbb{C}P^2$ we have

$$\tau \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \right) =\int_{\mathbb{C}P^2 \times\mathbb{C}P^2}L_{{2}} \left( p_{{1}} \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \right) ,p_{{2}} \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \right)  \right) =1 $$

$$\int_{\mathbb{C}P^2 \times\mathbb{C}P^2}(18b_1+9b_2)f_1^2f_2^2 =1$$

$$(18b_1+9b_2)\int_{\mathbb{C}P^2 \times\mathbb{C}P^2}f_1^2f_2^2 =1$$

$$18b_1+9b_2=1$$

Solving the equations for $b_1$ and $b_2$ we obtain

$$b_{{1}}=-\frac{1}{45}$$

$$b_{{2}}= \frac{7}{45}$$

then we have that

$$L_{{2}} \left( p_{{1}},p_{{2}} \right) =-\frac{1}{45}{p_{{1}}}^{2}+{\frac {7
}{45}}\,p_{{2}}$$

The Hirzebruch signature formula says that for $\mathbb{C}P^6 $ we have

$$\tau \left( \mathbb{C}P^6 \right) =\int_{\mathbb{C}P^6}L_{{3}} \left( p_{{1}} \left( \mathbb{C}P^6 \right) ,p_{{2}} \left(\mathbb{C}P^6 \right) ,p_{{3}} \left( \mathbb{C}P^6 \right)  \right)=1$$

$$\int_{\mathbb{C}P^6}(343c_1+147c_2+35c_3)h^6=1$$

$$(343c_1+147c_2+35c_3)\int_{\mathbb{C}P^6}h^6=1$$

$$343c_1+147c_2+35c_3=1$$

The Hirzebruch signature formula says that for $\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2$ we have

$$\tau \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2 \right) =\\ \int_{\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2}L_{{3}} \left( p_{{1}} \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2 \right) ,p_{{2}} \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2 \right) ,\\p_{{3}} \left( \mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2 \right)  \right)=1$$

$$ \int_{\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2}27\,{f_{{1}}}^{2}{f_{{3}}}^{2}{f_{{2}}}^{2} =1$$

$$27\left( 6\,c_{{1}}+3\,c_{{2}}+c_{{3}} \right)\int_{\mathbb{C}P^2 \times\mathbb{C}P^2 \times\mathbb{C}P^2}\,{f_{{1}}}^{2}{f_{{3}}}^{2}{f_{{2}}}^{2} =1$$

$$27\left( 6\,c_{{1}}+3\,c_{{2}}+c_{{3}} \right)=1$$

The Hirzebruch signature formula says that for $\mathbb{C}P^2 \times\mathbb{C}P^4$ we have

$\tau \left( \mathbb{C}P^2 \times\mathbb{C}P^4 \right) =\\\int_{\mathbb{C}P^2 \times\mathbb{C}P^4}L_{{3}} \left( p_{{1}} \left( \mathbb{C}P^2 \times\mathbb{C}P^4 \right) ,p_{{2}} \left( \mathbb{C}P^2 \times\mathbb{C}P^4 \right) ,p_{{3}} \left( \mathbb{C}P^2 \times\mathbb{C}P^4 \right)  \right)=1$

$$\int_{\mathbb{C}P^2 \times\mathbb{C}P^4}15\,{f}^{2}{g}^{4} \left( 15\,c_{{1}}+7\,c_{{2}}+2\,c_{{3}} \right) =1$$

$$15\left( 15\,c_{{1}}+7\,c_{{2}}+2\,c_{{3}} \right)\int_{\mathbb{C}P^2 \times\mathbb{C}P^4}{f}^{2}{g}^{4} =1$$

$$15\left( 15\,c_{{1}}+7\,c_{{2}}+2\,c_{{3}} \right)=1$$

Solving the equations for $c_1$, $c_2$ and $c_3$ we obtain

$$c_{{1}}={\frac {2}{945}}$$

$$  c_{{2}}=-{\frac {13}{945}} $$

$$c_{{3}}={\frac {62}{945}}$$

then we have that

$$ L_{{3}}={\frac {2}{945}}\,{p_{{1}}}^{3}-{\frac {13}{945}}\,p_{{1}}p_{{
2}}+{\frac {62}{945}}\,p_{{3}}$$

Comments

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

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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