12
$\begingroup$

This question is about the possible existence of a "compactified" algebraic De Rham cohomology in positive characteristic.

Namely, one knows that, for a smooth, but not proper, variety $U$ over a field of characteristic zero, one can recover the usual De Rham cohomology $H^{i}(U)$ by choosing a compactification - namely, a smooth proper variety $X$ containing $U$ such that $X-U = D$ is a normal crossings divisor - and then taking the log De Rham cohomology of $X$ with respect to $D$. In particular, the resulting log De Rham cohomology group is independent of the choice of $X$ and $D$.

Now suppose we are over a field of positive characteristic, $k$. For smooth $U$, one has the naively defined De Rham cohomology groups $H^i(U)$, but if $U$ is not proper these will in general be infinite dimensional over $k$. However, one can imagine trying to define a "compactified" De Rham cohomology of $U$, as the log De Rham cohomology of some compactification, if one exists (or, more generally, one could try to give a definition using alterations of some proper variety which contains $U$; which always exist). However, beyond the case of a curve, it seems non-obvious that this would give a well-defined object. If this was well defined, however, the resulting cohomology groups would be finite dimensional, and perhaps interesting.

Question: Does anyone know of any results pointing to the existence of such cohomology groups? Or, does anyone know compelling reasons to believe they don't exist; e.g., examples in positive characteristic of different compactifications giving different log De Rham cohomology groups?

More generally, one could replace in the above paragraphs "log De Rham cohomology" with "log Crystalline cohomology groups with coefficients in $W_n(k)$", the Witt vectors of length $n$ for any $n>1$, and I think the same question would be relevant.

Thanks,

Chris

Improve this question
$\endgroup$
6
  • $\begingroup$ Infinite dimensionality is not due, in positive characteristic, to non-properness but to the actual characteristic, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 19, 2016 at 0:53
  • 2
    $\begingroup$ Well, if $X$ is a proper scheme then the De Rham cohomology groups are finite dimensional; since they are computed as the hypercohomology groups of a finite complex of vector bundles. So in that sense it is "due to the non-properness." $\endgroup$
    – Chris Dodd
    Commented Jul 19, 2016 at 1:18
  • 1
    $\begingroup$ There is a well-defined theory called "rigid cohomology" due to Berthelot which gives finitely generated answer for arbitrary varieties. A drawback of it is that it is defined only rationally while crystalline cohomology has a structure of $W(k)$-module. As far as I know it is still an open problem to give rigid cohomology an integral structure. $\endgroup$
    – SashaP
    Commented Jul 19, 2016 at 18:42
  • $\begingroup$ Also, as you suggest rigid cohomology can be computed via the log-de Rham-Witt complex given a smooth compactification. Every such compactification gives an integral structure but it seems unclear to me how to identify these integral structures for different compactifications. $\endgroup$
    – SashaP
    Commented Jul 19, 2016 at 19:05
  • $\begingroup$ Hi Sahsa; thanks for the comment. This was one motivation for my question. As far as I can see, no one addresses this problem one way or the other in the literature; it seems like a strange gap. My feeling was that there must be some "folklore" thinking on the problem; we'll see if the MO community offers some. $\endgroup$
    – Chris Dodd
    Commented Jul 19, 2016 at 19:57

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .