2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $v = (v_x)_{x\in X}\in \ell^2(X)$.

$\endgroup$
2
  • $\begingroup$ 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 $C_0(X)$? $\endgroup$ Nov 10, 2020 at 7:24
  • $\begingroup$ 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) $\endgroup$
    – s.harp
    Nov 10, 2020 at 8:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.