connectedness of the complement of the zero set of a polynomial $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ 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!
 A: $SL(n,\mathbb{C})^N$ 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.
A: An answer to the more general question could be in three steps.
1) If $X$ is a complex algebraic variety, connected for the Zariski topology, then it is connected for the usual topology, an important and nontrivial fact that can be found in Mumford's Red Book, or in Shafarevich's one. 
2) The complement to a closed Zariski subset in an irreducible variety is connected for the Zariski topology.
3) The varieties $SL(n,{\bf C})$ and $SL(N,{\bf C})^N$ are irreducible 
A: Let $X_{j}$ be elements of $SL(n, \mathbb{C})$ with entries ${X_{j}^{k,l}}$.
Express the polynomial $P: SL(n,\mathbb{C})^N \rightarrow \mathbb{C}$ as a polynomial $Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C}$ in variables ${X_{j}^{k,l}}$
Similar express the polynomial
$$ Q''(X_1, \cdots, X_n)  = \prod\limits_{i=1}^N \left( \det(X_i) -1  \right)$$
in variables ${X_{j}^{k,l}}$. 
You can then  apply the well-known result for $\mathbb{C}^{N \cdot n^2}$ to $Q' \cdot Q''$, and get what you want.
Edit due the comment: Note that we have both $$\{ Q' Q'' = 0 \} \subset \mathbb{C}^{N  \cdot n^2}$$ is a connected subset, and that it is really a subset of $SL_n(\mathbb{C})^N$ 
$$\{ Q' Q'' = 0 \} \subset \{ Q'' =0 \} = SL_n(\mathbb{C})^N.$$
I am not sure, what topology you are working in, but it holds in any topology, the original results for $\mathbb{C}^N$ holds in;)
