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If $n=a+bi$ is the Cartesian decomposition of a complex number or a normal operator on a Hilbert space (in both cases $a=\frac{n+n^*}{2}$ and $b=\frac{n-n^*}{2i}$ and $a$ and $b$ commute), then $\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+sgn(b)\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}i$ is a square root of $n$. In the case where $n$ is an operator, the definition of the operator $sgn(b)$ proceeds by requiring that $sgn(b)\psi = b((b^2)^{1/2})^{-1}\psi$ for all $\psi \in \ker(b)^\perp$ while $sgn(b)\psi=\psi$ for all $\psi \in \ker(b)$. All in all, $sgn(b)$ commutes with every element $c$ that commutes with $b$ and $sgn(b)(b^2)^{1/2}=b$.

Now to my question: I don't know an elementary construction of $sgn(b)$ of a self-adjoint (or even positive) element $b$ in a general abstract $C^*$ algebra. Is an elementary definition of such an element possible? Is an elementary construction of a square root of a normal element possible?

5th decile
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  • Not every normal element of a $C^$ algebra admits a root. As an example consider the $C^$ algebra $C(S^1)$ and the element $z\mapsto z$. A root of this would be a continuous square root function on the unit circle, which doesn't exist. This problem of continuity can be lifted for example if you are looking at a von Neumann algebra. In that case you can apply a measurable (rather than just continuous) functional calculus to normal elements, and there exist many measurable roots. – s.harp Dec 27 '19 at 21:43
  • The expression I use selects the root with "positive real part"... I think you're missing something. – 5th decile Dec 27 '19 at 23:07
  • Anyway, I found the post https://math.stackexchange.com/questions/632521/detail-in-the-algebraic-proof-of-the-polar-decomposition?rq=1 which guided me to Sakai's book on Cstar algebras. He seems to give a construction of a $sgn(b)$-element e.g. for $b$ self-adjoint. I may write an answer to my own question at some point, using the insights found there. – 5th decile Dec 27 '19 at 23:10
  • Note that your goal is to construct a root of a normal element. What I'm saying is that this is not always possible (without increasing the size of the $C^*$-algebra). The root of a positive element $p$ always exists and can be given explicitly by the Riemann integral: $$\sqrt p = \frac1\pi \int_0^\infty d\lambda\ p(\Bbb1\lambda+p)^{-1}\lambda^{-1/2}$$ (Note that you need to pass to a unitisation of the algebra to evaluate this integral, but the integral itself lies in the original algebra.) – s.harp Dec 28 '19 at 05:44
  • Ok, it was me who was missing something. I think I need a unital $W^*$ algebra (i.e. the algebra is a dual Banach space to another Banach space) to construct/define $sgn(b)$ and the square root of a normal element. – 5th decile Dec 28 '19 at 11:55

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