$M \times N$ orientable if and only if $M, N$ orientable

Question

For two manifolds $M$ and $N$ I'm trying to prove that $M \times N$ is orientable if and only if $M$ and $N$ are orientable.

My attempt so far:

$\impliedby)$ Assume $M, N$ are orientable. Then $\widetilde{M}$ and $\widetilde{N}$ the orientation double covers of $M$ and $N$ are disconnected. But $\widetilde{M} \times \widetilde{N}$ is the orientation double cover for $M \times N$. Since $\widetilde{M}, \widetilde{N}$ are disconnected certainly $\widetilde{M} \times \widetilde{N}$ is disconnected. Therefore, $M \times N$ is orientable.

$\implies)$ Proceed by contradiction. Assume $M \times N$ is orientable and WLOG $M$ is nonorientable. Then the orientation double cover $\widetilde{M\times N}$ is disconnected. Now note that $\widetilde{M}$ is connected since $M$ is nonorientable. So $\widetilde{M} \times \widetilde{N}$ is connected since the product of a connected space and a disconnected space is connected. But $\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}$, so $\widetilde{M\times N}$ is connected. But this is a contradiction, therefore it must be that $M$ is orientable.

This completes our proof.

The validity of this proof hinges on two assumptions that I'm unsure about: 1. $\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}$ 2. The product of a connected space and a disconnected space is connected.

Are these two facts true? Is my proof valid? How do I go about proving these two necessary assumptions if they are indeed true? If I'm totally going the wrong way could someone send me in the proper direction?

Thanks

Answer 1

I'll use that a manifold $X$ is orientable if and only its first Stiefel-Whitney class vanishes: $$w_1(M)\stackrel {\text {def}}{=} w_1(TM)=0\in H^1(X,\mathbb Z/2)$$ Now back to our problem:
If $p_M:M\times N\to M$ and $p_N:M\times N\to N$ are the projections we have $T(M\times N)=p_M^*TM \oplus p_N^*TN$ which allows us to write $$w_1(M\times N)=(w_1M ,w_1M)\in H^1(M\times N,\mathbb Z/2)=H^1(M,\mathbb Z/2)\oplus H^1(N,\mathbb Z/2)$$ Thus $$w_1(M\times N)=0\iff w_1(M)=0 \operatorname {and} w_1(N)=0$$ which proves that $$ M\times N \operatorname {is orientable} \iff M \operatorname {and} N \operatorname {both are} .$$

Answer 2

Your method is flawed, as I mentioned in my comments.

Here's an outline:

1. Prove that if $M$ is not orientable, then $M \times \mathbb{R}^k$ is not orientable (any $k$).

   Hint: Show that $\widetilde{M \times \mathbb{R}^k} \cong \widetilde{M} \times \mathbb{R}^k$.

2. Prove that if $U$ is an open subset of a manifold $A$ (hence is a manifold of the same dimension) and $U$ is not orientable, then $A$ is not orientable.

3. Prove that $M \times N$ has an open subset diffeomorphic to $M \times \mathbb{R}^k$, for $k$ the dimension of $N$.

   (Equivalently $M \times B^k$ for $B^k$ the unit open ball, since that's diffeomorphic to $\mathbb{R}^k$.)

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A fancy perspective on what you were trying to do (ignore this if it doesn't make sense):

The space of orientations on $M$ is naturally a $\mathbb{Z}/2\mathbb{Z}$ principal bundle. That is to say, it's a manifold $\widetilde M$ admitting a projection $\pi: \widetilde M \rightarrow M$ such that the fiber over (i.e. pre image of) any point admits an action by $\mathbb{Z}/2\mathbb{Z}$, and such that every point $p \in M$ has an open neighborhood $U$ such that $\pi^{-1}(U) \cong U \times \mathbb{Z}/2\mathbb{Z}$ (such that here the action of $\mathbb{Z}/2\mathbb{Z}$ is by its own group operation.)

When we take $M \times N$, this admits the pullbacks $\pi_1^* \widetilde M$ and $\pi_2^* \widetilde N$, by the two projections $\pi_1: M\times N \rightarrow M$ and $\pi_2: M\times N \rightarrow N$, of these two $\mathbb{Z}/2\mathbb{Z}$ principal bundles.

Their fiber-wise product $\pi_1^* \widetilde M \times_{M\times N} \pi_2^* \widetilde N$ (which is diffeomorphic to $\widetilde M \times \widetilde N$ as a manifold) is not what you're interested in (it would have four points in every fiber).

Instead, the fibers of each admit an action by $\mathbb{Z}/2\mathbb{Z}$ and the simultaneous action of $\mathbb{Z}/2\mathbb{Z}$ on both should act trivially (swapping the orientation on each factor keeps the orientation the same), so we mod out by this action while taking the fiber product, getting only two points in each fiber, and call this $\pi_1^* \widetilde M \times_{\mathbb{Z}/2\mathbb{Z}} \pi_2^* \widetilde N$. This still admits an action of $\mathbb{Z}/2\mathbb{Z}$ (on one factor at a time) and is a $\mathbb{Z}/2\mathbb{Z}$ principal bundle over $M \times N$.

Then in fact $\widetilde{M\times N}$ is isomorphic (as principal bundle) to $\pi_1^* \widetilde M \times_{\mathbb{Z}/2\mathbb{Z}} \pi_2^* \widetilde N$.

Of course this doesn't prove your result; it's just an expression for the orientation double cover. You still need to do something like restrict to an open subset $M \times \mathbb{R}^k$ in this case, as suggested in the outline above, to complete the proof (which you can do without all this fanciness just about as easily).

Answer 3

You can define , for an m-manifold $M$ and an n-manifold $N$ the form $w_M \times w_N$ in $M \times N$ , with $w_M \times w_N(X,Y)=w_M(X)w_N(Y)$ as a top , nowhere-zero $(m+n)-$ form on $M \times N$. Conversely, since $T_{(x,y)}M \times N=T_xM (+)T_y N$ then $T(M\times N)=TM(+)TN$, do something for the $(m+n)$-th exterior power of the product bundle and work out the rest.