# factorization of homogeneous multivariable polynomials as sums of products of linear forms

Suppose we have $$n$$ variables $$x_{1\leq i \leq n }$$. Consider the vector space of degree-$$d$$ homogeneous polynomials of $$x_i$$ over the field $$\mathbb{C}$$. The dimension of this space is $$D_1 = (d+n -1)!/(n-1)!/d!$$.

Consider the special set of polynomials (which we shall refer to as decomposable polynomials) which can be written as the product of $$d$$ linear forms

$$\Omega = \prod_{j=1}^d \left( \sum_{i=1}^n a_{ij }x_i \right ) .$$

It is not hard to see that the degrees of freedom of $$\Omega$$ is $$D_2 = d(n-1) +1$$.

Now the question is, is it always possible to write an arbitrary $$d$$-degree homogeneous polynomial $$F$$ as the sum of $$M = \lceil D_1/D_2 \rceil$$ decomposable polynomials? Here $$\lceil \cdot \rceil$$ is the ceiling function.

• for generic polynomials one can do even better: arxiv.org/abs/1112.1371 (your question seems to be a generalisation of what's known as "Waring problem for polynomials" – Dima Pasechnik May 19 at 10:33
• Also at MSE. Please see here about cross-posting. – KReiser May 19 at 10:47
• @DimaPasechnik I just heard of Waring problem. But here I am not concerned with powers of linear forms. I am concerned with products of different linear forms. But anyway, I will have a look of the reference. – S. Kohn May 19 at 11:12
• the question does not forbid powers, right? – Dima Pasechnik May 19 at 12:19
• The question is about secants of the chow variety. There is literature about it, see e.g., arxiv.org/abs/2005.12436 and arxiv.org/abs/1602.04275 – Abdelmalek Abdesselam May 19 at 16:50

Yes, it appears to be always possible. Let $$0\neq v\in\mathbb{C}^n$$ be a zero of $$F$$, $$v\in V(F)$$. There are $$F_1,\dots,F_n\in R:=\mathbb{C}[x_1,\dots,x_n]$$ of degree $$d-1$$ each, such that $$v$$ is their only common zero, $$v\in V(F_1,\dots,F_n)\subset V(F)$$. Hence $$F\in (F_1,\dots,F_n)$$, which means that $$F=\sum_{k=1}^n \ell_k F_k, \quad\text{for \ell_t\in R, of degree 1, 1\leq t\leq n.}$$
By induction on $$d$$, we can assume that each $$F_k$$ is decomposed as required in the question, therefore this is a decomposition we're looking for. QED.
Edit: more precisely, one should flip the roles of $$\ell_k$$ and $$F_k$$ above. Then, as the ideal $$(\ell_1,...,\ell_n)$$ is radical, one can apply Nullstellensatz.
• certainly, I didn't claim it's a complete answer. However, for $d=2$ it appears to do the job, even beat the bound $M$ asked for (it suffices to take $n-1$ linear forms to start with, so you'd get $n-1$ terms). – Dima Pasechnik May 20 at 8:20
• For $d=2$ op’s $D_1=(n+1)n/2$, $D_2=2n-1$, so $D_1/D_2 \approx n/4$. That’s not achievable. The best is better than $n-1$, it’s $\lceil n/2 \rceil$, because a quadratic form of rank $r$ can be put in the form $x_1 x_2 + x_3 x_4+\dotsb +x_{r-1} x_r$ or $\dotsb + x_{r-2} x_{r-1} + x_r^2$ depending on parity. – Zach Teitler May 20 at 12:42