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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 $S(0,1)$?

Community
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Schüler
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  • 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
  • I don't think so, consider the operator on $\ell^2$ which maps an infinite number of coordinates to zero, while leaving an infinite number of them unchanged. – Not Mike Feb 15 '18 at 13:19
  • @Not Mike: The operator you are suggesting is not injective. – Jochen Glueck Feb 15 '18 at 13:25
  • @JochenGlueck Thanks for pointing that out. Clearly I need more coffee. – Not Mike Feb 15 '18 at 13:29
  • So $0$ is in the spectrum, but is not an eigenvalue. How about this example ... $F = L^2[0,1]$ and $M$ is multiplication, like this: $(Mx)(t) = tx(t)$ for $t \in [0,1]$. – Gerald Edgar Feb 15 '18 at 14:20
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    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
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    @JochenGlueck: You don't like coffee? Then what do you turn into theorems? – Nik Weaver Feb 15 '18 at 22:55
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    @NikWeaver: chocolate, definitely. – Jochen Glueck Feb 15 '18 at 23:05
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    Upvoted for chocolate! – Nik Weaver Feb 15 '18 at 23:21

1 Answers1

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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.

Taras Banakh
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    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
  • @NikWeaver Why is $M^{-1/2}$ continuous on $S$? For example, $M$ can be a compact strictly positive operator... – Taras Banakh Feb 15 '18 at 22:39
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    @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
  • @NikWeaver Thank you. Exactly this was written on StackExchange. Then I will remove this my Remark. – Taras Banakh Feb 15 '18 at 22:44
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    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
  • @NikWeaver By "strictly positive" I understand an operator $M$ such that $\langle Mx,x\rangle>0$ for all non-zero $x$. I hope this is stronger than just injectivity and positivity of $M$? Or it is the same? – Taras Banakh Feb 15 '18 at 22:48
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    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
  • If it is the same, then all my proof above strongly simplifies as $V$ is just $F\setminus{0}$! I asked exactly this and UserMaths answered that his operator is not strictly positive so I had to invent Lemma and Claim. Maybe instead of two conditions on $M$ in OP to write one: $\langle Mx,x\rangle>0$ for all non-zero $x$. This wil strongly simplify the answer. – Taras Banakh Feb 15 '18 at 22:56
  • @NikWeaver Thank you for the comment (concerning the Spectral Theorem). I have rewritten and simplified my answer. Now I am thinking, may be there exists some explicit homeomorphism between $S$ and $S_M$ (not involving Torunczyk's Theorem)? – Taras Banakh Feb 15 '18 at 23:10
  • @NikWeaver I have rewritten the answer. Too simple. Maybe something wrong? – Taras Banakh Feb 15 '18 at 23:17
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    That ... actually looks right. Nice! – Nik Weaver Feb 15 '18 at 23:33
  • @NikWeaver Then my initial proof could serve as an answer to this OP :) https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts – Taras Banakh Feb 15 '18 at 23:39
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    Thank you for your answer. I think we don't need spectral theorem. Since $M$ is postive and injective so $M$ is strictly positive. – Schüler Feb 16 '18 at 06:26
  • @UserMaths Why? Is it seen without spectral theorem? Why then your wrote 2 conditions on $M$ instead of one ($\langle Mx,x\rangle>0$?) – Taras Banakh Feb 16 '18 at 06:30
  • @UserMaths And look at your first comment (answering my question if your operator is strictly positive): you wrote that it is not. – Taras Banakh Feb 16 '18 at 07:00
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    @TarasBanakh: I think there has been some confusion over the term "strictly positive", which I believe you are using in a slightly nonstandard way. UserMaths is right, though: if $\langle Mx,x\rangle = 0$ for some nonzero $x$ then $|M^{1/2}x|^2 = \langle Mx,x\rangle = 0$, so that $M^{1/2}x = 0$ and therefore $Mx = M^{1/2}M^{1/2}x = 0$. That is, if $M$ is not strictly positive in your sense then it is not injective. So you don't really need the spectral theorem. – Nik Weaver Feb 16 '18 at 08:10
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    @UserMaths: if you find that the answer given is correct, don't forget to mark it as "accepted". – Nik Weaver Feb 16 '18 at 08:12
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    @NikWeaver Thank you for all these explanations. They were very helpful. Truly speaking, the last time I had contacts with Spectral Theory of Operators was on undergraduate studies in the university 30 years ago. On the other hand, in my Lviv University we still have a group of good specialists (including R.Hryniv and Ya.Mykytyuk), actievly working in this field, but I never had opportunity to realize what are they doing. This group is concentrated around a seminar in Functional Analysis supervised by (now dead) V.Lyantse who met Stefan Banach (being a student in 40-ies). – Taras Banakh Feb 16 '18 at 09:15
  • If $M$ is not injective, I hope to get an example under which $S(0,1)$ and $S_M(0,1)$ are not homeomorphic. And thank you very much. – Schüler Feb 17 '18 at 06:29
  • I try with the following: If we consider the $\ell^2$ complex Hilbert space, and the operator $$M \boldsymbol x=(x_1, x_2/4, x_3/9, \ldots, x_s/s^2,0,0,\ldots), $$ so that $$ S_M=\left{ \boldsymbol x\in \ell^2\ :\ \sum_{n=1}^s \frac{|x_n|^2}{n^2}=1\right}.$$ If we show that this set is compact so it cannot be homeomorphic to $S$. – Schüler Feb 17 '18 at 06:56
  • @UserMaths Your $S_M$ is not compact, but it is homeomorphic to the product of the sphere in $\mathbb R^n$ with $\ell^2$, so not homeomorphic to $S(0,1)$. – Taras Banakh Feb 17 '18 at 07:05
  • @TarasBanakh Thank you. I hope that we can include in your answer an example of $M$ which will make $S_M(0,1)$ compact and thus not homeomorphic to $S(0,1)$. – Schüler Feb 17 '18 at 07:08
  • @UserMaths It is seems that for positive operators $M$ on a Hilbert space it is possible to obtain a complete topological classification of the spaces $S_M(0,1)$: if $M$ has range of finite dimension $n$, then $S_M(0,1)$ is homeomorphic to ${x\in\mathbb R^n:|x|=1}\times F$; if $M$ has infinite-dimensional range, then $S_M(0,1)$ is homeomorphic to $F$ and hence to $S(0,1)$. – Taras Banakh Feb 17 '18 at 07:10
  • @TarasBanakh Thank you for your edit. I think that if $M$ has range of finite dimension $n$, then $S_M(0,1)$ is homeomorphic to ${x∈C^n:∥x∥=1}×F$ because $F$ is a complex Hilbert space – Schüler Feb 18 '18 at 08:51
  • @UserMaths Yes, you are right! Thank you for your comment. I will make a correction. – Taras Banakh Feb 18 '18 at 09:10
  • A little correction to your last statement: if $M$ has $1$-dimensional range then the homeomorphism also exists, since $S_M$ is a hyperplane in this case. – erz May 20 '18 at 22:42
  • @erz I am afraid that for 1-dimensional M the space $S_M$ consists of two parallel hyperplanes (so it is not connected). – Taras Banakh May 21 '18 at 06:06
  • Oh yes, sorry about that. – erz May 21 '18 at 06:43