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I have been trying to understand this question from Mathoverflow: https://mathoverflow.net/questions/34044/group-cannot-be-the-union-of-conjugates

The discussion is interesting, but the specific example is a bit beyond me. Could someone provide a proof that invertible upper triangular matrices have infinite index in $\text{GL}_2(\mathbf{C})$?

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Joe Corneli
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    It should be enough to note that they are a (complex) codimension 1 subgroup. This says the coset space is 1-dimensional, so in particular infinite. – john Apr 01 '17 at 01:09
  • OK, I believe you're saying that the cosets are $gH$ for $g$ in $\mathbf{C}$. And yeah this does really look like a matter of knowing the definition of "index" I guess. Wikipedia reminds me: "Formally, the index of [subgroup] $H$ in $G$ is defined as the number of cosets of $H$ in $G$." There should be one for each complex number, as you're telling me. Thanks, I'm obviously a bit rusty at this stuff. – Joe Corneli Apr 01 '17 at 01:29
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    Well, not $g$ in $\mathbb{C}$ but $g$ in $\mbox{GL}_2(\mathbb{C})$. I'm sort of appealing to the structure of $\mbox{GL}_2(\mathbb{C})$ as a manifold here but not too much. – john Apr 01 '17 at 01:35
  • Yes, right. But we could find a copy of $\mathbb{C}$ in there. The issue of whether the topological structure is needed comes up in the MO discussion. – Joe Corneli Apr 01 '17 at 13:48
  • @john could you make your appeal to the manifold structure explicit in an answer, so I can accept it? Are you using the regular value theorem? I've found a proof of the codimension-1 for SL group that uses that. – Joe Corneli Apr 01 '17 at 17:22

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Suppose $H,K\le G$ are subgroups. Since $[G:K]\ge [HK:K]=[H:H\cap K]$, to prove that $[G:K]$ is infinite it suffices to prove $[H:H\cap K]$ is infinite.

Can you find an infnite subgroup $H\le\mathrm{GL}_2(\mathbb{C})$ for which $H\cap U$ is finite? (Where $U$ stands for the subgroup of upper diagonal matrices.)

anon
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Turning my comment to an answer per Joe's request:

$\mbox{GL}_2(\mathbb{C})$ is an open 4-(complex)-dimensional submanifold (8-real-dimensions) of $\mbox{Mat}_{2\times 2}(\mathbb{C}) \cong \mathbb{C}^4$ (say $x_{11},x_{12},x_{21},x_{22}$ are complex coordinates) since it is the complement of the closed zero locus of the determinant function $x_{11}x_{22} - x_{21}x_{12}$.

The subgroup $U$ of invertible upper triangular matrices is a hyperplane section (namely, $\{x_{21} = 0\}\cap \mbox{GL}_2(\mathbb{C})$) and hence is a closed 3-(complex)-dimensional submanifold (6-real-dimensions). Importantly, we can think of it as a 6-real-dimensional Lie group.

$U$ acts on $\mbox{GL}_2(\mathbb{C})$ by left multiplication. The action (call it $\rho$) is smooth (the coordinate functions are polynomial), proper (it acts by diffeomorphisms which are of course proper) and free (all left-multiplication group actions are free). Smoothness, properness and freeness of the action are sufficient for the quotient by the action to also be a manifold (see here).

By a dimension count, the resulting manifold $M := \mbox{GL}_2(\mathbb{C})/\rho$ is 1-complex-dimensional (2-real-dimensions). It's points are in bijection with the orbits of the action $\rho$ but these orbits are exactly the left cosets of $U$. Having positive dimension means $M$ is in particular infinite. So the index of $U$ in $\mbox{GL}_2(\mathbb{C})$ is also infinite.

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john
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The subgroup of upper triangular matrices $B$ is the one invariating the subspace spanned by $e_1= (1,0)$. So the cosets of $B$ are in $1-1$ correspondence with the $1$-dimensional subspaces of $\mathbb{C}^2$ ( since $G = GL_2(\mathbb{C})$ acts transitively on it). Therefore $G/B \simeq \mathbb{P}^1(\mathbb{C})$.

That would work for any field. Now since $\mathbb{C}$ is infinite, we conclude that the cosets form an infinite set.

orangeskid
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