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 $X$ be proper Cohen-Macaulay scheme of pure dimension 1 over an algebraically closed field $k$. When $X$ is moreover reduced, Rosenlicht's theory of regular differential forms gives a beautiful explicit description of Grothendieck duality for $X$. I am wondering if there is an analogue of Rosenlicht's theory in the general context of proper CM curves.

More precisely, in the reduced case Rosenlicht defines a sheaf of $\cal{O}_X$-modules $\omega$

whose sections over an open $U$ in $X$ are those meromorphic differentials $\eta$ on the inverse image of $U$ in the normalization $X'$ of $X$ with the property that for each closed point $x\in U$, the sum of the residues of $f\eta$ over all points $y$ of $X'$ mapping to $x$ is zero for all $f\in \mathcal{O}_{X,x}$. He also defines a trace morphism $Tr:H^1(X,\omega)\rightarrow k$ in terms of "sums of residues" with the property that the pair $(\omega,Tr)$ is canonically isomorphic to the relative dualizing sheaf with its (Grothendieck) trace mapping (one proves this by showing that Roesnlicht's construction satisfies the right universal property).

Is there a similarly explicit (i.e. in terms of certain kinds of differential forms and residues) description of the sections of the dualizing sheaf and of the trace map in the general proper CM setting?

Things to note: A random normal proper and flat curve (=scheme of pure relative dimension 1) $\cal{X}$ over $W(k)$ will always have CM special fiber. However, this special fiber is "very often" not reduced, so there are many examples of non-reduced CM curves.

I asked a previous question on MO hinting at this one: Adjunction for underlying reduced subschemes

I'm happy to assume that $X$ is Gorenstein, if that is at all useful.

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.