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Given a smooth manifold $X$ with a very-ample line bundle $L$, we have an associated embedding $$ \phi_{|L|}:X \to \mathbb{P}(H^0(X,L)^\vee) .$$ We know that if an automorphism $\alpha:X\to X$ leaves invariant the class of $L$ inside the Picard group of $X$, then $\phi_{|L|}$ is equivariant with respect to the action of $\alpha$ on the right and on the left.

Consider now as a submanifold $X$ inside a Grassmannian $G(k,n)$, and $L$ the line bundle on $X$ given by the restriction of the Schubert class $\sigma_{1,1}$ to $X$. We have then that the map $\phi_{|L|}$ is the inclusion $X\hookrightarrow G(k,n)$.

My question is: if an automorphism $\alpha:X\to X$ leaves invariant the class of $L$ inside the Picard group of $X$, can we say that $\alpha$ is the restriction of an automorphism of $G(k,n)$, as we did for $\mathbb{P}(H^0(X,L)^\vee)$. It seems very reasonable to me, but I struggle to find out why.

Thank you!

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Both statements are wrong without extra hypotheses. Indeed, let $X$ be a zero-dimensional subscheme of length $N \gg 0$ (i.e., just $N$ points). Then the map $\phi_{|L|}$ is a map $X \to \mathbb{P}^{N-1}$ which does not factor through $Gr(k,n)$. Similarly, the automorphism group $\mathfrak{S}_N$ of $X$ is not embedded into the authomorphism group of $Gr(k,n)$.

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  • $\begingroup$ Ok, excuse me I think I dropped some assumption, my bad. If I consider a manifold $X$, inside the Grassmannian, which is not contained in any hyperplane section of the Grassmannian (seen through Pl\"ucker embedding) I think that the map $\phi_{|L|}$ does factor thorugh G(k,n), isn't it? The map $\phi_{|L|}$ is literally the inclusion inside $\mathbb{P}(\bigwedge^k \mathbb{C}^n)$, or I am missing something? $\endgroup$ – Nutella Warrior Jun 8 at 14:04
  • $\begingroup$ If $X$ is contained in a hyperplane it is not a problem, the problematic case is when the restriction map $H^0(Gr(k,n),\mathcal{O}(1)) \to H^0(X,L)$ is not surjective. $\endgroup$ – Sasha Jun 8 at 16:38
  • $\begingroup$ Ok, now I think I figure it out a little better, thank you. Do you have any reference, to have an overview? Also, the condition you give is sufficient to have the properties we are looking for? $\endgroup$ – Nutella Warrior Jun 8 at 19:11
  • $\begingroup$ Ok, sorry for the double comment. I can ask a new question about references and extra hypothesis, if it's the best way $\endgroup$ – Nutella Warrior Jun 9 at 14:53

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