# Prove the product of orientable manifolds is again orientable

1. May 1, 2012

### hatsoff

1. The problem statement, all variables and given/known data

Let M and N be orientable m- and n-manifolds, respectively. Prove that their product is an orientable (m+n)-manifold.

2. Relevant equations

An m-manifold M is orientable iff it has a nowhere vanishing m-form.

3. The attempt at a solution

I assume I would take nowhere vanishing m- and n-forms f and g on M and N, respectively, and use them to construct an (m+n)-form h on MxN. However I don't know how this construction would proceed. Any help would be much appreciated.

Last edited: May 1, 2012
2. May 1, 2012

### Office_Shredder

Staff Emeritus
I'm assuming that the word orientable in the problem statement is just missing. Given an m-form and an n-form there's only one real way to ever construct an m+n form. It might help to remember/prove that the cotangent bundle of MxN is TM*xTN*

3. May 1, 2012

### hatsoff

Thanks for the response. My first thought is to let

$$\varphi(m,n)(x_1,\cdots,x_{m+n})=f(m,n)(x_1,\cdots,x_m)+g(m,n)(x_{m+1},\cdots,x_{m+n})$$

so that each φ(m,n) is (m+n)-multilinear, and then apply the alternation mapping A to get the antisymmetric multilinear map h(m,n)=A(φ(m,n)), that is,

$$(h)(m,n)(x_1,\cdots,x_{m+n})=\frac{1}{(m+n)!}\sum_{ \sigma\in S_{m+n}}(\text{sgn }\sigma)\varphi(m,n)(x_{ \sigma(1)},\cdots,x_{ \sigma (m+n)})$$

But there is so much about that map which I wouldn't know how to prove. For instance, is h a diffeomorphism with its image? I know that it maps into the set of (m+n)-multilinear maps from R^{(m+n)(m+n)} into R, but is it really surjective like I need? And is it nowhere vanishing? If I knew in advance that the answers to these questions were all "yes," then I wouldn't mind spending a lot of time trying to prove it. But I don't know any of that, and it's very frustrating.

As to the cotangent bundle, I'm not sure how that would help. In fact I had to look it up on wikipedia, since I've never encountered it before.