# Rewriting a integral using a pullback between manifolds with different dimensions

Let $M$, $N$ be differentiable manifolds, let $f: M \to N$ be a smooth map. Let $\omega \in \Omega^{dim(N)}(N)$, a dim(N)-form on $N$. Consider the integral:

$$\int_N \omega$$

We know that in the case that $M$ and $N$ have the same dimensions and f is a diffeomorphism, we have:

$$\int_M f^*\omega = \int_N \omega$$

My question is the following: when $M$ and $N$ have different dimensions, say $dim(M) > dim(N)$ and $f$ is surjective, is there a way to express $\int_N \omega$ as a integral over $M$ using the pullback $f^*$ ?

• interesting question, I'm not sure I know, but, for the standard concept of integration of a form we'd need $f^* \omega$ is an $dim(M)$-form. So, how can we naturally add degree? If $M$ and $N$ were related by Hodge duality perhaps? I can trade two-forms on surfaces for one-forms along curves in three dimensions. I hope somebody answers your question. Jun 16, 2015 at 2:00
• @John: How does $\int_M f^* \omega$ even make sense when the degree of $\omega$ is not the dimension of $M$? Jun 16, 2015 at 8:49
• Perhaps one could take something like $f^* \omega \wedge \mu$ where $\mu \in \Omega^{m-n}(M)$ integrates to 1 along each fiber of $f$? Feels like it might work but I haven't actually computed anything. Will need some extra conditions on $f$ for it to work out - the picture in my head is probably a bundle but may require even more structure. Jun 16, 2015 at 9:04
• Small editorial remark: When $f$ is a diffeomorphism, you only know $\int_M f^*\omega = \int_N \omega$ (provided this makes sense) when $f$ is orientation-preserving. Jun 16, 2015 at 13:10
• @TedShifrin correct, will edit
– zzz
Jun 16, 2015 at 17:22

What does make sense when $\dim M>\dim N$ is the Gysin map, usually called integration over the fiber. Assuming everything is compact, for simplicity, and oriented, given a submersion $f\colon M\to N$ and any $\phi\in\Omega^k(M)$, one has $f_*\phi\in\Omega^{k-(\dim M-\dim N)}(N)$. When $k=\dim M$, in particular, you will have $$\int_M\phi = \int_N f_*\phi.$$
• How is defined the push-foward $f_*\phi$? Aug 8 at 18:24