# Objects with trivial automorphism group

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.

• Presumably for your $n$ punctures, $n\not=0,1,2$? Jan 30 '19 at 22:39
• Even for $n>2$ there can be nontrivial automorphisms (e.g. if each $s_i=i$). Jan 30 '19 at 23:02
• Yes, I am assuming $n > 2$ and that the punctures are distinct and ordered.
– user
Jan 31 '19 at 0:02
• 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. Jan 31 '19 at 2:53

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 $$n\geq 4$$.