What is the geometry behind psi classes in Gromov-Witten theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:49:08Z http://mathoverflow.net/feeds/question/24545 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24545/what-is-the-geometry-behind-psi-classes-in-gromov-witten-theory What is the geometry behind psi classes in Gromov-Witten theory? Simon Rose 2010-05-13T22:07:14Z 2010-07-20T18:16:08Z <p>Intuitively, Gromov-Witten theory makes perfect sense. Via Poincare duality, we look at the cohomology classes $\gamma_1, \ldots, \gamma_n$ corresponding to geometric cycles $Z_i$ on a target space $X$, pull them back and then the integral $$\langle \gamma_1 \cdots \gamma_n\rangle=\int_{\overline{\mathcal{M}}_{g,n}(X,\beta)^{vir}}ev_1^*\gamma_1 \smile \cdots \smile ev_n^*\gamma_n$$ should count the number of curves whose intersection with the given cycles is non-empty.</p> <p>However, we also have the &psi;-classes (or "gravitational descendants") arising from the moduli space $\overline{\mathcal{M}}_{g,n}$ which are the chern classes of the $i$-th cotangent line bundle to a given $(C, x_1, \ldots, x_n) \in \overline{\mathcal{M}}_{g,n}$.</p> <p>So what, geometrically, do these represent? The fact that they arise from $\overline{\mathcal{M}}_{g,n}$ means that the inclusion of a &psi;-class places restriction on the geometry of the curves which we count; that much is clear. What is this restriction?</p> <p>The reason that I am curious is that I am trying to evaluate the GW-invariants corresponding to maps which have components collapsing to an A<sub>1</sub> singularity (i.e. a $B\mathbb{Z}/2$), but such that not all of the curve collapses. It has been mentioned in passing that including a &psi;-class could help with this, and while the little I understand makes this sound plausible, I don't exactly see why.</p> <p>So what are &psi;-classes? Can I use them to split my curve up into parts so that a fixed component lands on my stacky point, while the rest of it does whatever else curves do?</p> http://mathoverflow.net/questions/24545/what-is-the-geometry-behind-psi-classes-in-gromov-witten-theory/32670#32670 Answer by ABayer for What is the geometry behind psi classes in Gromov-Witten theory? ABayer 2010-07-20T18:16:08Z 2010-07-20T18:16:08Z <p>The following answer is unfortunately not quite correct, but it may be useful anyway. I will of course be ignoring any virtual fundamental class issues.</p> <p>Imagine that you are computing a Gromov-Witten invariant where you require the i-th marked point to land at a specific point (i.e. your i-th insertion &gamma;<sub>i</sub> is the class of a point), and now lets add aditionally the i-th psi-class as an insertion. You can restrict to the subspace of maps with <code>$f(x_i) = x$</code> for some generic choice of $x \in X$. Fixing an arbitrary non-trivial map $\Phi \colon T_x \to k$ gives you by composition a map from the relative tangent bundle of the universal curve over <code>$M_{g, n}(X)$</code> at the section x<sub>i</sub> to the trivial line bundle, in other words a section $\phi$ of the relative cotangent bundle of the universal curve. It will vanish on curves which are tangent to a hypersurface through x with tangent direction matching the zero-locus of the map $\Phi$.</p> <p>So you can think of Gromov-Witten invariants with psi-classes as counting maps which additionally satisfy tangency conditions at the marked points.</p> <p>Why is this not correct? The zero locus of $\phi$ computes the Chern class of the relative cotangent bundle at $x_i$ over M<sub>g, n</sub>(X), which is not the same as the pull-back of the $\psi$-class from M<sub>g, n</sub>. Insertions of the former are sometimes called "gravitational ancestors", and the difference to gravitational descendants is described explicitly in alg-geom/9708024.</p>