4
$\begingroup$

The usual proof of the existence of an absolute value of a functional on a C*-algebra $A$ uses the polar decomposition of normal functionals on $A^{**}$, which relies on the compactness of the unit ball of $A^{**}$ in the weak*-topology.

Is it possible to derive the existence of an absolute value of a bounded linear functional on a C*-algebra via a compactness argument in $A^*$? For a definition of the absolute value of $\varphi \in A^*$, I mean a $\tau \in A^*_+$ such that $\|\tau\| = \|\varphi\|$ and $|\varphi(a)|^2 \leq \|\varphi\| \tau(a^* a)$.

Furthermore, is it possible to use this to show that $\varphi$ can be represented as $\varphi(a) = \langle \pi_\tau(a) \xi \,\vert\, \eta\rangle$ where $\pi_\tau$ is the GNS representation associated with $\tau$ and $\|\varphi\| = \|\xi\| \|\eta\|$?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I haven't worked out the details, but another way to think about absolute values is to use a $2\times 2$ matrix construction. What I have in mind is how you move from completely bounded maps to completely positive maps, by putting the cb map on the off diagonal, and using a Hahn-Banach argument to find cp maps on the diagonal which make the whole matrix CP. This allows you to use Stinespring to prove to representation result for CB maps. I think the same works for linear functionals, but I haven't written this out to check there is not a lurking circular argument... $\endgroup$
    – Matthew Daws
    Commented Jun 10, 2020 at 8:31
  • $\begingroup$ That does work, although you need to embed a proof that the CB norm of a linear functional is its norm, so it's not that much different than the full CB representation theorem. One other approach I tried that doesn't work is proving representation for self-adjoint functionals using Jordan Decomposition, and then trying to lift that to a self-adjoint functional in the 2x2 case. You do better than the naive estimate of $4 \|\varphi\|$, but you end up with a $\sqrt{2}$ factor in the norms of the vectors. $\endgroup$
    – Cameron Zwarich
    Commented Jun 10, 2020 at 14:02

1 Answer 1

Reset to default
2
$\begingroup$

This is maybe a "backwards" answer to what you might have been hoping for...

An affirmative answer to the 2nd question would give the 1st question as well. Indeed, if there is a $*$-representation $\pi:A\rightarrow B(H)$ and $\xi,\eta\in H$ with $\|\varphi\| = \|\xi\| \|\eta\|$ and $\varphi(a) = \langle \pi(a)\xi, \eta \rangle$, then by rescaling we may suppose that $\|\xi\|=\|\eta\|=\|\varphi\|^{1/2}$, and then $\tau:a\mapsto \langle \pi(a)\xi,\xi\rangle$ is a positive functional with $\|\tau\| = \|\xi\|^2 = \|\varphi\|$. Further, $$ \|\varphi\| \tau(a^*a) = \|\eta\|^2 \|\pi(a)\xi\|^2 \geq | \langle \pi(a)\xi, \eta \rangle|^2 = |\varphi(a)|^2. $$

I do not know a "compactness" argument which can show this. However, a while ago I wrote up an argument of how to prove Kaplansky Density using Arens products; see Notes on GitHub. To do this in a non-circular way, you need to use Hahn-Banach, and need to find a way to prove exactly this representation result, in a "simple" way. The way I did this was as follows.

Let $H$ be a Hilbert space. Let $B(H)_*$ be the trace-class operators on $H$, the predual of $B(H)$. By Hahn-Banach (Goldstine's Theorem) for $\mu\in B(H)^*$ there is a net $(\omega_i)$ in $B(H)_*$ converging weak$^*$ to $\mu$, and with $\|\omega_i\|=\|\mu\|$ for each $i$. Thus, for an ultrafilter $\mathcal U$ refining the order filter, the natural map $(B(H)_*)_{\mathcal U} \rightarrow B(H)^*$ (given by "take weak$^*$-limit") is a metric surjection. With $K = \ell^2(H)$, and $\pi_0:B(H)\rightarrow B(K)$ the "diagonal" map, for each $\omega\in B(H)_*$ we can find $\xi,\eta\in K$ with $\omega = \omega_{\xi,\eta}\circ\pi_0$. Now let $K = (\ell^2(H))_{\mathcal U}$ the ultrapower, a Hilbert space. We can find $\xi = (\xi_i), \eta=(\eta_i)\in K$ with $\omega_i = \omega_{\xi_i,\eta_i}\circ\pi_0$ for each $i$. Thus, with $\Pi:B(H)\rightarrow B(K)$ the diagonal of $\pi_0$, we have that $$ \langle \Pi(x)(\xi), \eta \rangle = \lim_{i\rightarrow\mathcal U} \langle \pi_0(x)\xi_i, \eta_i \rangle = \lim_{i\rightarrow\mathcal U} \omega_i(x) = \mu(x). $$

Given a $C^*$-algebra $A$, by the GNS construction, we can exhibit $A$ as a subalgebra of $B(H)$ for some $H$. For $\varphi\in A^*$ take a Hahn-Banach extension to $\mu\in B(H)^*$. Then the previous paragraph, with $\pi$ the restriction of $\Pi$ to $A$, gives the required representation.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I was going to update my question when I fully convinced myself I didn't understand, but there's a proof like the one I suggested in the middle of Theorem 1 in [ousar.lib.okayama-u.ac.jp/en/list/authors/T/Minoru,Tomita/item/… theory of operator algebras I) by Tomita. He says that since every bounded functional is a linear combination of states, you can find such a $\tau$, but maybe not one of the desired norm. Why can you find such a $\tau$ to begin with? It's also strange that he picks one of minimal norm, but never uses minimality... $\endgroup$
    – Cameron Zwarich
    Commented Jun 9, 2020 at 20:06
  • $\begingroup$ That aside, your proof seems like the morally correct one, so I'm going to accept it. There's a variant of this idea in Theorem 1.10.8 of Ilijas Farah's book Combinatorial Set Theory of C*-Algebras, which is actually attributed to Ozawa in the acknowledgements. However, it uses a Banach limit of functionals on the diagonal rather than a Hilbert space ultrapower, and hence needs a further step in the form of an inequality that is rather difficult to motivate. $\endgroup$
    – Cameron Zwarich
    Commented Jun 9, 2020 at 20:26
  • 1
    $\begingroup$ I don't know your motivation for an alternative proof. But maybe Bishop-Phelps theorem helps? $\endgroup$
    – Narutaka OZAWA
    Commented Jun 10, 2020 at 0:55

You must log in to answer this question.

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