Formal consequences of Riemann-Roch (multiple answers welcome) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:14:51Z http://mathoverflow.net/feeds/question/716 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome Formal consequences of Riemann-Roch (multiple answers welcome) Andrew Critch 2009-10-16T06:20:39Z 2010-09-28T21:10:53Z <p>This question aims to pin down what Riemann-Roch can tell us about a divisor on a curve, without any "geometric thinking". It can be annoying to wonder if there is some clever trick you're missing out on in a problem, and it would be nice to know the limits of Riemann-Roch formalism the way we know we can't solve a system of 2 linear equations in 3 variables.</p> <p>If someone prefers a different formalization of this question, I'd be happy to get an answer to that one instead. Here is mine: </p> <p>Situation:</p> <p>S1) <code>Div</code> is a free abelian group generated by an infinite set of letters <code>P</code>. [<em>like points</em>] <br> S2) <code>D</code> &isin; <code>Div</code> is "effective" if all its coefficients are non-negative. <Br> S3) <code>deg: Div -&gt; Z</code> is the sum of coefficients map. <Br> S4) <code>Prin</code> is a distinguished subgroup of ker(deg). [<em>like principal divisors</em>] <br> S5) <code>l: Div/Prin -&gt; N</code> is a function to the non-negative integers. [<em>like the dimension of global sections</em>] <br> S6) <code>K is an element of Div</code>. [<em>like a cannonical divisor</em>]</p> <p>Relations (<code>g:=l(K)</code>):</p> <p>R1) <code>l(D)-l(K-D) = deg(D) + 1 - g</code>. [<em>Riemann Roch</em>] <br> R2) <code>l(D+P) = l(D)</code> or <code>l(D)+1</code> for any generator P. <br> R3) <code>l(D)&gt;0</code> iff <code>D</code> is equivalent mod <code>Prin</code> to an effective divisor. <br> R4) If <code>l(D)&gt;0</code> and <code>deg(D)=0</code> then <code>D</code> &isin; <code>Prin</code>.</p> <blockquote> <p>Question A (hopefully manageable): What <em>exactly</em> can be inferred here about one of <code>l(D)</code> or <code>deg(D)</code>, given the other? </p> </blockquote> <p>Maybe someone already knows the answer to this, from experience with solving RR-related problems, or from literature. </p> <p>Awesomely, many other concepts can be reformulated in this context, and we can ask more...</p> <p>Optional definitions:</p> <p>O1) <code>D</code> is "free" if <code>l(D-P) = l(D)-1</code> for any generator <code>P</code>. <br>O2) <code>D</code> is "very ample" if <code>l(D-P-Q) = l(D)-2</code> for any generators <code>P,Q</code> (not nececesarily distinct) <br>O3) <code>D</code> is "ample" if <code>nD</code> is very ample for some <code>n&gt;0</code>. <br>O4) <code>D</code> is "big" if for some <code>c&gt;0</code> and all large <code>n</code>, <code>l(nD) ≥ cm^n</code></p> <blockquote> <p>Question B (partial answers welcome): What <em>exactly</em> can be inferred here about <code>l(D)</code>, <code>deg(D)</code>, effectiveness, freeness, (very) ampleness, and bigness of <code>D</code> given information about the others? </p> </blockquote> <p>Some examples (see Hartshorne chapter IV): </p> <ul> <li><code>l(0)=1</code></li> <li><code>deg(K) = 2g-2</code></li> <li>If <code>D</code> is very ample then <code>deg(D)&gt;0</code></li> <li>If <code>deg(D) ≥ 2g</code> then <code>D</code> is free</li> <li>If <code>deg(D) ≥ 2g+1</code> then <code>D</code> is very ample </li> <li><code>D</code> is ample iff <code>deg(D)&gt;0</code></li> </ul> <p>So, yeah! What's the deal with Riemann-Roch?</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/721#721 Answer by Lars for Formal consequences of Riemann-Roch (multiple answers welcome) Lars 2009-10-16T07:29:34Z 2009-10-16T07:29:34Z <p>The easiest thing that comes to mind is that if D is effective, then l(D) &le; deg(D) + 1, since l(K-D)-g &le; 0.</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/728#728 Answer by David Lehavi for Formal consequences of Riemann-Roch (multiple answers welcome) David Lehavi 2009-10-16T08:33:56Z 2009-10-16T08:33:56Z <ul> <li>deg D=2g-2, l(D)=g => D=K</li> <li>deg D=1, l(D) = 2 => g=0</li> </ul> <p>If you want a nice purely geometric fact, deduced from purely numerical arguments, look at Griffiths and Harris p. 258 where they show that a canonical genus 4 curve is the intersection of a quadric and a cubic in it's canonical system, by considering only h^0(nK) and h^0(P^3, n H), where H is a hyperplane and n=1,2,3.</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/744#744 Answer by Andy Putman for Formal consequences of Riemann-Roch (multiple answers welcome) Andy Putman 2009-10-16T14:53:58Z 2009-10-16T14:53:58Z <p>I don't have time to write up a detailed list of nice formal consequences, so I'll just give two references (neither of which seem to be as popular as they deserve to be). Probably the easiest place to read about the "standard" applications and tricks using Riemann-Roch is Rick Miranda's book "Algebraic Curves and Riemann Surfaces". It's written in such a way that an advanced undergraduate could read it, but it manages to cram in a lot of nice stuff. At a slightly higher level, you can't go wrong with Clemen's "Scrapbook of Complex Curve Theory".</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/39643#39643 Answer by roy smith for Formal consequences of Riemann-Roch (multiple answers welcome) roy smith 2010-09-22T18:15:38Z 2010-09-28T03:54:43Z <p>Riemann's original proof (as modified by Roch) showed that for D > 0 the space of differentials of functions in L(D) is the kernel of a linear period map from C^d--->C^g. Hence the dimension l(D)-1 of that kernel satisfies d-g ≤ l(D)-1 ≤ d. Equivalently, 1+d-g ≤ l(D) ≤ d+1. He also showed that deg(K) = 2g-2, and l(K) = g. It is trivial that d&lt;0 implies l(D) = 0.</p> <p>Roch showed that the cokernel of that map has dimension l(K-D), hence l(D) - l(K-D) = 1+d-g.</p> <p>The easy (formal) consequences are these:</p> <p>1) d = deg(D) > 2g-2 implies l(D) = 1+d-g, i.e. l(D) is minimal when deg(D) > deg(K).</p> <p>2) d = deg(D) > 2g-1 implies D is free, since l(D-P) must go down one, by 1).</p> <p>3) d = deg(D) > 2g implies D is very ample, since both l(D-P) and l(D-P-Q) go down.</p> <p>Since these consequences are easy, the interesting cases are for d = deg(D) ≤ 2g-2. The first one of these is that K is very ample unless there exist P,Q with l(P+Q) > 1.</p> <p>The next important, less formal result is that for 1 ≤ d ≤ 2g-1, one always has 2.l(D)-2 ≤ d, a generalization of the case for D = K. Moreover, equality holds if and only if either D = K or D is a multiple of a divisor P+Q as above with l(P+Q) > 1.</p> <p>Good sources for this are Griffiths - Harris for Riemann and Roch's proofs, and Arbarello - Cornalba- Griffiths - Harris for the refinements. I have written a longer article on the topic available free on my website at roy smith's web page at university of georgia math department. Let me recommend especially a fabulous book by the late george kempf, abelian integrals, available from the university autonoma de mexico. this is the best source for the famous "riemann kempf" singularities theorem. for the benefit of those who will not read it, I remark that the point, derived from insights of mumford, is that the singularities of a theta divisor are modeled by those of the discriminant locus of matrices of a given rank. I.e. if you understand singularities of rank loci of matrices then you also understand singularities of theta divisors. This explains the structure of the book by Arbarello, Cornalba, Griffiths, Harris, on Geometry of Algebraic Curves.</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/39695#39695 Answer by fcukier for Formal consequences of Riemann-Roch (multiple answers welcome) fcukier 2010-09-23T05:39:44Z 2010-09-23T05:39:44Z <p>To the sources given for RR in other answers, I would like to add Mumford's Complex Projective Varieties. He gives a nice proof, using the residue theorem plus basic linear algebra.</p> http://mathoverflow.net/questions/716/formal-consequences-of-riemann-roch-multiple-answers-welcome/40378#40378 Answer by stankewicz for Formal consequences of Riemann-Roch (multiple answers welcome) stankewicz 2010-09-28T19:52:28Z 2010-09-28T21:10:53Z <p>Dino Lorenzini has a preprint in which he considers "Riemann-Roch structures", roughly as you've laid them out. <a href="http://www.math.uga.edu/~lorenz/RRNovember11.dvi" rel="nofollow">http://www.math.uga.edu/~lorenz/RRNovember11.dvi</a></p> <p>Namely, he considers the situation of </p> <p>(S1) The free abelian group $\mathbf{Z}^n$ (okay, so not infinite... Riemann-Roch also works over a finite field)</p> <p>(S2) $R\in \mathbf{Z}^n$ an "effective" vector or divisor with coprime integer entries</p> <p>(S3) degree of a divisor $D$ with respect to $R$ is simply the dot product $R\cdot D$</p> <p>(S4) the "principal divisors" are chosen to be a lattice $\Lambda$ inside of $\Lambda_R$, the lattice of vectors or divisors perpendicular to $R$</p> <p>So that if (S6) a canonical divisor exists, there exists a function (S5) $h: \mathbf{Z}^n/\Lambda \to \mathbf{Z}_{\ge 0}$ which satisfies some of your relations and optional requirements called a "Riemann-Roch structure".(This is proposition 2.4)</p> <p>Of interest and proved by Baker and Norine is that if we take $\Lambda$ to be the image of the LaPlacian matrix of a graph with $n$ vertices and $m$ edges, we get such a structure.</p>