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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?

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    $\begingroup$ 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. $\endgroup$ Commented Mar 4 at 21:03
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    $\begingroup$ @TobiasFritz those seem to be spaces with trivial automorphism group, in which case any positive linear functional would do. $\endgroup$
    – Nik Weaver
    Commented Mar 4 at 21:12
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    $\begingroup$ 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. $\endgroup$
    – David Gao
    Commented Mar 5 at 19:20
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    $\begingroup$ 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. $\endgroup$
    – David Gao
    Commented Mar 5 at 19:23
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    $\begingroup$ @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. $\endgroup$
    – Bedovlat
    Commented Mar 7 at 12:37

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