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Inspired by this MSE question we ask the following question:

Is there a noncommutative $C^*$-algebra $A$ for which the following identity holds for all $x,y \in A$?

$$e^{(xy-yx)}= e^xe^y e^{-x}e^{-y}$$

That is $$e^{[x,y]}=[e^x,e^y]$$ where the bracket on the left-hand side is the algebra commutator, and the bracket on the right-hand side denotes the group commutator.

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    $\begingroup$ Thanks for the clarification. I have reworded the title in case anyone makes the same error as I did $\endgroup$
    – Yemon Choi
    Commented Oct 13, 2019 at 18:46
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    $\begingroup$ I would expect that the Baker-Campbell-Hausdorff formula could be used to say that $A$ is commutative; failing that, that for every $X, Y$ we have that $[X, Y]$ commutes with $X$ and $Y$. Have you tried it? $\endgroup$
    – user44191
    Commented Oct 13, 2019 at 21:05
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    $\begingroup$ Taylor at order 3 $[+O(\|x\|^4+\|y\|^4)]$ yields that a unital Banach algebra with this identity has to be 2-step nilpotent (that is $[x,y]$) is central for all $x,y$); indeed it says that $[x,[x,y]]$ is zero, which by polarization gives the result. $\endgroup$
    – YCor
    Commented Oct 13, 2019 at 21:12
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    $\begingroup$ @YCor I think it may be worth noting that the Jacobi identity is needed, that is, that polarization doesn't quite give the result directly. The derivation is straightforward, but not immediate. $\endgroup$
    – user44191
    Commented Oct 14, 2019 at 4:04
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    $\begingroup$ Conversely BCH for a unital Banach algebra satisfying $[x,[x,y]]=0$ identically writes as $\exp(x)\exp(y)=\exp(x+y+\frac12[xy])$ and this gives the commutator identity. So a unital Banach algebra satisfies the given commutator identity iff its underlying Lie algebra is 2-step nilpotent. What remains is to determine whether a $C^*$-algebra whose underlying Lie algebra is 2-step nilpotent is necessarily commutative. $\endgroup$
    – YCor
    Commented Oct 14, 2019 at 8:45

1 Answer 1

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Yes:

A $C^*$-algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ iff it is commutative.

This follows from two independent facts (I write $[x,y]=xy-yx$)

1) A (real/complex) unital Banach algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ $\Leftrightarrow$ it satisfies the identity $[x,[x,y]]=0$ $\Leftrightarrow$ it satisfies the identity $[x,[y,z]]$, i.e., its underlying Lie algebra is 2-step nilpotent.

2) A $C^*$-algebra satisfying the identity $[x,[x,y]]=0$ is commutative.

For (1), one obtains that the main implication, namely the identity implies $[x,[x,y]]=0$, by a 3rd order Taylor expansion: in every Banach algebra one has $$e^{xy-yx}=1+xy-yx+o(\|x\|^3+\|y\|^3);$$ $$e^xe^ye^{-x}e^{-y}=1+xy-yx+\frac12([x+y,[x,y]])+o(\|x\|^3+\|y\|^3),$$ so the commutator identity forces $[x+y,[x,y]]=0$ for all $x,y$, and hence $[x,[x,y]$ identically vanishes by homogeneity.

(As mentioned in the comments, a simple argument then implies $[x,[y,z]]=0$ for all $x,y,z$, but this is not needed to run the argument.)

Conversely, if $[x,[x,y]]$ is identically zero, the Baker-Campbell-Hausdorff formula reads as: for all $x,y$, one has $\exp(x)\exp(y)=\exp(x+y+(1/2)[x,y])$. The commutator identity follows.

Let us pass to (2). Let $A$ be a $C^*$-algebra satisfying the PI-identity $[x,[x,y]]=0$. First, if $A$ is finite-dimensional, it is a product of matrix algebras, and hence is commutative (since the given identity fails for $M_n(\mathbf{C})$ for $n\ge 2$). In general, I need the last emphasized statement in my answer here: $A$ has a family $(J_i)$ of finite-codimensional 2-sided closed $*$-ideals with trivial intersection (so $A$ embeds into the product $\prod_i A/J_i$). By the finite-dimensional case, $A/J_i$ is commutative for every $i$. Hence $A$ is commutative.

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  • $\begingroup$ I think your 3rd order Taylor expansion can't be right; there should be some symmetry between $x$ and $y$, which should force a term of the form $[x, [x, y]]$; this can also be seen by looking at the inverse (which takes the form $e^y e^x e^{-y} e^{-x}$). This can be dealt with by slightly more careful manipulation of the coefficients of $x$ and $y$, though. $\endgroup$
    – user44191
    Commented Oct 14, 2019 at 16:18
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    $\begingroup$ @user44191 Thanks! I had actually selected one of two homogeneous terms in my draft, then forgot the other one. It indeed yields $[x+y,[x,y]]$. Corrected. $\endgroup$
    – YCor
    Commented Oct 14, 2019 at 16:45

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