# Stiefel-Whitney Classes and

1. May 5, 2010

### slackersven

Stiefel-Whitney Classes and immersions

1. The problem statement, all variables and given/known data

I don't know whether this goes here or somewhere in the math section of these forums as I am brand new here.
My algebraic topology professor is rather cryptic and the other two guys in the class are just as stuck as I am. I have gone to him and he told me to use what i have derived and a few properties of cohomology to get the desired result. So here it goes.
Given two bundles V and W such that $$V \oplus W$$ is trivial. Let $$w_j(V)=w_j$$ and $$w_j(W)=w^j$$ we can show that if
$$w(V)=1+ w_1 +w_2+w_3+ \dotsc$$
then
$$w(W)= 1+(w_1+w_2+ \dotsc)+(w_1+w_2+ \dotsc)^2 +(w_1+w_2+ \dotsc)^3 + \dotsc$$

We have shown that when $$X=\mathbb{R}\textsf{P}^n$$,
$$w(TX)=(1+x)^{n+1} \in H^*(X)=\mathbb{Z}[x]/x^{n+1}$$

So the part I am having trouble with is the following:
If $$X=\mathbb{R}\textsf{P}^n$$ immerses in $$R^{n+c}$$ then $$\binom{-n-1}{j}$$ is even for c<j<=n.
The hint given is Show $$w^j=\binom{-n-1}{j}x^j$$ where V=TX

2. Relevant equations
Maybe:
$$w^k=w_1w^{k-1}+w_2w^{k-2}+w_3w^{k-3} +\dotsc w_{k-1}w^1 + w_k$$
$$w_1=w^1$$
$$w^2=(w_1)^2 + w_2$$
$$w^3=(w_1)^3 +w_3$$

3. The attempt at a solution
I have tried many ways but have not found any success. Just now while writing this i tried an induction:
$$w^k=\binom{n+1}{1} \binom{-n-1}{k-1} x^k + \binom{n+1}{2} \binom{-n-1}{k-2} x^k + \dotsc + \binom{n+1}{k-1} \binom{-n-1}{1} x^k + \binom{n+1}{k} \binom{-n-1}{0} x^k$$
but got stuck when this did not work out to anything nice:
$$\binom{n+1}{i} \binom{-n-1}{k-i}$$

Anybody know how to do this? Are there binomial identities that I am missing or is this misunderstanding entirely in the structure of the cohomology ring? This is the last homework of my undergrad and I really want to be done.

Last edited: May 6, 2010