# A variant on characteristic $p$ de Rham cohomology

I was thinking about de Rham cohomology in characteristic $$p$$, and in particular the recent question about Poincare residues, and I came up with the following construction.

Let $$k$$ be a perfect field of characteristic $$p$$ and let $$A$$ be a regular $$k$$-algebra. Let $$\Omega^j$$ be the Kahler $$j$$-forms, let $$Z^j$$ be the closed $$j$$-forms, $$B^j$$ the exact $$j$$-forms and $$H^j = Z^j/B^j$$. The inverse Cartier operator is the unique isomorphism $$C^{-1} : \Omega^j \to H^j$$ satisfying $$C^{-1}(\alpha \wedge \beta) = C^{-1}(\alpha) \wedge C^{-1}(\beta) \quad C^{-1}(f) = f^p \quad C^{-1}(df) = f^{p-1} df$$ for $$f \in A$$. (It is easy to see that there is at most one such map, a nice exercise to see that it is well defined, and not at all clear that it is an isomorphism.)

The inverse operator is an isomorphism $$Z^j/B^j \to \Omega^j$$, which we can also consider as a surjection $$Z^j \to \Omega^j$$. By abuse of notation, I'll write $$C$$ for the surjection $$Z^j \to \Omega^j$$ as well. We thus have two maps $$Z^j \to \Omega^j$$: The surjection $$C$$, and the obvious inclusion.

Define a differential form $$\alpha \in \Omega^j$$ to be forever closed if, for all $$i$$, we have $$C^i(\alpha) \in Z^j$$. Note that we must have $$C^{i-1}(\alpha) \in Z^j$$ for it to make sense to define $$C^i(\alpha)$$, so this condition spells out as "we impose that $$\alpha$$ is closed, and therefore $$C(\alpha)$$ is defined, and we impose that $$C(\alpha)$$ is closed, and therefore $$C^2(\alpha)$$ is defined, etcetera."

Define a forever closed form $$\alpha$$ to be "eventually exact" if $$C^k(\alpha)$$ is $$0$$ for $$k$$ sufficiently large. Note that exact forms are eventually exact, since the exact forms are the kernel of $$C$$. Define the eventual cohomology, $$EH^j$$, to be the forever closed forms modulo the eventually exact forms.

It looks like $$EH^{\bullet}$$ is always finite dimensional, and forms a graded ring. It does not appear that the dimension of $$EH^j$$ gives topological betti numbers -- it appears to give something like the multiplicity of the highest weight part of the cohomology.

Is this some object people have studied before?

• Old version had a statement about Meyer-Vietores which I no longer think is right. If $X$ is affine and $X = U \cup V$ with $U = \{ f \neq 0 \}$ and $V = \{ g \neq 0 \}$, then I thought before that any forever closed form $\gamma$ on $U \cap V$ was of the form $\alpha + \beta$ for $\alpha$ defined on $U$, $\beta$ defined on $V$ and both forever closed. But all I can really show is that, for any $N$, I can find $\alpha$ and $\beta$ with $\alpha+\beta=\gamma$ and $\alpha$ and $\beta$ each $N$-fold closed. – David E Speyer Jan 17 at 4:12

To be able to compute the iterates of the Cartier operator it is convenient to understand how $$C$$ interacts with the de Rham differential:

Cartier isomorphism induces an isomorphism of complexes $$(\Omega^{*}_A,d_{dR})\simeq (H^{*}(\Omega^{\bullet}_A),\beta)$$ where $$\beta$$ is the Bockstein differential provided by the distinguished triangle $$\Omega^{\bullet}_A\to R\Gamma_{cris}(A/W_2(k))\to \Omega^{\bullet}_A$$ It shows that for a closed form $$\alpha$$ the image $$C(\alpha)$$ is closed iff the class $$[\alpha]$$ is annihilated by the Bockstein homomorphism which in turn is equivalent to the liftability of $$\alpha$$ to class in $$H^i_{cris}(A/W_2(k))$$. Passing to cohomology in the above isomorphism, composing it with the Cartier isomorphism and iterating this procedure $$(i-1)$$ times we get an isomorphism $$(\Omega^{*}_A,d_{dR})\simeq (E_i^{(1-i)*,*},\beta_i)$$ of the de Rham complex with the complex appearing on the $$i$$-th page of the Bockstein spectral sequence associated to the crystalline cohomology complex.

These facts can be seen easily from the following description of the Cartier isomorphism: choose a lift $$\tilde{A}$$ of $$A$$ to a complete formally smooth algebra over $$W(k)$$ equipped with a lift $$\widetilde{Fr}$$ of the Frobenius endomorphism of $$A$$(the existence of such lift follows from the vanishing of the relevant obstruction groups which is implied by smoothness of $$A$$ over $$k$$). The Cartier operator applied to a form $$\omega\in \Omega^i_A$$ is then given by $$C(\omega)=\overline{\frac{\widetilde{Fr}^*(\tilde{\omega})}{p^i}}$$ where $$\tilde{\omega}$$ is any lift of $$\omega$$ to a form on $$\tilde{A}$$ and $$\overline{\cdot}$$ denotes the reduction.

By tracing through the construction of the Bockstein differentials we get the following

Lemma. For a closed form $$\alpha$$ the $$i$$-th iteration of the Cartier operator is defined and gives a closed form if and only if $$[\alpha]\in H^j(\Omega_A^{\bullet})$$ lifts to a class $$\widetilde{[\alpha]}$$ in the crystalline cohomology of $$A$$ over $$W_{i+1}(k)$$. The $$(i+1)$$-th iteration of the Cartier operator is zero if and only if the class $$p^i\widetilde{[\alpha]}\in H^j_{cris}(A/W_{i+1}(k))$$ vanishes.

Combining these conditions for all $$i$$ we get that a form is forever closed iff its class is in the image of the map $$H^j_{cris}(A/W(k))\to H^i_{dR}(A/k)$$ and it is eventually exact iff the class is in the image of $$H^j_{cris}(A/W(k))[p^{\infty}]\to H^i_{dR}(A/k)$$.

Crystalline cohomology of $$A$$ coincides with the cohomology of the $$p$$-adically completed de Rham complex of any lift of $$A$$ to $$W(k)$$ and for the purposes of computing the above invariants we can replace $$H^j_{cris}(A/W(k))$$ by the (non-complete) de Rham cohomology $$H^j_{dR}(\widetilde{A}/W(k))$$ where $$\widetilde{A}$$ is any lift. The quotient $$H^j_{dR}(\widetilde{A})/H^j_{dR}(\widetilde{A}){[p^{\infty}]}$$ is a $$W(k)$$-lattice in the finite-dimensional vector space $$H^j_{dR}(\widetilde{A}[1/p]/W(k)[1/p])$$(it is finite-dimensional e.g. by comparison with singular cohomology).

It indeed seems to follow that $$EH^j$$ is a finite-dimensional vector space over $$k$$ with dimension at most the $$j$$-th rational Betti number of any lift of $$A$$.

• Thanks! Am I reading correctly to say that $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[p^{\infty}]$ lives in $H_{dR}(\tilde{A}[1/p]/W(k)[1/p])$ as those de Rham classes of $\tilde{A}[1/p]$ which can be represented by differential forms in $\Omega^{\bullet}(\tilde{A})$ (i.e. without denominators)? And then $EH$ is the tensor product of this with $W(k)/p W(k) \cong k$? – David E Speyer Jan 2 at 1:21
• Wait, I'm suspicious of your removing the $p$-adic completion. Let $A$ be the coordinate ring of a supersingular elliptic curve with $1$ point removed and let $\tilde{A}$ be an elliptic curve mins a point over $\mathbb{Z}_p$ deforming $A$. I believe that every $1$-form is eventually effective, so $EH=0$. But let $\omega$ be the invariant differential form. I am pretty sure that $p^k \omega$ is not exact for any $k$. I suspect the resolution is that there is an element in the $p$-adic completion of $\tilde{A}$ whose differential is $\omega$, but that we need to work with this completion. – David E Speyer Jan 2 at 3:25
• I am now less confident in the last sentence. I took $y^2 = x^3+x$. The invariant form is $\omega = \tfrac{dy}{3x^2+1} = \tfrac{dx}{2y}$, which reduces to just $dy$ in characteristic $3$. So $\omega$ is exact modulo $3$, but is not exact on the $3$-adic lift. This seems to me to be a problem for your formula $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[3^{\infty}]$ (or for my understanding of it) because it seems to me that $\omega$ is a nonzero class in $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[3^{\infty}]$ . I thought that I could work with the $3$-adic completion of $\tilde{A}$ instead, (continued) – David E Speyer Jan 2 at 4:01
• @DES-SupportsMonicaAndTransfolk I don't think that the reduction of the module $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[p^{\infty}]$ is necessarily equal to $EH$. Rather, the reduction surjects onto $EH$ but there can be a non-torsion class in the de Rham cohomology over $W(k)$ that is congruent to a torsion class and it happens in your example: $dy/(3x^2+1)$ is equal to $dy+3\frac{-x^2}{x^2+1}dy$(this is just saying that $\omega$ reduces to an exact form, as you're saying). – SashaP Jan 2 at 10:19
• So the quotient of the de Rham cohomology by the torsion is a rank two module(because the cohomology of a punctured elliptic curve in characteristics zero is two-dimensional) that surjects onto one-dimensional $EH$ in characteristic $p$. I agree that the cohomology of complete elliptic curve does not contribute to the $EH$ but there is a log-form $dy/y$ invariant under Cartier operator that makes $EH$ one-dimensional, I think. – SashaP Jan 2 at 10:24