Suppose we have a $C^*$ algebra $A=\{(x_n)\in \prod M_n(\Bbb C),lim_n tr_n(x_n^*x_n)=0\}$.
If $B$ is any nonzero $C^*$-sub algebra of $A$,does there exist a finite dimensional irreducible representation of $B$?
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7$\begingroup$ That's not a C*-algebra ... $\endgroup$– Nik WeaverCommented Jan 5, 2019 at 15:11
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5$\begingroup$ @YemonChoi: it's not an algebra, in the same way that the set of zero trace matrices in $M_n$ isn't an algebra (the product of two matrices with zero trace could have nonzero trace). $\endgroup$– Nik WeaverCommented Jan 5, 2019 at 18:11
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$\begingroup$ @NikWeaver Oh of course, somehow my brain mentally put in the trace-norm of $x_n$ rather than the trace itself $\endgroup$– Yemon ChoiCommented Jan 5, 2019 at 20:13
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$\begingroup$ That makes more sense, and then I think you're right that it would be a C*-algebra. $\endgroup$– Nik WeaverCommented Jan 5, 2019 at 20:57
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1 Answer
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Yes. Irreducible representations are the building blocks. If there is a finite dimensional representation, there is a finite dimensional irrep. Any C*-subalgebra of $A$, indeed any C*-subalgebra of $\prod M_n$, has plenty of finite dimensional representations, namely evaluation on the $n$th factor, for any $n$.
answered Jan 6, 2019 at 5:21
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$\begingroup$ If $\phi$ is a finite representation of $A$,$B$ is a $C^*$ subalgebra of $A$,I think that there is a possibility that $\phi_{B}=0$ $\endgroup$ Commented Jan 6, 2019 at 6:36
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1$\begingroup$ If $B$ is not zero then it has to contain some element which is nonzero on some factor. Evaluation on that factor will not be zero on $B$. $\endgroup$ Commented Jan 6, 2019 at 13:43
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$\begingroup$ (Indeed, any nonzero element of $B$ will not be in the kernel of some finite dimensional irrep.) $\endgroup$ Commented Jan 6, 2019 at 15:48