Inspired by this MSE question we ask the following question:

Is there a noncommutative $C^*$ algebra $A$ for which the following identity hold 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.

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.

Inspired by this MSE question we ask the following question:

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

$$e^{(xy-yx)}= e^xe^y e^{-x}e^{-y}$$ that is $$e^{[x,y]}=[e^x,e^y]$$

Inspired by this MSE question we ask the following question:

Is there a noncommutative $C^*$ algebra $A$ for which the following identity hold 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.

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

$$e^{(xy-yx)}= e^xe^y e^{-x}e^{-y}$$ that is $$e^{[x,y]}=[e^x,e^y]$$

Inspired by this MSE question we ask the following question:

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

$$e^{(xy-yx)}= e^xe^y e^{-x}e^{-y}$$ that is $$e^{[x,y]}=[e^x,e^y]$$