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 the Waring rank homogeneous polynomials sub-multiplicative?
You asked:



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



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> Is $W(PQ) \leq W(P) W(Q)$ for two homogeneous polynomials $P$ and $Q$?
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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 $W(Q)=1$?
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My hunch is that the answer is probably negative, but I don't know. I haven't thought about it.


