24
$\begingroup$

It is known in characteristic $0$ that (algebraic) de Rham cohomology is a Weil cohomology theory. However, in characteristic $p > 0$ it isn't, if only because it has mod $p$ coefficients, whereas Weil cohomology theories should take values in characteristic $0$. It turns out that de Rham cohomology picks up torsion from Crystalline cohomology, so the dimensions are actually not the correct ones.

However, even if it's not a Weil cohomology theory, one could still try to prove some of the properties. The one I am interested in is Poincaré duality, including the construction of the cup product pairing, say for smooth proper varieties.

I think I know how to do this, but it is a bit technical. I was wondering if there is a place in the literature where it is carried out. It seems that most sources on algebraic de Rham cohomology immediately assume characteristic $0$, and don't prove the results for characteristic $p$ (presumably because de Rham cohomology is 'not the correct thing to consider').

Edit. Maybe I should have stated more clearly what I already know, and what it is that I am looking for. I am certainly convinced of the truth of the statement. For $X$ projective, there is a nice Čech-theoretic approach, worked out below by David Speyer. In general, there is a dévissage approach where we truncate the de Rham complex and induct on the number of terms. One place where one can read about this is these notes (p. 3-4) by Johan de Jong. In the notes, it is only carried out for $X$ projective, but the same approach should work for $X$ proper. (The notes assume characteristic $0$, but this is not used in the construction.)

However, the reason I posted this reference request is that I want to use the result in a paper. I prefer using peer reviewed sources to informal notes, and this is the type of result that should have been known for more than 30 years. I was hoping someone happens to know where to find it.

$\endgroup$
  • $\begingroup$ Are you claiming you can prove this always holds, without exceptions (e.g., including cases where the Hodge to dR spectral sequence doesn't degenerate)? $\endgroup$ – nfdc23 Nov 1 '15 at 1:17
  • $\begingroup$ I think so, yes, but I haven't gone over the whole argument in enough detail to be completely confident. $\endgroup$ – R. van Dobben de Bruyn Nov 1 '15 at 1:45
  • $\begingroup$ Did you eventually find a good reference? I came up with Bhatt's paper on $p$-adic derived de Rham cohomology, but it seems that no Poincaré duality is seriously covered. $\endgroup$ – Yai0Phah Nov 3 '19 at 8:56
22
$\begingroup$

$\def\dr{d_{\rightarrow}}\def\du{d_{\uparrow}}$ This is true. I don't know a reference, but here is a proof. I will show, more strongly, that, at every stage in the Hodge-de Rham spectral sequence, we have a perfect pairing $E^{pq}_r \times E^{(n-p)(n-q)}_r \to k$. In particular, $E^{pq}_{\infty}$ and $E^{(n-p)(n-q)}_{\infty}$ are dual and, since $H^k_{DR}(X)$ is filtered with associated graded $\bigoplus_{p+q=k} E^{pq}_{\infty}$, this shows that $H^k_{DR}$ and $H^{2n-k}_{DR}$ are dual.


We begin with a bunch of formal nonsense about spectral sequences.

Let $V_1$, $V_2$, $W_1$ and $W_2$ be $k$-vector spaces equipped with bilinear pairings $V_1 \times V_2 \to k$ and $W_1 \times W_2 \to k$. Let $\phi_1 : V_1 \to W_1$ and $\phi_2 : W_2 \to V_2$ be linear maps. We say that $\phi_1$ and $\phi_2$ are adjoint if $\langle \phi_1(v_1), w_2 \rangle = \langle v_1, \phi_2(w_2) \rangle$ for all $v_1 \in V_1$ and $w_2 \in W_2$. (Note that, as yet, we don't require the pairings to be perfect or anything to be finite dimensional.)

Let $A^{pq}$ be an $n \times n$ double complex of $k$-vector spaces, with $\dr: A^{pq} \to A^{(p+1)q}$ and $\du: A^{pq} \to A^{p(q+1)}$ the rightward and upward maps. Set $d = \dr+\du$. (I'll be sloppy about signs throughout, but I think my convention is that squares anti-commute.) Suppose that, for all $(p,q)$, we are given a bilinear pairing $A^{pq} \times A^{(n-p)(n-q)} \to k$; suppose that $\dr: A^{pq} \to A^{(p+1)q}$ and $\dr: A^{(n-p-1)(n-q)} \to A^{(n-p)(n-q)}$ are adjoint, as are $\du: A^{pq} \to A^{p(q+1)}$ and $\du: A^{(n-p)(n-q-1)} \to A^{(n-p)(n-q)}$. Let $E^{pq}_r$ be the corresponding spectral sequence, where we first take differentials in the $\du$ direction.

Our main results are

Theorem 1: The bilinear pairing between $A^{pq}$ and $A^{(n-p)(n-q)}$ descends to a bilinear pairing between the subquotients $E^{pq}_r$ and $E^{(n-p)(n-q)}_r$. With respect to this pairing, the differentials $E^{pq}_r \to E^{(p+r)(q+1-r)}_r$ and $E^{(n-p-r)(n-q-1+r)}_r \to E^{(n-p)(n-q)}_r$ are adjoint.

Theorem 2: In the above setting, if the vector spaces on the $r$-th page are finite dimensional and the pairings between them are perfect, then so are the pairings on every succeeding page.

To prove this, we need to recall how $E^{pq}_r$ is defined. We follow Vakil, section 1.7.7, except for the unfortunate point that he does the map in the $p$ direction first and we do the map in the $q$ direction first. This is forced on us because the notation $H^q(X, \Omega^p)$ is standard. This means that many of our coordinates are reversed from Vakil.

Set $S^{pq} = \bigoplus_{k \geq 0} A^{(p+k)(q-k)}$. Let $S^{pq}_r$ be $S^{pq} \cap d^{-1}(S^{(p+r)(q+1-r)})$. (So $S^{pq}_0=S^{pq}$, and increasing $r$ makes the condition more restrictive.) Then $$E^{pq}_r = \frac{S^{pq}_r}{S^{(p+1)(q-1)}_{r-1} + d S^{(p-r+1)(q+r-2)}_{r-1}}.$$ The term $S^{(p+1)(q-1)}_{r-1}$ is simply those elements in $S^{pq}_r$ which have no contribution from the $A^{pq}$ summand, so $S^{pq}_r/S^{(p+1)(q-1)}_{r-1}$ injects into $A^{pq}$ and $E^{pq}_r$ is a subquotient of $A^{pq}$.

Lemma: Let $p+p' = n-r$ and $q+q'=n+r-1$. Let $v \in S^{pq}_r$ and $w \in S^{p'q'}_r$. Then $$\langle v^{pq}, (dw)^{(p'+r)(q'+1-r)} \rangle = \langle (d v)^{(p+r)(q+1-r)}, w^{p' q'} \rangle.$$ Here $u^{pq}$ denotes the projection of $u$ onto the $A^{pq}$ summand.

Proof Sketch: We have $(dw)^{(p'+r)(q'+1-r)} = \dr (w^{(p'-1+r)(q'+1-r)} ) + \du (w^{(p'+r)(q'-r)})$. By our adjointness hypotheses, $\langle v^{pq}, (dw)^{(p'+r)(q'+1-r)} \rangle = \langle \dr(v^{pq}), w^{(p'-1+r)(q'+1-r)} \rangle + \langle \du(v^{pq}), w^{(p'+r)(q'-r)} \rangle$. But $\du(v^{pq})=0$ since $v \in S^{pq}_r$, so we only need to think about the first term. Now, since $v \in S^{pq}_r$, we have $\dr(v^{pq}) = - \du(v^{(p+1)(q-1)})$ and, using adjointness again, $\langle \du(v^{(p+1)(q-1)}), w^{(p'-1+r)(q'+1-r)} \rangle = \langle v^{(p+1)(q-1)}, \du (w^{(p'-1+r)(q'-r)}) \rangle$. Since $w \in S^{p'q'}_r$, we have $ \du (w^{(p'-1+r)(q'-r)}) = - \dr(w^{(p'-2+r, q'-r-1})$. Continuing in this manner, we eventually establish the result. $\square$

Proof Sketch of Theorem 1: We first check that the bilinear form descends to the quotient. In other words, if $v \in S^{(p+1)(q-1)}_{r-1} + d S^{(p-r+1)(q+r-2)}_{r-1}$ and $w \in S^{(n-p)(n-q)}_r$, then $\langle v^{pq} ,w^{(n-p)(n-q)} \rangle = 0$. If $v \in S^{(p+1)(q-1)}_{r-1}$ then $v^{pq}=0$, so this is immediate. Now, suppose that $v=du$ for $u \in S^{(p-r+1)(q+r-2)}_{r-1}$. Note that $S^{(p-r+1)(q+r-2)}_{r-1} \subseteq S^{(p-r)(q+r-1)}_r$. So the lemma shows that $\langle (du)^{pq}, w^{(n-p)(n-q)} \rangle = \langle u^{(p-r)(q+r-1)}, (dw)^{(n-p+r)(n-q-r+1)} \rangle$. But $u^{(p-r)(q+r-1)}=0$, since $u \in S^{(p-r+1)(q+r-2)}$. We have shown that the bilinear form descends.

We now recall the definition of the differential $d_r: E^{pq}_r \to E^{(p+r)(q+1-r)}_r$. Take $v \in E^{pq}_r$ and lift $v$ to $ \tilde{v} \in S^{pq}_r$. Then $d_r(v)$ is the class of $(d \tilde{v})^{(p+r)(q+1-r)}$ in the quotient $E^{(p+r)(q+1-r)}_{r}$. Take $v \in E^{pq}_r$ and $w \in E^{(n-p-r)(n-q-1+r)}_r$ and lift them to $\tilde{v}$ and $\tilde{w}$. We want to show that $\langle (d \tilde{v})^{(p+r)(q+1-r)}, \tilde{w}^{(n-p-r)(n-q-1-r)} \rangle = \langle \tilde{v}^{pq}, (d \tilde{w})^{(n-p)(n-q)} \rangle$. Again, this is the Lemma. $\square$

Proof Sketch of Theorem 2: Suppose that all the vector spaces on the $r$-th page are finite dimensional and the pairings between them are perfect. Write $V^{\vee}$ for the dual to a finite dimensional vector space $V$. We have complexes $\cdots \to E^{(p-r)(q+r-1)}_r \to E^{pq}_r \to E^{(p+r)(q-r+1)}_r \to \cdots$, and $E^{pq}_{r+1}$ is the cohomology of this complex. If two complexes of finite dimensional vector spaces are dual, then their cohomologies are also dual. "$\square$"


Now, we must explain why all of this applies to our setting. Let $X$ be smooth projective irreducible of dimension $n$ over a field $k$. We can extend the base field, and hence can assume the base field is infinite. Thus, by Bertini, we can find $n+1$ mutually transverse very ample divisors $D_0$, $D_1$, \ldots, $D_n$. Let $U(i_0, i_1, \ldots, i_q)$ be the affine open $X \setminus (D_{i_0} \cup \cdots \cup D_{i_q})$. Set $A^{pq} = \bigoplus_{0 \leq i_0 < i_1 < \cdots < i_q \leq n} \Omega^p(U(i_0, \ldots, i_q))$. We organize these into a double complex in the usual way, so $\dr$ is the de Rham differential and $\du$ is the Cech differential. $H^{\ast}_{DR}$ is the total cohomology of this complex.

We note that $A^{nn}$ is Cech co-chains for $H^n(X, \Omega^n)$ and, since we only have $n+1$ open sets, these co-chains are co-cycles. So we have a natural projection $A^{nn} \to H^n(X, \Omega^n) \cong k$, by Serre duality. We denote this map $A^{nn} \to k$ by $\int$.

We have a product $A^{pq} \times A^{p' q'} \to A^{(p+p')(q+q')}$ by $(\alpha \beta)(i_0, i_1, \ldots, i_{q+q'}) = \alpha(i_0, i_1, \ldots, i_q) \wedge \beta(i_q, \ldots, i_{q+q'})$. (Being sloppy about signs as usual, and omitting notation for restriction. It is standard (and also easy, if you neglect signs) that $\dr$ and $\du$ are both derivations for this multiplication. We define a pairing $A^{pq} \times A^{(n-p)(n-q)} \to k$ by $\langle \alpha , \beta \rangle = \int \alpha \beta$.

We wish to check that $\dr$ and $\du$ are self-adjoint. We have $\langle \du(\alpha), \beta \rangle = \int \du(\alpha) \beta$ and $\langle \alpha, \du(\beta) \rangle = \int \alpha \du(\beta)$. So our goal is to show that $\int \alpha \du(\beta) \pm \du(\alpha) \beta = 0$ or, in other words, that $\int \du(\alpha \beta)=0$. This just says that a Cech coboundary is zero in $H^n(X, \Omega^n)$, so it is true. When we do the same computation for $\dr$, we wind up needing to show that $\int \dr(\alpha \beta)=0$. The explicit description of $\int$ in terms of residues makes it clear that $\int d(\eta)=0$ for any $\eta \in A^{(n-1)n}$.

So the hypotheses of our general set up apply. Also, $E^{pq}_1 \cong H^q(X, \Omega^p)$, so the hypotheses of Theorem 2 apply for $r=1$ by Serre duality and the finiteness of cohomology of coherent sheaves. Theorem 2 then proves the result.

$\endgroup$
  • 1
    $\begingroup$ and thanks to permalinks we now also have a reference $\endgroup$ – pro Nov 2 '15 at 4:24
  • $\begingroup$ Thanks; this is great! I will take some time to parse all that is written. In the meantime, I have two questions. Are you assuming that $X$ is geometrically integral (to get $H^{nn} = k$, and not some finite extension)? (I am perfectly happy to stick to this assumption.) More importantly, do you think this could be generalised to the proper case? In that case we don't have access to a cover by $n+1$ opens (I think), so something else is needed. $\endgroup$ – R. van Dobben de Bruyn Nov 2 '15 at 8:37
  • $\begingroup$ I was using gemetrically integral, although that should be removable: Properly fromulated, the statement plays well with change of base field and with disjint union. Not so sure about the proper case. I wanted to have $\int$ defined on all of $A^{nn}$ in order to have the pairings defined on the whole double complex. My guess is yes, but I don't really know. $\endgroup$ – David E Speyer Nov 2 '15 at 14:01
  • $\begingroup$ I'll also mention that I had a much more conceptual way that I wanted to do it. Let $\Omega^{\leq p}$ be the de Rham complex truncated to the first $p$ terms, and define $\Omega^{\geq p}$ likewise. I wanted to show, by induction on $p$, that $\mathbb{H}^q(\Omega^{\leq p})$ and $\mathbb{H}^{n-q}(\Omega^{\geq n-p})$ are dual. The idea would be to use the short exact sequences of complexes $0 \to \Omega^p[p] \to \Omega^{\leq p} \to \Omega^{\leq p-1} \to 0$ and $0 \to \Omega^{\geq n-p+1}[1] \to \Omega^{\geq n-p} \to \Omega^{n-p} \to 0$. This gives two long exact sequences in hypercohomology ... $\endgroup$ – David E Speyer Nov 2 '15 at 14:05
  • $\begingroup$ Serre duality shows $1/3$ of the terms are dual, induction shows another $1/3$ of the terms are dual. If I could build compatible pairings between the sequences, I could deduce by formal linear algebra that the last $1/3$ are dual. But I couldn't figure out how to build the pairings without looking inside the definition of hypercohomology, at which point the above route became faster. If I really cared about the proper case, I might go back and try harder. $\endgroup$ – David E Speyer Nov 2 '15 at 14:07
2
$\begingroup$

Prompted by user Yai0Phah's comment, let me answer my question: I did indeed find a reference. It turns out the correct place to look for these things is any book on crystalline cohomology, because $$H^i_{\operatorname{dR}}(X) = H^i_{\operatorname{cris}}(X/k).$$ One canonical reference for Poincaré duality for smooth proper varieties in the more general setting of crystalline cohomology over $W_n(k)$ is [Berthelot, Ch. VII, Thm. 2.1.3]. The proof for $H^*_{\operatorname{cris}}(X/W_n(k))$ actually immediately reduces to the case of $H^*_{\operatorname{cris}}(X/k)$. This is then carried out by a careful dévissage on the successive truncations $\tau_{\geq i} \Omega_X^*$ and $\tau_{\leq n-i} \Omega_X^*$, which in the end deduces the statement from Serre duality. $\square$


References.

[Berthelot] Berthelot, Pierre, Cohomologie cristalline des schemas de caractéristique $p > 0$, Lecture Notes in Mathematics 407 (1974). ZBL0298.14012.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.