Cartan formula

Remarks

The same formulas hold for the cross product $H^*(X) \otimes H^*(Y) \to H^*(X \times Y)$, and the cup product case follows by pulling back along the diagonal $X \to X \times X$. The cross product case is checked in the construction of the operations, which are built from a product-compatible map on cochains.

In terms of the total square $Sq = \sum_{i \ge 0} Sq^i$, part 1 says that $Sq$ is a ring homomorphism of $H^*(X; \F_2)$.

Example: let $x$ generate $H^1(\R P^\infty; \F_2)$, so $H^*(\R P^\infty; \F_2) = \F_2[x]$. Since $Sq^0 x = x$ and $Sq^1 x = x^2$, one has $Sq(x) = x + x^2$. Then $Sq(x^n) = (x + x^2)^n$, so $Sq^i(x^n) = \binom{n}{i} x^{n+i}$.

Going deeper

For a student

With the Cartan formula and the values on degree-one or degree-two generators, one computes all Steenrod operations on the cohomology of $\R P^n$, $\C P^n$ and products of these. It also gives the Hopf algebra structure of the Steenrod algebra: the coproduct sends $Sq^k$ to $\sum_{i+j=k} Sq^i \otimes Sq^j$. Milnor used this coproduct to describe the dual of the Steenrod algebra.