# A connection between the Gelfand Naimark Theorem and the GNS construction

Let $$\mathcal{A}$$ be an abelian $$C^*$$ algebra. By the Gelfand Naimark theorem, we know $$\mathcal{A}$$ is isometrically $$*$$ isomorphic to $$\mathcal{C}_0(\Delta_{\mathcal{A}})$$ where $$\Delta_{\mathcal{A}}$$ is the space of non zero characters on $$\mathcal{A}$$. We have that $$\Delta_{\mathcal{A}}$$ is a locally compact Hausdorff space.

Now by the GNS construction, we know every $$C^*$$ algebra $$\mathcal{A}$$ admits a faithful representation into some $$\mathcal{B(H)}$$ where $$\mathcal{H}$$ is a Hilbert space i.e. $$\mathcal{A}$$ is a closed subalgebra of a $$\mathcal{B(H)}$$.

My question is whether for abelian $$C^*$$ algebras, can we say something more specific regarding the embedding $$\mathcal{B(H)}$$ using the Gelfand Naimark Theorem?

Note: I was thinking along the lines of getting the irreducible representations of $$\mathcal{A}$$ which are in direct correspondence with the pure states of $$\mathcal{A}$$ which are exactly the characters on $$\mathcal{A}$$ as it is abelian.

Thanks.

The characters of $$C_0(X)$$ are the maps $$\{ ev_x\mid x\in X\}$$ where $$ev_x: C_0(X)\to\Bbb C$$ is given by $$f\mapsto f(x)$$. As such the associated semi-definite inner-product $$(,)_{ev_x}$$ is given by: $$(f,g)_{ev_x}:= ev_x(f^* g) =\overline{f(x)}\cdot g(x)$$ And clearly $$C_0(X) /N_{ev_x}\cong\Bbb C$$ where $$N_{ev_x}$$ is the null space of $$(,)_{ev_x}$$. The isomorphism $$C_0(X)/N_{ev_x}\to \Bbb C$$ is given by $$[g]\mapsto g(x)$$ as you can check explicitly. Now the action of $$C_0(X)$$ on this Hilbert space is given by:
$$f\cdot [g] = [f\cdot g]$$ which, under the above isomorphism $$C_0(X)/N_{ev_x}\cong \Bbb C$$, corresponds to: $$f\cdot z = f(x)\cdot z$$
Now if you put all this together what you get is that the GNS space is: $$H= \bigoplus_{x\in X} \Bbb C = \ell^2(X)$$ and the representation is defined by: $$(\pi(f) v)_{x} = (f(x)\cdot v_x)$$ For some $$v = (v_x)_{x\in X}\in \ell^2(X)$$.
• I am sorry to ask this after a long time. You have taken the direct sum only over the characters of $C_0(X)$ which are just the pure states. For the universal representation, you need to take the direct sum over all the states. Now I agree that taking only the pure states gives you a faithful representation, but is there any way to characterize all the states of $C_0(X)$? Nov 10, 2020 at 7:24
• Yes, the states of $C_0(X)$ are the probability Radon measures on $X$. If you carry out the analogous argument as above you get that $$H=\bigoplus_{\mu \in P_1(X)} L^2(\mu)$$ (note that the pure states / characters correspond to the Dirac measures, and $L^2(\delta_x) \cong \Bbb C$, as a comptability statement with the above) Nov 10, 2020 at 8:56