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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.

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Schüler
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1 Answers1

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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 $\mathbb{Z}$.

erz
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  • 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 06:59
  • 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
  • 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 – Schüler Nov 16 '18 at 10:56
  • I don't know of any – erz Nov 16 '18 at 11:11
  • According to this link: https://en.m.wikipedia.org/wiki/Contractible_space The $1$-shere of $l^2$ is always contactible. On the other hand $S_M(0,1)$ is homeomorphic to $S^1\times l^2$. Could you please help me show that $S^1\times l^2$ is not contactible? – Schüler Nov 29 '18 at 14:36
  • It's not contractible because it's not simply connected. The latter claim is proven in the answer. – erz Nov 30 '18 at 01:12