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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
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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