This might be a very basic question . I am new to Lie Algebra. In general we know that it is not necessary that every subgroup of a connected matrix lie group is connected. For instance $GL(n,\mathbb{R})\leq GL(n,\mathbb{C})$ but $GL(n,\mathbb{C})$ is connected while $GL(n,\mathbb{R})$ is not connected. Is this undersstanding correct ? Now let $G\leq GL(n,\mathbb{C})$ be a matrix lie group. is there a way to show that $G$ is connected/ disconnected ? Is there any major misconception in my understanding ?
Connected matrix lie subgroups
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general-topology
lie-groups
1 Answers
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There is no general algorithm for checking that. However there are some tools to help you. For example the way you show $GL(n, \mathbb{R})$ is disconnected is by looking at $\det$ function:
$$\det:GL(n,\mathbb{R})\to\mathbb{R}\backslash\{ 0\}$$
And so $GL(n,\mathbb{R})=\det^{-1}(-\infty,0)\cup\det^{-1}(0, \infty)$. Note that both sets on the right are disjoint and open ($\det$ is continous) so as long as they are nonempty (and they are since $\det(I)=1$ and $\det(-I)=-1$) the $GL(n, \mathbb{R})$ group is disconnected.
But not every subgroup of $GL(n, \mathbb{R})$ is disconnected. Obviously for a topological group $G$ the connected component of $1$ is a connected subgroup of $G$. In this case that would be all matrices with positivie determinant. Another example would be the special linear group
$$SL(n,\mathbb{R})=\{M\in GL(n, \mathbb{R})\ |\ \det(M)=1\}$$
is connected (both as a nontrivial consequence of Gram-Schmidt).
Now the problem with $\mathbb{C}$ is that $\mathbb{C}\backslash\{0\}$ is still connected. So you can't really use this method in case $G