A cylinder in infinite-dimensional Hilbert space cannot be homeomorphic to a sphere

Question

Consider the $\ell^2$ complex Hilbert space.

Let $m\in \mathbb{N}^*$ be a fixed number, and set $$ S=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^m \frac{|x_n|^2}{n^2}=1\right\}.$$

I want to show that $S$ cannot be homeomorphic to $$ S(0,1)=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^\infty |x_n|^2=1\right\}.$$

I guess that $S$ is homeomorphic to $S_m\times \ell^2$ with $$S_m:=\{y\in \mathbb{C}^m; \|y\|=1\}$$

Answer 1

The sphere $S(0, 1)$ in the Hilbert space is contractible (Reference). I include the proof given by Loop Space for convenience:

1. $S(0, 1)$ is mapped onto its own equator $S(0, 1)\cap \{x_1=0\}$ by the forward shift $T$.
2. The map $T$ is homotopic to the identity map via the normalized straight-line homotopy $(tx + (1-t)Tx)/\|tx + (1-t)Tx\|$, $0\le t\le 1$.
3. The map $T$ is also homotopic to the constant map $Cx = (1, 0, 0, \dots)$, again via the normalized straight-line homotopy.

On the other hand, the set $S$ is homotopy equivalent to finite-dimensional sphere $S^m$ via the maps $$f:S\to S^m, \quad f(x) = (x_1/1, x_2/2, x_3/3, \dots, x_m/m)$$ $$g:S^m\to S, \quad g(x) = (1 x_1, 2x_2, 3x_3, \dots, mx_m, 0, 0, \dots)$$ Indeed, both $f$ and $g$ are continuous, $f\circ g$ is the identity on $S^m$, and $g\circ f$ is the projection $x\mapsto (x_1, \dots, x_m, 0, 0, \dots)$ which is homotopic to the identity via straight-line homotopy.

And a finite dimensional sphere is not contractible.

Answer 2

$S$ is homeomorphic to $S_m \times \ell^2$. Indeed, we consider the following map $$S\ni (x_1,\dots)\mapsto (x_1,2x_2,\dots,mx_m,x_{m+1},\dots)\in S_m \times \ell^2.$$