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 Post subject: Vector fields on projective spacePosted: Wed, 13 Jul 2011 03:20:10 UTC
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Question (Milnor/Stasheff again) is to prove that if with odd. then there do not exists vector fields on projcetive space which are everywhere linearly independent.

During the chapter we have showed that for the Stiefel-Whitney class is not equal to 1, and I want to use this somehow (this also gives it is not parallelizable) A manifold of dimension n is parallelizable iff there are vector fields which are everywhere linearly independent. So immediately we have there are not vector fields which are everywhere linearly independent.

Any hints on how to proceed?

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 Post subject: Re: Vector fields on projective spacePosted: Wed, 13 Jul 2011 05:28:36 UTC
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qwirk wrote:
Question (Milnor/Stasheff again) is to prove that if with odd. then there do not exists vector fields on projcetive space which are everywhere linearly independent.

During the chapter we have showed that for the Stiefel-Whitney class is not equal to 1, and I want to use this somehow (this also gives it is not parallelizable) A manifold of dimension n is parallelizable iff there are vector fields which are everywhere linearly independent. So immediately we have there are not vector fields which are everywhere linearly independent.

Any hints on how to proceed?

I assume you know the interpretation of Steifel-Whitney class as the (mod-2) obstruction to constructing linearly independent tangent vector fields to M on the i-th skeleton?

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 Post subject: Posted: Wed, 13 Jul 2011 06:50:38 UTC
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No, they don't really talk about the concept of an 'obstruction' until later.

(I know that if the bundle posses cross-sections which are nowhere linearly independent then the last Stiefel-Whitney classes are zero. )

I looked it up on mathworld and it says the -th Stiefel-Whitney class being nonzero implies that there do not exist everywhere linearly independent vector fields

Thus gives .

Thus it suffices to show that the i-th Stiefel Whitney class is non-zero, which implies the binomial coefficient is non-zero, so the question reduces to the fact that

Is this correct? (In which case, I will try harder to prove the binomial identity - it eluded my brief attempts - no spoilers if that is the case please!)

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 Post subject: Posted: Wed, 13 Jul 2011 10:44:58 UTC
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qwirk wrote:
No, they don't really talk about the concept of an 'obstruction' until later.

(I know that if the bundle posses cross-sections which are nowhere linearly independent then the last Stiefel-Whitney classes are zero. )

I looked it up on mathworld and it says the -th Stiefel-Whitney class being nonzero implies that there do not exist everywhere linearly independent vector fields

Thus gives .

Thus it suffices to show that the i-th Stiefel Whitney class is non-zero, which implies the binomial coefficient is non-zero, so the question reduces to the fact that

Is this correct? (In which case, I will try harder to prove the binomial identity - it eluded my brief attempts - no spoilers if that is the case please!)

Yes. Note that you don't need the (m-1) since .

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 Post subject: Posted: Wed, 13 Jul 2011 13:30:07 UTC
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Ahh thanks. I managed to prove it using Lucas' theorem using the base-2 expansion, in which case there is only one non-zero term. (Is there an easier/alternative way? I was trying to prove by induction, but it was a bit of a mess)

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 Post subject: Posted: Wed, 13 Jul 2011 13:51:32 UTC
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qwirk wrote:
Ahh thanks. I managed to prove it using Lucas' theorem using the base-2 expansion, in which case there is only one non-zero term. (Is there an easier/alternative way? I was trying to prove by induction, but it was a bit of a mess)

You could prove it directly: Pairing up with in

and note that the powers of 2 cancel exactly.

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 Post subject: Posted: Wed, 13 Jul 2011 14:09:37 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
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Because

for odd?

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 Post subject: Posted: Wed, 13 Jul 2011 14:23:40 UTC
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qwirk wrote:
Because

for odd?

That need not be an integer...

Rather, note that , so whichever power of 2 that divides k must not exceed and hence will also divide , and vice versa.

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