Is there an element in an infinite tensor power of a C*-algebra that is invariant under finite permutations?

Question

Let $A$ be a C*-algebra, and consider the infinite tensor power $A^{\otimes {J}}$, where $J$ is infinite (we consider the minimal or maximal tensor product). To any finite permutation, which is a bijection $\sigma\colon J\to J$ fixing all but finitely many elements of $J$, we can associate a *-homomorphism $\hat{\sigma}\colon A^{\otimes {J}} \to A^{\otimes J}$ which sends the $i$-th copy of $A$ (in $A^{\otimes J}$) to the $\sigma(i)$-th copy of $A$.

Is there an element $x \in A^{\otimes J}$, other than $\lambda 1$ with $\lambda \in \mathbb{C}$, such that $\hat{\sigma}(x)=x$ for every finite permutation?

($\lambda 1$ is the trivial case: $\hat{\sigma}(\lambda 1) = \lambda \hat{\sigma}(1) = \lambda 1$).

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From what I could do, it seems there is no such element, but I am unable to work out the details. Here are some inherent remarks.

1. If we consider $A^{\odot J}$, the pre-C*-algebra given by the algebraic infinite tensor power, every element $x$ is represented in some $A^{\odot F}$ with $F$ finite (i.e., it's an element of the form $x \odot 1 \odot 1 \odot\dots$). Therefore, any $x \ne \lambda 1$ cannot satisfy $\hat{\sigma}(x)=x$.
2. Let $J= \mathbb{N}$. Consider $x \in A$, and imagine to construct $x^{\otimes \mathbb{N}}$. If $\lVert x \rVert<1$, we notice that $x^{\otimes n} \to x^{\otimes \mathbb{N}}$ as $n$ increases, so $\lVert x^{\otimes n}\rVert \to \lVert x^{\otimes \mathbb{N}}\rVert$. However, $\lVert x^{\otimes n} \rVert= \lVert x\rVert^n \to 0$: we conclude that $x^{\otimes N} =0$, which is a trivial case.

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Some references:

Bruce E. Blackadar. Infinite tensor products of C*-algebras. Pacific Journal of Mathematics, 72(2):313–334, 1977.

Tobias Fritz and Eigil Fjeldgren Rischel. Infinite products and zero-one laws in categorical probability. Compositionality 2. 2020.

Answer 1

Thanks to the comments of @DiegoMartinez and @CalebEckhardt, I can answer my question.

Briefly, the answer is no: any $x$ satisfying the hypothesis is of the form $\lambda 1$.

Let us consider $x \in A^{\otimes J}$, where $J$ is infinite, such that $\hat{\sigma}(x) = x$ for every finite permutation $\sigma$. As suggested by @DiegoMartinez, we can proceed as follows. By definition of $A^{\otimes J}$, there is a sequence $a_n \in A^{\otimes F_n}\subset A^{\otimes J}$, where $F_n$ is a finite set, such that $a_n \to x$. For clarity, let us consider $\epsilon_n>0$ such that $\lVert x-a_n \rVert \le \epsilon_n$ and $\epsilon_n \to 0$.

We pick a finite permutation $\sigma$ such that $\sigma(F_n)\cap F_n=\emptyset$. Then $$\lVert a_n - \hat{\sigma}(a_n)\rVert\le \lVert a_n - x + \hat{\sigma} (x) - \hat{\sigma}(a_n) \rVert \le \lVert a_n - x \rVert + \lVert \hat{\sigma} \rVert \lVert x-a_n \rVert \le 2 \epsilon_n,$$ where we used $\hat{\sigma}(x) = x$ and $\lVert \hat{\sigma} \rVert=1$.

We now adopt a strategy similar to the one suggested by @CalebEckhardt. For any state $\Phi\colon A^{\otimes J} \to \mathbb{C}$, we notice that $$\lVert \Phi(\hat{\sigma}(a_n)) - \Phi(x)\rVert \le \epsilon_n$$ for any finite permutation $\sigma$ (because $x= \hat{\sigma}(x)$).

Let us fix $\Phi$. Using the CPU map $\Psi := \Phi_{\mid A^{\otimes \sigma(F_n)}} \otimes \operatorname{Id}_{A^{\otimes J\setminus \sigma(F_n)}} \colon A^{\otimes J} \to A^{\otimes J\setminus \sigma(F_n)}$, we obtain $$\lVert a_n - \Phi(\hat{\sigma}(a_n))1 \rVert=\lVert \Psi (a_n -\hat{\sigma}(a_n)) \rVert\le\lVert a_n -\hat{\sigma}(a_n) \rVert\le 2\epsilon_n.$$

Finally, $$ \lVert x- \Phi(x)1 \rVert \le \lVert x-a_n\rVert + \lVert a_n - \Phi(\hat{\sigma}(a_n))1 \rVert + \lVert \Phi(\hat{\sigma}(a_n))1 - \Phi(x)1 \rVert \le 4 \epsilon_n \to 0,$$ from which we conclude that $x=\Phi(x)1$.