Steenrod Squares over an Infinite Projective Space

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Let $u$ be a generator of $H^1(\mathbb{R} P^\infty; \mathbb{F}_2)$. Prove the relations $$\text{Sq}^i(u^n) =\binom{n}{i} u^{n+i}$$

 

[Infrared]

Spoiler

We can induct on $n.$ The base case is clear. Next, assuming the formula to be true for exponent $n$ (and for all $i$), we have:

$$\text{Sq}^i(u^{n+1})=\text{Sq}^i(u^n\cup u)=\sum_{a+b=i} Sq^a(u^n) \cup Sq^b(u).$$

Since $Sq^b(u)$ vanishes when $b>1$, the only terms in the sum are $\text{Sq}^i(u^n)\cup \text{Sq}^0(u)+\text{Sq}^{i-1}(u^n) \text{Sq}^1(u)=\binom{n}{i}u^{n+i+1}+\binom{n}{i-1}u^{n+i+1}.$ So we just need to verify that $\binom{n}{i}+\binom{n}{i-1}=\binom{n+1}{i}$ (in fact we only need to check that it is true mod 2, but it is true over the integers). The number of ways of picking $i$ items from $n+1$ items is the number of ways of picking $i$ items where the first item is included ($\binom{n}{i-1}$ ways) plus the number of ways of picking $i$ items from $n+1$ where the first item is not picked ($\binom{n}{i}$ ways).