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The commutative Gelfand-Naimark theorem tells us that every unital commutative C* algebra is isometrically isomorphic to the space of continuous functions on its maximal ideal space. The non- commutative GNS construction, on the other hand, tells us that every C* algebra is isometrically embedded into a * subalgebra of $\mathfrak{B}(\mathbb{H})$ for some Hilbert space $\mathbb{H}$.

My question is the following: how does the commutative case fit into the non commutative one?

I could see that given a commutative C* algebra $C(X)$, where $X$ is a locally compact Hausdorff space, one can find a non zero positive linear functional $F$ on $C(X)$ such that $F(e)=1$, where $e$ is the identity function on $C(X)$. Then using Riesz Representation theorem, one can find a corresponding Borel $\sigma$ algebra on $X$ and a unique probability measure $\mu$ such that $F(f)= \int f\,d\mu \quad \forall f\in C(X)$. From here it is easy to deduce that $C(X)$ is isometrically embedded into a * subalgebra of $\mathfrak{B}(\mathcal{L^2}(X,\mu))$ via multiplication operators.

My question is: Is there a canonical way of finding such a probability measure? Clearly if we choose a different positive linear functional, we get a different probability measure. So is there some unique/ natural way of doing this? Also, is this where the Hahn-Hellinger theorem comes in?

I would be grateful for some clarity on this, and also references if possible.

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  • $\begingroup$ You have to be more careful: assuming you mean $e$ is the constant function 1, $F$ could be an evaluation functional, so that $\mu$ is a point mass. Then $\mathcal{L}^2(X,\mu)$ is one-dimensional, and so is $\mathfrak{B}(\mathcal{L}^2(X,\mu))$, and you can't isometrically embed $C(X)$ into that. $\endgroup$ – Nate Eldredge Sep 30 '13 at 16:01
  • $\begingroup$ As another example, suppose $X$ is an uncountable set with the discrete topology. Any measure $\mu$ is supported on some countable set $S_\mu$, and so if $f \in C_0(X)$ is supported on $X \setminus S_\mu$, multiplication by $f$ is the zero operator on $\mathcal{L}^2(X,\mu)$. So no linear functional $F$ can give rise to an isometric embedding. $\endgroup$ – Nate Eldredge Oct 1 '13 at 3:14
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    $\begingroup$ I should have been more careful in phrasing the question. Every commutative algebra is isometrically isomorphic to $C(X)$ where $X$, its maximal ideal space, is compact and Hausdorff. In this case would it be true that we can find a positive linear functional that leads to an embedding into $\mathfrak{B}(\mathcal{L^2}(X,\mu))$ for some probability measure $\mu$? $\endgroup$ – Arundhathi Oct 1 '13 at 4:46
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The commutative case fits into the noncommutative case quite simply; in both cases there is not a canonical choice of linear functional which leads to an isometric representation of the C$^*$-algebra.

Let's consider the GNS representation more carefully.

It is not the case that there is always one linear functional $f$ whose GNS representation gives you an isometric representation of your C$^*$-algebra $A$ in some $B(H)$. If the linear functional $f$ is faithful, i.e., $f(x^*x)=x$ implies $x=0$, then the corresponding representation $\pi_f$ is isometric. If your $C^*$-algebra is separable, then there will always be such a faithful positive linear functional $f$, but there will be many different choices for $f$. Of course, the C$^*$-algebras $\pi_f(A)$ will all be isometrically isomorphic for all these different choices of faithful states $f$.

In general you will need to consider a direct sum of many different GNS representations using many different states in order to obtain the isometric inclusion of $A$ into some $B(H)$. For example, one can take the "universal representation" formed by taking the direct sum of the GNS representations for all positive linear functionals in $A^*$.

Similarly, one may need more than one measure to obtain an isometric representation of your commutative C$^*$-algebra. Instead of fixing one $F \in C(X)^*$, $F \geq 0$, one could take a universal representation formed by taking the direct sum over all such $F$. So there is not a canonical way to pick one such $F$ or corresponding measure $\mu$.

Also, if $X$ is a locally compact space that is not compact, then one should instead consider $C_0(X)$, the continuous functions on $X$ that vanish at infinity. In this case, the function $e$ in your question (given by $e(x)=1$ for all $x$) will not be an element of your C$^*$-algebra.

Yes, this lack of a canonical representation of self-adjoint operators by multiplication operators makes it more difficult to determine when two such operators are unitarily equivalent and it is for this reason that we need results like the Hahn-Hellinger theorem.

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In addition to the above response, perhaps I could also add this : Your goal is to faithfully represent a C* algebra on some object that is (hopefully) better understood. In order to do this, you are looking for a faithful representation.

The simple idea is to look for irreducible representations, ie. ones that cannot be decomposed further as the direct sum of representations. It turns out, in the commutative case, all these irreducible representations are one dimensional (they correspond to the evaluation maps from $C(X) \to \mathbb{C}$).

In the non-commutative case, however, one is not so lucky. For each irreducible reprentation, one gets a Hilbert space (which may or may not be $\mathbb{C}$). Adding all these representations up results in a possibly infinite dimensional Hilbert space. This is the GNS construction.

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  • $\begingroup$ If in the commutative case, every irreducible representation is one dimensional, would this not mean that $C(X)$ is isometrically embedded into a one dimensional space, which does not make sense, as Nate Eldredge had pointed out in his comment? Also, how do we show the same? Forgive me if I am missing out on something obvious. $\endgroup$ – Arundhathi Sep 30 '13 at 18:03
  • $\begingroup$ No. Irreducible reps. may not be faithful. For instance the only irreducible reps. of $C(X)$ are the evaluation maps (This is because the kernels of these reps. must be primitive ideals) $\endgroup$ – Prahlad Vaidyanathan Oct 1 '13 at 2:37

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