I want to prove that $U_n(\mathbb{C})$ is a Lie subgroup of $GL_n(\mathbb{C})$ and find its dimension (as a real subgroup). Okay, let's consider the following action of $U_n(\mathbb{C})$ on $Mat_n(\mathbb{C})$: $$(x, y) \longrightarrow xyx^{*}$$ If i take $y= 1$ I can see that $U_n(\mathbb{C})$ is its kernel, so it's a Lie subgroup. Is this proof correct? To prove that its dimension is $2n^2$ (is it true?) I can look at the tangent space at $1 \in U_n(\mathbb{C})$. So I have that $T_1 U_n(\mathbb{C}) = \{A: A + A^* = 0\}$. But how from this I can find that its dimension is $2n^2$? Moreover, is there any calculation of dimension which calculates the codimension?
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1Does this answer your question? Dimension of $SU(N)$ – José Carlos Santos Sep 19 '21 at 17:40
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I don't think the dimension of $U_n$ is $2n^2$ as a real Lie group. Since $U_n$ is the stabilizer of $1$ under the action you described, its dimension should be $2n^2$ minus the dimension of the orbit space $xx^{\ast} : x \in \operatorname{GL}_n(\mathbb C)$. – D_S Sep 19 '21 at 17:51
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@JoséCarlosSantos no, there is no proof that it's subgroup and more there is no proof of equvialence for first answer, i am interested more in some less direct proof – NeoFanatic Sep 19 '21 at 17:55
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2Every closed subgroup of a Lie group is a Lie subgroup. – José Carlos Santos Sep 19 '21 at 17:57
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@JoséCarlosSantos but why it's closed?Because it's a preimage of ${-1, +1}$ of $det$ map from $GL_n(\mathbb{C})$ to $\mathbb{C}^{*}$? – NeoFanatic Sep 19 '21 at 17:59
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2No. Because it is the preimage of ${\operatorname{Id}_n}$ with respect to the map $x\mapsto x^*x$. – José Carlos Santos Sep 19 '21 at 18:02