2
$\begingroup$

I have a simple question I've managed to get myself quite confused about.

Given a Hilbert space H, what do we know about the cardinality of

(a) the set $P(H)$ of projection operators onto $H$ (equivalently, the set of all closed subspaces of $H$),

and

(b) the set of all Boolean subalgebras of $P(H)$?

I imagine that if $H$ is separable, we can say more than we could otherwise.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.

So in finite dimension, $\mathrm{card}(P(H)) = \sum_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.

In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(P(H))=\mathrm{card}(U(H))=\mathrm{card}(B(H))=\max(\dim(H),\mathrm{card}(\mathbb{R}))$.

So regardless of dimension, $P(H)$ has cardinality $\max(\dim(H), 2^{\aleph_0})$.

Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $2^{2^{\mathrm{card}(P(H))}}$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks Chris. That's really helpful. But there's one thing I don't understand. By $\mathbb{R}$, do you mean the reals? Because it seems weird that for a finite dimensional Hilbert space, P(H) should be uncountable. Or have I misunderstood? $\endgroup$
    – King Kong
    Commented Jan 15, 2014 at 19:07
  • 2
    $\begingroup$ @BenEva A Hilbert space of finite dimension 2 or more has continuum many projection operators. For example, the Euclidean plane has a projection operator to each line through the origin (plus the zero and identity operators). $\endgroup$
    – Andreas Blass
    Commented Jan 15, 2014 at 22:30
  • $\begingroup$ Sorry, one more question. Dosen't this mean that if H is an infinite dimension seperable hilbert space, then $card(P(H))$ = $dim(H) = card(\mathbb{N})$, and so an infinite dimensional seperable hilbert space has fewer closed subspaces than a finite dimensional hilbert space? $\endgroup$
    – King Kong
    Commented Jan 16, 2014 at 16:18
  • $\begingroup$ Yes, sorry, I was being sloppy there, edited it now. $\endgroup$
    – Chris Heunen
    Commented Jan 16, 2014 at 16:53
  • $\begingroup$ ah, thanks, makes perfect sense now. $\endgroup$
    – King Kong
    Commented Jan 16, 2014 at 16:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .