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Let $G(k, n)$ be the Grassmannian of $k$-dimensional subspaces of $K^{n}$, $K$ a field, embedded in $\mathbb{P}^{N}$ by the Plücker embedding. In Harris' Algebraic Geometry, A First Course, Theorem 10.19 states that

$$\mathrm{Aut}(G(k, n)) = \mathrm{Aut}(G(k, n), \mathbb{P}^{N}),$$ where $\mathrm{Aut}(G(k, n), \mathbb{P}^{N}) := \{ T \in \mathrm{Aut}(\mathbb{P}^{N}) \mid T(G(k, n)) = G(k, n) \}$. In his proof, Harris says that it comes down to the assertion that every codimension $1$ subvariety of $G(k, n)$ is the intersection of $G(k, n)$ with a hypersurface in $\mathbb{P}^{N}$. In other words, $\mathrm{Pic}(G(k, n)) \cong \mathrm{Pic}(\mathbb{P}^{N})$, right? How such information can say about the automorphism group of these varieties? Is it true in general? That is, if $Y$ is a subvariety of a variety $X$ and $\mathbb{Pic}(Y) = \mathbb{Pic}(X)$, then every automorphism of $Y$ extends to an automorphism of $X$?

Thank you!

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In fact, one has the following result.

Proposition 1. Let $X \subset \mathbb{P}^n$ be a smooth subvariety such that $\textrm{Pic}(X)$ is generated by the hyperplane section $\mathscr{O}_X(1)$. Then every automorphism of $X$ is induced by an automorphism of $\mathbb{P}^n.$

The proof is very easy. Any automorphism of $X$ gives an automorphism of the Picard group $\textrm{Pic}(X)=\mathbb{Z} \cdot \mathscr{O}_X(1)$, and since effective curves go into effective curves it follows that such a automorphism is the identity. In other words, the hyperplane class is preserved and any automorphism sends hyperplane sections into hyperplane sections. This precisely means that $\textrm{Aut}(X)$ is induced by the projective linear group $\textrm{Aut}(\mathbb{P}^n)$.

Examples of the situation given in the Proposition 1 are Grasmannians and hypersurfaces $X \subset \mathbb{P}^n$ with $n \geq 4$ (in the last case the condition on the Picard group follows from Lefschetz Hyperplane Theorem).

Another situation where the automorphism group is induced by the ambient space is the following.

Proposition 2. Let $X \subset \mathbb{P}^n$ be a smooth subvariety which is either canonically or anti-canonically embedded (i.e., the embedding is the morphism associated with either $|K_X|$ or $|-K_X|$). Then every automorphism of $X$ is induced by an automorphism of $\mathbb{P}^n$.

In fact, the canonical (or anti-canonical) class is preserved by any automorphism. But this class is by assumption nothing but $\mathscr{O}_X(1)$, so we can conclude as before.

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  • $\begingroup$ It probably goes without saying, but this "another situation" is indeed strictly another -- the anticanonical class of the Grassmannian is ${\mathcal O}(n)$. $\endgroup$ Aug 21, 2013 at 17:49

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