# $S_M$ 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. May 20, 2018 at 15:00
• $M = 0$ satisfies your criteria, though it's probably not what you're looking for. May 20, 2018 at 15:52
• @ArunDebray Yes $M$ is assumed non zero operator May 20, 2018 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 May 23, 2018 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, 2018 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. Nov 16, 2018 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, 2018 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 Nov 16, 2018 at 10:56