$U(n)\cap\mathfrak{u}(n)$ as a submanifold of $\mathbb{C}^{n^2}$ This is a question from an exam on an extra-curricular course in differential geometry.
The task reads "Express $U(n)\cap\mathfrak{u}(n)$ as a submanifold of $\mathbb{C}^{n^2}$, where $U(n)$ and $\mathfrak{u}(n)$ are both considered as embedded submanifolds of $\mathbb{C}^{n^2}$". I am not sure what an answer would be, i.e. just how explicitly can this intersection be expressed. So any hint/idea will be appreciated.
What I came up with:
For $n=1$ the intersection consists of just 2 points $\pm i$, so topologically it is $S^0$.
For $n=2$ each unitary matrix can be expressed as \begin{bmatrix}
a & b\\
-e^{i\varphi}\overline{b} & e^{i\varphi}\overline{a}
\end{bmatrix}
with $\varphi\in\mathbb{R}$ and $a,b\in\mathbb{C}$ satisfying $|a|^2+|b|^2=1$. Since $u(n)$ constists of skew-hermitian matrices, then by imposing this condition on matrices of this form we get 2 cases:


*

*$e^{i\varphi}=1$, $a\in\mathbb{R}$. These are unitary skew-symmetric matrices with unit determinant. 
\begin{bmatrix}
a & b\\
-\overline{b} & \overline{a}
\end{bmatrix}
Since $a\in\mathbb{R}$, $a=\pm\sqrt{1-|b|^2}$. Topologically the set of such matrices seems to be diffeomorphic to $S^2$.


*

*$b=0$, $e^{i\varphi}=\pm1$, $a\in\mathbb{R}$, and there are 4 matrices of this form, of which only 2 are not covered in the previous case:


$$ \left[ \begin{array}{cc}
i & 0 \\
0 & -i
\end{array} \right] \text{and}
\left[ \begin{array}{cc}
-i & 0 \\
0 & i
\end{array} \right]
$$
In total, the intersection is $S^0\sqcup S^2$.
It would be cute if $U(n)\cap\mathfrak{u}(n)\cong\coprod\limits_{k=0}^{n-1} S^{2k}$ were true in general, but I have no idea how to prove it at the moment.
The only expression I found is through a solution to a system of equations
$$U(n)\cap\mathfrak{u}(n)=\{S\in\mathfrak{u}(n)|S^2-I=0\}.$$ But I have no idea how to get anything explicit from this.
 A: s.harp is right that the space is diffeomorphic to $\coprod_{k=0}^n U(n)/(U(k)\times U(n-k))$. Call $X:=U(n)\cap\mathfrak{u}(n)$.
Proof:  Consider the $U(n)$ action on $C^{n^2} = M_n(\mathbb{C})$ by conjugation.  This action preserves $U(n)\subseteq M_n(\mathbb{C})$ becuase $U(n)$ is a subgroup.  It also preserves $\mathfrak{u}(n)\subseteq M_n(\mathbb{C})$ because this is nothing but the adjoint action.
It follows that the $U(n)$ conjugation action maps $X$ to itself.  Further, since every element in $U(n)$ is conjugate to a diagonal element, every orbit contains a diagonal matrix.  Further, each of the permutation matrices is in $U(n)$, so we may further conjugate the diagonal matrix to put the diagonal elements in any order we like.
As noted in the comments, the eigenvalues must be $\pm i$.  (One way to see this:  the eigenvalues of an element of $U(n)$ are all norm $1$, while the eigenvalues of an element in $\mathfrak{u}(n)$ are purely imaginary.)  Since conjugation preserves the eigenvalues, and because every element of $U(n)$ is conjugate to a diagonal matrix we have proven:

Under the $U(n)$ action on $X$, for each element $A$ of $X$, there is a unique integer $k$ with $0\leq k \leq n$ for which $A$ is conjugate to $E_k:=diag(\underbrace{i,...,i}_{k \text{ copies}}, \underbrace{-i,...,-i}_{n-k \text{ copies}}).$

(So far, all of this was essentially in the comments.)
Now, by compactness, the $U(n)$ orbits through distinct $E_k$s have a positive distance between them.  So $X$ is given as a disjoint union of orbits through the $E_k$.
Finally, we use this wonderful fact that for a compact Lie group action, the orbit $U(n)\cdot E_k$ is canonically diffeomorphic to $U(n)/U(n)_{E_k}$, where $U(n)_{E_k}$ denotes the stabilizer at $E_k$.  That is, $U(n)_{E_k} = \{B\in U(n): BE_k B^{-1} = E_k$.}
So, we need only show that $U(n)_{E_k} = U(k)\times U(n-k)$.  Given $B\in U(n)$, write $B$ in the block form $B = \begin{bmatrix}C & D\\ E & F\end{bmatrix}$ with $C$ a $k\times k$ matrix and $F$ an $(n-k)\times (n-k)$ matrix.
Then \begin{align*} B\in U(n)_{E_k}&\iff BE_k = E_k B\\ &\iff iD = -iD\text{ and } -iE = iE\\ &\iff D = 0\text{ and } E = 0\\ &\iff B\in U(k)\times U(n-k).\end{align*}
$\square$
The spaces $U(n)/(U(k)\times U(n-k))$ are called complex Grassmannians and have the geometric interpretation of being the set of all $k$-dimensional subspaces in $\mathbb{C}^n$.  For certain values of $n$ and $k$, they have other descriptions.
For example, for any $n$, $U(n)/U(0)\times U(n)$ and $U(n)/U(n)\times U(0)$ are both points (as you saw in the $n=1$ and $n=2$ case).  Further, for any $n$, $U(n)/U(1)\times U(n-1)$ and $U(n)/U(n-1)\times U(1)$ are both diffeomorphic to $\mathbb{C}P^{n-1}$.  Note that when $n=2$, $\mathbb{C}P^1 = S^2$, as you saw.
