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Let $A$ be a $C^*$-algebra. Does there exist a non-trivial positive linear functional $\nu\in A^*$ which is $\mathrm{Aut}(A)$-invariant? That is, $\nu\circ\alpha=\nu$ for all $\alpha\in\mathrm{Aut}(A)$. Note that $\nu$ does not have to be bounded.

Question: Is there a reference where this question is addressed for wide classes of infinite-dimensional $C^*$-algebras?

I am specifically looking for references. Thank you.

Discussion: If $A=M_n$ (matrices of dimension $n\in\mathbb{N}$) then $\nu(a)=\mathrm{tr}[a]$ is one such. If $A$ is commutative then $A\simeq C_0(X)$ for a locally compact Hausdorff space $X$, and the question boils down to a Radon measure on $X$ which is $\mathrm{Homeo}(X)$-invariant. If $X$ is discrete then the counting measure does the job. If $X\simeq(a,b)\subset\mathbb{R}$ then the answer is no. What happens in between?

Bedovlat
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    The answers to Hausdorff spaces with trivial automorphism group give examples in the commutative case, since some of the spaces mentioned there are also compact. – Tobias Fritz Mar 04 '24 at 21:03
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    @TobiasFritz those seem to be spaces with trivial automorphism group, in which case any positive linear functional would do. – Nik Weaver Mar 04 '24 at 21:12
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    There are $C^\ast$-algebras with unique traces. For example any UHF algebra, or any II$_1$ factor. So traces on such algebras would be examples. – David Gao Mar 05 '24 at 19:20
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    For an unbounded example, the canonical trace on $B(H)$ for an infinite-dimensional $H$ would work. Or if $M$ is a II$_1$ factor with trivial fundamental group, then any trace on $M \mathbin{\bar{\otimes}} B(l^2)$ would do. – David Gao Mar 05 '24 at 19:23
  • @YemonChoi I did not say that $\nu$ cannot be unbounded. I said quite the contrary: $\nu$ can be unbounded. Such is the case when $A$ is not unital. – Bedovlat Mar 06 '24 at 09:58
  • @Bedovlat I made a typo in my previous comment. What I meant to say is: since $\nu$ can be unbounded, what does it mean for it to be Aut($A$)-invariant? Are you requiring the ideal of definition of $\nu$ to be Aut($A$)-invariant? (perhaps this comes for free) – Yemon Choi Mar 06 '24 at 20:28
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    @YemonChoi All right, if you view $\nu:A^+\to[0,+\infty]$ as a $(0,+\infty)$-linear functional, then its $\mathrm{Aut}(A)$-invariance does yield the invariance of the ideal. – Bedovlat Mar 07 '24 at 12:37

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