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I am applying the GNS construction to the algebra $M_{N \times N}$ of all $N\times N$ matrices on $\mathbb{C}$. I have the state $\omega$ given by $$\omega(A) = Tr(\rho A),$$ for $A\in M_{N\times N}$, with $\rho = \rho^* \in M_{N\times N}$ and $\rho = diag(\lambda_1, \ldots, \lambda_N)$, $\sum_i \lambda_i =1, \ \lambda_i \geq 0$.

I have deduced (please correct me if I'm wrong), but the kernel of the above state: $\ker(\omega):= \{A\in M_{N \times N} : \forall B \in M_{N \times N} \implies \omega(B^*A) = \omega(A^*B)=0 \}$ is the set: $$\ker(\omega) = \{A\in M_{N \times N} : A^2 = 0\}.$$

By the GNS construction, we then look to the quotient space given by $M_{N \times N}/ \ker (\omega)$. Let $\xi_A$ denote the equivalence class of $A$, then we need to construct a representation $\pi_{\omega}$ such that we have:

$$\pi_{\omega}(A) \xi_B = \xi_{AB}.$$

I'm a little unsure of how to do this, but I have two ideas:

  • $\pi_{\omega}(A)B = AB,$
  • $\pi_{\omega}(A)B = \sqrt{A}B.$

I'm not sure about the accuracy of either of these answers, but I need a little guidance on this technique.

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You are looking for the subspace $$ J=\{A:\ \omega(A^*A)=0\}, $$ which is not the kernel of $\omega$. If $\lambda_j>0$ for all $j$, this will be just $\{0\}$. You have $$ 0=\omega(A^*A)=\sum_j \lambda_j (A^*A)_{jj}=\sum_{\lambda_j>0}\lambda_j (A^*A)_{jj}. $$ So, for each $\lambda_j>0$, the $j,j$ entry of $A^*A$ has to be zero. This is equivalent to $A$ having its $j^{\rm th}$ row and column equal to zero. So, if $F=\{j:\ \lambda_j>0\}$ and $m=|F|$, then $$ M_N(\mathbb C)/J=M_m(\mathbb C). $$ The representation is indeed $$ \pi_\omega(A)B=AB, $$ but this needs to understood in the quotient. That is, as you wrote yourself, if $\xi_B$ denotes the class of $B$ in $M_N(\mathbb C)/J$, then $$ \pi_\omega(A)\xi_B=\xi_{AB}. $$

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  • $\begingroup$ Not sure where you are going with that. You can always make any degenerate representation non-degenerate by just cutting the irrelevant part of the codomain. If you are doing the canonical GNS construction, the representation is non-degenerate by construction. $\endgroup$ – Martin Argerami Nov 7 '18 at 20:23

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