This question arose by reading the paper [1], in particular, the remark at p. 737:

As an example, consider a non-isotrivial smooth projective morphism $f \colon X \to Y$ from a smooth projective surface to a smooth projective curve. Suppose that there exists a $y \in Y$ such that the Kodaira-Spencer map $$\rho_{f, \, y} \colon T_{Y, \, y} \to H^1(X_y, \, T_{X_y})$$ is zero. Then $\Omega_X|_{X_y}=\mathcal{O}_{X_y} \oplus \Omega_{X_y}$, so $\Omega_X|_{X_y}$ is not ample.

Smooth, non-isotrivial fibrations from a surface to a curve are called Kodaira fibrations (see [2] for a good introduction to the subject) and the usual way to construct them is taking suitable ramified covers of a product of two curves.

Now, I do not see how to perform the construction in such a way that the above condition on the Kodaira-Spencer map is satisfied. In the paper [3] Kodaira shows, by means of a local computation, that in all his original examples the Kodaira-Spencer map is everywhere non-vanishing (p. 212-213). On the other hand, I am not aware of any example in which it vanishes at some point (in fact, I am not aware of any other example of Kodaira fibration for which the Kodaira-Spencer map has been explicitly computed).

So let me ask the

Question. What are examples of non-isotrivial smooth projective morphisms $f \colon X \to Y$ from a smooth projective surface to a smooth projective curve such that the Kodaira-Spencer map vanishes at some $y \in Y$?


[1] Jabbusch, Kelly, Positivity of cotangent bundles, Mich. Math. J. 58, No. 3, 723-744 (2009). ZBL1186.14037.

[2] Catanese, Fabrizio, Kodaira fibrations and beyond: methods for moduli theory, Jpn. J. Math. (3) 12, No. 2, 91-174 (2017). ZBL1410.14010.

[3] Kodaira, Kunihiko, A certain type of irregular algebraic surfaces, J. Anal. Math. 19, 207-215 (1967). ZBL0172.37901.

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    $\begingroup$ This is vanishing at a point $y$, not in a neighborhood of $y$. The fact that the derivative a function vanishes at a point does not imply that the function is constant in a neighborhood of that point... $\endgroup$ Jan 25, 2022 at 12:31
  • $\begingroup$ @PiotrAchinger: In fact, I checked Kodaira's original reference, and he does prove (by means of a local computation) that the KS-map in his examples is non-zero at every point. So I undeleted and edited the question, asking for examples with KS-map vanishing at some point. $\endgroup$ Jan 25, 2022 at 12:57
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    $\begingroup$ Any ramified base change of a smooth, non-isotrivial fibration should give an example (the Kodaira map being trivial for the ramification points). $\endgroup$ Jan 25, 2022 at 13:14
  • $\begingroup$ @NikolasKuhn: That's nice. Please write the explicit computation as an answer, and I will be glad to accept it. $\endgroup$ Jan 25, 2022 at 16:05

1 Answer 1


Any ramified base change of a smooth, non-isotrivial fibration should give an example.

You can give a simple proof using one of the various characterizations of the Kodaira spencer map: https://en.wikipedia.org/wiki/Kodaira%E2%80%93Spencer_map#In_scheme_theory

Namely, if $f:X\to Y$ is a smooth fibration over a curve $Y$, then the Kodaira spencer map at a point $\operatorname{Spec} k = y\in Y$ is obtained by restricting $f$ to a first order thickening $\overline{f}:\mathfrak{X}_{y}\to \operatorname{Spec} k[\varepsilon]$ of $y$ and considering the associated sequence of differentials $$0\to \overline{f}^*\Omega_{k[\varepsilon]/k}\to \Omega_{\mathfrak{X}_y}\to \Omega_{\mathfrak{X}_y/k[\varepsilon]}\to 0 $$ By smoothness of the morphism, this sequence is exact and the last nonzero term is locally free as an $\mathcal{O}_{\mathfrak{X}_y}$-module. In particular, the sequence remains exact when you restrict to the reduced fiber $X_y$. By the base change properties of differentials, you get an exact sequence $$0\to f^*T_{Y,y}^{\vee}\to \Omega_{\mathfrak{X}_y}|_{X_y} \to \Omega_{X_y}\to 0$$

The associated element in $T_{Y,y}^{\vee}\otimes \operatorname{Ext}^1(\Omega_{X_y},\mathcal{O}_X)$ is equivalent to the data of a map $T_{Y,y}\to H^1(X_y,T_{X_y})$ which is the Kodaira spencer map.

Now if $h:Y'\to Y $ is a morphism of smooth curves that is ramified at $y'\in Y'$ such that $h(y')=y$, we get a diagram $$ \begin{array}{ccc} \mathfrak{X}_{y'} & \rightarrow & \mathfrak{X}_y \\ \downarrow & & \downarrow \\ \operatorname{Spec}k[\varepsilon] & \rightarrow & \operatorname{Spec}k[\varepsilon] \end{array}$$ in which the lower horizontal map is given by the dual map on algebras with $\varepsilon\mapsto 0$. By functoriality of differentials, it follows that we have a morphism of exact sequences $$ \begin{array}{ccccccccc} 0 & \rightarrow & f_y^*T_{Y,y}^{\vee} & \rightarrow & \Omega_{\mathfrak{X}_y}|_{X_y}&\rightarrow &\Omega_{X_y}& \rightarrow & 0\\ & &\downarrow & & \downarrow& &\downarrow & & \\0 & \rightarrow & h^*f_{y'}^*T_{Y',y'}^{\vee} & \rightarrow & \Omega_{\mathfrak{X}_y}|_{X_y}&\rightarrow &\Omega_{X_y}& \rightarrow & 0 \end{array}$$ Here, the first vertical map is the differential of $h$, which vanishes when we have ramification, while the last vertical map is the identity. Finally, it follows that we have a commutative diagram of Kodaira-Spencer maps $$ \begin{array}{ccc} T_{Y',y'} & \rightarrow & H^1(X_{y'},T_{X_{y'}}) \\ \downarrow & & \downarrow \\T_{Y,y} & \rightarrow & H^1(X_y,T_{X_y})\end{array}$$ where the first vertical map is zero, and the last vertical map an isomorphism. This implies that the upper horizontal map is zero.

  • $\begingroup$ Very clean explanation. Morally, since the fibres of a Kodaira fibrations have genus at least $2$ (actually, at least $3$) there is a moduli space $\mathcal{M}$ for them. If $y \in Y$, there is a germ of deformation $\mathcal{X} \to U$ of the fibre $X_y$, where $U$ is a small neighborhood of $y$ in $Y$. Then the KS map is the differential of the corresponding moduli map $U \to \mathcal{M}$, and composing with a base change of $U$ ramified at $y$ we obtain a map whose differential vanishes at $y$ (by the chain rule). Thank you for pointing this out to me. $\endgroup$ Jan 25, 2022 at 20:42

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