Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$ 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.

share|improve this question
When would you ever need to take the absolute value of a determinant of a complex matrix? –  Qiaochu Yuan Sep 20 '12 at 2:37
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." –  Ben Webster Sep 20 '12 at 2:44
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, –  Emerton Sep 20 '12 at 14:57

3 Answers 3

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.

share|improve this answer
In particular, as a Lie group it is not that interesting. Its Lie algebra, for example, is a direct product of two simple factors. –  Mariano Suárez-Alvarez Sep 20 '12 at 2:40
(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. ;) ) –  Qfwfq Sep 20 '12 at 8:51
@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. –  Jim Conant Feb 27 '13 at 0:02

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).

share|improve this answer

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 $\lbrace\text{diag}(z,z^{-1},1,...,1) \mid z \neq 0\rbrace$.

share|improve this answer
Concerning compactness: Of course one can directly argue that $G$ isn't bounded. –  Ralph Sep 20 '12 at 4:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.