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I need help on this one:

In Chriss & Ginzburg book on representation theory and complex geometry I came across the following statement:

maximal compact (in analytic topology) subgroup G_comp of reductive group G is dense in Zariski topology of G.

If this is true, what about the case of S^1 inside C*. Since it is closed it doesn't look dense to me.

Answers appreciated!

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    $\begingroup$ You are talking about Zariski topology. For example, the set of integers is closed in the Euclidean topology on the complex plane but is Zariski dense in $\mathbb C$. (Easy exercise). $\endgroup$
    – Venkataramana
    Commented Sep 25, 2017 at 12:59
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    $\begingroup$ The non-trivial content is that $\mathbf{C} \otimes_{\mathbf{R}} {\rm{Lie}}(G_{\rm{comp}}) \to {\rm{Lie}}(G)$ is an equality. The Zariski closure of any subgroup of $G(\mathbf{C})$ is a Zariski-closed $\mathbf{C}$-subgroup, necessarily smooth, so for the above Lie algebra reason the Zariski closure of $G_{\rm{comp}}$ has full dimension and hence is $G$ if $G$ is Zariski-connected. If $G$ is not assumed to be Zarski-connected then you need $G_{\rm{comp}}$ to meet each Zariski-component of $G$. It even meets each connected component of $G(\mathbf{C})$ for the analytic topology (requires proof). $\endgroup$
    – nfdc23
    Commented Sep 25, 2017 at 13:20

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When talking about Zariski-topology it is important to specify which field you are working over! Over $\mathbb{C}$ the cirkel is not closed in the Zariski topology on the one-dimensional reductive complex algebraic group $\mathbb{C}^*$, because it is not the zero set of a polynomial in one variable over $\mathbb{C}$. However, if we view $\mathbb{C}^*$ as a two-dimenional Lie group over $\mathbb{R}$ (i.e. isomorphic to the cylinder $S^1 \times \mathbb{R}_{>0}$) then the circle subgroup is Zariski-closed and your objection applies. So perhaps check if the book makes any assumptions about the Lie groups in question being reductive algebraic groups over $\mathbb{C}$.

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