Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $x,y,z$.

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.

  • Is it possible to characterize $\phi$ concretely in terms of the polynomial $f$?

If you want, feel free to restrict to the hyperelliptic case:

$f(x,y,1;\epsilon):=y^2-\Pi_{i=1}^{2g+2}(x-\lambda_i(\epsilon))$,

(which I think has to be desingularized though)

in which case a basis of $H^0(X,K_X)$ is given by $\frac{x^{k}}{y}dx$ for $k=0 \cdots g-1$.

share|improve this question
add comment

2 Answers

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.

  • Some Generalities on $\phi$: Any deformation of an affine hyperelliptic curve such as

$$ y^2 = \prod (x - \lambda_i(\epsilon)) $$

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: the Kodaira-Spencer class comes from deforming the gluing data NOT from deforming the equations.

  • Computation of $\phi$: 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

$$ U_1 := \operatorname{Spec}(k[x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)), $$ $$ U_2 := \operatorname{Spec}(k[x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)), $$ 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.

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.

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 $$ \operatorname{Spec}(k[\epsilon, x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)), $$ $$ \operatorname{Spec}(k[\epsilon, x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)). $$

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 $q(x_2) = \prod_{i=1}^{6} (1-r_i x_2)$.

The expression for the image of $y_1$ is quite complicated, but it hopefully is just $g(y_1/x_1 \frac{\partial y_1}{\partial x_1})$.

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 $\phi$.

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 $H^{1}(X, TX)$.

share|improve this answer
    
Thank you, I will have a look to this answer later! –  Qfwfq May 12 '10 at 17:30
    
@unknown, given a curve X on a smooth, projective surface S, you can deform the curve by deforming the equation of the curve and these deformations are computed by a map H^{0}(X,O_{X}(X)) \to H^1(X,O). After typing the answer above, I realized this might be closer to your original question. Let me know if you are interested in the details –  jlk May 13 '10 at 5:30
    
@jlk: I was interested in a 1-parameter first-order deformation arising moving the Weierstrass points of the curve, hence (I thought) deforming the (desingularized curve corresponding to the) above equation. I don't know if this is related to embedded deformations of curves inside projective surfaces; if you think yes, it'd be interesting to see the details. –  Qfwfq May 13 '10 at 10:35
    
(continued) Also, could you please explain what do you mean by "you can deform the curve by deforming the equation"? Isn't it that in this specific case [as every h.ell. curve can be obtained as a branched cover of P^1 via the above equation] all such deformations are obtained by "deforming the equation" (in that specific way)? –  Qfwfq May 13 '10 at 10:38
    
@unknown: By "deforming the equation" I meant deforming a curve X on a surface S by deforming it as a closed subscheme (i.e. finding a k[e]-flat closed subscheme in X \times k[e] that is equal to X when e=0). If you write everything out, then these deformations are constructed by deforming the equations of X in S. There is a natural map H^{0}(O(X)|){X}) \to H^{1}(O), and the image of this is the deformations of X that can be realized as embedded deformations. However, I don't think this is what you are interested in. –  jlk May 13 '10 at 14:38
show 2 more comments

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...

(I had posted another answer, but then I realized that what I wrote was nonsense...)

share|improve this answer
    
I was trying to compute the cocycle in H^1(X,T_X) as explained, e.g., in the book by Claire Voisin. You cover the (total space of the family) with affine open sets $V$ so small that you have an isomorphism between each of them and a cartesian product $V_{red}\times \Delta$, then the epsilon components of the "transition functions" give derivations on the coordinate rings of the double intersections: that's the Cech cocycle. - The problem is that it's not obvious (to me) what this trivializing open affines should be in this simple concrete case. –  Qfwfq Apr 23 '10 at 15:48
    
I guess the problem with the computation may be due to the fact that the dual of $H^1(X,T_X)$ is NOT $H^0(X,K_X)$, BUT $H^0(X,2K_X)$! –  Sasha Apr 24 '10 at 3:48
    
Haha! ;) No, that was just a misprint! I've edited now. –  Qfwfq Apr 24 '10 at 18:14
    
Actually, I was trying genus 2, in which case $H^0(X,K_X^2)=S^2H^0(X,K_X)$, so you can use the (tensor products of pairs of elements of the) base I indicated. –  Qfwfq Apr 24 '10 at 18:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.