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](https://mathoverflow.net/questions/293025/are-the-following-subsets-of-a-hilbert-space-always-homeomorphic) $S\_M(0,1)$ 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.



$S_M$ is not always homeomorphic to the 1-sphere of $F$
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 $\mathbb{Z}$.


