Given a nonsingular, projective variety $X$ of dimension $n$ over an algebraically closed field $k$. Over $k=\mathbb{C}$, the top chern class $c_n(T_X)$ of the tangent sheaf is the Euler characteristic of the associated complex manifold. Is there some kind of (geometric) intuition or well-known formulas for the value of $c_n(T_X)$ in the case of $p=\mathrm{char}(k)>0$?
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$\begingroup$ There are several choices of what you mean by "top chern class". I don't know enough to know if they all give the same answer, but in positive characteristic there isn't just one obvious cohomology theory people mean. Should we pick an $\ell \neq p$ and use $\ell$-adic cohomology, or crystalline cohomology, etc? $\endgroup$– MattCommented Nov 11, 2011 at 18:52
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4$\begingroup$ I don't think this is too much of a problem. The top Chern class can be understood as an element of the Chow group $CH^n(X)=CH_0(X)$, which maps to $\mathbb{Z}$ by degree. The Chern classes with values in the other cohomologies factors through this as well. $\endgroup$– Donu ArapuraCommented Nov 11, 2011 at 19:05
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$\begingroup$ Ah, thanks. I should have thought of that. $\endgroup$– MattCommented Nov 11, 2011 at 21:57
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The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course cannot use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension.
Addendum: It's all coming back to me, a third possibility is to use that the self-intersection of the diagonal is on the one hand the Euler characteristic (as it is gives the trace of the identitity map), on the other hand that self-intersection is given by the top Chern class of the normal bundle of the diagonal which is exactly the cotangent bundle.
answered Nov 11, 2011 at 18:57
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$\begingroup$ Just to make sure I understand, this Euler characteristic would be defined via, e.g., étale or $\ell$-adic cohomology groups. However, we only have a Grothendieck topology on $X$ and not a canonical, classical topology whose singular homology groups yield the same number. $\endgroup$ Commented Nov 12, 2011 at 1:20
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$\begingroup$ That's right, though you could also use crystalline cohomology which could be defined as the cohomology of a complex of sheaves in the Zariski topology. $\endgroup$ Commented Nov 12, 2011 at 5:11