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For which simple unital $C^*$ algebras does the following functional equation have a solution:

$$ d^2=0,\;{(d+d^*)}^2=1$$

The Calkin algebra and $M_{2n}(\mathbb{C})$ are some examples. It is not solvable in $M_{2n+1}(\mathbb{C})$. What infinite dimensional simple $C^*$ algebra does not admit an element $d$ with the above property?

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    $\begingroup$ @BranimirĆaćić but is not k theiry stable not depending of the size matrices? $\endgroup$ – Ali Taghavi Apr 30 '18 at 22:24
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    $\begingroup$ OK, let me clean up my stupid mess to avoid confusion. Here's a cheap observation. If such a $d$ exists in $A$, then $d^\ast d$ and $dd^\ast$ are Murray–Von Neumann equivalent projections, such that $d^\ast d + dd^\ast = 1_A$. Hence, a necessary condition for $d$ to exist is that $[1_A] \in K_0(A)$ is divisible by $2$. For example, by the isomorphism $\operatorname{Tr} : K_0(M_N(\mathbb{C})) \to \mathbb{Z}$, we find $d$ exists in $M_N(\mathbb{C})$ only if $[1_{M_N(\mathbb{C})}] \in K_0(M_N(\mathbb{C}))$ is divisible by $2$, if and only if $\operatorname{Tr}([1_{M_N(\mathbb{C})}]) = N$ is even. $\endgroup$ – Branimir Ćaćić May 1 '18 at 0:43
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    $\begingroup$ It is an exercise to see that the complex $*$-algebra generated by an element $d$ satisfying the relations above is $M_2(\mathbb C)$. Hence, any $*$-algebra contains such an element if and only if it contains a unital copy of $M_2(\mathbb C)$. $\endgroup$ – Andreas Thom May 1 '18 at 10:52
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    $\begingroup$ @AliTaghavi: The universal $C^*$-algebra generated by a contraction $d$ with $d^2=0$ can be understood by observing that $Z=dd^*+d^*d$ is central in the $*$-algebra generated by $d$. Hence, in any irreducible representation $Z$ will act as a scalar $t \in [0,2]$. This and the observation for $t=1$ implies that the universal $C^*$-algebra (which can be decomposed over its center) is the non-unital algebra $M_2(C_0((0,1]))$ where the generator $d$ corresponds to the matrix with one non-zero entry $t \in C_0((0,1])$ in the upper right corner. $\endgroup$ – Andreas Thom May 1 '18 at 21:14
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    $\begingroup$ In particular, the spectrum of $dd^*+d^*d$ is a subset of $[0,1]$ if $d$ is a contraction with $d^2=0$. $\endgroup$ – Andreas Thom May 1 '18 at 21:16
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If such a $d$ exists in $A$, then $d^\ast d$ and $d d^\ast$ are Murray–Von Neumann equivalent projections, such that $d^\ast d + d d^\ast = 1_A$. Hence, a necessary condition for $d$ to exist is that $[1_A] \in K_0(A)$ is divisible by $2$. For instance, this rules out $M_{2n+1}(\mathbb{C})$, since $\operatorname{Tr}([1_{M_{2n+1}(\mathbb{C})}]) = 2n+1$ is odd. With regard to your actual question, this also rules out any UHF algebra $A$ such that $\tfrac{1}{2} \notin K_0(A) \leq (\mathbb{Q},+)$ (i.e., the inductive limit of a sequence $M_{k_1}(\mathbb{C}) \to M_{k_2}(\mathbb{C}) \to \cdots$, where each $k_i$ is odd) or any irrational rotation algebra: in either case, you have have a simple unital $C^\ast$-algebra $A$ together with a normalised trace $\tau$, such that

  1. $K_0(\tau) : K_0(A) \to \mathbb{R}$ is injective;
  2. $K_0(\tau)([1_A]) = 1$ (since $\tau$ is normalised);
  3. $\tfrac{1}{2} \notin K_0(\tau)(K_0(A))$.
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