# Bézout and products in algebraic groups

Let $$G$$ be a linear algebraic group -- be it a Lie group or a group of Lie type. Let $$V$$, $$W$$ be subvarieties of $$G$$. Of course, $$V\cap W$$ is also a variety (not necessarily irreducible) and $$V\cdot W^{-1}$$ is a constructible set. It is easy to see that $$\dim(\overline{V\cdot W^{-1}}) = \dim(V)+\dim(W)-\dim(V\cap a W)$$ for $$a\in V\cap W$$ generic. By Bézout, we know that the degree $$\deg(V\cap a W)$$ of $$V\cap a W$$ (meaning the sum of the degrees of its components) is at most $$\deg(V)\cdot \deg(W)$$.

Is it always the case that $$\deg(V \cap a W) \cdot \deg(\overline{V\cdot W^{-1}}) \leq \deg(V)\cdot \deg(W)$$, for $$a\in \overline{V\cdot W^{-1}}$$ generic?

If not, is that the case under some sort of generic conditions, or, say, for $$W = g V g^{-1}$$ and $$g\in G$$ generic?

Reason why such a thing might be plausible: if $$\dim(V) + \dim(W)\geq \dim(G)$$, then, generically, $$\overline{V\cdot W}$$ is the entire group, and, while $$\deg(G)$$ is not in general $$1$$ for $$G$$ seen as a subvariety of $$M(n)\sim \mathbb{A}^{n^2}$$, the degree of $$G$$ "relative" to itself is in some sense $$1$$ (of course we would need to formalise that notion; the point is that that's how it works out in further applications of Bézout's theorem, in a trivial sense, in so far as, for $$V\subset G$$, $$\deg(V\cap G) = \deg(V) \cdot 1$$). If $$\dim(V)+\dim(W)<\dim(G)$$, then, generically, the intersection $$V\cap W$$ is empty, and non-generically, Bézout may not be tight.

• The claim that $\dim(\overline{V\cdot W})$ has dimension $\dim(V)+\dim(W)-\dim(V\cap W)$ is not true. For instance, typically when $V=W$ (say, for a general enough curve in the plane), we have $\dim(V\cdot V)=2\dim(V)$ (and not $=\dim(V)$). For the rest of the question, I'm not sure what you call degree.
– YCor
Oct 12, 2018 at 16:00
• Hm, you are right. For the rest of the question - let us go with "sum of the degrees of irreducible components of maximal dimension" as the definition of the degree of reducible variety. Oct 12, 2018 at 16:05
• What {\em is} the case is that $\dim(\overline{V\cdot W})$ has dimension $\dim(V)+\dim(W)-\dim(V\cap a W^{-1})$ for $a\in \overline{V\cdot W}$ generic (or for $a$ in the image of a generic point of $V\times W$ under the multiplication map, which is the same). Oct 12, 2018 at 16:07
• What is the degree? Oct 12, 2018 at 19:02
• Laurent Moret-Bailly's question is still pertinent. Into which affine or projective space are you embedding your varieties? "Degree" is not defined without choosing such an embedding. Oct 12, 2018 at 20:48