# Dimension of a general partial derivative of a linear subspace of polynomials

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: $$\binom{n+d-1}{d}\dim(\partial U) \geq \binom{n+d-2}{d-1} \dim(U)$$

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

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.
• 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?
• 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. Nov 7, 2022 at 21:36
• 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. Nov 8, 2022 at 16:36