MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 '14 at 16:33

I came to this question hoping to learn more, but I can record what I've learned. I find the Cartier operator much more natural in a larger context. This is a combination of material from Brion and Kumar's book, Section I.3, and things I saw asserted in papers and worked out for myself. Of course, there is a risk that any of this is wrong.$\def\cF{\mathcal{F}}$ $\def\cC{\mathcal{C}}$

Notation: Let $k$ be a perfect field of characteristic $p$ and $A$ a $k$-algebra. Let $\Omega^r$ be the $r$-th wedge power of the Kahler differentials of $A/k$. Let $Z^r = \mathrm{Ker}(d: \Omega^r \to \Omega^{r+1})$ (the de Rham co-cycles) and $B^r = \mathrm{Im}(d: \Omega^{r-1} \to \Omega^r)$ (the de Rham co-boundaries). Let $H^r = Z^r/B^r$. Let $A^p$ denote the ring of $p$-th powers in $A$. Note that $d$ is a map of $A^p$ modules and hence $Z^r$, $B^r$ and $H^r$ are all $A^p$ modules.

The inverse of the Cartier operator is more basic than the Cartier operator.

Theorem There is a unique map $\cF: \Omega^r \to H^r$ such that

  • $\cF(f) = f^p$ for $f \in \Omega^0 = A$.

  • $\cF(\alpha+\beta) = \cF(\alpha)+\cF(\beta)$.

  • $\cF(d u) = u^{p-1} du$ for $u \in A$.

  • $\cF(\alpha \wedge \beta) = \cF(\alpha) \wedge \cF(\beta)$.

The only hard part to check is that $(u+v)^{p-1} (du+dv)$ is equivalent to $u^{p-1} du+ v^{p-1} dv$ modulo $B^1$. To this end, note that $$(u+v)^{p-1} (du+dv) - u^{p-1} du - v^{p-1} dv = d \sum_{k=1}^{p-1} \frac{1}{p} \binom{p}{k} d(u^k v^{p-k}). \quad (\ast)$$ Here $\frac{1}{p} \binom{p}{k}$ must be interpreted as an integer which we reduce modulo $p$ to make an element of $k$. (Morally, equation $(\ast)$ comes from the relation $d(u+v)^p = \sum_{k=0}^{p} \binom{p}{k} d(u^k v^{p-k})$. We can't deduce $(\ast)$ directly from this because we can't divide by $p$, but $(\ast)$ is still true in characteristic $p$.)

If $A$ has a sufficiently nice deformation $\tilde{A}$ over the ring of Witt vectors $W(k)$, and $F: \tilde{A} \to \tilde{A}$ is a lift of Frobenius, then $\cF$ for $\omega \in \Omega^r$ is given by reduction modulo $p$ of $\frac{1}{p^r} (F^{\ast})^r \omega$. I don't know exactly what "sufficiently nice" means, but $\tilde{A}$ smooth over $W(k)$ is certainly enough. This is my favorite way to think of $\cF$.

$\cF$ respects localization, and thus makes sense on schemes. It also respects etale extension and completion, for those who prefer other topologies.

We now have Theorem I.3.4 in Brion and Kumar (presumably not original, but they don't cite it): If $A$ is regular, than $\cF$ is an isomorphism $\Omega^{\bullet} \to H^{\bullet}$. Proof sketch: The claim is that a map of finitely generated $A^p$ modules is an isomorphism; this can be checked on completions. So we are reduced to proving the claim in a power series algebra. This is a combinatorial exercise, very similar to the proof of the Poincare lemma for formal power series in characteristic zero.

Note that this is interesting even for $r=0$: It says that $df=0$ for $f \in A$ if and only if $f$ is a $p$-th power.

For $A$ regular, the Cartier operator $\cC: H^{\bullet} \to \Omega^{\bullet}$ is the inverse to $\cF$.

In particular, if $A$ regular of dimension $1$, then we can take the composition $\Omega^1 \to H^1 \to \Omega^1$ to get the definition you gave.

$\cC$ on a curve is analogous to residue in several senses:

  • It is the composition of $\Omega^1 \to H^1$ and an isomorphism between $H^1$ and an explicit free $H^0$-module of rank $1$. In the case of residue, working in the Laurent series ring $k((x))$, we have $H^0 = k$ and we have a canonical isomorphism $H^1 \cong k$; the residue map is $\Omega^1 \to H^1 \to k$. In the Cartier case, $H^0 = A^p$ and $H^1$ is a free $A^p$-module of rank $1$. It doesn't have a canonical generator, but what you can do canonically is use $\cC$ to turn $A^p$ into $A$ and $H^1$ into $\Omega^1$.

  • We have $\cC(df) = 0$ and $\cC(du/u) = du/u$, for $u$ a unit of $A$.

  • Concretely, $\mathrm{res}(\cC(\omega)) = \mathrm{res}(\omega)^{1/p}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.