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Let $A$ be a $C^*$-algebra, $\phi$ a state on $A,$ and $\pi$ the GNS representation associated with $\phi.$

Can it be said in general that $\sup_{a\in A}\frac{\|\pi(a)\|^2}{\phi(a^*a)}=M<\infty$?

If not, can we find increasing, continuous $u:[0,\infty)\rightarrow[0,\infty)$ s.t. $u(0)=0$ and $\|\pi(a)\|\leq u(\phi(a^*a))$ for all $a\in A$? Notice the above is the case $u(t)=Mt.$

This is a pointed version of this question. I'm interested because I want to show that $\phi(x^*x)$ is a weakly stable predicate in this language I'm working with (model theory nonsense).

Miles Gould
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  • Also, just out of curiosity, what are weakly stable predicates? I’ve never seen that terminology when I was studying the model theory of operator algebras before. – David Gao Jun 20 '25 at 05:44
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    @DavidGao It’s a notion from the model theory of metric structures, and its main use is proving that zero-sets of certain predicates are definable sets. In this case, I wanted $L_\phi$ to be definable in the language of $C^*$-algebras with states. This would have been sufficient to axiomatize “$\phi$ is pure.” I haven’t been on MSE in a while, this is what I’ve been learning. Note: It’s very closely related to weakly stable relations if you’ve heard of those. – Miles Gould Jun 20 '25 at 14:44
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    I haven’t heard of weakly stable relations either, but I have learned continuous model theory before. So the goal is, just to spell it out, to show the operator norm distance from $L_\phi$ can be uniformly approximated by formulas in the language of $C^\ast$-algebras together with an additional predicate for the state $\phi$, correct? – David Gao Jun 20 '25 at 15:15
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    (If $L_\phi$ is definable, then $\sup_x |d_\infty(x, L_\phi) - \phi(x^\ast x)^{1/2}|$ would axiomatize $\phi$ being pure. I suppose that’s the main goal?) – David Gao Jun 20 '25 at 15:41
  • @DavidGao That wasn't the predicate I was thinking of, but that one is probably better! And yes, that is the goal. – Miles Gould Jun 20 '25 at 15:59
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    I can write down more details if you post a question specifically about this, but just for a quick answer: Using the fact that $\phi$ being pure is equivalent to $d_\infty(x, L_\phi) = \phi(x^\ast x)^{1/2}$ and the ultraproduct characterization of axiomatizability, one can indeed show $\phi$ being pure is axiomatizable, though it’s not clear to me what exactly are the axioms achieving this. $L_\phi$ is not definable in general, though. For example, you can have a sequence of projections whose $2$-norm converge to $0$ w.r.t. a faithful trace. – David Gao Jun 20 '25 at 16:37
  • @DavidGao That’s alright, I already showed that this class is axiomatizable by showing it’s closed under isomorphisms, ultraproducts, and ultraroots. I specifically was hoping to find definable predicates which axiomatize it. – Miles Gould Jun 20 '25 at 16:45
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    I see. Well, as I mentioned, $L_\phi$ is not definable, so that won’t work. But I think I have figured out an axiom that probably works:$$\sup_x\inf_y(||y|^2-\phi(x^\ast x)|+\phi((y-x)^\ast(y-x)))$$I haven’t checked all the details (and don’t have the time to do so right now), but I’m fairly confident this should work. – David Gao Jun 20 '25 at 17:35
  • @DavidGao I’m not sure why this would be equivalent to φ pure, any hints? – Miles Gould Jun 21 '25 at 16:27
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    Suppose $\phi$ is pure. Then the natural map $A/L_\phi\to H_\phi$ is an isometric isomorphism. So, for any $\epsilon>0$, $x\in A$ of norm at most $1$, there exists $y\in A$ of norm at most $1$ s.t. $\hat{x}=\hat{y}$ and $||y|^2-\phi(x^\ast x)|<\epsilon$. This shows the axiom holds for pure $\phi$. – David Gao Jun 23 '25 at 02:21
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    For the converse, suppose $\phi$ is not pure. There there exists a nontrivial projection $p\in\pi(A)’$. Consider the vector $p\hat{1}$, which must have norm strictly between $0$ and $1$ (since $\hat{1}$ is cyclic for $\pi$). For any $\epsilon>0$, there exists $x\in A$ s.t. $|\hat{x}-p\hat{1}|<\epsilon$. If the axiom holds, then there exists $y\in A$ with $|\hat{y}-\hat{x}|<\epsilon$ and $||y|-|\hat{x}||<\epsilon$. This means $|y|<|p\hat{1}|+2\epsilon$ and $|\hat{y}-p\hat{1}|<2\epsilon$. But $p\hat{y}=\pi(y)p\hat{1}$, so $|y|\geq|\pi(y)|\geq\frac{|p\hat{y}|}{|p\hat{1}|}$… – David Gao Jun 23 '25 at 02:35
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    … We also have $|p\hat{y}-p\hat{1}|=|p(\hat{y}-p\hat{1})|\leq|\hat{y}-p\hat{1}|<2\epsilon$, so,$$|p\hat{1}|+2\epsilon>|y|\geq\frac{|p\hat{y}|}{|p\hat{1}|}>\frac{|p\hat{1}|-2\epsilon}{|p\hat{1}|}$$But then letting $\epsilon\to0$ yields $|p\hat{1}|\geq1$, a contradiction. – David Gao Jun 23 '25 at 02:40
  • @DavidGao Sadly the converse doesn't work. In fact, "$\phi$ is pure" is not axiomatizable. If it were, an ultrapower $(B,\psi)$ of such a pair $(A,\phi)$ would also have $\psi$ pure. However, take $(A,\phi)=(\mathbb{C},\iota)$ where $\iota$ is the identity map. Then $\psi$ is faithful on $B$ (and satisfies your axiom) but $\mathbb{C}$ is the only $C^*$-algebra with a faithful pure state, a contradiction. It appears my proof that the class is closed under ultraproducts fails somewhere. – Miles Gould Jun 30 '25 at 03:46
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    It does not fail. All ultrapowers of $(\mathbb{C},\iota)$ are just itself. There is no contradiction. (Also, even if the ultrapower were not $\mathbb{C}$, I don’t see what is the contradiction you’re claiming. That being pure is axiomatizable certainly does not imply being faithful and pure is axiomatizable - it happens to be just because being $(\mathbb{C},\iota)$ happens to be axiomatizable, but that’s more of an accident. Just being faithful is certainly not axiomatizable.) – David Gao Jun 30 '25 at 05:58
  • @DavidGao I see now, I was terribly mistaken about ultrapowers of $\mathbb{C}$. – Miles Gould Jun 30 '25 at 07:09

3 Answers3

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Can it be said in general that $\sup_{a\in A}\frac{\|\pi(a)\|^2}{\phi(a^*a)}=M<\infty$

No, not even in finite dimension. Let $A=M_2(\mathbb C)$ and $\phi(a)=a_{11}$. Then your $M$ is not defined on $\mathbb C\,E_{22}$. And since $\pi$ is faithful (because $A$ is simple) you have $$ \frac{\|\pi(a)\|^2}{\phi(a^*a)}=\frac{\|a^*a\|}{\phi(a^*a)}. $$ If you take $a=\varepsilon E_{11}+E_{22}$ then $$ \frac{\|\pi(a)\|^2}{\phi(a^*a)}=\frac{1}{\varepsilon^2}, $$ and you can do this for any $\varepsilon>0$. This example will also fail with any continuous non-negative $u$ such that $u(0)=0$, as you will get $$ \frac1{u (\varepsilon^2)}\xrightarrow[\varepsilon\to0]{}\infty. $$

Martin Argerami
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  • Thank you for you answer! I decided to reward the problem to David for going above and beyond (as usual), but I love a good explicit example! – Miles Gould Jun 20 '25 at 14:50
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This basically never happens. $\phi(a^\ast a) \leq \|\pi(a)\|^2$ is clear, so what you want (even the weaker version - which is really equivalent to the a priori stronger version by normalizing $a$) is just the two seminorms $\|\pi(\cdot)\|$ and $a \mapsto \phi(a^\ast a)^{1/2}$ being equivalent. But this implies two things:

  1. $\pi(A)$ is linearly homeomorphic to a Hilbert space, namely $H_\phi$;
  2. $\phi$ is faithful on $\pi(A)$, i.e., $\pi(a) \neq 0$ implies $\phi(a^\ast a) \neq 0$.

The first condition can never happen unless $\pi(A)$ is finite-dimensional. Indeed, $\pi(A)$ being linearly homeomorphic to a Hilbert space will imply it is a dual space and thus a von Neumann algebra. An infinite-dimensional von Neumann algebra necessarily contains a copy of $\ell^\infty$, but this cannot happen for a space linearly homeomorphic to a Hilbert space.

Even if $\pi(A)$ is finite-dimensional, the second condition needs not hold, since there are plenty of non-faithful states on, say, matrix algebras, as in the example in Martin’s answer. Of course, if $\pi(A)$ is finite-dimensional and $\phi$ is faithful on $\pi(A)$, then the desired result does hold since $\pi(a) \mapsto \phi(a^\ast a)^{1/2}$ would be a norm on $\pi(A)$ in that case and any two norms on a finite-dimensional space are equivalent. So, your desired result is equivalent to $\pi(A)$ being finite-dimensional and $\phi$ being faithful on $\pi(A)$.

David Gao
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In the GNS representation, we have $\phi(a^*a) = \|\pi(a)\xi_\phi\|^2$ where $\xi_\phi$ is the cyclic vector. So the question I think becomes: is there a uniform bound on $\frac{\|\pi(a)\|^2}{\|\pi(a)\xi_\phi\|^2}$?

Part 1: Linear bound fails

This we can do away with using a counterexample. Consider $A = C([0,1])$ with the integration state $\phi(f) = \int_0^1 f(t) dt$. The GNS representation has $L^2([0,1])$ as the Hilbert space, $\pi(f)$ acts as multiplication by $f$, and the cyclic vector is essentially the constant function $1$.

For small $\epsilon > 0$, define $f_\epsilon$ to be a continuous function that equals $1/\sqrt{\epsilon}$ on $[0,\epsilon]$ and $0$ elsewhere. Then:

  • $\|\pi(f_\epsilon)\| = \|f_\epsilon\|_\infty = 1/\sqrt{\epsilon}$
  • $\phi(f_\epsilon^*f_\epsilon) = \int_0^1 |f_\epsilon(t)|^2 dt = \int_0^\epsilon \frac{1}{\epsilon} dt = 1$

Therefore: $$\frac{\|\pi(f_\epsilon)\|^2}{\phi(f_\epsilon^*f_\epsilon)} = \frac{1/\epsilon}{1} = \frac{1}{\epsilon} \to \infty$$

This shows no uniform bound $M$ exists.

Part 2: General function bound also fails

From the construction above, as $\phi(a^*a) \to 0^+$, we can make $\|\pi(a)\|$ arbitrarily large. More precisely, by scaling the previous example, for any $t > 0$ we can find elements $a$ with $\phi(a^*a) = t$ but $\|\pi(a)\| \geq C/\sqrt{t}$ for arbitrarily large constants $C$.

Any continuous function $u$ with $u(0) = 0$ satisfying $\|\pi(a)\| \leq u(\phi(a^*a))$ would need $u(t) \geq C/\sqrt{t}$ near $t = 0$. But this contradicts $u(0) = 0$ and continuity at $0$.

WTF is going on

I think the problem is that cyclicity of $\xi_\phi$ doesn't control individual operator norms. While individual operators can have large norms in directions "orthogonal" to how they act on $\xi_\phi$.

Since both questions have negative answers in general, for weak stability of $\phi(x^*x)$ you may need:

  1. Restricted classes: The bounds might hold for specific C*-algebras (e.g., finite-dimensional ones).
  2. Alternative approaches: Work with the universal representation where $\|a\| = \sup_{\psi \text{ state}} \|\pi_\psi(a)\|$.
  3. Reformulation: Consider that both $\|\pi(a)\|$ and $\sqrt{\phi(a^*a)}$ are seminorms with the same null space, which might give the definability properties you need without explicit bounds.
arjo
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