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 $\dim V \leq \dim \Bbb{P}+1$.

  • $\begingroup$ I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $\sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y \subsetneq X$? $\endgroup$
    – user44191
    Mar 19, 2019 at 20:57
  • $\begingroup$ Sure. Is that different from what I wrote? $\endgroup$
    – abx
    Mar 19, 2019 at 21:00
  • $\begingroup$ I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written). $\endgroup$
    – user44191
    Mar 19, 2019 at 21:04
  • 1
    $\begingroup$ Oh, right. $\ell\leq n$ would be fine for me, though this can be certainly weakened. $\endgroup$
    – abx
    Mar 19, 2019 at 21:43
  • 3
    $\begingroup$ $\ell \leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent. $\endgroup$
    – user44191
    Mar 19, 2019 at 21:48

2 Answers 2


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.

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.

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 $$\text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))\otimes V \geq \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p+d)) + \text{dim} H^0(\mathbb{P},\mathcal{O}_{\mathbb{P}}(p))^2.$$

  • $\begingroup$ Great, thanks a lot! $\endgroup$
    – abx
    Mar 20, 2019 at 10:35
  • $\begingroup$ You are welcome. $\endgroup$ Mar 20, 2019 at 10:57

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.

  • 1
    $\begingroup$ Thanks, but could you explain why the Koszul complex gives that? $\endgroup$
    – abx
    Mar 19, 2019 at 21:03
  • 1
    $\begingroup$ Let $E$ the cokernel of $\ldots \rightarrow \mathcal{O}(p-d)^{\binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$. $\endgroup$
    – Libli
    Mar 19, 2019 at 22:20
  • 1
    $\begingroup$ Hence, if you assume that $l \leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha. $\endgroup$
    – Libli
    Mar 19, 2019 at 22:22
  • $\begingroup$ @abx: I added an explanation of the spectral sequence argument. $\endgroup$
    – Sasha
    Mar 20, 2019 at 6:35
  • $\begingroup$ Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation. $\endgroup$
    – abx
    Mar 20, 2019 at 8:14

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