2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
3
$\begingroup$

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 $Up=qU$.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.