What is the easiest/quickest way to see the following?



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> If $n + 1 = 2^rm$ with $m$ odd, then there do not exist $2^r$ vector fields on the projective space $\mathbb{P}^n$ which are everywhere linearly independent.
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Thanks!



$n + 1 = 2^rm$ with $m$ odd $\implies$ do not exist $2^r$ vector fields on $\mathbb{P}^n$ that are everywhere linearly independent?
The standard way to treat statements like this one is to analyze the Stiefel-Whitney classes of $\mathbb P^n$. The full Stiefel-Whitney class of the tangent bundle is
$$w(T\mathbb P^n)=(1+u)^{n+1},$$
where $u\in H^1(\mathbb P^n,\mathbb Z\_2)$ is the generator.


In your particualr case $(n+1)=2^rm$ and let $p=2^r(m-1)$. Then $w\_p(T\mathbb P^n)=\binom{n+1}{p}u^p=u^p\neq 0$. Here we used the fact $\binom{2^rm}{2^r}=1 \pmod 2$.


On the other hand, if there are $k$ linearly independent vector fields on $\mathbb P^n$, then $T\mathbb P^n=E\oplus \mathbb R^k$ and, therefore, $w\_q(T\mathbb P^n)=w\_q(E)=0$ for $q>n-k$. Combining with the fact $w\_p(T\mathbb P^n)\neq 0$, we conclude $k<n-(p-1)=2^r$.


