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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 $\partial U \subseteq \mathbb{C}[x_1,\dots, x_n]_{d-1}$ obtained by applying $\partial$ to every element of $U$ satisfies the following dimension bound: $$ \binom{n+d-1}{d}\dim(\partial U) \geq \binom{n+d-2}{d-1} \dim(U) $$

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    $\begingroup$ 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. $\endgroup$ Oct 15, 2022 at 23:04
  • $\begingroup$ I removed two off-topic tags (algebraic geometry and symmetric spaces). $\endgroup$ Oct 16, 2022 at 11:10

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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: $x_1^d, x_1^{d-1} x_2, \dotsc$.

If $\dim U \leq \binom{n+d-2}{d-1}$ 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 lex-leading terms, so they are linearly independent). In this case $\dim(\partial U) = \dim(U)$.

Otherwise, $\dim U > \binom{n+d-2}{d-1}$, but $\partial U$ consists of all the degree $d-1$ forms: as before, the first $\binom{n+d-2}{d-1}$ basis elements of $U$ map injectively, hence surjectively (the other basis elements have no $x_1$, and are annihilated). In this case $\dim(\partial U) = \binom{n+d-2}{d-1}$. Since $\dim(U) \leq \binom{n+d-1}{d}$, the inequality you seek is satisfied.

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    $\begingroup$ 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? $\endgroup$
    – Ben
    Nov 7, 2022 at 5:55
  • $\begingroup$ Ah, there is something to think about here. A theorem of Galligo-Bayer-Stillman (Theorem 15.20 in Eisenbud's book) shows that the generic initial ideal of the ideal generated by $U$ is Borel-fixed. However a Borel-fixed 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 Borel-fixed 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 Borel-fixed 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

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