Let $U \subseteq \mathbb{C}[x_1,\dots, x_n]_d$ be a linear subspace of the vector space of homogeneous degree$d$ polynomials (including zero). I would like a proof or counterexample of the claim that for a general linear partial derivative $\partial=\alpha_1 \frac{\partial}{\partial{x_1}}+\dots + \alpha_n \frac{\partial}{\partial{x_n}}$, it holds that the linear subspace $\partial U \subseteq \mathbb{C}[x_1,\dots, x_n]_{d1}$ obtained by applying $\partial$ to every element of $U$ satisfies the following dimension bound: $$ \binom{n+d1}{d}\dim(\partial U) \geq \binom{n+d2}{d1} \dim(U) $$

1$\begingroup$ Have you tried letting $\partial=\partial/\partial x_1$, and tak $U$ general? Then a reduced basis for $U$ in lex order will start with distinct monomials… should be possible to count dimensions. $\endgroup$– Zach TeitlerOct 15, 2022 at 23:04

$\begingroup$ I removed two offtopic tags (algebraic geometry and symmetric spaces). $\endgroup$– Vladimir DotsenkoOct 16, 2022 at 11:10
1 Answer
Change coordinates, or act by a linear transformation, so that $U$ is a general subspace and we are differentiating by $\partial = \frac{\partial}{\partial x_1}$. Since $U$ is general, it has a basis whose leading monomials, in lexicographic order, are an initial segment: $x_1^d, x_1^{d1} x_2, \dotsc$.
If $\dim U \leq \binom{n+d2}{d1}$ then every basis element of $U$ has a leading term containing an $x_1$. Then $\partial$ is injective on $U$ (the derivatives of basis elements are nonzero and have pairwise distinct lexleading terms, so they are linearly independent). In this case $\dim(\partial U) = \dim(U)$.
Otherwise, $\dim U > \binom{n+d2}{d1}$, but $\partial U$ consists of all the degree $d1$ forms: as before, the first $\binom{n+d2}{d1}$ basis elements of $U$ map injectively, hence surjectively (the other basis elements have no $x_1$, and are annihilated). In this case $\dim(\partial U) = \binom{n+d2}{d1}$. Since $\dim(U) \leq \binom{n+d1}{d}$, the inequality you seek is satisfied.

1$\begingroup$ I am confused about your first line... do you claim that $GL(n) \cdot U$ is Zariski open dense in the Grassmannian variety of $\dim(U)$dimensional subspaces of the polynomial ring? This is false by dimension considerations. So I guess you mean that $GL(n) \cdot U$ is ``general enough" so that one of its elements admits such a basis? If so, why is this true? $\endgroup$– BenNov 7, 2022 at 5:55

$\begingroup$ Ah, there is something to think about here. A theorem of GalligoBayerStillman (Theorem 15.20 in Eisenbud's book) shows that the generic initial ideal of the ideal generated by $U$ is Borelfixed. However a Borelfixed collection of monomials in degree $d$ is not quite the same thing as a lex initial segment. $\endgroup$ Nov 7, 2022 at 21:36

$\begingroup$ I'm not sure if it's salvageable. Sure, if the basis elements of $U$ have leading terms involving $x_1$, then what I wrote earlier is fine, but that's not guaranteed. For example if $n=4$ and $U$ is spanned by $\{x_2 x_3, x_1 x_3, x_2^2, x_1 x_2, x_1^2\}$ then it is Borelfixed but missing the monomial $x_1 x_4$. So... I don't know. Worse: I'm now questioning whether we have enough generality to get the Borelfixed property while still fixing $\partial = \partial/\partial x_1$. My hunch is that it's okay but I'm not sure. $\endgroup$ Nov 8, 2022 at 16:36