# Are the following subsets of a Hilbert space always homeomorphic?

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 $$S_M(0,1)=\{x\in F:\;\langle Mx, x\rangle=1\}.$$ Is $$S_M(0,1)$$ always homeomorphic to the 1-sphere $$S(0,1)$$?

• Just a vague idea: of course the question is only non-triivial if $0$ is a spectral value of $M$. If we would like to show that $S(0,1)$ and $S_M(0,1)$ are not homeomorphic, we need a topological invariant. The space $S(0,1)$ is a Baire space, but I would suspect that $S_M(0,1)$ is not. Yet, I haven't found a rigorous argument for this intuition yet... (nor am I sure that my intuition is correct) Feb 15, 2018 at 13:13
• My first comment is of course non-sense. $S_M(0,1)$ is a closed subspace of a complete metric space, and thus complete itself; in particular, $S_M(0,1)$ is a Baire space. It seems that I need more coffee, too... (but unfortunately I don't like coffee) Feb 15, 2018 at 14:21
• @JochenGlueck: You don't like coffee? Then what do you turn into theorems? Feb 15, 2018 at 22:55
• @NikWeaver: chocolate, definitely. Feb 15, 2018 at 23:05
• Upvoted for chocolate! Feb 15, 2018 at 23:21

The topological equivalence of the set $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ and the unit sphere $S:=\{x\in F:\|x\|=1\}$ can be proved as follows.

The assumptions on $M$ and the spectral theorem (or just the equality $\langle Mx,x\rangle=\langle \sqrt{M}x,\sqrt{M}x\rangle$) imply that $\langle Mx,x\rangle>0$ for any non-zero vector $x\in F$. Then the map $$h:S\to S_M,\;h:x\mapsto \frac{x}{\sqrt{\langle Mx,x\rangle}},$$ is a homeomorphism with inverse $$h^{-1}:S_M\to S,\;\;h^{-1}:y\mapsto \frac{y}{\|y\|}.$$

Acknowledgement. I would like to thank Nik Weaver for his very helpful comments, which allowed to simplify the initial answer (which infolved a powerful machinery of infinite-dimensional topology) to the present (almost) trivial form.

Added after comments of @MathUsers: For a positive (but not necessarily injective) opeartor $M$ on a Hilbert space $F$ the set $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ is homeomorphic to the product of the sphere in a Hilbert space and a Hilbert space (of suitable dimensions). This can be shown as follows.

Using the Spectral Theorem, show that the Hilbert space $F$ admits an orthogonal projector $P:F\to Y$ onto its subspace $Y\subset F$ such that $M=M\circ P=P\circ M$ and $\langle My,y\rangle>0$ for every $y\in Y\setminus\{0\}$. Write $F$ as the orthogonal sum $F=X\oplus Y$ where $X=P^{-1}(0)$ is the kernel of the projector $P$. Let $S_Y=\{y\in Y:\|y\|=1\}$ be the unit sphere in the Hilbert space $Y$.

Theorem. The space $S_M:=\{x\in F:\langle Mx,x\rangle=1\}$ is homeomorphic to $X\times S_Y$.

Proof. The map $$h:X\times S_Y\to S_M,\;\;h:(x,y)\mapsto x+\frac{y}{\sqrt{\langle My,y\rangle}}$$is a homeomorphism with the inverse $$h^{-1}:z\mapsto (z-Pz,\tfrac{Pz}{\|Pz\|}).$$

Corollary. The space $S_M$ is homeomorphic to the unit sphere $S$ in $F$ if and only if the positive opeartor $M$ has infinite-dimensional range.

• If $M$ is strictly positive the problem is trivial: $M^{-1/2}$ takes $S$ homeomorphically onto $S_M$. Feb 15, 2018 at 20:23
• @TarasBanakh: about your last remark, if $\langle Av,v\rangle$ is real for all $v$ then $A$ is self-adjoint. This is a consequence of the polarization identity. Feb 15, 2018 at 22:42
• I assumed "strictly positive" meant that $0$ is not in the spectrum. If all you mean is that it has no kernel, that is the same as saying it is injective, which the OP did assume. Feb 15, 2018 at 22:44
• It's the same. (Use the spectral theorem to assume $M$ is a multiplication operator.) But that's good, it means you're free to assume this. Feb 15, 2018 at 22:52
• @UserMaths: if you find that the answer given is correct, don't forget to mark it as "accepted". Feb 16, 2018 at 8:12