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Example 107.7.2 (Smooth positive dimensional automorphism group). Let $k$ be an algebraically closed field. If $X$ is a smooth genus $0$, resp. $1$ curve, then the automorphism group has dimension $3$, resp. $1$. Namely, in the genus $0$ case we have $X \cong \mathbf{P}^1_ k$ by Algebraic Curves, Proposition 53.10.4. Since

\[ \mathit{Aut}(\mathbf{P}^1_ k) = \text{PGL}_{2, k} \]

as functors we see that the dimension is $3$. On the other hand, if the genus of $X$ is $1$, then we see that the map $X = \underline{\mathrm{Hilb}}^1_{X/k} \to \underline{\mathrm{Pic}}^1_{X/k}$ is an isomorphism, see Picard Schemes of Curves, Lemma 44.6.7 and Algebraic Curves, Theorem 53.2.6. Thus $X$ has the structure of an abelian variety (since $\underline{\mathrm{Pic}}^1_{X/k} \cong \underline{\mathrm{Pic}}^0_{X/k}$). In particular the (co)tangent bundle of $X$ are trivial (Groupoids, Lemma 39.6.3). We conclude that $\dim _ k H^0(X, T_ X) = 1$ hence $\dim \mathit{Aut}(X) \leq 1$. On the other hand, the translations (viewing $X$ as a group scheme) provide a $1$-dimensional piece of $\text{Aut}(X)$ and we conlude its dimension is indeed $1$.

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