# Isotrivial families with non-zero Kodaira spencer map

Let $S$ be a smooth quasi-projective curve over the complex numbers. Let $P$ be a closed point in $S$. Let $f:\mathcal X \to S$ be a polarized family of smooth projective connected varieties. To this family and the point $P$ we can associate the Kodaira-Spencer map

$$Tan_S(P) \to H^1(X,T_X).$$

Here $X$ is the fibre of $f$ over $P$.

Of course, if $f$ is a product/trivial family (so $\mathcal X = X \times S$) then the Kodaira-Spencer map is zero.

On the other hand, I can't prove that the same holds if all closed fibres of $f$ are isomorphic (i.e. $f$ is isotrivial). Therefore, I suspect that there might be examples of non-trivial isotrivial families with non-zero Kodaira-Spencer map.

Are there examples of such families of varieties? (Note that such a family if it exists is not trivial, locally for the fppf topology.)

• If all fibers are isomorphic, then by the Fischer-Grauert theorem the family is locally trivial. Then, the Kodaira-Spencer map -being "local"- should be zero, isn't it? – diverietti Jul 30 '15 at 11:42
• In addition to Fischer-Grauert, if you are working with projective schemes (which you are), there are also purely algebro-geometric proofs. Over $S'=S\times S$ with its two pullback families, $\mathcal{X}'_1 =\text{pr}_1^*\mathcal{X}$ and $\mathcal{X}'_2 = \text{pr}_2^*\mathcal{X}$, you can form the relative Isom scheme $\text{Isom}_{S'}(\mathcal{X}'_1,\mathcal{X}'_2)\to S'$ using existence of the Hilbert scheme. Your hypothesis implies that this is surjective over $S'$, thus generically flat over $S$. Base change, and then apply your result for a product family. – Jason Starr Jul 30 '15 at 11:58
• ... not just generically flat, but smooth (on a dense open of some irreducible component dominating $S'$) because the characteristic is $0$. – Jason Starr Jul 30 '15 at 12:18
• @JasonStarr I agree that the Isom scheme is generically flat (even smooth). But it might fail to be so over $P$. So I think your argument only shows that the family $f$ is locally trivial over some dense open of $S$. (This dense open might not contain the point $P$ though.) Or am I missing something obvious? – Pancho Jul 30 '15 at 13:30
• A morphism of tosion-free coherent sheaves on a reduced scheme $S$, e.g., your Kodaira-Spencer map $T_S\to R^1 f_* T_{\mathcal{X}/S}$, is identically zero if and only if it is zero on a dense open. – Jason Starr Jul 30 '15 at 13:34

## 1 Answer

The most general (most useful) formulation of the Kodaira-Spencer map uses the triangle of transitivity of cotangent complexes and relative duality. Even for such mildly singular varieties as nodal curves, the following is not sufficient (I recommend Section 6 of K. Behrend and B. Fantechi, "The Intrinsic Normal Cone", for a quick introduction to Kodaira-Spencer in the case of singular, projective varieties). For a proper, smooth morphism of relative dimension $n$, $f:\mathcal{X}\to S$, with $S$ a finitely presented $k$-scheme, the natural exact sequence $$f^*\Omega_{S/k} \to \Omega_{\mathcal{X}/k} \to \Omega_{\mathcal{X}/S} \to 0,$$ is left exact (thus a short exact sequence), and the cokernel $\Omega_{\mathcal{X}/S}$ is locally free of rank $n$. Denote the dual locally free sheaf by $T_{\mathcal{X}/S}$. Tensoring by this locally free $\mathcal{O}_{\mathcal{X}}$-module gives a short exact sequence, $$0 \to f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to \Omega_{\mathcal{X}/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to \Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to 0.$$ This short exact sequence gives rise to a long exact sequence of higher direct image sheaves with respect to $f$. The relevant part here is the first connecting map, $$\delta: f_*\left(\Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right) \to R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right).$$ The identity map on $\Omega_{\mathcal{X}/S}$ gives a natural global section, $\text{Id}:\mathcal{O}_S \to f_*\left(\Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right)$. Thus there is a global section, which is a version of the Kodaira-Spencer map, $$\delta'_{\mathcal{X}/S}:\mathcal{O}_S \to R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right)$$

If $S$ is smooth over $k$, then $\Omega_{S/k}$ is locally free. Then, by the projection formula, the natural map $$\Omega_{S/k}\otimes_{\mathcal{O}_S} R^1 f_*\left( T_{\mathcal{X}/S} \right) \to R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right),$$ is an isomorphism. Thus, $\delta'_{\mathcal{X}/S}$ is naturally an $\mathcal{O}_S$-module homomorphism, $$\delta'_{\mathcal{X}/S} : \mathcal{O}_S \to \Omega_{S/k}\otimes_{\mathcal{O}_S} R^1 f_*\left( T_{\mathcal{X}/S} \right).$$ Denoting by $T_{S/k}$ the dual locally free sheaf to $\Omega_{S/k}$, this is equivalent to another $\mathcal{O}_S$-module homomorphism (closer to the usual definition), $$\delta_{\mathcal{X}/S} : T_{S/k} \to R^1 f_*\left( T_{\mathcal{X}/S} \right),$$ where $T_{S/k}$ is the dual of the locally free sheaf $\Omega_{S/k}$.

By Grauert's theorem, Corollary III.12.9, p. 288, of R. Hartshorne, "Algebraic Geometry", $R^1f_*\left(T_{\mathcal{X}/S}\right)$ is a locally free $\mathcal{O}_S$-module compatible with base change to reduced schemes (we only need closed points) if for every point $s$ of $S$ (closed points suffice), $h^1(\mathcal{X}_s,T_{\mathcal{X}_s})$ is constant. In particular, if all fibers of $f$ are isomorphic as $k$-schemes, this is true. In that case, $\delta$ is a map of locally free $\mathcal{O}_S$-modules. On a reduced scheme, a morphism of torsion-free quasi-coherent $\mathcal{O}_S$-modules is zero if and only if it is zero at the generic point if and only if it is zero on some dense open subscheme. Thus, in what follows, it suffices to restrict to a dense open subscheme of $S$.

Let $s_0$ be a $k$-point of $S$. Let $X_0$ denote the fiber of $f$ over $s_0$. As a corollary of existence of the Hilbert scheme (really, algebraic space if $f$ is merely a proper, flat morphism of algebraic spaces), the Isom functor is representable, and there is an Isom scheme, $$\pi: \text{Isom}_S(X_0\times S,\mathcal{X}) \to S.$$ By hypothesis, this morphism is surjective on $k$-points. Now assume that $k$ is characteristic $0$ and uncountable (under the hypotheses above, $\delta$ is compatible with base change, so we can base change to make $k$ uncountable).

Since $\pi$ is surjective, and since $k$ is uncountable, there exists an irreducible component $I$ of $\text{Isom}_S(X_0\times S,\mathcal{X})$ such that the restion $\pi|_I$ is dominant. Denote this restriction by $$\rho:I\to S.$$ Since $k$ is algebraically closed, there exists a dense open subscheme of $I_{\text{red}}$ that is a smooth $k$-scheme. Replace $I$ by this dense open subscheme of $I_{\text{red}}$. Since $\rho$ is a dominant morphism of smooth $k$-schemes, and since $k$ has characteristic $0$, by generic smoothness there exists a dense open subscheme of $S$ over which $\rho$ is smooth. Replace $S$ by this open subscheme and assume that $\rho$ is smooth and surjective. By construction of the Kodaira-Spencer maps, there is a commutative diagram of $\mathcal{O}_I$-modules, $$\begin{array}{ccc} \rho^*\mathcal{O}_S & \xrightarrow{\rho^*\delta_{\mathcal{X}/S}} & \rho^*(\Omega_{S/k})\otimes_{\mathcal{O}_I} \rho^*R^1f_* T_{\mathcal{X}/S} \\ \downarrow & & \downarrow \\ \mathcal{O}_I & \xrightarrow{\delta_{\mathcal{X}_I/I}} & \Omega_{I/k}\otimes_{\mathcal{O}_I} R^1f_{I,*} T_{\mathcal{X}_I/I} \end{array}.$$ Since $\rho$ is faithfully flat, to prove that $\delta_{\mathcal{X}/S}$ is zero, it suffices to prove that $\rho^*\delta_{\mathcal{X}/S}$ is zero. By Grauert's theorem, the natural map $\rho^* R^1f_* T_{\mathcal{X}/S} \to R^1f_{I,*} T_{\mathcal{X}_I/I}$ is an isomorphism. Since $\rho$ is smooth, the morphism $$d\rho^\dagger: \rho^*\Omega_{S/k} \to \Omega_{I/k},$$ is injective with locally free cokernel. Thus, the second vertical arrow above is injective (the first vertical arrow is an isomorphism). So to prove that $\rho^*\delta_{\mathcal{X}/S}$ is zero, it suffices to prove that $\delta_{\mathcal{X}_I/I}$ is zero.

By construction, there is an isomorphism over $I$ of the base change $\mathcal{X}_I = I\times_S \mathcal{X}$ with the constant family $X_0\times I$. That isomorphism of $I$-schemes identifies the Kodaira-Spencer map of $\mathcal{X}_I/I$ with the Kodaira-Spencer map of $X_0\times I/ I$. Of course since $X_0\times I/I$ is the pullback of $X_0/\text{Spec}(k)$ by the constant $k$-morphism $p:X\to \text{Spec}(k)$, there is another commutative diagram, $$\begin{array}{ccc} p^*\mathcal{O}_{s_0} & \xrightarrow{p^*\delta_{X_0/s_0}} & \rho^*(\Omega_{k/k})\otimes_{k} H^1(X_0,T_{X_0/k}) \\ \downarrow & & \downarrow \\ \mathcal{O}_I & \xrightarrow{\delta_{X_0\times I/I}} & \Omega_{I/k}\otimes_{\mathcal{O}_I} R^1f_{I,*} T_{X_0\times I/I} \end{array}.$$ Since $\Omega_{k/k}$ is the zero sheaf, it follows that $\delta_{X_0\times I/I}$ is the zero homomorphism. Therefore also $\delta_{\mathcal{X}/S}$ is zero.

• This is beautifully clear. Thank you. One small comment: Somewhere one should mention that the Isom-scheme satisfies some finiteness properties, i.e., the morphism $I\to S$ is of finite type. If not, one can't use all the "generic" arguments, right? – Pancho Jul 30 '15 at 18:54
• @JasonStarr You wrote "Since k is algebraically closed, there exists a dense open subscheme of I that is k-scheme." Did you mean to write "Since $k$ is algebraically closed and of characteristic zero, there exists a smooth dense open subscheme of $I$ over $k$"? – Ariyan Javanpeykar Aug 17 '15 at 14:58
• @AriyanJavanpeykar: "Did you mean to write ..." Yes, you are absolutely right. Thank you for catching the error. I will correct it now. – Jason Starr Aug 17 '15 at 15:01
• @JasonStarr How can one get rid of the assumption on the characteristic of $k$? It seems to me that if, for instance, the morphism $\pi$ were finite type, we don't need $k$ to be of characteristic zero. Does that seem right to you? – Ariyan Javanpeykar Aug 17 '15 at 15:57