This question is Exercise 2.10 in Rordam's K-Theory for C$^{*}$-Algebras book. Let $p,q$ be projections in an infinite-dimensional separable Hilbert space.

Show that $p$ and $q$ are unitarily equivalent (i.e., $\exists$ unitary $u\in H$ with $q=upu^{*}$) if and only if $\dim(p(H))=\dim(q(H))$ and $\dim(p(H)^{\perp})=\dim(q(H)^{\perp})$.

This question was asked once before here, but no answer was given.

I have solved the first part of the problem, which asks to show that $p$ and $q$ are equivalent in the Murray von-Neumann sense iff $\dim(p(H))=\dim(q(H))$. Since in a C$^{*}$-algebra unitary equivalence implies Murray von-Neumann equivalence, the forward direction reduces to showing $\dim(p(H)^{\perp})=\dim(q(H)^{\perp})$. But I am stuck in excluding the case where $\dim(p(H))$ and $\dim(q(H))$ are co-finite with $\dim(H)\setminus\dim(p(H))\not=\dim(H)\setminus\dim(q(H))$ I have thought about this question for hours, but I am at a loss as to how to proceed. I would really appreciate any help with completing the forward implication and with obtaining the reverse implication.

Thank you.

  • $\begingroup$ These are hermitian projections, right? $\endgroup$
    – s.harp
    May 17 '17 at 17:41
  • $\begingroup$ Yes. They are hermetian. $\endgroup$
    – ervx
    May 17 '17 at 17:41
  • $\begingroup$ Find Hilbert basises and of $p(H)$, $q(H)$ and extend these to Hilbert basises of all of $H$. Can you see how to define a sensible $U$? $\endgroup$
    – s.harp
    May 17 '17 at 17:44
  • $\begingroup$ I thought about doing this, but did not figure out how to define $U$. $\endgroup$
    – ervx
    May 17 '17 at 17:51
  • $\begingroup$ How to prove $p,q$ are equivalent iff $dim(p(H))=dim(q(H))$? $\endgroup$
    – math112358
    Feb 4 '19 at 17:02

Let $\{e_i\mid i\in I_1\}$ be a Hilbert basis of $p(H)$ and $\{b_i\mid i\in I_1\}$ be a Hilbert basis of $q(H)$, note that the index set is the same, this follows from $\mathrm{dim}\,p(H)=\mathrm{dim}\,q(H)$.

Extend these to Hilbert basises $\{e_i\mid i\in I_2\}$ and $\{b_i\mid i\in I_2\}$. Again the index sets are the same, which follows from $\mathrm{dim}\,(1-q)(H) = \mathrm{dim}\,q(H)^\perp=\mathrm{dim}\,p(H)^\perp=\mathrm{dim}\,(1-p)(H)$.

The map $$U:H\to H,\quad \sum_{i\in I_2} x_i e_i\mapsto \sum_{i\in I_2} x_i b_i$$ is unitary. Note that: $$Up\left(\sum_{i\in I_2} x_i e_i\right) = U\left(\sum_{i\in I_1} x_i e_i\right)=\sum_{i\in I_1}x_ib_i = q\left(\sum_{i\in I_2}x_i b_i\right)=qU\left(\sum_{i\in I_2}x_i e_i\right) $$

so $Up=qU$.

  • $\begingroup$ Thank you very much. Might you also tell me how to prove that remaining part of the forward implication; i.e., that if $p\sim_{u} q$, then $\dim(p(H))^{\perp}=\dim(q(H))^{\perp}$? $\endgroup$
    – ervx
    May 17 '17 at 18:42
  • $\begingroup$ I was thinking of maybe defining an isomorphism from $(1-q)(H)\to(1-p)(H)$, using the fact that the two projections are unitarily equivalent, but I'm not sure how to do this, or if this is the way to go. $\endgroup$
    – ervx
    May 17 '17 at 18:43
  • 1
    $\begingroup$ From the hermiticity of $p$ and $q$ you know that $(1-q)(H)=q(H)^\perp$ and the same with $p$. Further then $U(1-p)(H)=(U-Up)(H)=(U-qU)(H)=(1-q)U(H)$ so the unitary $U$ also sends $p(H)^\perp$ to $q(H)^\perp$. Unitary maps preserve dimension of subspaces. $\endgroup$
    – s.harp
    May 17 '17 at 19:51
  • $\begingroup$ Thank you so much for all your help. I just figured that last part out a minute before you typed it! $\endgroup$
    – ervx
    May 17 '17 at 19:53

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