# $S_M$ is not always homeomorphic to the 1-sphere of $F$

Let $F$ be a infinite-dimensional complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$, the norm $\|\cdot\|$, the 1-sphere $S(0,1)=\{x\in F;\;\|x\|=1\}$ and let $\mathcal{B}(F)$ be the algebra of all bounded linear operators on $F$.

Let $M\in \mathcal{B}(F)$ be a bounded operator. Suppose

• that $M\in \mathcal{B}(F)^+$, i.e., $\langle Mx,x\rangle\geq0$ for all $x\in F$, and

• that $M$ is an injective operator on $F$.

Consider $$S_M(0,1)=\{x\in F:\;\langle Mx, x\rangle=1\}.$$

According to this answer $S_M(0,1)$ is always homeomorphic to the 1-sphere $S(0,1)$.

If $M$ is not injective ($M\ne 0$), I want to find an example such that $S_M(0,1)$ is is not homeomorphic to the 1-sphere of $F$ denoted $S(0,1)$.

I think if $F$ is an infinite-dimensional complex Hilbert space and if we find an operator $M$ such that $S_M(0,1)$ is compact then $S_M(0,1)$ is not homeomorphic to $S(0,1)$. Indeed $S(0,1)$ is compact iff $F$ is finite-dimensional.

• If $M$ is e. g. a projection, then if I am not mistaken $S_M(0,1)$ is the cartesian product of the 1-sphere for the subspace you project onto and the orthogonal complement of that subspace. – მამუკა ჯიბლაძე May 20 '18 at 15:00
• $M = 0$ satisfies your criteria, though it's probably not what you're looking for. – Arun Debray May 20 '18 at 15:52
• @ArunDebray Yes $M$ is assumed non zero operator – Schüler May 20 '18 at 16:04

Taras Banakh's answer to your original question essentially answers this one too. Take $F=l^2$ and take $M$ to be the projection on the first to coordinates. Then $S_M(0,1)=\{(a_1,a_2,a_3,...)\in l^2,|a_1|^2+|a_2|^2=1\}$, which is homeomorphic to $S^1\times l^2$, where $S^1$ - the usual circle.
Then the unit sphere of $l^2$ and $S^1\times l^2$ are not homeomorphic, since the former is simply connected (easy to see), and the latter is not: its fundamental group is the product of the fundamental groups of $S^1$ and $l^2$ and is therefore isomorphic to $\mathbb{Z}$.
• Could you please explain me why the unit sphere of $l^2$ and $S^1\times l^2$ are not homeomorphic?Thanks a lot – Schüler May 23 '18 at 14:35
• I kind of do explain in the body of the answer: the unit sphere of $l^2$ is simply connected, while $S^1\times l^2$ is not. – erz May 23 '18 at 23:44
• Could you please explain me why the unit sphere of $l^2$ is simply connected however $S^1\times l^2$ is not? Thanks a lot. – Schüler Nov 16 '18 at 6:59
• The explanation of the second claim is already written in the answer. The unit sphere of $l^{2}$ is simply connected, because of stereographic projection: draw a loop, take a point not on that loop, stereographically project the sphere on $l^2$ with respect to that point, deform the obtained loop to a point, then project the homotopy back. I hope this explanation makes sense. – erz Nov 16 '18 at 10:23