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.

[this answer]: https://mathoverflow.net/questions/293025/are-the-following-subsets-of-a-hilbert-space-always-homeomorphic