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Let $C$ be a smooth curve over a finite field of characteristic $p$. Let $t$ be a local parameter at a point. If $f$ is a regular function on a neighbourhood of the point, one can write uniquely $$f = \sum_{i=0}^{p-1} f_i^p t^i $$ for functions $f_i$. Using this decomposition one can define an operator on differential forms on $C$ via $$ f \mathrm d t \mapsto f_{p-1} \mathrm d t.$$ Amazingly (?) this does not depend on the choice of coordinate, producing a map of sheaves $\Omega_C \to \Omega_C$. This is the Cartier operator.

Question: What is so special about $\Omega_C$ here? Is there a similar construction on quadratic differentials? On spin curves, i.e. on a square root of $\Omega_C$?

I am told that the Cartier operator is analogous to the residue of a logarithmic differential form. So I guess a similar question (that I don't know how to answer either) is what makes the residue tick.

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I see them as analogous because both are about the failure of $d$ to be onto. On Laurent polynomials in $t$, in characteristic zero, every $t^k$ is a derivative except $t^{-1}$, and the residue looks at the coefficient of that. On honest polynomials in $t$, in characteristic $p$, all the $t^k$s are derivatives except when $p|k+1$, and the Cartier looks at the (now infinitely many!) coefficients of those. –  Allen Knutson Jun 19 '13 at 19:00
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BTW my favorite application of the Cartier operator is this. Given a nonzero anticanonical section $\sigma$, look at $f \mapsto \sigma$Cartier$(f/\sigma)$. Since $f/\sigma$ is a (meromorphic) top form, it's closed, so we can apply Cartier. When we contract it with $\sigma$ again, it becomes a well-defined function. So $\sigma$ gives a Frobenius-inverse-linear map from functions to functions, which for good $\sigma$ is a Frobenius splitting that compatibly splits $\{\sigma = 0\}$. –  Allen Knutson Jun 19 '13 at 20:00
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To see why the Cartier operator is independent of the choice of the coordinate, you can check out Exe4.13 in the book 'Algebraic function fields and codes' by Stichtenoth. In short, it tells you that the map defined by certain properties of the Cartier operator has to be Cartier. –  Yujia Qiu Jun 12 at 16:33

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