Relations between homogeneous polynomials

Question

Edit: The formulation of my question was incorrect, for several reasons. Here is what I hope to be the correct formulation:

Let $\mathbb{P}$ be a projective space, and $V$ a general linear subspace of $H^0(\mathcal{O}_{\mathbb{P}}(d))$ (that is, a general point in the corresponding Grassmannian). Then for $p<d$ the multiplication map $$H^0(\mathcal{O}_{\mathbb{P}}(p))\otimes V\rightarrow H^0(\mathcal{O}_{\mathbb{P}}(p+d))$$ is of maximal rank, i.e. either injective or surjective.

Is this true? Known? Sasha's answer shows that it is true when $\dim V \leq \dim \Bbb{P}+1$.

Answer 1

I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.

MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)

This special case is mostly solved by work of Gleb Nenashev.

MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf

Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$\text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))\otimes V \leq \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p+d)) - \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))^2,$$ and surjectivity whenever $$\text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))\otimes V \geq \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p+d)) + \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))^2.$$

Answer 2

I think this can be controlled as follows. Let $Z \subset \mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution $$ \dots \to \mathcal{O}(-2d)^{\binom{\ell}{2}} \to \mathcal{O}(-d)^\ell \to \mathcal{O} \to \mathcal{O}_Z \to 0. $$ Twisting it by $\mathcal{O}(d+p)$ we obtain $$ \dots \to \mathcal{O}(p-d)^{\binom{\ell}{2}} \to \mathcal{O}(p)^\ell \to \mathcal{O}(d+p) \to \mathcal{O}_Z(d+p) \to 0.\tag{*} $$ Your question is equivalent to injectivity of the induced map $$ H^0(\mathcal{O}(p)^\ell) \to H^0(\mathcal{O}(d+p)). $$ If $n \ge \ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?

EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term $$ E_1^{i,j} = H^j\left(\mathcal{O}(d+p+id)^{\binom{\ell}{-i}}\right),\qquad i \le 0 $$ and converges to $E_\infty^k = H^k(\mathcal{O}_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^{n-1}$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is $$ E_1^{-\ell,n-1} = H^{n-1}\left(\mathcal{O}(d+p-\ell d)\right) $$ is in the total grading $-\ell + n - 1 \ge -1$, hence all differentials from it go to terms of total grading $\ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is $$ E_1^{-1,0} = H^0\left(\mathcal{O}(p)^{\ell}\right) $$ is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential $$ d_1^{-1,0} \colon H^0(\mathcal{O}(p)^\ell) \to H^0(\mathcal{O}(d+p)). $$ is nonzero, it survives in the spectral sequence and gives a contribution to $E_\infty^{-1} = H^{-1}(\mathcal{O}_Z(d+p)) = 0$, which is absurd.