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  "body": "\n## Statement\n\nLet $X$ be a space, let $p$ be a prime, and let $x, y \\in H^*(X; \\F_p)$ be homogeneous classes, with $\\abs{x}$ the degree of $x$.\nWrite $\\smile$ for the [[cup product|JX47MS1R1H]].\n\n1. For $p = 2$, the [[Steenrod squares|VR7RHP41M0]] satisfy, for every $k \\ge 0$,\n$$Sq^k(x \\smile y) = \\sum_{i + j = k} Sq^i x \\smile Sq^j y.$$\n2. For $p$ odd, the [[Steenrod reduced powers|EZYHSE1E8P]] satisfy, for every $k \\ge 0$,\n$$P^k(x \\smile y) = \\sum_{i + j = k} P^i x \\smile P^j y.$$\n3. For $p$ odd, the [[Bockstein homomorphism|H7PCM776GR]] $\\beta$ of $0 \\to \\Z/p \\to \\Z/p^2 \\to \\Z/p \\to 0$ is a derivation:\n$$\\beta(x \\smile y) = \\beta x \\smile y + (-1)^{\\abs{x}} x \\smile \\beta y.$$\n\n## Remarks\n\nThe 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 $X \\to X \\times X$.\nThe cross product case is checked in the construction of the operations, which are built from a product-compatible map on cochains.\n\nIn terms of the total square $Sq = \\sum_{i \\ge 0} Sq^i$, part 1 says that $Sq$ is a ring homomorphism of $H^*(X; \\F_2)$.\n\nExample: let $x$ generate $H^1(\\R P^\\infty; \\F_2)$, so $H^*(\\R P^\\infty; \\F_2) = \\F_2[x]$.\nSince $Sq^0 x = x$ and $Sq^1 x = x^2$, one has $Sq(x) = x + x^2$.\nThen $Sq(x^n) = (x + x^2)^n$, so $Sq^i(x^n) = \\binom{n}{i} x^{n+i}$.\n\n## Explanations\n\n### L1 human\n\nThe Cartan formula says how a Steenrod operation acts on a product of two classes.\nThe answer looks like the product rule for derivatives, with a sum over all ways to split the operation between the two factors.\nIt turns many computations of Steenrod operations into simple algebra.\n\n### L2 human\n\nWith 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.\nIt also gives the Hopf algebra structure of the Steenrod algebra: the coproduct sends $Sq^k$ to $\\sum_{i+j=k} Sq^i \\otimes Sq^j$.\nMilnor used this coproduct to describe the dual of the Steenrod algebra.\n",
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