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Suppose that $k$ is an algebraically closed field. GL$(n,k)$ is the general linear group. It can be considered as $\{x \in \mathbb{A^{n^2}}: det(x) \neq 0\}$. Clearly, this is an open subset of $\mathbb{A^{n^2}}$. But why is it closed? and why is it connected?

Let T$(n,k)$ denotes the subgroup of GL$(n,k)$ consists of diagonal matrices. It is clearly closed. But how to prove its connectedness?

Many thanks~

  • 6
    $\text{GL}_n(k)$ is _not_ closed in $\mathbb{A}^{n^2}$; it's Zariski dense. It _is_ closed in $\mathbb{A}^{n^2 + 1}$; you actually want to embed it as $\{ x \in \mathbb{A}^{n^2}, y \in \mathbb{A}^1 : \det(x) y = 1 \}$.2011-05-21
  • 3
    (In any case, being closed is not an intrinsic property of a variety, so it is mildly misleading to treat it linguistically the same way as being connected, which _is._)2011-05-21
  • 3
    Dear ShinyaSakai, More generally, any open subset of an irreducible topological space is irreducible (hence connected). Since $\mathbb{A}_k^{n^2}$ is irreducible, so is $GL_n(k)$. For the second question, note that $T(n, k)$ is isomorphic to $k^* \times \dots \times k^*$.2011-05-21

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