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.

  • $\begingroup$ 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" $\endgroup$ – Dima Pasechnik May 19 at 10:33
  • $\begingroup$ Also at MSE. Please see here about cross-posting. $\endgroup$ – KReiser May 19 at 10:47
  • $\begingroup$ @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. $\endgroup$ – S. Kohn May 19 at 11:12
  • $\begingroup$ the question does not forbid powers, right? $\endgroup$ – Dima Pasechnik May 19 at 12:19
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    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ This shows the existence of decompositions. But OP asked for decompositions with a certain number of terms. $\endgroup$ – Zach Teitler May 19 at 21:33
  • $\begingroup$ 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). $\endgroup$ – Dima Pasechnik May 20 at 8:20
  • $\begingroup$ 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. $\endgroup$ – Zach Teitler May 20 at 12:42
  • $\begingroup$ right, I didn't realise the author was too optimistic. $\endgroup$ – Dima Pasechnik May 20 at 13:36

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