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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?

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  • $\begingroup$ 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. $\endgroup$ – David E Speyer Jan 17 at 4:12
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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 $A$.

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  • $\begingroup$ 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 $W(k)/p W(k) \cong k$? $\endgroup$ – David E Speyer Jan 2 at 1:21
  • $\begingroup$ 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. $\endgroup$ – David E Speyer Jan 2 at 3:25
  • $\begingroup$ 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) $\endgroup$ – David E Speyer Jan 2 at 4:01
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    $\begingroup$ @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). $\endgroup$ – SashaP Jan 2 at 10:19
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    $\begingroup$ 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. $\endgroup$ – SashaP Jan 2 at 10:24

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