$U(n)$ is a subgroup of $SO(2n)$ How can I show that $U(n)$ is a subgroup of $SO(2n)$? I can see how we can identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ by mapping $a+ib\mapsto (a,b)$, but after that I'm a bit confused. In particular, the determinant of a matrix in $U(n)$ has to be nonzero, whereas in $SO(2n)$ it has to be 1. How do we make this transition?
 A: Hint: Think of the complex entry $a+bi$ as the $2\times2$ submatrix
$$
\pmatrix{a&-b\\b&a}
$$
More rigorously, consider the mapping that makes this replacement, and note that it is an injective, continuous homomorphism.
A: A determinant of a matrix in $U(n)$ is not merely nonzero, it is of absolute value 1.
Start with a trivial example. $U(1)$ is just a complex number of absolute value 1, so geometrically it sits on the unit circle. You can represent it as $cos(\theta) +i* sin(\theta)$. What's the real $2\times 2$ matrix that represents a rotation by the angle $\theta$? Is it in $SO(2)$?
Once you understand this example, extend to $U(n)$. Replace every complex element in the matrix by what you think is the right real counterpart, so you get a $2n \times 2n$ real matrix, and see if you can show it's in $SO(2n)$.
A: Let me write the answer, this is a good exercise.
All you have to do is to show the inclusion $U(n)\subset O(2n)$, which makes sense after identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ by for instante $(z_1,\cdots,z_n) = (x_1+iy_1,\cdots,x_n+iy_n)\longleftrightarrow (x_1,\cdots,x_n,y_1,\cdots,y_n)$.
Consider then $U\in U(n)$, that is, $U \in Mat_{n\times n}(\mathbb{C})$ such that $UU^\dagger = U^\dagger U = 1$. Such matrices acts on $\mathbb{C}^n$ by $$(z_1,\cdots,z_n) \longmapsto (U_{1j}z_j,\cdots,U_{nj}z_j)$$
Using our identification, we will discover how $U(n)$ acts on $\mathbb{R}^{2n}$. To do so we decompose the matrices of $U(n)$ in real and imaginary parts: $$ U = U^R + iU^I $$ with $U^R,U^I \in Mat_{n\times n}(\mathbb{R})$ and
$$ UU^\dagger =  (U^R + iU^I)(U^R + iU^I)^\dagger = U^R (U^R)^T + U^I (U^I)^T + i\left( U^I(U^R)^T - U^R(U^I)^T \right)=1 $$
$$ U^\dagger U =  (U^R + iU^I)^\dagger(U^R + iU^I) =  (U^R)^TU^R +  (U^I)^TU^I + i\left( (U^R)^TU^I - (U^I)^TU^R \right)=1 $$
The condition follows as: $$U^R (U^R)^T + U^I (U^I)^T = (U^R)^TU^R +  (U^I)^TU^I = 1 \ \ (*)$$
$$ U^I(U^R)^T - U^R(U^I)^T = (U^R)^TU^I - (U^I)^TU^R = 0 \ \ (**)$$
With this notation:
$$ (z_1,\cdots,z_n) \longmapsto (U^R_{1j}z_j + iU^I_{1j}z_j,\cdots,U^R_{nj}z_j+iU^I_{nj}z_j) $$
writing in terms of $z_i=x_j+iy_j$  : $$ \left( (U^Rx)_1 - (U^Iy)_1 + i(U^Ix)_1+i(U^Ry)_1,\cdots, (U^Rx)_n - (U^Iy)_n + i(U^Ix)_n+(U^Ry)_n) \right) $$
Therefore, expliciting the identification: $$ (x_1,\cdots,x_n,y_1,\cdots,y_n)\longmapsto \left( (U^Rx)_1 - (U^Iy)_1, \cdots, (U^Rx)_n - (U^Iy)_n,(U^Ix)_1+(U^Ry)_1,\cdots, (U^Ix)_n+i(U^Ry)_n \right) $$
In matrix notation:
$$U = \begin{bmatrix}U^R& -U^I\\ U^I & U^R\end{bmatrix} \in Mat_{2n\times 2n}(\mathbb{R}) $$
To see that this matrix is on $SO(2n)$, we have to check the ortogonal condition: $UU^T=1$. This follows immediately by $(*)$ and $(**)$ :
$$ UU^T = \begin{bmatrix}U^R& -U^I\\ U^I & U^R\end{bmatrix}\begin{bmatrix}(U^R)^T& (U^I)^T\\ -(U^I)^T & (U^R)^T\end{bmatrix}  $$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   = \begin{bmatrix}U^R(U^R)^T + U^I(U^I)^T& U^R(U^I)^T-U^I(U^R)^T\\ U^I(U^R)^T-U^R(U^I)^T & U^I(U^I)^T+U^R(U^R)^T\end{bmatrix}= \begin{bmatrix}1& 0\\ 0 & 1\end{bmatrix}$$
ps: Repeated indices always denotes sum over the indices $\left(a_jb_j=\sum_ja_jb_j\right)$
