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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 $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ ?


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Could you give us some more details? Which topologies are you using? What exactly is a polynomial $P\colon SL(N,\mathbb{C})^n \to \mathbb{C}$? – Mark Grant Apr 27 '12 at 10:34
Do you mean the restriction of a polynomial in $N^2n$ variables to the special linear group? – Mark Grant Apr 27 '12 at 13:18
Sorry. It is more ambiguous than I thought. I assume a matrix representation of the elements of $SL(N,\mathbb{C})$. In this way I define the sum as the sum of matrices. The pruduct is the one of the group. So, the polynomial are defined. Yes, I guess this is equivalent to polynomial in $N^2 n$ variables restricted to the special linear group. As for the topology, I was assuming the topology inherited from $\mathbb{C}^{n(N^2−1)}$ via the exponential parametrization of the group from the algebra. But other topologies in which the set is connected might be useful. – Luigi Scorzato Apr 27 '12 at 13:36
@Luigi: I was interpreting things as in Neil's answer (I think). However, your comment (and also that of Mrc Plm) makes it sound like you want to take a polynomial in $n$ variables (with $\mathbb{C}$ coefficients, say) and plug in $n$ matrices. But this would give back a matrix, right? – Mark Grant Apr 27 '12 at 13:54
No, my comment was non-sense. I was also thinking in my answer about a polynomial to $\mathbb{C}$, and I worked with the entries as well. – Marc Palm Apr 27 '12 at 14:00
up vote 4 down vote accepted

$SL(n,\mathbb{C})^N$ 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.

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thanks! could you please give me a reference for the standard fact about transversality of the path? – Luigi Scorzato Apr 27 '12 at 13:54
@Neil: Just use Alexander duality to show that a subcomplex of dimension $<k-1$ cannot separate ${\mathbb R}^{k}$. In this particular case, complex subvariety in ${\mathbb C}^n$ of complex dimension $<n$ admits a triangulation (Lojasiewicz), so we can regard it as a simplicial complex of real dimension $<2n-1$. The same thing works locally. – Misha Apr 27 '12 at 14:01

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

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thanks! About point 2) I have probably imprecise definitions in mind and a quick googling does not help me. Isn't the set $\{x\in \mathbb{R}^n| x_1=0 \}$ a closed Zariski subset in an irreducible variety? Because its complement in $\mathbb{R}^n$ is not connected. What am I missing? – Luigi Scorzato Apr 28 '12 at 9:16
R^n isn't a complex variety – Ostap Chervak Apr 28 '12 at 12:37
interesting! So, in that case it is point 1) that fails? i.e. the complement of $\{ x\in \mathbb{R}^n |x_1=0\}$ is Zariski connected? Really interesting... – Luigi Scorzato Apr 28 '12 at 12:52
@Luigi: Yes, considering $\mathbf R$ or $\mathbf C$ makes a great difference. The affine line minus the origin over a field $k$ is connected (it is affine with ring of functions given by $k[T,T^{-1}]$ is an integral domain). If you look at $\mathbf C$-points, you get the complement to the origin in~$\mathbf C$ (a punctured plane, hence a connected topological space), but if you look at $\mathbf R$-points, you get the complement to the origin in $\mathbf R$ which is obviously not connected. – ACL Jun 6 '12 at 20:28

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;)

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Sorry, but I still do not see the conclusion: Your argument shows that the complement of the zero set of $Q'\cdot Q''$ is connected in $\mathbb{C}^{N n}$ but this does not mean that it is connected in $SL_n(\mathbb{C})$. Does it? – Luigi Scorzato Apr 27 '12 at 13:15
Clear now?..... – Marc Palm Apr 27 '12 at 13:48
I apologize for my misleading fomulation – Luigi Scorzato Apr 28 '12 at 8:57

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