How does multiplication affect degrees?

Question

Let $M(n) \sim \mathbb{A}^{n^2}$ be the space of $n$-by-$n$ matrices, seen as an affine space over a field $K$, and endowed with the usual matrix multiplication. Let $V$ and $W$ be subvarieties of $M(n)$. The product $V\cdot W$ is a constructible set (by Chevalley's theorem); write $\overline{V\cdot W}$ for its Zariski closure.

Is it the case that $\deg(\overline{V\cdot W}) \leq \deg(V) \cdot \deg(W)$?

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Note: this is a less ambitious version of a question I previously asked (Bézout and products in algebraic groups).

Answer 1

This is false. Let $G$ be the open subset $\textbf{GL}_2$ in the affine space $\textbf{Mat}_{2\times 2} = \mathbb{A}^4$ with affine coordinates $x_{1,1}$, $x_{2,1}$, $x_{1,2}$ and $x_{2,2}$. Let $V$ be the one-dimensional subvariety, $$ V = \text{Zero}(x_{2,1},x_{2,2}-1,x_{1,1}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & s \\ 0 & 1 \end{array} \right] : s\in \mathbb{A}^1 \right\}. $$ Similarly, let $W$ be the following one-dimensional subvariety, $$ V = \text{Zero}(x_{2,1},x_{1,1}-1,x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} 1 & t \\ 0 & t \end{array} \right] : t\in \mathbb{A}^1 \right\}. $$ Each of $V$ and $W$ is a one-dimensional, affine linear variety. In particular, each has degree $1$. Yet the product set $\overline{V\cdot W}$ is the following two-dimensional variety, $$ W = \text{Zero}(x_{2,1},2x_{1,1}x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & 2st \\ 0 & t \end{array} \right] : (s,t)\in \mathbb{A}^2 \right\}. $$ This has degree $2$.

Answer 2

I was about to say that, yes, the bound is false. Since the counterexample I saw is not the same (though it is strangely similar; are all such examples in low dimension related?), I'll post it here.

Let $n=3$. Let $$V=\left\{\left(\begin{matrix} r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{matrix}\right): r\in \mathbb{A}^1\right\},\;\;\;\; V=\left\{\left(\begin{matrix} 1 & 0 & 0 \\ 0 & s & 0 \\ 0 & 0 & s\end{matrix}\right): s\in \mathbb{A}^1\right\}.$$ Both $V$ and $W$ are linear varieties. Their product is $$V\cdot W=\left\{\left(\begin{matrix} r & 0 & 0 \\ 0 & r s & 0 \\ 0 & 0 & s\end{matrix}\right): r, s\in \mathbb{A}^1\right\}.$$ Since $V\cdot W$ intersects the linear variety given $a_{11}=a_{33}$, $a_{22}=1$ at exactly two points, it cannot be of degree $1$ (and indeed is of degree $2$). Hence $\deg(V\cdot W)\leq \deg(V) \deg(W)$ does not hold.

The next question suggests itself: for given $n$, what is the smallest $c_n$ such that $$\deg(\overline{V \cdot W}) \leq c_n \deg(V) \deg(W)$$ for arbitrary subvarieties $V, W\subset M(n)$? The bound is true with $c_n = 2^d$, where $d = codim(\overline{V\cdot W})\leq n-1$: the number of points of intersection of $\overline{V\cdot W}$ with a generic linear variety $L$ of dimension $d$ is bounded by the degree of the intersection of $\phi^{-1}(L)$ with $V\times W$, where $\phi:M(n)\times M(n)\to M(n)$ is the multiplication map. By Bézout's theorem, $$\begin{aligned}\deg(\phi^{-1}(L)\cap (V\times W) &\leq \deg(\phi^{-1}(L)) \deg(V\times W)\\ &\leq\deg(\phi^{-1}(L)) \deg(V) \deg(W),\end{aligned}$$ and it is easy to see that $\deg(\phi^{-1}(L))\leq 2^d$.

It seems simple to exclude the case of $\dim(\overline{V\cdot W})=1$, and thus obtain $c_n\leq 2^{n-2}$ rather than $c_n\leq 2^{n-1}$. Can one do much better? Is there, say, a polynomial bound or a linear bound on $c_n$?