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).