If the Product Manifold MxN is orientable, so are M,N.

[WWGD]

Hi, this is just a review exercise. Let M,N be n- and m- manifolds respectfully , so that the product manifold MxN is orientable. I want to show that both M,N are orientable.

I could do some computations with product open sets of $\mathbb R^n$ , or work with orientation double-covers, but I am trying to work with top/orientation differential forms, so I am thinking of doing the pullback of the top (nowhere-zero) form $w_{n+m}$on MxN into N along the inclusion map, and showing that the pulled-back form is nowhere-zero:

$i^{ *}w_{n+m} : T_{(p,q)}N \times M \rightarrow T_p N: i^{*}(w_{n+m} (X_1,...,X_{n+m}))=w_{n+m}(i(X_1),...,i(X_n))$ , where,

$i: N \rightarrow N \times M : (x_1,...,x_n) \rightarrow (x_1,..,x_n, 0,..,0)$

How do we show this pulled-back form is nowhere-zero on N ? I guess formally, this is the
restriction of an (n+m)-linear map to the first n arguments, where the last (m-n)-arguments are set to 0. Is this clearly a nowhere-zero form?

 

[WWGD]

I saw an argument somewhere in terms of the Stiefel-Whitney class, which is the obstruction to orientability. The argument is that if the Stiefel-Whitney class of the product MxN is zero, it must be the case , using pullbacks by each of the projection maps we show that the Stiefel-Whitney class of each of M,N must also be zero. I will reproduce it in more detail soon.

 

[homeomorphic]

I like thinking about orientation in terms of a preferred equivalence class of basis at each point. The tangent space breaks down into a direct product of the tangent spaces, so just project a preferred basis to each factor to get a preferred basis at each point of M and N. I think that should work, and the argument for other definitions ought to be "homotopic" to the argument for the basis definition, if you know how the definitions are equivalent.

 

[WWGD]

Good idea, homeomorphic, I was thinking of doing something of that sort, where the Jacobian of the chart maps (fxg )(m,n) , where f is a chart map for M and g is a chart map for n (so that the Jacobian of the coordinate change is always non-zero ) can somehow be split into a Jacobian for f or g. It would be great if you could de-wedge the top $(n+m)-$ orientation form into two n- and m- orientation forms respectively for N,M . I guess this is sort-of what I was trying to do when pulling back the top form by the inclusion , but I don't see how that would produce nowhere-zero n- , m- forms.

It may be possible to restrict the determinant of the larger (m+n)- matrix to determinants for the (nxn) and (mxm) sub-matrices and show these submatrices have positive determinant, but I can't see now how to do it. .

 

[lavinia]

Hi, this is just a review exercise. Let M,N be n- and m- manifolds respectfully , so that the product manifold MxN is orientable. I want to show that both M,N are orientable.

I could do some computations with product open sets of $\mathbb R^n$ , or work with orientation double-covers, but I am trying to work with top/orientation differential forms, so I am thinking of doing the pullback of the top (nowhere-zero) form $w_{n+m}$on MxN into N along the inclusion map, and showing that the pulled-back form is nowhere-zero:

$i^{ *}w_{n+m} : T_{(p,q)}N \times M \rightarrow T_p N: i^{*}(w_{n+m} (X_1,...,X_{n+m}))=w_{n+m}(i(X_1),...,i(X_n))$ , where,

$i: N \rightarrow N \times M : (x_1,...,x_n) \rightarrow (x_1,..,x_n, 0,..,0)$

How do we show this pulled-back form is nowhere-zero on N ? I guess formally, this is the
restriction of an (n+m)-linear map to the first n arguments, where the last (m-n)-arguments are set to 0. Is this clearly a nowhere-zero form?

This is the right idea. Use the Kunneth Theorem. Over the reals you should get the orientation forms on M and N.

Over Z/2Z you can use it for your Stiefel Whitney class argument. Note that the tangent bundle of MxN is the Whitney sum of the tangent bundles of the pull backs of the tangent bundles of M and N.