Canonical orientation on level set submanifolds?

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quasar987
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There's a problem here from past years qualifiers exam that says

"Show that if M is an oriented manifold,F:M-->N is smooth, and c is a regular values of F, then S:=F-1(c) is an orientable submanifold of M."

I know that in the case where N=R, then there is a canonical transverse vector field to S given by grad(F), so this determines an orientation on S. But what about the general case? To generalize the above "argument", we would need dim(N) independant vector fields along S (or sections of the normal bundle of S if you prefer) that are transverse to S... where do we find those??
 

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quasar987
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Mmh, I think I have figured this out.... Basically, the key in the "classical argument" (the case N=R) is that grad(F)|_S gives a global section of the 1-dimensional normal bundle of S... that is to say, a trivialization of nu_S.

More generally, if S is a submanifold (embedded or immersed) of an oriented manifold M, then S is orientable whenever the normal bundle of S is trivial. (Just take a global frame (E_1,...,E_k) of the normal bundle and we define the orientation of T_pS to be the orientation determined by a basis wich completes the frame (E_i(p)) to a positively oriented basis of T_pM.)

In particular, in the case where S=F-1(c), c a regular value, then [tex]F_*:\nu_S\rightarrow T_cN[/tex] is a bundle map, hence triviality of the normal bundle..
 
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lavinia
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There's a problem here from past years qualifiers exam that says

"Show that if M is an oriented manifold,F:M-->N is smooth, and c is a regular values of F, then S:=F-1(c) is an orientable submanifold of M."

I know that in the case where N=R, then there is a canonical transverse vector field to S given by grad(F), so this determines an orientation on S. But what about the general case? To generalize the above "argument", we would need dim(N) independant vector fields along S (or sections of the normal bundle of S if you prefer) that are transverse to S... where do we find those??


Your idea seems right. I think the inverse image of the tangent bundle at c is trivial so the normal bundle over S is trivial and is thus orientable. I am just thinking that the inverse image of any bundle over a point is trivial and the derivative is a bundle map from the normal bundle of S to the tangent bundle at c. I guess a global basis would be the inverse image of a basis for the tangent bundle at c.
 
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quasar987
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I guess a global basis would be the inverse image of a basis for the tangent bundle at c.
Indeed.


The question then follows up with asking if the result remains true when the ambiant manifold M is not orientable. In other words, is every level set submanifold orientable? I assume the answer is no, but what is a counterexample?

I tried RP^2 in RP^4 as the preimage of 0 by the map f:RP4-->R:[x1:...:x5]-->1-(x12+...+x32), but 0 is not a regular value of that map!

Any idea?
 
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lavinia
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Indeed.


The question then follows up with asking if the result remains true when the ambiant manifold M is not orientable. In other words, is every level set submanifold orientable? I assume the answer is no, but what is a counterexample?

I tried RP^2 in RP^4 as the preimage of 0 by the map f:RP4-->R:[x1:...:x5]-->1-(x12+...+x32), but 0 is not a regular value of that map!

Any idea?

I'm not sure what you are asking but here is an example to show you why I am confused.

Take a non-orientable manifold Cartesian product the circle and map this Cartesian product onto the circle by projection. Then each level set is a non-orientable manifold.
 
  • #6
quasar987
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lavinia, recall that the original question was

"Show that if M is an oriented manifold,F:M-->N is smooth, and c is a regular values of F, then S:=F-1(c) is an orientable submanifold of M."

The question then asks

"Is this still true if M is not orientable?"

By your clever counter-example, the answer is no! :) Thanks!
 

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