Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:30:10Z http://mathoverflow.net/feeds/question/23607 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23607/are-the-arithmetic-genera-of-cohen-macaulay-curves-in-a-fixed-homology-class-boun Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? David Steinberg 2010-05-05T18:15:47Z 2010-07-24T01:03:04Z <p>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). </p> <blockquote> <p>I want to show that if you fix a curve class &beta; in H<sub>2</sub>(X), then there is a bounded interval which contains the genus of every CM curve whose homology class is &beta;.</p> <p>Do you know of a reference for this result?</p> </blockquote> <p>Motivation: I am trying to prove a certain class of sheaves forms a bounded family (namely, the collection of sheaves underlying <a href="http://arxiv.org/abs/0707.2348" rel="nofollow">stable pairs</a>). 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.</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/23607/are-the-arithmetic-genera-of-cohen-macaulay-curves-in-a-fixed-homology-class-boun/23611#23611 Answer by damiano for Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? damiano 2010-05-05T18:45:10Z 2010-05-06T14:42:16Z <p>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.</p> <p>Choose an embedding of <em>X</em> 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 <strong>reduced</strong>) 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 <strong>above</strong>, you conclude that the arithmetic genera of your curves are bounded <strong>above</strong>.</p> http://mathoverflow.net/questions/23607/are-the-arithmetic-genera-of-cohen-macaulay-curves-in-a-fixed-homology-class-boun/23613#23613 Answer by Angelo for Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? Angelo 2010-05-05T19:33:51Z 2010-05-06T08:05:22Z <p>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 $-d-1$.</p> <p>[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.</p> http://mathoverflow.net/questions/23607/are-the-arithmetic-genera-of-cohen-macaulay-curves-in-a-fixed-homology-class-boun/33157#33157 Answer by ABayer for Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? ABayer 2010-07-24T01:03:04Z 2010-07-24T01:03:04Z <p>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:</p> <p>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).</p> <p>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.</p>