Suppose I am working with a category of objects such that each object $X$ in the category has as a trivial automorphism group. An example of such an object would be genus zero curves with $n$-punctures, i.e., $(\mathbb{P}_k^1, (s_i)_{i=1}^n \in k )$.

Let $F(S)/{\sim}$ represent families of these objects over an affine scheme $S=\operatorname{Spec} A$ where $A$ is a $k$-algebra and $k$ an algebraically closed field. As an example, consider families of genus zero curves with $n$-punctures over $S$ and fibers $X= X_s \cong (\mathbb{P}_k^1, (s_i: k \to \mathbb{P}_k^1)_{i=1}^n) $. Then $H^1(X, TX)=0$ and $Aut(X)=0$.

Motivation for question: It seems that since both $H^1(X, TX)=0$ and $Aut(X)=0$ we can conclude that every family of genus zero curves with $n$-punctures is trivial, i.e any family over $S$ is isomorphic to the trivial family $(\mathbb{P}_A^1, (s_i: A \to \mathbb{P}_A^1)_{i=1}^n)$.

Question: Now going back to the beginning, suppose my family of objects over $S$ is such that all fibers are isomorphic to some objet $X$, $H^1(X, TX)=0$ and $Aut(X)=0$, does this imply that all families are trivial?

Or even more generally, suppose $H^1(X, TX)=n$ and $Aut(X)=0$, does this imply that there exists $n$ distinct families $F(S)/{\sim}$?

I know there are probably many conditions on $X$ I should include, however, just assume that my object $X$ has any property that makes it "nice" to work with.

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    $\begingroup$ Presumably for your $n$ punctures, $n\not=0,1,2$? $\endgroup$
    – David Roberts
    Commented Jan 30, 2019 at 22:39
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    $\begingroup$ Even for $n>2$ there can be nontrivial automorphisms (e.g. if each $s_i=i$). $\endgroup$ Commented Jan 30, 2019 at 23:02
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    $\begingroup$ Yes, I am assuming $n > 2$ and that the punctures are distinct and ordered. $\endgroup$
    – user
    Commented Jan 31, 2019 at 0:02
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    $\begingroup$ I think you are using the wrong criterion to establish triviality of deformations. Even though $H^1(X,TX)=0$, this does not account for the marked points, which can move around. $\endgroup$
    – S. Carnahan
    Commented Jan 31, 2019 at 2:53

1 Answer 1


No. An example of a non-trivial family would occur with $S=\mathbb{A}^1\setminus\{0,1\}$ and $X_s$ given by $\mathbb{P}^1$ marked at the points $(0,1,\infty,s)$. I think the issue you run into is basically what S. Carahan noted; if $X$ were a curve, its infinitesimal deformations would be given by $H^1(X,TX)$, but you've also bundled in the data of the points. So the first-order deformation space is instead $H^1(X,T\mathbb{P}^1\otimes \mathcal{O}(-p_1-\ldots-p_n))$, which is nonzero for $n\geq 4$.

I'm not quite sure what your second question is asking.


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