# Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded?

Let X be a smooth projective variety over the complex numbers. Recall that a Cohen-Macaulay curve is a one-dimensional closed subscheme without embedded or isolated points (fat components are allowed).

I want to show that if you fix a curve class β in H2(X), then there is a bounded interval which contains the genus of every CM curve whose homology class is β.

Do you know of a reference for this result?

Motivation: I am trying to prove a certain class of sheaves forms a bounded family (namely, the collection of sheaves underlying stable pairs). The above result will allow me to reduce to the case where the support is a fixed CM curve, and from there, I know how to finish the proof.

I am aware that Le Potier (who built the moduli space of stable pairs, among others things, in "System Coherents et Structures de Niveau" --- pardon my lack of accents) has proven a much stronger result. However, for my purposes, it would be very helpful to have a proof (hopefully much less technically demanding than Le Potier's) that is no more general than what I need.

EDIT: in hindsight, and in light of the negative answer below, I realize that Le Potier does not claim to prove quite what I claimed he claimed.

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I don't think this is true. Take $X = \mathbb P^1 \times \mathbb P^2$. Let $C$ be $\mathbb P^1 \times 0$, and let $C_1$ be the first infinitesimal neighborhood of $C$. The curve $C_1$ is the relative spectrum of $\mathcal O_{\mathbb P^1} \oplus O_{\mathbb P^1}^{\oplus 2}$, where $\mathcal O_{\mathbb P^1}^{\oplus 2}$ is a square-zero ideal. Any sheaf $\mathcal O_{\mathbb P^1}(d)$, with $d \ge 0$, is a quotient of $\mathcal O_{\mathbb P^1}^{\oplus 2}$. If $C(d)$ denotes the relative spectrum of $\mathcal O_{\mathbb P^1} \oplus O_{\mathbb P^1}(d)$, then $C(d)$ is contained in $C_1$ for all $d \ge 0$; hence it is embedded in $X$ with fundamental class $2[C]$. But the arithmetic genus of $C(d)$ is $-d-1$.

[Added later] For a surface, a Cohen-Macaulay curve is a divisor, and the adjunction formula shows that the arithmetic genus is determined by the cohomology class, so the answer is positive. I believe that the answer is negative for all $X$ of dimension at least three.

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EDIT: As Angelo mentions, the argument below has a problem in the case of non-reduced curves. I am not sure that an upper bound for the arithmetic genus is impossible to show, but certainly the lower bound for Cohen-Macaulay curves is false, as Angelo's examples show. The argument below shows that there is an upper bound (and in fact there is also a lower bound) for the arithmetic genus of a reduced subscheme of pure dimension one in projective space, whether Cohen-Macaulay or not. I tried to play a little with the Cohen-Macaulay condition to prove that there is an upper bound, but with little success.

Choose an embedding of X in projective space. Since your curves are all in the same homology class, they all have the same degree: this is simply the intersection number of the ample class with the homology class of the curves. Generic projection to a plane tells you (since the curves are reduced) that the curves you are interested in are partial normalizations of plane curves with bounded degree. Since the arithmetic genus decreases under (partial) normalizations, and since the arithmetic genus of a plane curve of bounded degree is bounded above, you conclude that the arithmetic genera of your curves are bounded above.

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Does this work for non-reduced curves? –  Angelo May 5 '10 at 19:10
I strongly believe that there should be an upper bound for Cohen-Macaulay (notice that reduced curves are automatically Cohen-Macaulay). Of course for proving boundedness this is not good enough, but if the OP cares I might try to write up a proof. –  Angelo May 6 '10 at 15:07
Why isn't this enough to prove boundedness? The class beta (i think) is enough to fix the dimension of H^0, and an upper bound on genus fixes H^1, so they curves live in a finite number of Hilbert schemes. –  David Steinberg May 7 '10 at 17:50
The component of the Hilbert scheme is determined by degree and genus; infinitely many genera means infinitely many components. –  Angelo May 7 '10 at 18:37

Hi David, there is a two-line proof for the bound from above, if I am allowed to use the boundedness of the Hilbert scheme:

If there is a curve with degree $\beta$ and $\chi = m-d$, then the Hilbert scheme of curves of degree $\beta$ and Euler characteristic $\chi = m$ has dimension at least $d \cdot \dim X$, as I can always add $d$ floating points to my given curve. But the Hilbert scheme for $\beta, \chi = m$ has finite dimension. So $\chi$ is bounded from below (and the genus from above).

I also think this bound for $\chi$ from below should be enough to prove the boundedness of sheaves appearing in stable pairs that you need.

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Great! Thank you, Arend. –  David Steinberg Aug 4 '10 at 17:52