Kodaira-Spencer map in a concrete instance - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:04:58Z http://mathoverflow.net/feeds/question/22345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22345/kodaira-spencer-map-in-a-concrete-instance Kodaira-Spencer map in a concrete instance Qfwfq 2010-04-23T12:59:28Z 2010-05-13T03:47:45Z <p>Let $\pi:X_{\epsilon} \rightarrow \Delta$ be a family of (say smooth) projective plane curves parametrized by $\Delta:=\operatorname{Spec}(k[\epsilon])$, and let $X=X_0$ be the closed fiber. Suppose that $X_\epsilon$ is given by a polynomial $f(x,y,z;\epsilon)$ homogeneus in $x,y,z$.</p> <p>Let $\phi=\operatorname{ks}(\pi)=\operatorname{ks}(\partial/\partial\epsilon) \in H^1(X,T_X)=H^0(X,K_X^2)^{*}$ be the Kodaira-Spencer image of the above family.</p> <ul> <li>Is it possible to characterize $\phi$ concretely in terms of the polynomial $f$?</li> </ul> <p>If you want, feel free to restrict to the hyperelliptic case: </p> <p>$f(x,y,1;\epsilon):=y^2-\Pi_{i=1}^{2g+2}(x-\lambda_i(\epsilon))$,</p> <p>(which I think has to be desingularized though)</p> <p>in which case a basis of $H^0(X,K_X)$ is given by $\frac{x^{k}}{y}dx$ for $k=0 \cdots g-1$.</p> http://mathoverflow.net/questions/22345/kodaira-spencer-map-in-a-concrete-instance/22362#22362 Answer by Kevin Lin for Kodaira-Spencer map in a concrete instance Kevin Lin 2010-04-23T14:50:15Z 2010-04-23T14:50:15Z <p>One way to concretely realize the Kodaira-Spencer map is via Cech cohomology. Take a vector field downstairs, lift it to a vector field upstairs, and apply the Cech differential to the lifted vector field...</p> <p>(I had posted another answer, but then I realized that what I wrote was nonsense...)</p> http://mathoverflow.net/questions/22345/kodaira-spencer-map-in-a-concrete-instance/23973#23973 Answer by jlk for Kodaira-Spencer map in a concrete instance jlk 2010-05-09T01:06:32Z 2010-05-13T03:47:45Z <p>Here is an attempt. Based on your comment to Kevin Lin's post, I think that you know the first part of what I have written, but I included this for the sake of completeness.</p> <ul> <li><em><strong>Some Generalities on $\phi$:</em></strong> Any deformation of an <em>affine</em> hyperelliptic curve such as</li> </ul> <p>$$y^2 = \prod (x - \lambda_i(\epsilon))$$</p> <p>is trivial and hence corresponds to the zero cohomology class. Indeed, any deformation of a smooth, affine scheme (separated and of finite type over a field?) is trivial. Given a deformation $X_{\epsilon} \to \Delta$ as you describe, the Kodaria-Spencer map is computed by fixing an open affine cover $U_i$ of $X_0$ and isomorphisms $\phi_i \colon X_{\epsilon}|_{U_i} \to U_{i} \otimes k[\epsilon]$ of the restriction of $X_{\epsilon}$ to $U_i$ with the trivial deformation of $U_{i}$. The automorphism $\phi_{i} \circ \phi_{j}^{-1}$ of determines an explicit Cech cocyle that represents a class in $H^{1}(X_0, TX_0)$, and one checks that this class is independent of the choices made. The main point: <em>the Kodaira-Spencer class comes from deforming the gluing data NOT from deforming the equations.</em> </p> <ul> <li><em><strong>Computation of $\phi$:</em></strong> As you wrote, it is not clear from that description how everything works in a concrete cases. Here is how it works out in the case of a general genus $2$ hyperelliptic curve. Working over the field $k$, this curve can be described as the curve obtained by gluing the two affine schemes</li> </ul> <p>$$U_1 := \operatorname{Spec}(k[x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)),$$ $$U_2 := \operatorname{Spec}(k[x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)),$$ over the usual opens via the isomorphism $g$ defined by the rules $$x_1 \mapsto x_2^{-1},$$ $$y_1 \mapsto y_2 x_2^{-3}.$$ Here $r_1, \dots, r_6$ are general scalars.</p> <p>Associated to the affine open cover ${U_1, U_2}$ is the usual Cech complex, and we can use this complex to compute $H^{1}(X, TX)$. Some elements of this cohomology group are given by the Cech cocycles $$y_1/x_1 \frac{\partial}{\partial x_1}, y_1/x_1^{2} \frac{\partial}{\partial x_1}, y_1/x_1^{3} \frac{\partial}{\partial x_1} \in H^{0}(U_{12}, TX).$$ Here $U_{12}$ denotes the intersection of $U_1$ and $U_2$. Note: one needs to check that these vector fields are regular on $U_{12}$. The vector field $\frac{\partial}{\partial x_1}$ has simple poles at ramification points of the degree $2$ to $\mathbb{P}^1$, and the $y_1$ terms are needed to cancel these poles. I think these elements form a basis, but you just asked for an example so I guess we don't care about this.</p> <p>Let's compute the 1st order deformation of $X$ associated to $D:= y_1/x_1 \frac{\partial}{\partial x_1}$. To construct the deformation, we take the trivial deformations of $U_1$ and $U_2$ and deform the gluing automorphism. The trivial deformations are $$\operatorname{Spec}(k[\epsilon, x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)),$$ $$\operatorname{Spec}(k[\epsilon, x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)).$$</p> <p>The general rule is that the deformed gluing map $\tilde{g}$ is given by $\tilde{g}(a) = g(a) + \epsilon \cdot g(D(a))$. For our particular choice of $D$, I think this yields: $$x_1 \mapsto x_2^{-1} + y_2 x_2^{-2} \epsilon,$$ $$y_1 \mapsto y_2 x_2^{-3} + y_2 x_2^{-2} \frac{-x_2^{-1} q'(x_2) + 6 x_2^{-2} q(x_2)}{2 y_2} \epsilon.$$ Here $q(x_2) = \prod_{i=1}^{6} (1-r_i x_2)$.</p> <p>The expression for the image of $y_1$ is quite complicated, but it hopefully is just $g(y_1/x_1 \frac{\partial y_1}{\partial x_1})$.</p> <p>One can work our a similar description for the deformations coming from the other cohomology classes that I wrote down. Assuming these form a basis, this completely describes the map $\phi$.</p> <p>It is easy to reverse this construct as well. Every deformation arises by deforming the map $g$ to a map $\tilde{g}$ as we have done. The associated cohomology class can be described by writing $\tilde{g} = g + \epsilon \cdot D$ for some function $D$. One can show that $D$ defines a regular vector field on $U_{12}$ and hence represents an element of $H^{1}(X, TX)$.</p>