# A tale of two maps into a Grassmannian

I suspect that the answer to this question is well-known to the experts. However, I was not able to find it in the literature, so let me ask here.

Setup. In the sequel, $$X$$ is a compact complex surface and $$E$$ is a rank $$2$$ vector bundle on $$X$$. We call $$\pi \colon \mathbb{P}(E) \to X$$ the projective bundle in the sense of Hartshorne. In my applications, $$X$$ is of general type and $$E=\Omega_X^1$$, but let me ask the question in generality.

We assume that $$S^nE$$ is globally generated, namely, that the evaluation of sections $$H^0(X, \, S^nE) \otimes \mathcal{O}_X \to S^nE$$ is surjective.

If $$V$$ is a vector space, we denote by $$G(r, \, V)$$ the Grassmannian of $$r$$-dimensional subspaces of $$X$$, and by $$G(V, r)$$ the Grassmannian of the $$r$$-dimensional quotients of $$V$$, so that $$G(r, \, V) \simeq G(V^*, \, r)$$. Moreover, we denote by $$\mathbb{G}(r-1, \, \mathbb{P}(V))$$ and $$\mathbb{G}(\mathbb{P}(V), \, r-1)$$ the corresponding projective Grassmannians.

We are now going to construct two morphisms $$X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*)$$.

The first morphism. Take a point $$x \in X$$. The fibre $$\pi^{-1}(x)$$ is the projective line $$\mathbb{P}(E(x))$$, and $$|\mathcal{O}_{\mathbb{P}(E)}(n)|$$ cuts on it the complete linear system $$\mathcal{O}(n)$$. Identyfying the global sections of $$\mathcal{O}_{\mathbb{P}(E)}(n)$$ with those of $$S^nE$$, we see that $$\pi^{-1}(x)$$ is embedded in $$\mathbb{P}H^0(X, \, S^nE)^*$$ as a rational normal curve $$C_x$$ of degree $$n$$. There is exactly one $$n$$-plane $$L_x$$ containing $$C_x$$, and we get a morphism $$\mathscr{G} \colon X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*), \quad x \mapsto L_x.$$

The second morphism. Take a point $$x \in X$$. There is a surjection $$H^0(X, \, S^nE) \to S^nE(x) \to 0,$$ where $$S^nE(x)$$ is the fibre of $$S^nE$$ over $$x$$, namely, a $$n+1$$-dimensional vector space. This provides by definition an element $$s_x \in G(H^0(X, \, S^nE), \, n+1) \simeq G(n+1, \, H^0(X, \, S^nE)^*) \simeq \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*),$$ and we can define a morphism $$\mathscr{H} \colon X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*), \quad x \mapsto s_x.$$

Question. How are the morphisms $$\mathscr{G}$$ and $$\mathscr{H}$$ related? Are they the same?

Every answer and/or reference to the relevant literature will be greatly appreciated.

I guess, when you say "in the sense of Hartshorne" you mean the projective spectrum of $$\oplus S^kE$$.
Yes, the morphisms are the same, and to see this just note that there is a natural (relative Veronese) embedding $$\mathbb{P}_X(E) \to \mathbb{P}_X(S^nE),$$ such that every fiber of the (former) $$\mathbb{P}^1$$-bundle becomes a rational normal curve in the (latter) $$\mathbb{P}^n$$-bundle. Moroever, the composition $$\mathbb{P}_X(E) \to \mathbb{P}_X(S^nE) \to \mathbb{P}(H^0(X,S^nE)^*)$$ (where the second arrow is induced by the dual of the evaluation morphism $$H^0(X,S^nE) \otimes \mathcal{O}_X \to S^nE$$) coincides with the morphism induced by the line bundle $$\mathcal{O}_{\mathbb{P}_X(E)}(n)$$. This shows that the two maps coincide.