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In the first reply to this post, I made a comment, but nobody answered me yet. Link: Orientability of a product of smooth manifolds implies orientability of each factor

My problem is in the return of the affirmation $(\Leftarrow)$.

Problem statement:

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.

First reply from rmdmc89:

$1)$ If $M^m,N^n$ are both orientable, there are volume forms (i.e., non-zero top-forms) $\omega\in\Omega^m(M)$ and $\sigma\in\Omega^n(N)$. Define $\eta\in\Omega^{n+m}(M)$ as: $$\eta(X_1,...,X_n,Y_1,...,Y_m):=\omega(X_1,...,X_n)\sigma(Y_1,...,Y_m)$$

which is a volume form on $M\times N$, so $M\times N$ is orientable.

$2)$ Conversely, if $M\times N$ is orientable, there is a volume form $\eta\in\Omega^{n+m}(M\times N)$. Fix a point $q\in N$ and a basis $\left\{\left.\frac{\partial}{\partial y_1}\right|_q, ...,\left.\frac{\partial}{\partial y_n}\right|_q\right\}$ a for $T_qN$. Define $\omega\in\Omega^m(M)$ as: $$w(X_1,...,X_n):=\eta_{(\cdot,q)}\left(X_1,...,X_n,\left.\frac{\partial}{\partial y_1}\right|_q,...,\left.\frac{\partial}{\partial y_m}\right|_q\right)$$

which is a non-zero top-form on $M$, so $M$ is orientable. We can make a similar construction for $N$.

My comment on the first answer:

"I think it's wrong at the beginning of "Define $η∈Ω^{n+m}(M)$" must be "Define $η∈Ω^{n+m}(M\times N)$". And even then, I think the product is not well defined. In $Λ^{m+n}(M×N)$ has more vectors than ωσ 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. Am I correct?"

How to correct?

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  • $\begingroup$ You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment. $\endgroup$ – Saucy O'Path Jun 10 '18 at 20:00
  • $\begingroup$ @SaucyO'Path Yes! $\endgroup$ – Mancala Jun 10 '18 at 20:01
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You’re right that $\eta \in \Omega^{n+m} (M \times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $\omega \in \Omega^m(M)$ and $\sigma \in \Omega^n(N)$ be two volume forms. Let $\pi_M : M\times N \to M$ and $\pi_N : M\times N \to N$ be the canonical projection maps. Then let $\eta = \pi_M^\ast(\omega) \wedge \pi_N^\ast(\sigma)$. This is a volume form on $M \times N$ and is more or less the $\eta$ defined in the original response.

Edit: As Shifrin points out below, the form $\eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.

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  • $\begingroup$ Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :( $\endgroup$ – Mancala Jun 10 '18 at 20:30
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    $\begingroup$ Why do you think the product isn’t well defined? As for $\bigwedge^{m+n}(M \times N)$, this accepts $m+n$ vector (fields) as does $\omega \sigma$. Where do you see a mismatch? $\endgroup$ – Osama Ghani Jun 10 '18 at 20:32
  • $\begingroup$ I'm going to write what I do not understand: $\endgroup$ – Mancala Jun 10 '18 at 20:35
  • $\begingroup$ No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $M\times N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product. $\endgroup$ – Ted Shifrin Jun 10 '18 at 20:47
  • $\begingroup$ Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula. $\endgroup$ – Osama Ghani Jun 10 '18 at 21:04

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