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$$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$ This seems to me to be a perfectly natural group to study; it is easy to define and contains $U(n), SL_n$, and all the torsion. Is there any good reason why this group isn't among the usual classical groups that are so well-understood and thoroughly discussed? I understand that most of those are studied/defined by looking at groups preserving particular inner products, but it still surprises me that I've never heard of any interesting results/properties of this group. The only guess I currently have is it's not compact.

I could ask the same question with $\mathbb{R}$ but then "morally" the group is just two copies of $SL_n(\mathbb{R})$ so I understand why it's less interesting.

A perfectly acceptable answer is that I'm totally misinformed and this group is perfectly understood, classical, named, etc., in which case any reference would be appreciated.

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    $\begingroup$ When would you ever need to take the absolute value of a determinant of a complex matrix? $\endgroup$ Sep 20, 2012 at 2:37
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    $\begingroup$ Most groups, especially Lie groups, are interesting to people because they're the symmetry groups of objects people care about. I guess the group you suggest is the group of complex linear transformations that preserve the real volume form, but I'm hard pressed to take that seriously as a "structure." $\endgroup$
    – Ben Webster
    Sep 20, 2012 at 2:44
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    $\begingroup$ Dear Jon, This group and groups like it come up all the time in the work of Arthur and others on the trace formula, where they use various notations (maybe $G^1$, if $G$ were denoting $GL_n$; but I could be misremembering). On the other hand, this is not an algebraic group, which is perhaps why it has a slightly different status than the usual classical groups. Incidentally, determining its representation theory, etc., is a fairly trivial exercise given the known results about $GL_n$ and $SL_n$. Regards, $\endgroup$
    – Emerton
    Sep 20, 2012 at 14:57

3 Answers 3

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I guess one answer is there's an isomorphism between your group and

$$SL_n \mathbb C \times_{\mathbb Z /n\mathbb Z} SO_2$$

My notation means take the product and mod out by the diagonally embedded copy of $\mathbb Z/n \mathbb{Z}$.

The embedding of $\mathbb Z/n\mathbb Z$ in $SO_2$ is as the cyclic subgroup of order $n$, and the embedding in $SL_n \mathbb C$ is the matrices of the form $\lambda I$ where $\lambda \in S^1$ is an $n$-th root of unity and $I$ is the identity matrix in $SL_n \mathbb C$.

So it's almost a direct product of two fairly nice groups.

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    $\begingroup$ In particular, as a Lie group it is not that interesting. Its Lie algebra, for example, is a direct product of two simple factors. $\endgroup$ Sep 20, 2012 at 2:40
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    $\begingroup$ (Edited notation. Sorry, but I'm a member of the International Committee for the Abolition of the Notation $\mathbb{Z}_n$ to Denote Cyclic Groups Because it Conflicts with the Notation for p-Adic Integers. ;) ) $\endgroup$
    – Qfwfq
    Sep 20, 2012 at 8:51
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    $\begingroup$ @Qfwfq: but why do p-adics have primacy? (Pun intended.) I could make a similar case that we need to go around changing every $\mathbb Z_p$ for $p$-adics to a different notation. $\endgroup$
    – Jim Conant
    Feb 27, 2013 at 0:02
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Denote the group in question by $G$. Then there is a split extension $$1 \to SL_n(\mathbb{C}) \to G \xrightarrow{\text{det}} S^1 \to 1$$ where the splitting is given by $S^1 \to G,\; z \mapsto \text{diag}(z,1,...,1)$. Hence, from the group theoretical point of view $G$ is just the semi-direct product $$G = SL_n(\mathbb{C}) \ltimes S^1$$

Added: Your guess that $G$ isn't compact (in Euclidean topology) if $n>1$ is correct. For, suppose $G$ is compact. Then, the closed subgroup $SL_n(\mathbb{C}) = \text{det}^{-1}(1)$ is also compact, in contradiction to the fact that it contains the unbounded subset $\lbrace\text{diag}(z,z^{-1},1,...,1) \mid z \neq 0\rbrace$.

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  • $\begingroup$ Concerning compactness: Of course one can directly argue that $G$ isn't bounded. $\endgroup$
    – Ralph
    Sep 20, 2012 at 4:17
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This is just a guess: If you're going to sacrifice complex analyticity by allowing complex conjugation in the defining equations, then you should take full advantage of that by using lots of complex conjugates and gaining something important in return, like compactness (as in the case of the unitary group).

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