Consider two differentiable manifolds $M^m,N^n$ and the associated set of differential forms for order $k \in \{0,\ldots,m\}$ and $q \in \{0,\ldots,n\}$, respectively $\Omega^k(M^m)$ and $\Omega^q(N^n)$.

Now consider the differentiable manifold $M^m \times N^n$ and the associated set of differential forms for order $l \in \{0,\ldots,m+n\}$, namely $\Omega^l(M^m \times N^n)$. Let's denote the canonical projections from the product space to the components by $\pi_M$ and $\pi_N$.

Well, there are some relations between the $\Omega$'s being considered above.

Notably, we have that $$\begin{matrix} \times :& \Omega^k(M^m) \times \Omega^q(N^n) & \rightarrow & \Omega^{k+l}(M^m \times N^n) \\ & (\omega,\eta) & \mapsto & \omega \times \eta := {\pi_M}^*\omega \wedge {\pi_N}^*\eta \end{matrix}$$ is an injective map.

Even though we can't hope to have this map $\times$ inverted (see these posts (1) and (2)), as it fails to be surjective, my question is if we can put something in the opposite direction (analogous to taking the marginal of a measure on a product space):

Given $l \in \{0,\ldots,m+n\}$, let's fix a decomposition $l = k_0 + q_0$ with $k_0 \in \{0,\ldots,m\}$ and $q_0 \in \{0,\ldots,n\}$. Can we define some natural way a map $$\begin{matrix} \Pi_M: \Omega^l(M^m \times N^n) & \rightarrow & \Omega^{k_0}(M^m) \end{matrix}$$ so that, in particular, $\Pi_M (\omega \times \eta) = \omega$ for any $(\omega, \eta) \in \Omega^{k_0}(M^m) \times \Omega^{q_0}(N^n)$? If so, how?

Well, if so, we should have something similar in the $N$ component. And, just to be clear: we are not hoping that $\xi = \Pi_M(\xi) \times \Pi_N(\xi)$ for every $\xi \in \Omega^l(M^m \times N^n)$.

Also, we can't define something in this spirit but getting rid the decomposition choice $(k_0,q_0)$ above, right?

Thanks in advance!


1 Answer 1


If $N$ is compact and oriented, then for each $k$ there is a linear map $\Pi_M: \Omega^{k+n}(M\times N) \to \Omega^k(M)$ called the pushforward or integration along the fiber. Here's how that works. For any $\alpha\in \Omega^{k+n}(M\times N)$, define a form $\beta= \Pi_M\alpha\in \Omega^k(M)$ as follows: for arbitrary $x\in M$ and $v_1,\dots,v_k\in T_xM$, $$ \beta_{x} (v_{1}, \dots, v_{k}) = \int _{\{x\}\times N} {\tilde v}_{1} \ \lrcorner\ {\tilde v}_{2} \ \lrcorner\ \dots \ \lrcorner\ {\tilde v}_{k} \ \lrcorner\ \alpha, $$ where $\tilde v_j|_{(x,y)} = v_j \oplus 0 \in T_x M\oplus T_y N \cong T_{(x,y)}(M\times N)$. In particular, if $\alpha = \omega\times\eta$, then \begin{align*} \beta_{x} (v_{1}, \dots, v_{k}) &= \int _{\{x\}\times N} \big( ​{\tilde v}_{1} \ \lrcorner\ {\tilde v}_{2} \ \lrcorner\ \dots \ \lrcorner\ {\tilde v}_{k} \ \lrcorner\ \pi_M^*\omega\big)\pi_N^*\eta\\ &= \int _{\{x\}\times N} \big({v}_{1} \ \lrcorner\ {v}_{2} \ \lrcorner\ \dots \ \lrcorner\ {v}_{k} \ \lrcorner\ \omega_x\big)\pi_N^* \eta\\ &=\omega_x(v_1,\dots,v_k)\int _{N} \eta, \end{align*} so $\Pi_M(\omega\times\eta) = \Big(\int_N\eta\Big)\omega$.

You can use this operation to define a left inverse for $\pi_M^*\colon \Omega^k(M\times N)\to \Omega^k(M)$: just choose $\eta\in \Omega^n(N)$ such that $\int_N\eta=1$ (such a form exists because $N$ is compact and oriented), and define $\rho\colon \Omega^k(M\times N) \to \Omega^k(N)$ by $\rho\alpha = \Pi_M(\alpha\wedge \pi_N^*\eta)$. The calculation above shows that $\rho(\pi_M^*\omega) = \omega$.

An important feature of these maps is that they both commute with exterior differentiation: $\Pi_M(d\alpha) = d(\Pi_M\alpha)$ and $\rho(d\alpha) = d(\rho\alpha)$. The first equation can be proved by using a partition of unity to break up $\alpha$ into pieces that are compactly supported in product coordinate charts and doing the computation there; and the second one follows from the first because $d(\pi_N^*\eta)=\pi_N^*(d\eta)=0$. Thus both maps descend to maps on de Rham cohomology.

  • $\begingroup$ I get it! Many thanks for your answer! As we are integrating on $\{x\} \times N$, which is very natural, there consequences on the orders. Eg, say $m=n=10$. If $\omega$ and $\eta$ are 3-forms, we can do nothing. If $\omega$ and $\eta$ are 6-forms we can do something, but we won't ever recover any of them. If $\omega$ is a 2-form and $\eta$ a 10-form, we can recover $\omega$ (modulo a constant factor) but not $\eta$. Only if both are 10-forms we'll be able to recover both. I am interpreting this correclty? Also, in your first $=$, you want to use $\alpha$ instead of $\beta$ in the RHS, right? $\endgroup$
    – LuaLua
    Commented Apr 11, 2021 at 0:54
  • $\begingroup$ Yes, you're interpreting it correctly. The only pushforward map (as far as I know) takes $(k+n)$-forms to $k$-forms, where $n=\dim N$. And thanks for the correction -- fixed now. $\endgroup$
    – Jack Lee
    Commented Apr 11, 2021 at 4:15

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