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 $S(0,1)$?

  • $\begingroup$ 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) $\endgroup$ Feb 15, 2018 at 13:13
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    $\begingroup$ 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) $\endgroup$ Feb 15, 2018 at 14:21
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    $\begingroup$ @JochenGlueck: You don't like coffee? Then what do you turn into theorems? $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 22:55
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    $\begingroup$ @NikWeaver: chocolate, definitely. $\endgroup$ Feb 15, 2018 at 23:05
  • 4
    $\begingroup$ Upvoted for chocolate! $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 23:21

1 Answer 1


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.

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    $\begingroup$ If $M$ is strictly positive the problem is trivial: $M^{-1/2}$ takes $S$ homeomorphically onto $S_M$. $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 20:23
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    $\begingroup$ @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. $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 22:42
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    $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 22:44
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    $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Feb 15, 2018 at 22:52
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    $\begingroup$ @UserMaths: if you find that the answer given is correct, don't forget to mark it as "accepted". $\endgroup$
    – Nik Weaver
    Feb 16, 2018 at 8:12

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