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This is the Problem 8-B form the characteristic classes by John W. Milnor and James D. Stasheff.

[Problem 8-B]. If the total Stiefel-Whitney class $w(\xi) \neq 1$, show that the smallest $n>0$ with $w_n(\xi) \neq 0$ is a power of 2. (Use the fact that $\binom{x}{k}$ is odd whenever $x$ is an odd multiple of $k=2^r$.)

In the problem, $\xi$ is a vector bundle and $w_n(\xi)$ is the $n$th Stiefel-Whitney class of $\xi$.

[Problem 8-A] is the problem about proving Wu's explicit formula

$Sq^k(w_m) = w_k w_m + \binom{k-m}{1} w_{k-1}w_{m+1} + \ldots + \binom{k-m}{k}w_0 w_{m+k}$

where $Sq^k$ is the Steenrod squaring operation.

I try to use the problem 8-A to solve the problem 8-B. However, it does not work yet.

Can anybody give me a hint or a proof?

Thank you.

Eric Wofsey
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ljh8372
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1 Answers1

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Let $n>0$ be minimal such that $w_n(\xi)\neq 0$, and suppose $n$ is not a power of $2$. Let $k=2^d$ be the greatest power of $2$ that divides $n$, and let $m=n-k$. Then Wu's formula gives $$Sq^k(w_m)=w_k w_m + \binom{k-m}{1} w_{k-1}w_{m+1} + \ldots + \binom{k-m}{k}w_0 w_{m+k}.$$

By minimality of $n$ and the fact that $n$ is not a power of $2$ (so $0<k<n$ and $0<m<n$), the left-hand side of this equation vanishes, as does every term on the right-hand side except the last term. Thus we get that $0=\binom{k-m}{k}w_0 w_{m+k}=\binom{k-m}{k}w_n$. But $k-m=2k-n$ is an odd multiple of $k$ since $n$ is an odd multiple of $k$, so this $\binom{k-m}{k}$ is odd. This means $w_n=0$, which is a contradiction. Thus $n$ must have been a power of $2$.

Eric Wofsey
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