Unitary group is a properly embedded Lie subgroup of $GL(n, \mathbb C)$ I would like to show that $U(n)$, the unitary group of degree $n$, is a properly embedded Lie subgroup of $GL(n, \mathbb C)$ of dimension $n^2$, and find the matrices that are in the tangent space at the identity.
To show the first part, I think I can argue it is a kernel $\Phi^{-1}(I_n)$, where $\Phi(A) = \overline {A^T}A$. However, I am not sure how to show its dimension, and how to identify matrices that are in the tangent space at the identity.
 A: To figure out what the tangent space to $U(n)$ at the identity, use the trajectories of curves definition of the tangent space. Let the tangent space to the identity of $U(n)$ be denoted $\mathfrak{u}(n)$. Let $A\in \mathfrak{u}(n)$ and let $\gamma:(-\varepsilon,\varepsilon)\to U(n)$ so that $\gamma(0)=I_n$ and $\gamma'(0)=A$. By definition, as $\gamma(t)\gamma(t)^\dagger=I_n$, and so differentiating at $t=0$ we get
$$
0=\frac{d}{dt}\bigg|_{t=0}\gamma(t)\gamma(t)^\dagger=\gamma'(0)\gamma(0)^\dagger+\gamma(0)\gamma'(0)^\dagger=A+A^\dagger.
$$
You can use this idea to argue that $\mathfrak{u}(n)$ consists of the set of $n\times n$ skew-Hermitian matrices. I.e.
$$
\mathfrak{u}(n)=\{A\in \mathfrak{gl}(n,\Bbb{C}):A=-A^{\dagger}\}.
$$
Next, you need to come up with a basis for this space and hence count its dimension. For $n=2$, a basis over $\Bbb{R}$ is given by
$$
\begin{bmatrix}
i&0\\
0&0
\end{bmatrix},
\:\:
\begin{bmatrix}
0&0\\
0&i
\end{bmatrix},\:\:
\begin{bmatrix}
0&1\\
-1&0
\end{bmatrix},\:\:
\begin{bmatrix}
0&i\\
i&0
\end{bmatrix}.
$$
I leave the generalization of this to you.
