comparison of de Rham cohomology and etale cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:11:58Z http://mathoverflow.net/feeds/question/47950 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47950/comparison-of-de-rham-cohomology-and-etale-cohomology comparison of de Rham cohomology and etale cohomology shenghao 2010-12-01T21:47:46Z 2010-12-02T14:03:12Z <p>I have a basic question concerning comparison of different cohomology theories. Let $X$ be a projective smooth (or just proper smooth) variety over a separably closed field $k$ of characteristic $p,$ which is not liftable to characteristic zero. Is there any relation between the de Rham cohomology $H^n(X,\Omega^{\bullet}_X)$ and the $\ell$-adic cohomology? For example, do they have the same dimension (over $k$ and $Q_l$ resp.)?</p> http://mathoverflow.net/questions/47950/comparison-of-de-rham-cohomology-and-etale-cohomology/47966#47966 Answer by Hunter Brooks for comparison of de Rham cohomology and etale cohomology Hunter Brooks 2010-12-01T23:42:53Z 2010-12-01T23:42:53Z <p>I believe the answer is no, that these two spaces need not have the same vector space dimension. Grothendieck here cites an example of Serre in a footnote on the last page; unfortunately, I don't have access to Serre's original paper at the moment.</p> <p>29_95_0">http://www.numdam.org/item?id=PMIHES_1966_<em>29</em>_95_0</p> http://mathoverflow.net/questions/47950/comparison-of-de-rham-cohomology-and-etale-cohomology/47967#47967 Answer by Minhyong Kim for comparison of de Rham cohomology and etale cohomology Minhyong Kim 2010-12-01T23:50:16Z 2010-12-02T14:03:12Z <p>Often, yes. What always has the same dimension as $H^n_{et}(X,Q_l)$ is the rational crystalline cohomology $H^n_{cr}(X)\otimes K$ with coefficients in the fraction field $K$ of the Witt vectors $W$ of $k$. $H^n_{cr}(X)$ itself will have coefficients in $W$, and of course, have rank equal to the dimension of $H^n_{cr}(X)\otimes K$. But it might have torsion in general. On the other hand, there is an exact sequence $$0\rightarrow H^n_{cr}(X)\otimes_W k\rightarrow H^n(X,\Omega_X^{\cdot})\rightarrow H^{n+1}_{cr}(X)[p]\rightarrow 0$$ as in the universal coefficient theorem. This is because crystalline cohomology can be taken with any of the torsion coefficients $W/p^n$, and when you take it with coefficients in $W/p=k$, you get exactly De Rham cohomology. (One of the most important things to learn at the beginning about crystalline cohomology with $W/p^n$ coefficients is that it can be computed using the divided power De Rham complex associated to a smooth embedding over $W/p^n$, which reduces to the De Rham complex of $X$ itself when the coefficients are $W/p$.)</p> <p>So you will get the same dimensions you want if enough of crystalline cohomology is torsion-free. All this is explained in introductory books, such as the one by Berthelot and Ogus, except the comparison with \'etale cohomology. That is perhaps explained in a paper by Katz and Messing from the 70's.</p>