Why has to be M,N be orientable?

[Canavar]

Hello,

I want to solve this Problem:

If M,N are manifolds then MxN is orientable iff M,N is orientable.



I have solved the direction "<=" This was no problem.

But i have a lot of problems to solve the other direction!

Let us assume that MxN are orientable.
Why has to be M,N be orientable?

We have defined orientable in different ways. Once per atlases and also with a differential non-vanishing form...
But i could't see a connection.

Can you please help me?

Regards

 

[fzero]

If MxN is orientable, it has a nonvanishing volume form. Can you relate this form to forms on M and N? Presumably the way you proved the other part of theorem would shed some light on this.

 

[Canavar]

Hello,

There are different äquivalent criteria for a manifold to be orianted.
A manifold is oriented if it has an atlas, s.d. the differential of the coordinate changes have positiv determinant.
Therefore the product MxN has a canonically atlas, which also satisfy this criterion.

I hope you understand me.
We can assume that MxN has an atlas A, s.t. forall maps x,y in A we have
det d(x\circ y^{-1})>0. That is MxN is orientable.

But why has M to be orientable, if we assume that MxN has to be?

Regards

 

[fzero]

You have to use the product structure of MxN in your proof. Is there any way to write an atlas on MxN in terms of an atlas on M and an atlas on N?

 

[Canavar]

Yes of course there is! But not any atlas has to be in that form! This is a real problem.

For example, if (U_i,f_i) is a atlas for M and (V_i,g_i) is a atlas for N => (U_i x V_i, f_i x g_i) is a atlas for MxN.
But it has not to be in this form. That is f,g can depend on both, the element in M and N.

I mean in general the maps have the form (U_i x V_i, h) with h(x,y).

What can i do?

 

[fzero]

A single atlas satisfying the correct conditions should be enough to show orientabliltiy.

If it's too hard to construct, you might try using differential forms, which seems simpler.

 

[Canavar]

Oh i think it is a misunderstanding. You are right, we inly need a single atlas, which is oriented, to show that a manifold is oriented.
But I want to show something else. i want to show tha MxN is not orientable, if M is not. That is i have to show that any atlas is not orientable of MxN!

I hope the problem is now clear.
Once again:
I want to show:

If M is not orientable =>MxN is not orientable

I have no idea how i can show this. Neither with atlases nor with differential forms.

Regards

 

[fzero]

There is a relationship between the volume form on MxN and those on M and N. You can use that.