# Gelfand-Naimark-Segal representations and pure states

In my functional analysis course, we proved the following theorem.

Theorem (Gelfand-Naimark-Segal construction). Let $$A$$ be a $$C^*$$-algebra and $$\rho \in S(A)$$ a state on $$A$$. Then there exists a Hilbert space $$L^2 (A, \varphi)$$ and a unique (up to equivalence) representation $$\pi : A \to \mathcal{B} (L^2 (A, \varphi))$$ and a unit cyclic vector $$1_{\varphi}$$ such that $$\varphi(x) = \langle \pi(x) 1_{\varphi} , 1_{\varphi} \rangle,\quad \forall x \in A.$$

Then we had the following proposition.

Proposition. A state $$\varphi \in S(A)$$ is pure iff a GNS representation $$\pi_{\varphi}: A \to \mathcal{B} (L^2 (A, \varphi))$$ with a cyclic vector $$1_{\varphi}$$ is irreducible.

I need help understanding the proof of the left implication in the proposition.

Proof $$(\Leftarrow)$$ We assumed that a representation $$\pi_{\varphi}: A \to \mathcal{B} (L^2 (A, \varphi))$$ is irreducible and $$\varphi = \frac{1}{2} \varphi_1 + \frac{1}{2} \varphi_2$$ for some states $$\varphi_1, \varphi_2 \in S(A)$$. Our goal is to show that one of the states $$\varphi_1, \varphi_2$$ is zero. First we defined a linear map $$U: L^2 (A, \varphi) \to L^2 (A, \varphi_1) \oplus L^2 (A, \varphi_2),\quad \pi_\varphi(x) 1_\varphi \mapsto \frac{1}{\sqrt{2}} \pi_{\varphi_1}(x) 1_{\varphi_1} \oplus \frac{1}{\sqrt{2}} \pi_{\varphi_2} (x) 1_{\varphi_2}.$$ Next, we showed that this $$U$$ preserves scalar product: $$\langle \pi_\varphi (x) 1_\varphi, \pi_\varphi (y) 1_\varphi \rangle = \langle U \pi_\varphi (x) 1_\varphi, U \pi_\varphi (y) 1_\varphi \rangle$$ and it intertwines: $$U \pi_{\varphi}(x) (\pi_{\varphi} (y) 1_{\varphi}) = \left(\pi_{\varphi_1} (x) \oplus \pi_{\varphi_2} (x)\right) U (\pi_{\varphi} (y) 1_{\varphi}).$$ This is all good so far. Then we defined an orthogonal projection $$p_1 : L^2 (A, \varphi_1) \oplus L^2 (A, \varphi_2) \to L^2 (A, \varphi_1) \oplus L^2 (A, \varphi_2)$$ onto the first factor. Then my professor claimed that the map $$U^* p_1 U \in \mathcal{B}(L^2 (A, \varphi))$$ commutes with $$\pi_{\varphi} (x)$$ for any $$x \in A$$ and that this is trivial. This is where I need help. To prove this, I think we need to prove that the map $$U U^*$$ commutes with $$p_1$$, which I'm unable to do.

Any help is very appreciated. I know the notation is burdensome and varies greatly from author to author, so I'll be glad to clear it up. Thanks in advance.

Ok, so you proved that all $$x\in A$$ $$U\pi_\varphi(x)=(\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x))U\tag{1}$$ Starring both sides of $$(1)$$ gives $$(\pi_\varphi(x))^* U^* = U^*(\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x))^*\tag{2}$$ but $$\begin{split} (\pi_\varphi(x))^* =\pi_\varphi(x^*)\\ (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x))^* = (\pi_{\varphi_1}(x))^*\oplus(\pi_{\varphi_2}(x))^*=\pi_{\varphi_1}(x^*)\oplus\pi_{\varphi_2}(x^*) \end{split}$$ So from $$(2)$$ $$\pi_\varphi(x^*)U^* = U^*(\pi_{\varphi_1}(x^*)\oplus\pi_{\varphi_2}(x^*)\tag{3})$$ and $$(3)$$ holds for all $$x\in A$$, so substituting $$x$$ for $$x^*$$ we get $$\pi_\varphi(x)U^* = U^*(\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x))\tag{4}$$ for all $$x\in A$$.
Next, for all $$x\in A$$ and all $$(s,t)\in L^2(A,\varphi_1)\oplus L^2(A,\varphi_2)$$ $$\begin{split} (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) p_1 (s,t) = (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) (s,0) = (\pi_{\varphi_1}(x)s, 0)\\ p_1 (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) (s,t) = p_1 (\pi_{\varphi_1}(x) s, \pi_{\varphi_2}(x)(t)) = (\pi_{\varphi_1}(x)s, 0) \end{split}$$ so $$(\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) p_1 = p_1 (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x))\tag{5}$$ Finally, using $$(1), (4)$$ and $$(5)$$ \begin{align} \pi_\varphi(x)U^* p_1 U&=U^*(\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) p_1 U\\ &=U^* p_1 (\pi_{\varphi_1}(x)\oplus\pi_{\varphi_2}(x)) U\\ &=U^* p_1 U\pi_\varphi(x) \end{align}