# Is the Waring rank homogeneous polynomials sub-multiplicative?

For a homogeneous degree $d$ polynomial $P$, the symmetric or Waring rank $W(P)$ is the minimum $r$ such that $P = \sum_{j=1}^r l_j^d$, where $l_j$s are linear forms. Now, is the Waring rank sub-multiplicative, i.e. for two homogeneous degree $d$ polynomials $P$ and $Q$, $W(P Q) \leq W(P) \, W(Q)$?

Is $W(P \otimes Q) \leq W(P) W(Q)$ for two homogeneous polynomials $P$ and $Q$?

First, I think you have to be a little bit careful about the difference between tensor product and multiplication of polynomials. If you are just using the ordinary tensor product, then the resulting $P \otimes Q$ is no longer a symmetric tensor. For example, if $P = P(x,y) = xy$ and $Q = Q(z,w) = zw$, then as tensors, $P = x \otimes y + y \otimes x$, $Q = z \otimes w + w \otimes z$, and $$P \otimes Q = x \otimes y \otimes z \otimes w + \dotsb + y \otimes x \otimes w \otimes z,$$ a sum of $4$ terms. But it is not a symmetric tensor because it is missing terms such as $x \otimes w \otimes y \otimes z$.

So perhaps a better question would be:

Is $W(PQ) \leq W(P) W(Q)$ for two homogeneous polynomials $P$ and $Q$?

Here the answer is no. For a simple example, let's say... $P=xy$, $Q=zw$. Then $W(P)=W(Q)=2$, but $W(PQ)=W(xyzw)=8$, see for example https://arxiv.org/abs/1104.3648.

And in fact one should not expect $W(PQ) \leq W(P)W(Q)$ anyway. If we write $P=\sum_{i=1}^r \ell_i^d$ and $Q=\sum_{j=1}^s m_j^e$ (where $r=W(P)$, $s=W(Q)$, $\ell_i$ and $m_j$ are linear forms, $d=\deg(P)$, and $e=\deg(Q)$), then we get $$PQ = \sum \ell_i^d m_j^e,$$ a sum of $rs$ terms. Naively we would expand each $\ell_i^d m_j^e$ into a sum of powers. This needs $W(x^d y^e) = \max(d+1,e+1)$ terms for each term in the above summation. This proves that $$W(PQ) \leq W(P) W(Q) \max(\deg(P)+1, \deg(Q)+1).\tag{*}$$

The inequality can be strict. For example let, once again, $P=xy$ and $Q = zw$, so $W(P)=W(Q)=2$, $\deg(P)=\deg(Q)=2$, and $W(PQ) = 8$, strictly less than the naive $W(P) W(Q) W(\ell^2 m^2) = 2 \cdot 2 \cdot 3 = 12$.

Also equality is possible. For example if $W(P)=W(Q)=1$, and $P=x^d$, $Q=y^e$ for independent variables $x,y$ (as opposed to $P=x^d$, $Q=x^e$ for the same variable), then equality occurs. Off the top of my head, that's the only example I can think of where equality occurs in $(*)$. Slightly more generally say $P=x^d$, $Q=y_1^{a_1} y_2^{a_2} \dotsm y_k^{a_k}$, $1 \leq a_1 \leq \dotsb \leq a_k$, $d \geq \deg(Q) = a_1+\dotsb+a_k$. Then $W(P)=d+1$ and $W(Q)=(a_2+1)\dotsm(a_k+1)$ (see https://arxiv.org/abs/1112.3474). So $(*)$ provides the bound $W(PQ) \leq 1 \cdot W(Q) \cdot (d+1)$. Since $PQ$ is a monomial we know that $W(PQ) = (a_2+1)\dotsm(a_k+1)(d+1)$, which matches the bound.

But if $1 \leq d < a_1 \leq a_2 \leq a_k$ then we are back to strict inequality: $W(PQ) = (a_1+1)\dotsm(a_k+1) = (a_1+1) W(Q)$, which is strictly less than $W(P)W(Q)\max(\deg(P)+1,\deg(Q)+1)$, since that would be $(a_1+\dotsb+a_k)W(Q)$. So $W(P)=1$ or $W(Q)=1$ is certainly not a sufficient condition for equality in $(*)$.

Is it a necessary condition?

Suppose $P$ and $Q$ are homogeneous polynomials such that $$W(PQ) = W(P) W(Q) \max(\deg(P)+1,\deg(Q)+1).$$ Does it follow that $W(P)=1$ or $W(Q)=1$?

My hunch is that the answer is probably negative, but I don't know. I haven't thought about it.