Is there a simple $C^*$ algebra $A$, not isomorphic to $M_2(\mathbb{C})$, such that for every commutator element $x=ab-ba$, $x^2$ is an scalar element?
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1$\begingroup$ Well, over $\mathbb{R}$ the quaternions qualify, but if you complexify them you'll get $M_2(\mathbb{C})$ which you already ruled out. $\endgroup$– VincentCommented May 9, 2018 at 8:44
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$\begingroup$ @Vincent Yes I mean complex $C^*$ algebra. In particular I wonder whether there is an infinite dimensional one? $\endgroup$– Ali TaghaviCommented May 9, 2018 at 9:03
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3$\begingroup$ Such an algebra satisfies the identity $[[X,Y]^2,Z]$ and hence is a PI-algebra (Polynomial Identity). Simplicity probably implies that it's primitive (a little doubt, because of the topological vs purely algebraic setup). In encyclopediaofmath.org/index.php/PI-algebra it is said that a primitive PI-algebra is a matrix algebra over a skew-field, which would entail a negative answer to your question. $\endgroup$– YCorCommented May 9, 2018 at 10:17
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1$\begingroup$ @YCor I learn a lot from your great comment. however the topological simplicity is not equivalent to the algebraic one. I know a non unital example: The algebra of compact operators is a simple $C^*$ algebra but it has dense ideal of finite rank operator or trace class operatores or Hilbert Schmidt operators. $\endgroup$– Ali TaghaviCommented May 10, 2018 at 9:13
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1$\begingroup$ Yes I know, this is why my comment is rather a sort of expectation. Still, topological simplicity implies the existence of a (topologically) irreducible faithful Hilbert module. Then one would need a substitute for the Jacobson density theorem, it does not sound hopeless to me. $\endgroup$– YCorCommented May 10, 2018 at 9:31
1 Answer
The answer is no:
A simple C*-algebra in which $[x,y]^2$ is central for all $x,y$ is isomorphic to $M_n(\mathbf{C})$ for $n\in\{1,2\}$.
Let $A$ be a C*-algebra. Let $H$ be an irreducible representation of $A$. By irreducibility its commutant is reduced to scalars, and hence its double commutant is equal to the whole algebra $B(H)$, and the double commutant is knows (von Neumann's bicommutant theorem) to be the strong operator topology ($\mathsf{SOT}$) closure.
Recall that an associative algebra $A$ is a PI-algebra if it satisfies some nontrivial identity $\forall x\in A^n,$ $F(x_1,\dots,x_n)=0$, where $F$ is nonzero in the free associative algebra $\mathbf{C}\langle X_1,\dots,X_n\rangle$ - $F$ can be then chosen homogeneous.
In the above setting, we have a continuous algebra homomorphism with dense image $A\to (B(H),\mathsf{SOT})$. If $A$ satisfies the identity $F$ (say of degree $d$) so is $B(H)$. This implies (by a theorem of Kaplansky, see §2.1 in Procesi's book) that $H$ has dimension $\le d/2$. Note that in the precise case in consideration, we have $F(x,y,z)=[[x,y]^2,z]$, which is clearly not an identity in $B(H)$ when $\dim(H)\ge 3$.
Now if $A$ is a simple C*-algebra, not reduced to scalars, then $A$ admits a nontrivial irreducible representation (cf §2.5 in Dixmier's book), in a Hilbert space $H$. Since $A$ is simple, this is a faithful representation. If $A$ is a PI-algebra, this forces, by the above, $H$ to be finite-dimensional, so $A$ is a matrix algebra.
This yields the complement result:
A simple C*-algebra is a PI-algebra if and only if it is isomorphic to a matrix algebra.
This follows from Kaplansky's theorem, but also directly follows if one can embed the free associative $\mathbf{C}$-algebra on countably many variables into $B(H)$ for $H$ separable infinite-dimensional.
Actually from Dixmier's book, §2.9, the intersection of kernels of irreducible representations is reduced to 0. Hence one gets a stronger result:
A C*-algebra that is a PI-algebra admits a family of finite-dimensional representations of bounded dimension, whose intersection of kernels is trivial.
(In the case of the identity $[[x,y]^2,z]$ these are of dimension $\le 2$. I'm not sure if one can strengthen the latter conclusion to a more explicit description, but in any case this is beyond what is asked, since the simple case is clear-cut now.)
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$\begingroup$ Thank you for your great answer. BTW, according to your first statement, I think "central" is equivalent to "scalar" since the algebra is simple. $\endgroup$ Commented May 13, 2018 at 19:49
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1$\begingroup$ Possibly but "central" is enough and is the good point of view to develop the argument. $\endgroup$– YCorCommented May 13, 2018 at 21:10