Vector fields on projective space

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Question (Milnor/Stasheff again) is to prove that if n+1=2^rm

with

odd. then there do not exists 2^r

vector fields on projcetive space $\mathbb{R}P^n$ which are everywhere linearly independent.

During the chapter we have showed that for n+1=2^rm

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 $n \quad (=2^rm-1)$ vector fields which are everywhere linearly independent. So immediately we have there are not 2^rm-1

vector fields which are everywhere linearly independent.

Any hints on how to proceed?

outermeasure · **Posted:** Wed, 13 Jul 2011 05:28:36 UTC

qwirk wrote:

Question (Milnor/Stasheff again) is to prove that if n+1=2^rm

with

odd. then there do not exists 2^r

vector fields on projcetive space $\mathbb{R}P^n$ which are everywhere linearly independent.

During the chapter we have showed that for n+1=2^rm

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 $n \quad (=2^rm-1)$ vector fields which are everywhere linearly independent. So immediately we have there are not 2^rm-1

vector fields which are everywhere linearly independent.

Any hints on how to proceed?

I assume you know the interpretation of Steifel-Whitney class w_i(M)

as the (mod-2) obstruction to constructing n-i+1

linearly independent tangent vector fields to M on the i-th skeleton?

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

No, they don't really talk about the concept of an 'obstruction' until later.

(I know that if the bundle $\xi$ 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 (n-i+1)

everywhere linearly independent vector fields

Thus n-i+1=2^r

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

$\binom{2^rm}{2^r(m-1)} \equiv 1 (\text{mod} 2)$

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!)

outermeasure · **Posted:** Wed, 13 Jul 2011 10:44:58 UTC

qwirk wrote:

No, they don't really talk about the concept of an 'obstruction' until later.

(I know that if the bundle $\xi$ 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 (n-i+1)

everywhere linearly independent vector fields

Thus n-i+1=2^r

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

$\binom{2^rm}{2^r(m-1)} \equiv 1 (\text{mod} 2)$

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 $\displaystyle\binom{2^rm}{2^r(m-1)}=\binom{2^rm}{2^r}$ .

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

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)

outermeasure · **Posted:** Wed, 13 Jul 2011 13:51:32 UTC

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 2^r(m-1)+k