Let $M$ be a compact, connected and oriented smooth manifold of dimension $m$ and let $\pi_1,\pi_2:M\times M\rightarrow M$ be the projections to each factor. Given $\alpha\in H^m(M;\mathbb{R})$, I would like to compute $$ \int_{M\times M}\pi_1^*\alpha\smile\pi_2^*\alpha $$ in terms of $\int_M\alpha$.

My approach is as follows: $$ \int_{M\times M}\pi_1^*\alpha\smile\pi_2^*\alpha=\int_M\alpha\smile\pi_1^!\pi_2^*\alpha=\pi_1^!\pi_2^*\alpha\int_M\alpha, $$ where $\pi_1^!$ denotes integration along the fibre and hence $\pi_1^!\pi_2^*\alpha\in H^0(M;\mathbb{R})\simeq\mathbb{R}$ can be thought as a number. Now let $\iota:F_p\hookrightarrow M\times M$ be the inclusion of the fibre over a point $p\in M$ under the projection $\pi_1$. Then $\pi_1^!\pi_2^*\alpha=\int_{F_p}\iota^*\pi_2^*\alpha=\int_M\alpha$, where the second equality follows from the fact that $\pi_2\circ\iota$ is a diffeomorphism between $F_p$ and $M$.

In conclusion I got $$ \int_{M\times M}\pi_1^*\alpha\smile\pi_2^*\alpha=\left(\int_M\alpha\right)^2. $$

Everything looks correct to me, but I'll appreciate a double check.

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    $\begingroup$ You're using machine guns to kill a fly. It's just Fubini's Theorem (which essentially goes into integration over the fiber in the general case). $\endgroup$ – Ted Shifrin Sep 19 '15 at 19:56
  • $\begingroup$ Thank you, Ted. You are right. The case I asked about is just a base for a more involved computation I have to carry out and in the meantime I may have lost a bit of perspective. $\endgroup$ – A. Bellmunt Sep 20 '15 at 10:31
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    $\begingroup$ It might be worth saying that your machine gun fire was accurate. $\endgroup$ – user98602 Sep 20 '15 at 22:27

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