The following set is connected or not? Let $E$ be a complex Hilbert space and $S\in \mathcal{L}(E)^+$ (i.e.$S^*=S$ and $\langle Sx\;, \;x\rangle\geq0$ for all $x\in E$).

Let $U=\{x\in E; \langle Sx\;, \;x\rangle=1\}$. Is $U$ a connected set?

Thank you.
 A: The answer for real Hilbert spaces is no.
Let $S : \ell^2 \to \ell^2$ be defined as
$$S(x_1, x_2, x_3, \ldots) = (x_1, 0, 0, \ldots, ) = x_1e_1$$
$S$ is bounded and positive. We have:
$$1 = \langle Sx,x\rangle = x_1^2$$
which implies $x_1 \in \{-1, 1\}$.
Therefore, $U = \{(x_1, x_2, \ldots) \in \ell^2 : x_1 \in \{-1, 1\}\}$.
Note that $U$ is closed since $U = \langle S\cdot, \cdot\rangle^{-1}(\{1\})$.
The sets $\{(-1, x_2, x_3, \ldots ) \in \ell^2\}$ and $\{(1, x_2, x_3, \ldots ) \in \ell^2\}$ are disjoint and both closed in $\ell^2$ so they are in particular closed in $U$.
A: Let $F = (\ker S)^{\perp}$. Then $S\lvert_F$ is positive definite, and the function $r \colon M \to (0,+\infty)$ given by $r(x) = \sqrt{\langle Sx,x\rangle}$, where $M$ is the unit sphere of $F$, is continuous.
Hence $W = U\cap F = \{ x/r(x) : x \in M\}$ is homeomorphic to $M$, and since unit spheres in complex normed spaces are connected, $W$ is connected [with the caveat that some people consider $\varnothing$ disconnected, for them we would need to exclude $S = 0$]. And since
$$\langle S(x+y),x+y\rangle = \langle Sx,x\rangle$$
for $y \in \ker F$, it follows that $U$ is homeomorphic to $(\ker S) \times W$, hence connected.
A: Hint (for finite dimensional spaces)
Using the spectral theorem, you can prove that $U$ is homeomorphic to a product of circles. As a circle is path connected and a product of path connected sets is path connected, $U$ is path connected. Hence connected.
