# How can I determine the Steenrod Square $Sq^2$ for complex projective space?

I am trying to learn about Steenrod Squares for algebraic varieties so that I can compute examples of complex topological K-theory using the Atiyah-Hirzebruch Spectral Sequence (AHSS).

One of the key points about this sequence is that the third differential $d_3$ is the steenrod squaring operation $Sq^3$ on integral cohomology. This is defined as the composition $$\beta \circ Sq^2 \circ r$$ where $r$ is the reduction mod $2$ cohomology and $\beta$ is the connecting morphism associated to $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0$$ Since these operations play well with functoriality, I will only need to determine the Steenrod operations in $\mathbb{CP}^n$ in specific cases: If I have a smooth projective 3-hypersurface $X$ then the only non-trivial differential factors through $H^2(\mathbb{CP}^4)$. How can I determine this cohomology operation?

The cohomology ring $$H^*(\mathbb{C}P^n;R)\cong R[x]/(x^{n+1})$$, $$|x|=2$$, can be calculated for any coeffient ring $$R$$ using the Gysin sequence of the fibration $$S^1\hookrightarrow S^{2n+1}\rightarrow \mathbb{C}P^{n}$$. In particular it is a truncated polynomial ring on a single degree 2 class $$x\in H^2(\mathbb{C}P^n;R)$$. Now use the fact that for any non-negative integer $$r$$ and any space $$X$$, $$Sq^r$$ acts on $$H^r(X;\mathbb{Z}_2)$$ as the squaring operation $$y\mapsto y^2$$. That is

$$Sq^ry=y^2,\quad y\in H^r(X;\mathbb{Z}_2).$$

Now on the complex projective plane $$\mathbb{C}P^1\cong S^2$$ we obviously have

$$Sq^2x=0$$

since it is homeomorphic to $$S^2$$. For $$n\geq 2$$ we have that

$$Sq^2x=x^2\in H^4(\mathbb{C}P^n;\mathbb{Z}_2)\cong \mathbb{Z}_2$$

is a generator. This calculates the action of $$Sq^2$$ on $$x$$. To proceed from here we use the fact that the total Steenrod square $$Sq=\sum_{r\geq 0}Sq^r=1+Sq^1+Sq^2+\dots$$ is an endomorphism of the graded ring $$H^*(X;\mathbb{Z}_2)=\oplus_{r\geq}H^r(X;\mathbb{Z}_2)$$, together with the facts that $$Sq^0y=y$$, and $$Sq^ry=0$$ if $$|y|, and the observation that $$H^*(\mathbb{C}P^n;\mathbb{Z}_2)$$ vanishes in odd degrees to get, on the one hand

$$Sq(x^k)=(Sq\,x)^k=(x+x^2)^k=\sum_{i}\binom{k}{i}(x^2)^{i}x^{k-i}=\sum_i\binom{k}{i}x^{k+i},$$

recalling that the binomial coefficients are calculated mod 2, and on the other

$$Sq(x^k)=x^k+Sq^2x^k+\dots Sq^rx^k+\dots=\sum_iSq^{2i}x^k.$$

Now match up the degree of the elements in each expression to get

$$Sq^{2r}x^k=\binom{k}{r}x^{k+r}.$$

In particular on $$\mathbb{C}P^n$$ we have

$$Sq^2x^k=\binom{k}{1}x^{k+1}=k\cdot x^{k+1}$$

as long as $$k and $$Sq^2x^n=0$$ for degree reasons.

If you only care about the action of $\operatorname{Sq}^2$ on $H^*(\mathbb{C}P^n; \mathbb{Z}/2)$, you can just recall that $\operatorname{Sq}^2$ acts by squaring on degree 2 elements in cohomology, and that $H^*(\mathbb{C}P^n; \mathbb{Z}/2)$ is generated as an algebra in degree 2. So additivity and the Cartan formula will tell us everything about the action of $\operatorname{Sq}^2$ on $H^*(\mathbb{C}P^n; \mathbb{Z}/2)$.

In general, you may want to know the action of the other Steenrod squares. As mentioned above, the cohomology of complex projective space $\mathbb{C}P^n$ is a truncated polynomial algebra on the first Chern class, and the action of Steenrod operations on characteristic classes are given by Wu's formula. See remark 6.4 here for a reference and proof.