# Automorphisms of a manifold inside a Grassmannian

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!

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 $$Gr(k,n)$$.
• 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? – Nutella Warrior Jun 8 at 14:04
• 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. – Sasha Jun 8 at 16:38