signature of the topological manifold $M= \mathbb{C}P^6\times \mathbb{C}P^6$ (using homology, cohomology)

Question

I want to prove that the signature $\operatorname{sig}(M)$ of the topological manifold $M= \mathbb{C}P^6\times \mathbb{C}P^6$ is nonzero.

First of all, $M$ is a compact 24-dimensional manifold without boundary and $M$ is $\mathbb{Z}$-orientable, because $H_{24}(M;\mathbb{Z})\neq 0$ (here $H_{24}(M;\mathbb{Z})$ denotes the 24-th singular homology group with coefficients in $\mathbb{Z}$). We defined $\operatorname{sig}(M)$ as follows ( https://en.wikipedia.org/wiki/Signature_%28topology%29) :

For a $4k$-dimensional $\mathbb{Z}$-orientable compact manifold without boundary $M$ it is the signature of the symmetric bilinear form $$H^{2k}(M;\mathbb{R})\otimes _{\mathbb{R} }H^{2k}(M;\mathbb{R})\to \mathbb{R}$$ $$\alpha\otimes \beta \mapsto (\alpha\cup\beta )(\mu_M),$$where $\mu_M\in H_{4k}(M;\mathbb{Z})$ is an orientation.

Now $M= \mathbb{C}P^6\times \mathbb{C}P^6$ is a $4k$-manifold with $k=6$ (and the other conditions are satisfied as well, see above), such that $\operatorname{sig}(M)$ is defined. I have two different appoaches:

1.I already know that $\operatorname{sig}(\mathbb{C}P^{2k})=1$ for a fixed, but arbitrary $k\in\mathbb{N}$. Can I conclude from this that $\mathbb{C}P^{2\cdot 3}\times \mathbb{C}P^{2\cdot 3}$ has nonzero signature (and is it possible to determine the signature of $M$ exactly in this case)? If yes, how to do it exactly?

2.I already calculated the homology- and cohomology groups of $M$ with coefficients in $\mathbb{Z}$ and $\mathbb{R}$ (I used the kuenneth-formula) , but I think this could be overkill. And if I have done nothing wrong, $V=H^{12}(M;\mathbb{R})\cong \mathbb{R}^7$ hence $V$ is a 7-dimensional $\mathbb{R}$-vector space, i.e $V$ has odd dimension. I would say from this I can conclude that the signature of $M$ is nonzero ( but I'm not sure), or is it wrong?

Best.

Answer 1

Let $\alpha$ and $\beta$ be the generator in second cohomology of two copies $\mathbb CP^6$, respectively. Then $H^{12}(M)$ is generated by $\alpha^0\beta^6, \alpha^1\beta^5\ldots,\alpha^6\beta^0$. In such basis the matrix of cup-product takes the form $$\left( \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) $$ It is easy exercise that such quadratic form has signature $(4,3)$.

Answer 2

The signature is a ring homomorphism $\Omega_\ast\to\mathbb{Z}$ on the oriented Cobordism ring. In particular one can show that $sig(M\times N) = sig(M)\cdot sig(N)$ holds by using that over the reals $H^\ast(M\times N) = H^\ast(M) \otimes H^\ast(N)$ holds (tensor product of graded commutative rings, i.e. $(a\otimes b)(c\otimes d):=(-1)^{|b| |c|} ac \otimes bd$). $sig(\mathbb{C}P^6\times\mathbb{C}P^6) = 1$ follows immediately.