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I am trying to understand a proof in this paper (specifically theorem 5.4). In it, a fact is used that every element of the $W^*$-algebra $A$ is a linear combination of exponential unitaries.

I've tried to reason this fact out, but I don't follow where this fact comes from or why it necessarily holds true. I do know that any element can be written as a sum of four unitaries (as explained here), but I can't seem to find any information about being a sum of specifically exponential unitaries, which is important to the logic of the proof.

Why is this true, and is this known specifically about $W^*$-algebras or is it a more general fact? Any information on this would be appreciated.

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    $\begingroup$ isn't every unitary $u$ also an exponential unitary $u=e^{ih}$, for some self-adjoint $h$? $\endgroup$
    – Carlo Beenakker
    Jan 25, 2023 at 9:19
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    $\begingroup$ It is true in any unital $C^*$-algebra. Any element $x$ can be written as a linear combination of self-adjoints, by $x = \tfrac{1}{2}(x+x^*) + \tfrac{1}{2i} i(x-x^*)$. Moreover, if $y$ is a self-adjoint element with $\| y\| \leq 1$, and $\arccos \colon [-1,1] \to [0,\pi]$, then $u = \exp(i \arccos(y))$ is an exponential unitary such that $y = \tfrac{1}{2} (u+u^*)$. By combining these two results one gets the desired fact. $\endgroup$
    – Jamie Gabe
    Jan 25, 2023 at 9:51
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    $\begingroup$ @JamieGabe Nice argument! Please post this as an answer so the question can be marked as solved. $\endgroup$
    – Robert Furber
    Jan 25, 2023 at 21:32
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    $\begingroup$ Carlo Beenakker's suggestion works in any W$^*$-algebra by the Borel functional calculus for normal elements, but interestingly it is not the case that for any C$^*$-algebra $A$, every unitary $u \in A$ is of the form $e^{ih}$ for a self-adjoint $h$. Consider the C$^*$-algebra $C(S^1)$, where $S^1$ is the unit complex numbers. For a self-adjoint $h \in C(S^1)$, the function $t \mapsto e^{i(1-t)h} : [0,1] \to C(S^1)$ defines a homotopy from $e^{ih}$ to $1$, so $e^{ih}$ has winding number zero. Therefore the unitary $z : S^1 \hookrightarrow \mathbb{C}$ is not $e^{ih}$ for any self-adjoint $h$. $\endgroup$
    – Robert Furber
    Jan 25, 2023 at 22:41

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It is true in any unital $C^\ast$-algebra. Any element $x$ can be written as a linear combination of self-adjoints, by $x=\tfrac{1}{2}(x+x^\ast)+\tfrac{1}{2i}i(x−x^\ast)$. Moreover, if $y$ is a self-adjoint element with $\| y\| \leq 1$, and $\arccos\colon [−1,1]\to [0,π]$, then $u=\exp(i \arccos(y))$ is an exponential unitary such that $y=\tfrac{1}{2}(u+u^\ast)$. By combining these two results one gets the desired fact.

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