I am trying to show that following:

Let $M$ be an oriented smooth manifold of dimension $m$, and $N$ be an oriented smooth manifold of dimension $n$. Then $M\times N$ is orientable.

Let $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$ be canonical projections.

Since $M$ and $N$ are orientable, we can find non-vanishing forms $\omega \in \Omega^m(M)$ and $\eta\in \Omega^n(N)$.

I think that $\pi_M^*(\omega)\wedge \pi_N^*(\eta)$ should then be a non-vahinsghing $(m+n)$-form on $M\times N$.

I am unable to show this.

Let us write $\pi_M^*\omega$ as $\omega^*$ and $\pi_N^*\eta$ as $\eta^*$. One of the difficulties I am facing is that $\omega^*\wedge \eta^*$ is $\binom{m+n}{n}\sum_{\sigma\in S_{m+n}} {^\sigma}(\omega^*\otimes \eta^*)$.

I need to operate this thing on a basis of $T_{(p,q)}(M\times N)$, where $p\in M$ and $q\in N$. I know that $T_{(p,q)}(M\times N)\cong T_pM\oplus T_qN$ via the isomorphism $Z\mapsto (d\pi_MZ, d\pi_NZ)$.

Can somebody see what to do from here?

  • 1
    $\begingroup$ Nonvanishing is a condition you can check in local coordinates. Then you are more or less showing that if $f(x_1, \dots x_m) dx_1 \dots dx_m$ is nonvanishing at the origin and $g(y_1, \dots y_n) dy_1 \dots dy_m$ is nonvanishing at the origin then so is their wedge product $f(x_1, \dots x_m) g(y_1, \dots y_m) dx_1 \dots dx_m dy_1 \dots dy_n$. But this is clear. $\endgroup$ – Qiaochu Yuan Apr 9 '15 at 17:47
  • $\begingroup$ Thank you. this odes help. But can you see a way to show that $\omega^*\wedge \eta^*$ (as defined in the post) is nonvanishing? I am unable to deal with the expansion of the wedge using the "Alt" operator. $\endgroup$ – caffeinemachine Apr 10 '15 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.