$w(\mathbb{R}P^q) = 1$ if and only if $q = 2^k - 1$ for some $k$

Question

How do I show that $w(\mathbb{R}P^q) = 1$ if and only if $q = 2^k - 1$ for some $k$?

Answer 1

We will draw on the computation in Chapter 23, Section 3 of May's A Concise Course in Algebraic Topology that$$w(\mathbb{R}P^q) = \sum_{0 \leq i \leq q} \binom{q+1}{i} \alpha^i,$$where $$H^*(\mathbb{R}P^q ; \mathbb{Z}_2) \approx \mathbb{Z}_2[\alpha]/\alpha^{q+1}.$$Suppose that $q = 2^k - 1$. Then note that$$\binom{2^k}{i} \equiv 2 \text{ }(\text{mod }2)$$ if $i \neq 0,\, 2^k$. Additionally, note that $\alpha^{2^k} = 0$ because the relation in the cohomology ring is given by $\alpha^{2^k} = 0$. Therefore, $w(\mathbb{R}P^q) = 1$ if $q = 2^k - 1$.

We now need to show that the converse also holds. This reduces to showing that$$\binom{n}{m}\text{ } (\text{mod }2)$$is $1$ if and only if the binomial expansion of $m$ is a subset of the binomial expansion of $n$. Then, if $q \neq 2^k - 1$, then there will be an $1 < m < n$ such that$$\binom{n}{m} \equiv 1\text{ }(\text{mod } 2)$$and $w(\mathbb{R}P^q) \neq 1$.