# Local surjectivity of word maps

Let $$F$$ be a free group on 2 generators and $$G = \operatorname{SL}(d,\mathbb{C})$$. A word $$w \in F$$ induces the word map $$\mathrm{ev}_w: G \times G \to G$$.

Can we find some (generic) conditions on $$A,B \in G$$ such that, for all $$w \in F$$, the differential of $$\mathrm{ev}_w$$ is surjective at $$(A,B)$$ ?

• Are you asking whether it's generic, or whether one can provide explicit generic conditions (while you know it's generic)?
– YCor
Commented Oct 23, 2020 at 11:15
• I'm asking whether it's generic, but an explicit generic condition would be even better.
– FMB
Commented Oct 23, 2020 at 11:21
• So, this is equivalent to asking whether $\mathrm{ev}_w$ is essential for every $w$ (that is, has Zariski-dense image, which by standard algebraic geometry is the same as having image containing a dense Zariski-open subset).
– YCor
Commented Oct 23, 2020 at 13:18
• I believe that a theorem of Borel says that $\text{ev}_w$ is dominant for every non trivial $w$ (see arxiv.org/abs/math/0211302 ). Could you explain how this is equivalent ?
– FMB
Commented Oct 23, 2020 at 15:07
• If $f:X\to Y$ is dominant ($X,Y$ irreducible complex smooth) there's a Zariski-dense open $U$ subset of $X$ on which the differential has maximal rank $r$, and $U\to Y$ is still dominant. At the (metric) neighborhood of any point of $U$ there's a neighborhood whose image is locally (up to local holomorphic change) a copy of $\mathbf{C}^r$ in $\mathbf{C}^{\dim(Y)}$, so if $r<\dim(Y)$ the image is meager, thus can't contain a Zariski-open subset of $Y$. So $r=\dim(Y)$, i.e. the $df(x)$ is surjective for every $x\in U$.
– YCor
Commented Oct 23, 2020 at 15:36

Since a theorem of Borel (which you quote in the comment) tells you that this regular map is dominant for every $$w$$, its differential is surjective on a Zariski-dense open subset $$U_w$$.

Hence there intersection (over all $$w\neq 1$$) of these subsets $$U_w$$ is a $$\mathrm{G}_\delta$$-dense subset of full measure.