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.

  • 1
    $\begingroup$ 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. $\endgroup$ May 20, 2018 at 15:00
  • $\begingroup$ $M = 0$ satisfies your criteria, though it's probably not what you're looking for. $\endgroup$ May 20, 2018 at 15:52
  • $\begingroup$ @ArunDebray Yes $M$ is assumed non zero operator $\endgroup$
    – Schüler
    May 20, 2018 at 16:04

1 Answer 1


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 $\mathbb{Z}$.

  • $\begingroup$ Could you please explain me why the unit sphere of $l^2$ and $S^1\times l^2$ are not homeomorphic?Thanks a lot $\endgroup$
    – Schüler
    May 23, 2018 at 14:35
  • $\begingroup$ 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. $\endgroup$
    – erz
    May 23, 2018 at 23:44
  • $\begingroup$ 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. $\endgroup$
    – Schüler
    Nov 16, 2018 at 6:59
  • $\begingroup$ 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. $\endgroup$
    – erz
    Nov 16, 2018 at 10:23
  • $\begingroup$ Thank you. Is there a reference to cite it?i.e. A reference where i find the resultat that explain why the unit sphere of a hilbert space is simply connected. Thanks $\endgroup$
    – Schüler
    Nov 16, 2018 at 10:56

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