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Let $U(n)$ denote the unitary group. That is,

$$ U(n) = \{A \in GL_n(\mathbb C)\mid A^\ast A = I\}$$

Wikipedia states:

"The unitary group $U(n)$ is a real Lie group of dimension $n^2$. "

There seem to be two typos: one, unitary matrices are complex. So it should be a statement about complex Lie group. The other typo is $n^2$: $n^2$ is the dimension of $GL_n$ and $U(n)$ is a proper subgroup so clearly its dimension must be smaller than $n^2$.

What's the dimension of $U(n)$?

Adam Hughes
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learner
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1 Answers1

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The unitary group $U(n)$ is a real Lie group of $GL(n, \mathbb{C})$. We can forget about the complex structure on $GL(n, \mathbb{C})$ and hence view it as a real Lie subgroup of (real) dimension $2 n^2$, which is greater than $\dim U(n) = n^2$.

Note, by the way, that $GL_+(n, \mathbb{R}) := \{A \in GL(n, \mathbb{R}) : \det A > 0\}$ is a proper subgroup of $GL(n, \mathbb{R})$ but $\dim GL_+(n, \mathbb{R}) = n^2 = \dim GL(n, \mathbb{R})$.

Travis Willse
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    I'm sorry but I don't believe that $U(n)$ is not a (complex) subgroup of $GL_n$. Can you prove it? As far as I can tell $U(n)$ contains $I$ and is closed with respect to multiplication so it's a group. Also, it is the inverse image of the $|\det|$ map of ${1}$ so it's closed... or am I missing something? – learner Jan 29 '15 at 07:40
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    It's certainly a subgroup of $GL(n, \mathbb{C})$, but it's not a complex subgroup. The defining equation of $U(n)$ is $A^* A = I$, which in components is $\sum_{i = 1}^n \bar{a}{ki} a{kj}$, but these are not analytic conditions, because they involve conjugation. (For odd $n$, note that $\dim U(n) = n^2$ is odd, whereas every complex group has even real dimension.) – Travis Willse Jan 29 '15 at 07:48
  • But in Tapp's book the definition of a matrix group is given as any topologically closed subgroup of $GL_n$. What is your idea with "analytic condition" for a matrix group? – learner Jan 30 '15 at 00:35
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    Again, $U(n)$ is a perfectly good real (matrix) Lie group, but it is not a complex Lie group: the latter requires that the group have a (1) complex-analytic manifold structure, and that (2) the multiplication and inversion operations on the group are holomorphic maps. This case is perhaps initially confusing because the underlying set of $U(n)$ is often taken to be a set of complex matrices, but this is an altogether separate condition. – Travis Willse Jan 30 '15 at 00:47
  • You're welcome, I'm glad you found it helpful. – Travis Willse Jan 30 '15 at 01:52
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    Doesn't $GL(n,\mathbb{C})$ have complex dimension $n^2$? I thought its real dimension would be $2n^2$. – Dhanush Giriyan Mar 02 '22 at 17:50
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    @DhanushGiriyan Here $n^2$ refers to the real dimension of $U(n)$, not of $GL(n, \Bbb C)$. I'll adjust the wording of the sentence to clarify. – Travis Willse Mar 02 '22 at 18:06