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Let K be a field of characteristic 0 and X a smooth projective variety over K of dimension n. Then it is well-known that there is a trace map $tr: H^{2n}(X) = H^n(X,\Omega^n_X) \longrightarrow K$ which is an isomorphism.

I would like to know whether there is an explicit formulation of this isomorphism when H stands for algebraic de Rham cohomology, in terms of hypercocycles afforded by a Cech resolution.

Let me explain in more detail my question and what kind of answer I expect. Let $\mathcal U = \{ U_0, ..., U_n \}$ be an open affine cover of X. Then

$H^n(X,\Omega^n_X) = \frac{\Omega^n(U_0 \cap ... \cap U_n)}{\sum^n_{i=0} \Omega^n(U_0 \cap ... \overset{\hat{}}{U_i} ... \cap U_n)}$

and therefore we can represent elements of $H^n(X,\Omega^n_X)$ by classes of differential n-forms on X which are regular on the intersection $U_0 \cap ... \cap U_n$ of the n+1 chosen open affine subsets.

Question: Is there an explicit formulation of the isomorphism tr on this space?

If n=1 the answer is clear: $U_0=X\setminus \{p\}$, $U_1=X\setminus \{q\}$ where p and q are points on X and an easy application of Riemann-Roch shows that any differential 1-form on $X\setminus \{p,q\}$ can be represented (up to adding to it 1-forms which have a single pole at either p or q) by a form $\omega$ which has log poles at p and q. By the residue formula we have $res_p(\omega)+res_q(\omega)=0$ and therefore we can define

$tr(\omega) = (1/2)ยท(res_p(\omega)-res_q(\omega))$

which yields an isomorphism.

Can you give me a similar formulation for higher-dimensional varieties? (Or better yet, so that I don't bother you much, just give me a precise reference for this, which should be very classical?)

My impression is that one could do the following (so I am asking if each of the steps is true):

1) Prove that the open subsets $U_i$ can be chosen as $U_i=X\setminus D_i$ where $\sum_i D_i$ is a divisor with normal crossings.

2) Prove that any n-form on $U_0 \cap ... \cap U_n$ can be represented by a n-form $\omega$ with log poles at the divisors $D_i$.

3) Prove that the recursive rule $tr(\omega) = (1/n+1) \sum_i (-1)^i tr(res_{D_i}(\omega))$ yields the desired isomorphism.

Can you prove it explicitly at least for algebraic surfaces, using Riemann-Roch?

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Dear vic, Have you looked at the discussion in Griffiths and Harris? Regards, Matthew – Emerton Nov 3 '11 at 0:18
Yes, although I might have overlooked something, of course. I looked at: Chapter 3, section 4 on Spectral sequences and applications. Chapter 5, section 1 on local duality and section 4 on global duality, which seem to be the parts of the book most relevant to my question, but they don't address it. Note that I ask for an explicit formulation of the trace isomorphism directly in terms of hypercocycles, not by invoking a comparison theorem to other cohomology theories. I also looked at Harsthorne's "Residues and duality", Chapter III, section 3, where he works it out beautifully for X=P^n. – vic Nov 3 '11 at 3:59
Dear vic, What I had in mind is the part near the end where they discuss residues and duality, and differentials of the various (first, second, third) kinds, first for curves, and then for higher dimensional varieties. I don't recall precisely what is in there, so if you looked carefully, you probably haven't missed anything; it's just one of the few references I can think of which says something explicit about residues and duality. Best wishes, Matthew – Emerton Nov 3 '11 at 4:21
Yes, they have a nice discussion about differentials of 1st and 2nd kind. I would be satisfied if one could describe H^p(X) via these differentials, but this is only the case for p=1: H^1(X)={1-forms of the second kind}/{exact forms} and partially the case for p=2: {2-forms of the second kind}/{exact forms} is a quotient of H^2(X), so they don't capture the whole thing. For p>2 differentials of the second kind are less well-behaved. But ok, H^p(X) is explicitly algebraically described in terms of hypercocycles, which is fine. I mainly ask if my suggestion (3) for the trace map is correct. – vic Nov 3 '11 at 13:51

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