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Let $\Sigma_g$ denote the compact Riemann surface of genus $g$. I'm trying to compute $c_1(\Sigma_g\times \Sigma_h)$. The tangent bundle $T(\Sigma_g\times \Sigma_h)$ is isomorphic to $p_1^*T\Sigma_g \oplus p_2^* T\Sigma_h$, where $p_1,p_2$ are projections. So $c_1(\Sigma_g\times \Sigma_h)=p_1^*c_1(\Sigma_g)+p_2^*c_1(\Sigma_h)$. Let $[\Sigma_g]\in H_2(\Sigma_g;\Bbb Z)$ be the fundamental class so that $\langle \alpha_g, [\Sigma_g] \rangle=1$, where $\alpha_g$ is the Poincare dual of the class $[pt]\in H_0(\Sigma_g)$. Since $c_1(\Sigma_g)=e(\Sigma_g)$, we must have $c_1(\Sigma_g)=\chi(\Sigma_g)\alpha_g=(2-2g)\alpha_g$ and similarly $c_1(\Sigma_h)=(2-2h)\alpha_h$. Now I am only left to compute $p_1^*\alpha_g$ and $p_2^*\alpha_h$, but I can't see how to do this. How can we express these classes as a linear combination of the Poincare duals of $[\Sigma_g \times pt]$ and $[pt\times \Sigma_h]$?

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The Poincare dual of $[\Sigma_g \times pt]$ is $p_2^*(\alpha_h)$ and similarly the Poincare dual of $[pt \times \Sigma_h]$ is $p_1^* (\Sigma_g)$. To see this we can just compute the intersection number of one submanifold with the other and with itself and then use the Kunneth theorem to decide which cohomology class the Poincare dual is.

There is only a single intersection of $pt \times \Sigma_h$ and $\Sigma_g \times pt$ which occurs at $pt \times pt$. This intersection is transverse since the tangent planes together span the entire tangent space at that point. Of course the question of if this intersection is counted positively or negatively must be asked, and I say this should be counted positively due to how the orientation on a product of oriented manifolds is defined.

Now we compute the intersections with themselves. Let us perturb $pt \times \Sigma_h$ so that it intersects its original position transversely. We do this by seeing that the normal bundle of this submanifold is the vector bundle that sums with the tangent bundle of $\Sigma_h$ to get the restriction of the tangent bundle of $\Sigma_g \times \Sigma_h$ to $pt \times \Sigma_h$. From this description we see that the normal bundle has fiber over every point equal to $T_{pt}\Sigma_g$, i.e. it is trivial. Since the normal bundle is trivial, we pick a section of it and translate $pt \times \Sigma_h$ along it. This perturbation has no intersection with the original submanifold. Similarly for $\Sigma_g \times pt$.

From these two calculations we see that the Poincare dual of $pt \times \Sigma_h$ evaluates on $\Sigma_g \times pt$ to 1 and on $pt \times \Sigma_h$ to 0. From the Kunneth formula for the (co)homology of a product, we see that this is exactly the cohomology class of $p_1^* (\alpha_g)$. Similarly, the Poincare dual of $\Sigma_g \times pt$ is $p_2 ^* (\alpha_h)$.

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  • $\begingroup$ Thanks. I see that the Poincare dual of $pt \times \Sigma_h$ evaluates on $\Sigma_g\times pt$ to 1 and on $pt\times \Sigma_h$ to 0, but how does the Kunneth formula implies that the Poincare dual of $pt\times \Sigma_h$ is $p_1^*\alpha_g$? $\endgroup$
    – user302934
    Mar 31 at 15:20
  • $\begingroup$ @user302934 In the torsion free case (which we are in), the Kunneth theorem says that the cohomology of a product is the tensor product of the cohomologies and that the homology of the product is the tensor product of the homologies. What is more, the homology/cohomology pairing is given by the tensor product of the original pairings. This last piece is what is used here. $\endgroup$ Mar 31 at 15:25
  • $\begingroup$ The homology/cohomology pairing is given by the tensor product of the original pairings. Does this means that $\langle \alpha \otimes \beta, a\otimes b\rangle=\langle \alpha,a\rangle \cdot \langle \beta,b\rangle$? $\endgroup$
    – user302934
    Mar 31 at 15:31
  • $\begingroup$ @user302934 Yes, this is what I mean. $\endgroup$ Mar 31 at 15:46

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