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:


(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 $k=0 \cdots g-1$.


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

  • $\begingroup$ @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 $\endgroup$ – jlk May 13 '10 at 5:30
  • $\begingroup$ @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. $\endgroup$ – Qfwfq May 13 '10 at 10:35
  • $\begingroup$ (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)? $\endgroup$ – Qfwfq May 13 '10 at 10:38
  • $\begingroup$ @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. $\endgroup$ – jlk May 13 '10 at 14:38
  • $\begingroup$ (continued): I think I understand your question better now. The hyperelliptic curve admits a degree 2 map to P^1, and deforming this map is closely related to deforming its branch locus (i.e. the Weierstrass points). Every deformation of the map to P^1 induces a deformation of X; you'd like to see these deformations written down in such a way that you can compare them to the deformations of the Weierstrass points. Is that a reasonable interpretation of your question? If so, I think this can be done. I am traveling, but I can try to write something down when I get back next week. $\endgroup$ – jlk May 13 '10 at 14:41

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...)

  • $\begingroup$ 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. $\endgroup$ – Qfwfq Apr 23 '10 at 15:48
  • $\begingroup$ 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 $H^0(X,2K_X)$! $\endgroup$ – Sasha Apr 24 '10 at 3:48
  • $\begingroup$ Haha! ;) No, that was just a misprint! I've edited now. $\endgroup$ – Qfwfq Apr 24 '10 at 18:14
  • $\begingroup$ 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. $\endgroup$ – Qfwfq Apr 24 '10 at 18:18

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