# Orientability of a product of smooth manifolds implies orientability of each factor

I've been learning a bit about orientability on smooth manifolds. I'm having torubles with this exercise:

Given two smooth manifolds $M$ and $N$, show that the product manifold $M \times N$ is orientable if and only if $M$ and $N$ are orientable.

Using the orientability of $M$ and $N$ one can obtain an oriented atlas for each manifold and the construct an oriented atlas for the product formed by the product charts (i.e. charts of the form $(U \times V, \phi \times \psi)$ with $(U, \phi)$ a chart of $M$ and $(V, \psi)$ a chart of $N$).

I'm stuck on proving the converse. Given an oriented atlas for $M \times N$ one can obtain another oriented atlas formed by charts with basic open sets as domains. But from this atlas I don't know how to extract an atlas for $M$.

I was given the following hint: If $M \times N$ is orientable then $M \times \mathbb{R}^n$ is orientable where $n$ is the dimension of $N$. I don't know either how to prove it nor how to use it.

• Do you know anything about deRham cohomology?
– user99914
Jul 20, 2014 at 3:34
• No, I don't. Is there a reasonable proof without using deRahm cohomology? Jul 20, 2014 at 4:07
• I guess there is but I cannot figure it out. To reduce to the case $M \times \mathbb R^n$, find an open set $U$ in $N$ diffeomorphic to $\mathbb R^n$. Thus $M \times U$ is open in $M\times N$ and thus is orientable (every open set in an orientable manifold is orientable).
– user99914
Jul 20, 2014 at 4:24
• So the point is to show: $M$ is orientable if and only if $M \times \mathbb R$ is orientable.
– user99914
Jul 20, 2014 at 4:25
• Possible duplicate of Product of manifolds & orientability
– user279515
Dec 14, 2018 at 19:46

$$(\Rightarrow)$$ If $$M^m\times N^n$$ is orientable, there is a volume form $$\eta\in\Omega^{n+m}(M\times N)$$. For any fixed point $$q\in N$$, take a basis $$\{w_1,...,w_n\}$$ for $$T_qN$$. Now define $$\omega\in\Omega^m(M)$$ by: $$\omega(X_1,...,X_m):=\eta_{(\cdot,q)}\left(X_1,...,X_m,w_1,...,w_n\right)$$

We prove that $$\omega$$ is a volume form. For a fixed $$p\in M$$, take a basis $$\{v_1,...,v_m\}$$ for $$T_pM$$. Using the identification $$T_{(p,q)}M\times N\equiv T_pM\oplus T_qN$$, then $$\{v_1,...,v_m,w_1,...,w_n\}$$ is a basis for $$T_{(p,q)}M\times N$$. Since $$\eta$$ is a volume form, $$\omega_p(v_1,...,v_m)=\eta_{(p,q)}(v_1,...,v_m,w_1,...,w_n)\neq 0$$, which means that $$\omega$$ is a volume form, so $$M$$ is orientable. By a similar argument, $$N$$ is orientable.$$_\blacksquare$$

$$(\Leftarrow)$$ If $$M,N$$ are orientable, there are volume forms $$\omega\in\Omega^m(M), \sigma\in\Omega^n(N)$$. For the natural projections $$\pi_M:M\times N\to M$$ and $$\pi_N:M\times N\to N$$, define: $$\eta:=\pi_M^*(\omega)\wedge \pi_N^*(\sigma)\in\Omega^{n+m}(M\times N)$$

We prove that $$\eta$$ is a volume form. For a fixed $$x=(p,q)\in M\times N$$, take basis $$\{v_1,...,v_n\}$$ for $$T_pM$$ and $$\{w_1,...,w_n\}$$ for $$T_qN$$. Because $$\omega$$ and $$\sigma$$ are volume forms, we have $$\omega_p(v_1,...,v_m)\neq 0$$ and $$\sigma_q(w_1,...,w_n)\neq 0$$. Since $$\{v_1,...,v_m,w_1,...,w_n\}$$ is a basis for $$T_x(M\times N)$$, it's enough to check that $$\eta_x(v_1,...,v_m,w_1,...,w_n)\neq 0$$. Indeed, noticing that $$(d\pi_M)_x(v_i)=v_i$$, $$(d\pi_M)_x(w_j)=0$$, $$(d\pi_N)_x(v_i)=0$$ and $$(d\pi_N)_x(w_j)=w_j$$, we have: $$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$

• I think it's wrong at the beginning of "Define $\eta\in\Omega^{n+m}(M)$" must be "Define $\eta\in\Omega^{n+m}(M\times N)$". And even then, I think the product is not well defined. In $\Lambda ^{m+n}(M\times N)$ has more vectors than $\omega \sigma$ can actually receive. Am I right? Think of the case $m=n=1$ you will see. I have a problem similar to posted, my idea was to do as you but I found this problem. How to correct? Jun 10, 2018 at 16:35
• Indeed, $\eta\in\Omega^{n+m}(M\times N)$ is defined by $\eta = \pi_1^*\omega\wedge\pi_2^*\sigma$. But your definition is wrong, as vector fields on $M\times N$ needn't necessarily come as vector fields on $M$ or vector fields on $N$. You really need to project vector fields down to the respective factors and you need to use the complete skew symmetry of the wedge product. Jun 10, 2018 at 20:45
• @TedShifrin, you're absolutely right. I've tried to develop your idea, hope it's correct. Jun 11, 2018 at 0:21
• @Mancala, I've followed TedShifrin's suggestion. Thanks for noticing the mistake Jun 11, 2018 at 0:21
• (⇒) How do we know whether $\omega$ is smooth? $w_i$ seem not to be fixed vectors, but rather their images under identifications $T_{(p,q)}M\times N \simeq T_pM \oplus T_qN$ (and the map is different at each point). Nov 10, 2019 at 16:52

A manifold is orientable iff it admits a volume form (a nowhere vanishing top degree alternating differential form). It follows that an open submanifold of an orientable manifold is orientable. Take an open subset $U\subset N$ diffeomorphic to $\mathbb R^n$. Then $M\times U$ is an open submanfold of $M\times N$, hence orientable. Now an orientation on $U$ and $M\times U$ defines an orientation on $M$.