how does one understand GRR? (Grothendieck Riemann Roch) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:32:43Z http://mathoverflow.net/feeds/question/63095 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch how does one understand GRR? (Grothendieck Riemann Roch) roy smith 2011-04-27T00:44:22Z 2011-04-29T03:28:31Z <p>I tried to answer an earlier question as to uses of GRR, just from my reading, although i do not understand GRR. Today i tried to understand the possible idea behind GRR. After editing my answer accordingly, it occurred to me i was asking a question instead of giving an answer. My question is roughly whether the following speculation is in the ball park as to the purpose of GRR.</p> <p>I've been thinking about Riemann Roch today, and reading Riemann. After dealing with a fixed divisor D, Riemann observes that his result proves every divisor of degree g+1 dominates the pole divisor of a non constant meromorphic function. Then he says that it may be possible to find a special divisor of even lower degree that dominates the poles of a non constant function. I.e. he begins to vary the divisor. By a rank calculation he shows one cannot expect a non constant function unless the pole divisor has degree at least (g/2)+1.</p> <p>Now following his lead, we are led to vary the curve instead of the divisor. E.g. we might consider the family of curves over the moduli space. Then a good Riemann Roch theorem should let us relate the riemann roch theorem for the curve fibers, to a conclusion for a related sheaf on the base space., like a kunneth type formula, relating cohomology of base space total space and fiber.</p> <p>I.e. a nice divisor like the canonical divisor on a curve, should be cut out on each curve fiber by a divisor on the total space, by intersecting it with each curve. (e.g. we could restrict the sheaf O(1) on the plane, to every curve of degree 4.) then we can push this sheaf from the total space down to the base space, i.e. the moduli space of curves. A good relative riemann roch theorem would then relate the universal canonical sheaf on the total space, to the canonical sheaves on the curve fibers, and the cohomology of the push down of the universal sheaf to the base space, the moduli space of curves.</p> <p>Ideally such a relation would let one compute invariants of sheaves on the moduli space that do arise by pushing down sheaves on the total space of curves. Hopeful applications might include finding ample sheaves on Mg, hence proving projectivity, and computing invariants of the canonical sheaf on Mg, hence potentially estimating the kodaira dimension.</p> <p>Now this is all speculation since i do not understand even the statement of the GRR, and have not read the paper of Harris-Mumford in which the application i cited above is made. Moreover I have never seen any proof of kodaira dimension of Mg using this method. Perhaps someone more knowledgable will comment on these speculative applications? </p> <p>Is this roughly the idea behind GRR and Mumford's applications of it? I.e. is the idea of GRR to understand the cohomology of a sheaf on a base space which arises as a push down, by restricting it to the fibers of the map? and how helpful is this in practice?</p> <p>specific question: if chi(O) is constant on fibers, does GRR allow one to determine chi(O) of either total space or base space from the other?</p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63121#63121 Answer by Sasha for how does one understand GRR? (Grothendieck Riemann Roch) Sasha 2011-04-27T06:25:42Z 2011-04-27T06:25:42Z <p>Let $f:X \to Y$ be a proper morphism of algebraic varieties. The idea behind GRR is that the class in the Grothendieck group $K_0(Y)$ of the derived pushforward $Rf_*(F)$ of $F \in D^b(coh X)$ depends only on the class of $F$ in $K_0(X)$. The GRR itself gives an explicit formula in terms of Chern characters.</p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63135#63135 Answer by Leo Alonso for how does one understand GRR? (Grothendieck Riemann Roch) Leo Alonso 2011-04-27T09:16:59Z 2011-04-27T09:16:59Z <p>Given a proper morphism of smooth algebraic varieties $f \colon X \to Y$ we have on one hand $K_0$ groups and on the other a "cohomology" theory $H^*$, you make take for this Chow groups, singular homology (in the complex case) or the associated graded ring to the $\gamma$ filtration on K-groups. In any case every cohomology theory with <em>rational</em> coefficients. There is a natural transformation:</p> <p>$$ch \colon K_0 \to H^*$$</p> <p>that relies both theories. It is a sort of categorified exponential in the sense that, as generalized cohomology theories, K-theory is associated to the multiplicative formal groups while ordinary cohomology theory is associated to the additive formal group. If we combine the Chern character $ch$ with pull-back, then both are compatible in the sense that $f^\ast ch = ch f{}^\ast$. But these cohomology theories are also covariant for proper morphisms. In this case, the expected formula does not hold and one has to put as extra factors the <em>Todd classes</em> of $X$ and $Y$. So, in a sense, multiplying by these classes repairs the lack of covariant functoriality. By the way, the correct formula reads:</p> <p>$$ch(f_* e) \cdot Td(Y) = f_*( ch(e) \cdot Td(X))$$</p> <p>with $e$ a class in $K_0(X)$.</p> <p>Of course there are generalizations and to remove the non singularity hypothesis requires as usual a lot of machinery. If $e$ comes from a vector bundle, you may think about the formula as a way to compare the Chern classes of the direct image (usually a perfect complex, but there is no trouble in defining its Chern invariants) with the direct image of its Chern classes. And the comparison is mediated with fundamental classes of $X$ and $Y$. In case $Y$ is a point we are talking about computing the Euler-Poincaré characteristic of $e$. To me, the lack of compatibility of the Chern character with push forward shows that there are subtle relationships between the exponential and the intrinsic classes of the varietis, but perhaps this last comment is much to vague.</p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63136#63136 Answer by Sándor Kovács for how does one understand GRR? (Grothendieck Riemann Roch) Sándor Kovács 2011-04-27T09:19:36Z 2011-04-27T09:32:45Z <p><strong>Disclaimer</strong>: I would not be surprised if I committed some mistakes below, especially in the computational part. Hopefully the errors are indeed limited to computation and not the concepts.</p> <p>So, let's try to understand GRR in the context you are interested in:</p> <p>Let $f:X\to Y$ be a smooth family of projective curves and let $\mathscr L$ be a line bundle on $X$.</p> <p>GRR says that two things are equal, so let's understand each side. The left hand side is the <em>exponential Chern character</em> <code>$\mathrm{ch}(\_)$</code> of <code>$f_!(\mathscr L)$</code>.</p> <p>Since $f$ is flat of relative dimension $1$, we have that <code>$$f_!(\mathscr L)= f_*\mathscr L- R^1f_*\mathscr L.$$</code> The subtraction takes place in $K(Y)$, but it is not important at the moment. If you accept that we can temporarily think of this as something reasonable, then everything is dandy. (At least so far).</p> <p>The exponential Chern character is at the end a recipe to take a polynomial of various Chern classes. I find it fascinating that it can be given in such a simple form yet it is nearly impossible to remember, but again, this is not important at the moment.</p> <p>Now for the sake of understanding what's going on assume $\mathscr L$ behaves nicely with respect to $f$ in the sense that $h^0(X_y, \mathscr L_y)$ and $h^1(X_y, \mathscr L_y)$ are constant. This <strong>is</strong> the case if $\mathscr L=\omega_{X/Y}^{\otimes m}$.</p> <p>Anyway, if $\mathscr L$ is such, then both <code>$f_*\mathscr L$</code> and <code>$R^1f_*\mathscr L$</code> are locally free. </p> <p>Perhaps at this point we could also assume that $Y$ is a smooth projective curve. In that case $\mathrm{ch}(\mathscr E)=r+c_1(\mathscr E)$ where $\mathscr E$ is a locally free sheaf of rank $r$. Then we have <code>$$\mathrm{ch}(f_!(\mathscr L))= \mathrm{ch}(f_*\mathscr L)- \mathrm{ch}(R^1f_*\mathscr L)= \chi(\mathscr L_y) + c_1(f_!(\mathscr L)).$$</code></p> <p>Let's see how far we get with the right hand side of GRR.</p> <p>First we need the exponential Chern character of $\mathscr L$ on $X$, but this is easy: $$\mathrm{ch}(\mathscr L)=1+c_1(\mathscr L).$$</p> <p>Then we need the <em>Todd class</em> of the relative tangent sheaf of $f$. That's not too hard either. $\mathscr T_f=\omega_{X/Y}^{-1}$ and if we keep the assumption that $Y$ is a curve, then $X$ is a surface, so the Todd class is relatively manageable: <code>$$\mathrm{td}(\mathscr T_f)=1-\dfrac 12 c_1(\omega_{X/Y}) + \dfrac 1{12} c_1^2(\omega_{X/Y})$$</code> (Actually, this would be the same even if $Y$ is a surface. If $Y$ is a threefold, then there is an additional term with $c_1^4$ and so on.)</p> <p>Now we have to take the intersection product of these two classes and get <code>$$\mathrm{ch}(\mathscr L)\cdot \mathrm{td}(\mathscr T_f) = 1 + \left(c_1(\mathscr L) - \dfrac 12 c_1(\omega_{X/Y})\right) + \dfrac 12 c_1(\mathscr L)\cdot c_1(\omega_{X/Y}) + \dfrac 1{12} c_1^2(\omega_{X/Y}).$$</code></p> <p>The right hand side of GRR is <code>$f_*$</code> of this. Remember that <code>$f_*$</code> is zero if the image of the cycle is lower dimensional than the cycle. In particular $f_*(1)=0$ and $f_*D=D\cdot X_y=\deg D_y$ for a divisor $D$ and an arbitrary $y\in Y$. For points the push forward is just the image points with the appropriate coefficients.</p> <p>So we have that </p> <p><code>\begin{multline} f_*(\mathrm{ch}(\mathscr L)\cdot \mathrm{td}(\mathscr T_f)) =\\ \left(\deg\mathscr L_y - \dfrac 12 \deg\omega_{X_y}\right) + \dfrac 12 f_*\left(c_1(\mathscr L)\cdot c_1(\omega_{X/Y})\right) + \dfrac 1{12} f_*c_1^2(\omega_{X/Y}) \end{multline}</code></p> <p>Now if we compare this to the computation for the left hand side, we see that the first term (in parentheses) equals $\chi(X_y)$, so the original RR is embedded in here and the other two terms are equal to $c_1(f_!(\mathscr L))$ as defined above. Taking the degree of these divisors on $Y$ gives us a numerical equality: <code>$$\deg f_*\mathscr L -\deg R^1f_*\mathscr L = \dfrac 12 c_1(\mathscr L)\cdot c_1(\omega_{X/Y}) + \dfrac 1{12} c_1^2(\omega_{X/Y}).$$</code></p> <p>So one may say that GRR is a relative version of RR (or HRR) where you get the original one on the fibers and you get some information about the "horizontal" variation of relative cohomology of your line bundle. </p> <p>Then if you feel adventurous you may try to understand more complicated morphisms. I believe that the case of the universal curve over the moduli space is pretty much the same as here except that the computation on $Y$ is a little more complicated and so is the Todd class on $X$.</p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63145#63145 Answer by Dan Petersen for how does one understand GRR? (Grothendieck Riemann Roch) Dan Petersen 2011-04-27T10:18:40Z 2011-04-27T21:32:41Z <p>Here is how I think about G-R-R in the context of moduli of curves. I realize now that I wrote something quite long. </p> <p>Let me recall first the definition of the tautological ring. As a consequence of the results on the birational geometry of $\overline M_g$ that there is no hope of understanding the whole Chow ring of $\overline M_g$ -- for instance, unlike in misleading low genus examples, the Chow ring will in general be infinite-dimensional. In David Mumford's "Towards an enumerative geometry..." he he introduces a finite-dimensional subring of the Chow ring which contains all "geometrically natural" classes in the Chow ring and proposes studying it instead, and this subring is called the tautological ring. Let me quote:</p> <p>"Whenever a variety or topological space is defined by some universal property, one expects that by virtue of its defining property, it possesses certain cohomology classes called tautological classes. The standard example is a Grassmannian [...] by its very definition, there is a universal bundle $E$ on Grass of rank $k$, and this induces Chern classes $c_l(E)$ in both the cohomology ring and Chow ring of Grass."</p> <p>Let me expand on the meaning of "geometrically natural". There are several possible definitions of the tautological ring. The one used by Mumford is that it is the subring generated by the so-called $\kappa$-classes, which is not really the right one: you should also for instance consider the boundary divisors as tautological classes (but this is implicit already in Mumford's paper). A nice definition is the one of Faber and Pandharipande, which defines the tautological ring for all spaces $\overline M_{g,n}$ simultaneously: it is the minimal system of subrings which contains all fundamental classes, is closed under all gluing morphisms, and is closed under all forgetting points-morphisms. </p> <p>Morally what this means is that: (i) for any "natural" bundle you can write down directly in terms of the moduli functor, its Chern classes are going to be tautological; (ii) any sort of "natural" gluing procedure on curves is going to keep you inside of the tautological ring. For example, the $\lambda$-classes (Chern classes of the Hodge bundle) are tautological, the $\psi$-classes are tautological (the line bundles given by the cotangent line at a marked point), and the $\kappa$-classes are tautological. </p> <p>OK, so let us return to G-R-R. Let $f \colon X \to Y$ be a proper morphism. On one side of the equation you have the Chern character of the derived pushforward $Rf_\ast F$. On the other side you have the pushforward of the Chern character of $F$ and the Todd class of the relative tangent sheaf $T_f$. The point is that both $F$, $Rf_\ast F$ and $T_f$ can all be made sense of by working locally/fiberwise: we don't need to know <em>anything</em> about the global structure of $Y$ to apply G-R-R to $f$ and $F$. But this is also how the tautological ring was set up: the classes in the tautological ring are exactly those that can be defined by pushing around classes of "fiberwise" defined bundles, which means that these are exactly the classes that can be defined without making any reference to any "global" structure of the moduli space. </p> <p>So in hindsight Grothendieck-Riemann-Roch seems tailor made for the study of tautological rings. On the other hand, this is also a limitation of G-R-R: it will produce lots of relations and identities relating tautological classes to each other, but it will never prove any "global" statement about any of them. </p> <p>As an example, it is possible to algorithmically compute the intersection number on $\overline M_{g,n}$ for any polynomial in boundary strata and $\lambda$-, $\psi$- and $\kappa$-classes. First you express the $\kappa$-classes as pushforwards of $\psi$-classes, then G-R-R can be used to express the $\lambda$-classes in terms of pushforwards of $\psi$-classes, which will finally reduce your computation to an intersection number only involving $\psi$-classes. All this was completely formal, but sooner or later you are going to need to use some global geometric property of $\overline M_{g,n}$ to find an actual number, and this is where it comes in: the Witten conjecture/Kontsevich's theorem tells you how to compute any intersection of $\psi$-classes.</p> <p>So let me finally talk a bit about the article of Harris and Mumford. The first application of G-R-R in their article is to derive the formula $K_{\overline{M}_g} = 13\lambda_1 - 2\delta_0 - 3\delta_{1} - \ldots - 2\delta_{n}$ in the tautological ring. This is done by applying GRR to the projection from the universal curve and truncating after the first term. Incidentally, if you don't truncate after the first term, you get Mumford's formula (derived in "Towards an enumerative geometry...") expressing the Chern character of the Hodge bundle in terms of $\kappa$-classes and pushforwards of $\psi$-classes from the boundary strata. </p> <p>But again, GRR will not tell you any global geometric information like if a class is big or ample. The idea is then to find an effective divisor $D$ such that $mK_{\overline M_g} = D + a\lambda_1$ with $a > 0$. It turns out that this is possible for $D$ equal to the locus of $k$-gonal curves, where they pick $g = 2k-1$. They describe in the article how they came up with this particular choice of $D$ by trying to generalize the work of Freitag on the Kodaira dimension of $A_g$ for $g$ large, in particular I think that there should be a Siegel modular form whose pullback to $M_g$ conjecturally would have $D$ as its vanishing locus. I don't know if this was actually worked out in later work. Then $nK_{\overline M_g}$ for large enough $n$ defines a birational map using the fact that $\lambda_1$ is ample on $A_g$, ultimately because the Satake compactification is the Proj of the ring of Siegel modular form, i.e. the sections of powers of the determinant of the Hodge bundle. (However $\lambda_1$ is not ample on $\overline M_g$!)</p> <p>This part is clarified by the later article of Cornalba and Harris showing that a linear combination $a\lambda - b\delta$ is ample if and only if $a > 11b$. The rational Picard group of $\overline M_g$ is generated by $\lambda_1$ and the boundary divisors, so any effective divisor has an expression of the form $a\lambda - \sum b_i \delta_i$, so estimating the Kodaira dimension of $\overline M_g$ really comes down to finding effective divisors such that the <em>slopes</em> $a/b_i$ are small. </p> <p>Anyway, the second application of GRR in their article is to show that on the open part $M_g$, the $k$-gonal locus is a multiples of $\lambda_1$. Actually, this part uses even more crucially Porteous's formula: once they express $k$-gonality in terms of a morphism of bundles having lower than expected rank, the class of the $k$-gonal locus can be expressed in terms of Chern classes of the two bundles, i.e. in terms of tautological classes. It follows then that $D$ is the sum of a multiple of $\lambda_1$ and an integral linear combination of boundary divisors. Finally these integers are determined by evaluating the divisors on suitable "test curves". They conclude that $\overline M_g$ is of general type for big $g$. </p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63157#63157 Answer by Paul Johnson for how does one understand GRR? (Grothendieck Riemann Roch) Paul Johnson 2011-04-27T12:25:14Z 2011-04-27T12:25:14Z <h2>Introduction: GRR gives relations in the tautological ring</h2> <p>I can't speak directly to the potential applications you had in mind as far as Kodaira dimension, but I can say something about Mumford's application of GRR to the moduli space of curves. It seems that it's quite close to what you imagine, and in fact it's very important in the study of (a certain part of) the cohomology ring of the moduli spaces $\mathcal{M_g}$, and its relatives such as Deligne-Mumford space $\overline{\mathcal{M}}_{g,n}$ (now the curve has $n$ marked points, and we compactify by adding certain nodal curves should the points try to collide or the complex structure of the curve degenerate), and even further into Gromov-Witten theory. I'm going to give an overview of this story, building out of your question (I hope).</p> <p>The part of the cohomology ring of $\mathcal{M_g}$ I'm talking about is called the <em>tautological ring</em>, and a gentle survey-introduction is Ravi Vakil's <a href="http://math.stanford.edu/~vakil/files/cimemay26.pdf" rel="nofollow">The moduli space of curves and Gromov-Witten theory</a>. Those notes do not explicitly mention GRR, but they do quote a result that comes directly from Mumford's GRR calculation, and what I'm going to try to do is explain this a little bit. </p> <p>I'd also like to mention that, as far as I understand it, this direction of application is essentially what Mumford had in mind. The paper he does this calculation is, after all, entitled <a href="http://books.google.com/books?hl=en&amp;lr=&amp;id=df_X80mKSi0C&amp;oi=fnd&amp;pg=PA235&amp;dq=mumford+twoard+an+enumerative&amp;ots=IKtif2znLe&amp;sig=CD36-cUyAAx9Aj48G4QEdi5PDVY#v=onepage&amp;q&amp;f=false" rel="nofollow">"Toward an Enumerative Geometry of the Moduli Space of Curves"</a>.</p> <h2>Warm-up: Grassmannian</h2> <p>You can see in the first paragraph of Mumford's paper that he is explicitly modeling what he's doing after the cohomology of the Grassmannian, so I'm going to spend a paragraph on them, to motivate what's coming in the moduli space of curves. </p> <p>On the one hand, we have the schubert cycles, given by the loci of planes that intersect the a fixed flag with given dimensions. On the other hand, since each point represents a vector space, these vector spaces fit together to give a tautological vector bundle, and we can take cycles representing the chern classes of this bundle, and get different classes -- it's not necessarily clear at all that these different tautological cycles should be related, but they are. </p> <p>Mumford, and many after him, are trying to find similar relations between different tautological classes in $\mathcal{M}_g$. Mumford ends his first paragraph with "Moreover, it appears that many geometrically natural classes are expressible in terms of a small number of basic classes" -- this is akin to that description of the Grassmannian, and it's what GRR will give us.</p> <h2>We start with your basic idea</h2> <p>Rather than get into all the technical details of it, I just want to point out that he proceeds essentially exactly as you imagined here: </p> <blockquote> <p>A good relative riemann roch theorem would then relate the universal canonical sheaf on the total space, to the canonical sheaves on the curve fibers, and the cohomology of the push down of the universal sheaf to the base space, the moduli space of curves.</p> </blockquote> <p>Rephrased slightly differently: let $\pi:\mathcal{M_{g,1}}\to\mathcal{M_g}$ be the map from a moduli space of curves with one marked point to the moduli space of curves with no marked points -- it turns out this is exactly the universal family. Then we have the universal canonnical sheaf $\omega_\pi$ on $\mathcal{M_{g,1}}$ that you were discussing. We can use this to get cohomology classes in $H^*(\mathcal{M_g})$ in two different ways: first take its chern character, and then push down to $\mathcal{M_g}$, or first push down to $\mathcal{M_g}$, and then take the chern character. </p> <p>These two alternatives give rise to a priori very different looking cohomology classes on $\mathcal{M_g}$, but GRR says that, after fiddling with Todd classes, they are the same.</p> <h2>Chern then pushforward</h2> <p>Let's see what happens when we take the first path. Since $\omega_\pi$ is one dimensional, taking the chern character is simply exponentiating $c_1(\omega_\pi)$. The class $c_1(\omega_\pi)$ is called the psi class $\psi$. Note that usually this is defined as the first chern class of the tangent bundle to the curve at the marked point, but through the identification of the universal curve with $\mathcal{M_{g,1}}$, these are equivalent. But I should warn you that this is no longer quite true if we start adding more marked points or nodes to our curves. So taking the Chern character of $\omega_\pi$ gives powers of $\psi$ on $\mathcal{M_{g,1}}$, and now if we push these forward we get the Morita-Mumford-Miller kappa classes $\kappa_i=\pi_*(\psi^{i+1})\in H^{2i}(\mathcal{M_g})$ -- indeed, this is the definition of $\kappa_i$.</p> <h2>Pushforward then Chern</h2> <p>Now, what happens if we go in the other direction? To pushforward $\omega_\pi$, we take the cohomology of $\mathcal{M_g}$ -- since $h^0(C, \omega_C)=g$, independent of the curve $C$, we have $\pi_*(\omega_\pi)=\mathbb{E}$, where $\mathbb{E}$ is a dimension $g$ vector bundle on $\mathcal{M_g}$ known as the hodge bundle. More simply, the fiber of $\mathbb{E}$ over a curve $C$ are the sections of the canonical bundle of $C$. The chern <em>classes</em> of the $\mathbb{E}$ are known as the $\lambda$ classes: $\lambda_i=c_i(\mathbb{E})$. Taking the Chern character of the $\mathbb{E}$ then would then give us a bit of a mess of polynomials in the $\lambda$ classes.</p> <h2>Comparing them</h2> <p>So taking the two different paths from $K(\mathcal{M_{g,1}})$ to $H^*(\mathcal{M_g})$ gives two different looking types of tautological classes, the $\kappa$ classes and the $\lambda$ classes. Since we set this up with GRR in mind, we should now see a relation between them. </p> <p>In this case it turns out that this relationship cleans up rather nicely if we package it in a generating function, and a better answer might explain how, but I'll just note that since we were working with the relative cotangent bundle to begin with, the relative Todd class can be manipulated into just giving us more $\kappa$ classes, and then with more mucking around with characteristic classes, it turns out we can express this relationship very beautifully in terms of generating functions:</p> <p>$$\sum_{i=0}^\infty \lambda_i t^i=\exp\left(\sum_{j=i}^\infty \frac{B_{2j}\kappa_{2j-1}}{2j(2j-1)}t^{2j-1}\right).$$</p> <p>Here $B_{2j}$ are the Bernoulli numbers, coming from the Todd class. This is the formula in Ravi's notes I alluded to earlier: he cites Faber for this particular expression.</p> <h2>Extensions</h2> <p>I've done this just for $\mathcal{M_g}$ for simplicity, but I want to indicate here that you can get a lot more gas out of the same basic idea.</p> <p>First, you can add marked points and boundary points and it essentially goes through the same. The universal curve is still just adding another marked, and then forgetting it. What gets a little complicated is that the relative dualizing sheaf $\omega_\pi$ stops being equal to just the cotangent line at the extra point when our curve becomes singular, but we can understand how they differ, and so we get some additional contributions involving the boundary strata -- I think Mumford already started to deal with this, and Faber and Pandharipande certainly dealt with it.</p> <p>Also, you can extend this to Gromov-Witten theory, and consider moduli spaces of stable maps, and consider a curve with marked points together with a map $f:C\to X$, and play the same game there. Or better, we can first pull back a bundle from $E\to X$, and play the game described above with $f^*(E)$. In case $E$ is a line bundle, $s:X\to E$ a line bundle, and $Y=s^{-1}(0)$ the vanishing set, this can give relationships between the Gromov-Witten invariants of $X$ and $Y$ in terms of the chern classes of $E$. And in case $E$ is a negative, this can express the Gromov-Witten invariants of the total space of $E$ in terms of the Gromov-Witten invariants of $X$ and the chern classes of $E$. This is worked out in Tom Coates's thesis, and in his joint Annals paper with advisor, Givental: <a href="http://annals.math.princeton.edu/2007/165-1/p02" rel="nofollow">Quantum Riemann–Roch, Lefschetz and Serre</a>, and is very important to GW theory, as its the method by which we can understand $X$ that is a complete intersection in a toric variety $Y$ -- this method relates the GW theory of $X$ to that of $Y$, and since $Y$ is toric we can localize with respect to the torus action to compute its Gromov-Witten invariants. </p>