Pairing of cotangent and tangent bundles

Question

I am reading the survey paper: "The de-Rham Witt complex and Crystalline cohomology" by Luc Illusie.

In math line (2.1.12), Illusie considers the pairing $\langle-,-\rangle:\Omega_{X/S}^1\times T_{X/S}\longrightarrow \mathcal{O}_X$ of the tangent and cotangent bundles on a scheme $X$ of characteristic $p$, relative to a morphism $X\longrightarrow S$. ($S$ also has characteristic $p$).

Earlier we defined a bunch of maps, which would be necessary to describe in order for me to ask my question. The first one is the Cartier operation, $C$, which sends a closed form $w\in \Omega_{X/S}^1$ to a 1-form $Cw \in \Omega_{X^{(p)}/S}^1$, where $X^{(p)}$ is the base change of $X\longrightarrow S$ with respect to the absolute Frobenius on $S$. Denote by $W$ the canonical projection $W:X^{(p)}\longrightarrow X$.

Illusie then wants to say something about the pairing $$\langle Cw,W^*D\rangle$$ which happens on $\Omega_{X^{(p)}/S}^1\times T_{X^{(p)}/S}\longrightarrow \mathcal{O}_{X^{(p)}}$. Namely, he gives the identity $$\langle Cw,W^*D\rangle^p = \langle w,D^p\rangle - D^{p-1}\langle w,D\rangle.$$ My questions are:

1. What does the $p$-th power of a tangent vector, $D^p$, mean? Does the tangent space have a ring structure?

2. The pairing apriori should have values in $\mathcal{O}_{X^{(p)}}$, once it is raised to the $p$'th power we regard it as having values in $\mathcal{O}_X$ as this is. For this reason it seems that Illusie is regarding $D^{p-1}$ as an element of $\mathcal{O}_X$, which is consistent with math line (2.1.13), why is that? I guess this goes back to the first question.

3. Illusie also writes $D_i^p = 0$, where $D_i = \partial/\partial x_i$, for some etale basis $(x_i)$, why is that true? I guess that this should all be clear once I understand this notation.

Thanks in advance!

Answer 1

For (1), recall that if $R$ is a ring, then a derivation $D: R \to R$ satisfies the Leibniz rule, which by induction on $n$ implies that if $D^n$ denotes the $n$-fold iterate of $D$, then $$D^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i(f) D^{n-i}(g).$$ Since $\binom{p}{i} \equiv 0$ for $0<i<p$, this implies that if $R$ is an $\mathbf{F}_p$-algebra, then $D^p(fg) = D^p(f) g + f D^p(g)$. In other words, $D^p$ is a derivation.

Let me answer (3) before (2). If you work locally, i.e., consider the derivation $\partial_x$ on $\mathbf{F}_p[x]$, then $(\partial_x)^p x^n$ is zero for $n<p$, and is $p! \binom{n}{p} x^{n-p}$ for $n\geq p$, which is zero. So the derivation $(\partial_x)^p$ is identically zero.

Let's now discuss (2); since everything is local on $X$ and $X$ is smooth, we can assume that $X$ is etale over $\mathbf{A}^n_S$ with basis $x_1, \cdots, x_n$. Let $\omega$ be a closed $1$-form on $X$. Because of the Cartier isomorphism $\mathfrak{C}: \mathcal{H}^i(F_\ast \Omega^\bullet_{X/S}) \xrightarrow{\sim} \Omega^i_{X^{(p)}/S}$, we can write $\omega = df + \sum_{i=1}^n F^\ast(g_i) x_i^{p-1} dx_i$ for some functions $g_i$ and $f$ on $X$. Both $\langle \mathfrak{C} \omega, D\rangle^p$ and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle$ kill $df$, so by Frobenius semilinearity, we can assume that $\omega = x_i^{p-1} dx_i$. We'll also just assume $n=1$ and write $x$ instead of $x_1$.

Then $\langle \mathfrak{C} \omega, D\rangle = \langle dx^{(p)}, D\rangle = D(x)$, and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx)$. So we need to prove that $$(Dx)^p = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx),$$ which is an identity due to Hochschild. See https://joshuamundinger.github.io/assets/notes/hochschild-identity.pdf for a cute argument; it reduces to using the multinomial analogue of the Leibniz rule for $D^p(x^p)$ and a multinomial analogue of the binomial coefficient vanishing.

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Clarification from comments: let's again just consider a single variable $x$ and assume $X = \mathbf{A}^1_{\mathbf{F}_p}$. The most general closed $1$-form is $\omega = df + g(x^p) x^{p-1} dx$. Then $\mathfrak{C}(df) = 0$, so we can take $a_i$ to be $g(x^p) x^{p-1}$. Now, $\partial_x$ is $\mathbf{F}_p[x^p]$-linear, so $$\partial_x^{p-1} (g(x^p) x^{p-1}) = g(x^p) \cdot (\partial_x^{p-1} x^{p-1}) = (p-1)! g(x^p) = -g(x^p).$$ This means that $\mathfrak{C}(g(x^p) x^{p-1} dx) = g(x^{(p)}) dx^{(p)}$ can be written as $-\partial_x^{p-1} (g(x^p) x^{p-1}) dx^{(p)}$, as desired.