# Unitary Equivalence of GNS representations

Consider a unital $C^*$-algebra $\mathfrak A$ and a state $\omega$ on it. Let $\mathcal N := \{A\in\mathfrak A : \omega(A^*A)=0 \}$. Construct a Hilbert space $\mathcal H:= \overline{\mathfrak A /\mathcal N}$ with the inner product defined by: $\langle \psi_A,\psi_B \rangle \equiv \omega(A^*B)$ where $\psi_A, \psi_B \in \mathcal H$. Construct further a GNS representation $\pi:\mathfrak A \to \mathcal L(\mathcal H)$ defined by: $\pi(A)\psi_B\equiv\psi_{AB}.$ The vacuum state is defined as $\Omega := \psi_1$ where $1 \in \mathfrak A$ is the unit.

Given two such GNS representations $(\mathcal H_i,\pi_i,\Omega_i)$ with $i=1,2$ of $\mathfrak A$, I want to prove that the map $U: \mathcal H_1 \to \mathcal H_2$ defined by $U\pi_1(A)\Omega_1:=\pi_2(A)\Omega_2 \ \forall A \in \mathfrak A$ is unitary. Note that $U\pi_1(A)\Omega_1=\pi_2(A)\Omega_2$ is the same as saying $U\psi^{(1)}_A = \psi^{(2)}_A$ where $\psi^{(1)}_A \in \mathcal H_1$ and $\psi^{(2)}_A \in \mathcal H_2 \, .$

If I could only prove that $\psi^{(1)}_A = U^*\psi^{(2)}_A,$ my job is done. How do I do that?

How would you prove the unitary equivalence of these representations?

## NOTE:

Plugging $A=1$ in $U\psi^{(1)}_A = \psi^{(2)}_A$ implies that $\Omega_2 = U \Omega_1$ which in turn implies from the original condition that $(U\pi_1(A)-\pi_2(A)U)\Omega_1=0 \Rightarrow U\pi_1(A)=\pi_2(A)U$. Then my question becomes very similar to this one. However, there is one crucial difference. I do not assume irreducibility of the representations at any point.

• I suggest not using $i$ as an index when dealing with C$^*$-algebras (or anything Hilbert spaces), since we already use $i$ for a certain complex number. – Martin Argerami May 8 '17 at 2:49

That $U$ is linear, isometric, and surjective, is trivial. Irreducibility has nothing to do with it, but note that one requires that the two GNS representations satisfy that $\pi_i(\mathfrak A)\Omega_i$ is dense in $\mathcal H_i$.
The only nontrivial issue is whether $U$ is well-defined. That is, you want to show that if $\pi_1(A)\Omega_1=\pi_1(B)\Omega_1$, then $\pi_2(A)\Omega_2=\pi_2(B)\Omega_2$. This requires that the two GNS representations are minimal, that is, that $\pi_i(\mathfrak A)\Omega_i$ is dense in $\mathcal H_i$.
We have, if $\pi_1(A)\Omega_1=\pi_1(B)\Omega_1$, \begin{align} \|\pi_2(A)\Omega_2-\pi_2(B)\Omega_2\|^2&=\langle \pi_2(A)\Omega_2,\pi_2(A)\Omega_2\rangle+\pi_2(B)\Omega_2,\pi_2(B)\Omega_2\rangle-2\text{Re}\,\langle \pi_2(A)\Omega_2,\pi_2(B)\Omega_2\rangle\\ \ \\ &=\langle\pi_2(A^*A)\Omega_2,\Omega_2\rangle+\langle\pi_2(B^*B)\Omega_2,\Omega_2\rangle-2\text{Re}\,\langle\pi_2(B^*A)\Omega_2,\Omega_2\rangle\\ \ \\ &=\omega(A^*A)+\omega(B^*B)-2\text{Re}\,\omega(B^*A)\\ \ \\ &=\|\pi_1(A)\Omega_1-\pi_1(B)\Omega_1\|^2\\ \ \\ &=0. \end{align} So $\pi_2(A)\Omega_2=\pi_2(B)\Omega_2$, and so $U$ is well-defined.
• I wanted to prove the unitarity of $U$ which I believe follows directly from construction: $\langle \psi^{(2)}_A,\psi^{(2)}_B \rangle = \omega(A^*B) = \langle \psi^{(1)}_A,\psi^{(1)}_B \rangle.$ Anyway, the representations are minimal by construction as well: $\pi_i(\mathfrak A)\Omega_i = \{ \psi^{(i)}_A : A \in \mathfrak A\}$ is a pre-Hilbert space. But where did you use this denseness property in the above analysis? – Nanashi No Gombe May 9 '17 at 8:06
• From the construction you get that $U$ is a symmetry; you still need to show that it is surjective. And the representations are not minimal "by construction": if both were constructed as in your first paragraph, there would be nothing to prove; the point is to show that two representations where the inner product is given by $\omega$ are unitarily equivalent, and you need the minimality for that. – Martin Argerami May 9 '17 at 12:31
• Your "unitary" is, by definition, an isometry. But you need it to be surjective. If the representation is not minimal, there is no reason for your $U$ to be surjective. – Martin Argerami May 9 '17 at 15:39