# 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) – Jochen Glueck Feb 15 '18 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) – Jochen Glueck Feb 15 '18 at 14:21
• @JochenGlueck: You don't like coffee? Then what do you turn into theorems? – Nik Weaver Feb 15 '18 at 22:55
• @NikWeaver: chocolate, definitely. – Jochen Glueck Feb 15 '18 at 23:05
• Upvoted for chocolate! – Nik Weaver Feb 15 '18 at 23:21

## 1 Answer

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 $S_M$. – Nik Weaver Feb 15 '18 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. – Nik Weaver Feb 15 '18 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. – Nik Weaver Feb 15 '18 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. – Nik Weaver Feb 15 '18 at 22:52
• @UserMaths: if you find that the answer given is correct, don't forget to mark it as "accepted". – Nik Weaver Feb 16 '18 at 8:12