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Nov 21, 2018 at 17:16 comment added inkspot In the projective space $\mathbb P(V_e^\vee)$ there is a unique closed orbit, the orbit of the highest weight vector $\partial^e/\partial x_1^e$. This is contained in the closure of every orbit. Let $U$ be the locus of points $D$ in $\mathbb P(V_e^\vee)$ for which $V_{d+e}\times\{D\}$ is surjective. Then $U$ is Zariski open and $GL_n$-invariant, so it is enough to show that $U$ contains $\partial^e/\partial x_1^e$. For this the map is obviously surjective.
Nov 21, 2018 at 14:38 comment added abx Same as @fedja: why is it enough to check surjectivity for one particular orbit?
Nov 21, 2018 at 14:34 comment added fedja I'm not sure I understand the last two sentences, What is obvious (to me) from what was written before is that it is enough to get $x_1^d$ in the image, but that's not what you are saying. Can you elaborate a bit?
Nov 21, 2018 at 14:18 history answered inkspot CC BY-SA 4.0