I know that the complement of the zero set of a polynomial $P: \mathbb{C}^n \rightarrow \mathbb{C}$ is connected in $\mathbb{C}^n$ (by the way, can anybody suggest a reference?).


Is it possible to extend the proof also to polynomials
 $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ ?


Thanks!



connectedness of the complement of the zero set of a polynomial $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$
$SL(n,\mathbb{C})^N$ is a smooth and irreducible variety, and thus a manifold. The zero set $Z$ of a nonzero polynomial is a subvariety with $\dim\_{\mathbb{C}}(Z)\leq Nn^2-1$ and so $\dim\_{\mathbb{R}}(Z)\leq 2Nn^2-2$. If $a$ and $b$ are points in $Z^c$ then we can choose a smooth path $\mathbb{R}\to SL(n,\mathbb{C})^N$ joining them, then perturb it slightly to make it transverse to $Z$. As $\dim\_{\mathbb{R}}(\mathbb{R})+\dim\_{\mathbb{R}}(Z)<\dim\_{\mathbb{R}}(SL(n,\mathbb{C})^N)$, transversality just means that the path does not meet $Z$, so we get a path in $Z^c$ as desired.


For this we need a result telling us that the path can be made transverse. This is a standard fact in differential topology if $Z$ is a submanifold, which is the generic case. Essentially the same proof should work even if $Z$ is not smooth, though it might be harder to find a convenient reference.


