In non-commutative geometry, Gelfand duality is the construction of multiplicative linear functionals of a commutative C*-algebra, which can be viewed as the space of all its irreducible complex representations.

When encountered with a non-commutative C*-algebra, we can speak of the space of pure states, which is the generalization of multiplicative linear functionals in commutative cases. The GNS construction can then be viewed as a map from the space of states to the space of all representations, and also from the space of pure states to the space of all irreducible representations.

My question is, is the "GNS map" surjective, or injective? What should be viewed as the non-commutative topological space, the space of all irreducible representations, or the space of pure states? And why?


On surjectivity: No, not every representation comes from a state; only the cyclic ones. Every nondegenerate representation of a C*-algebra is a direct sum of cyclic representations (Zorn), and every cyclic representation comes from a GNS construction. But yes, every irreducible representation (which is also cyclic) comes from a GNS construction.

On injectivity: To even consider this question you probably want to identify representations if they are unitarily equivalent. But no, because for example two unit cyclic vectors in the Hilbert space of a representation will often yield different vector states, but these vector states will yield the same representation.

Which space? That would depend on your applications, and I don't have any absolute answers to that. Instead I'll tell you what comes to mind in the hope that it will help orient you. (Then I'll wait with you to be enlightened by another answer.)

The space of (unitary) equivalence classes of irreducible representations is often called the spectrum of a C*-algebra; a closely related space is the set of kernels of irreducible representations, called the primitive ideal space. The primitive ideal space is given the hull kernel topology and is a quotient of the spectrum. I recommend the book by Raeburn and Williams; see Appendix A to learn about these spaces, and see the rest of the book for how they're used.

As for the space of pure states, I'm more accustomed to the idea of studying the space of all states, in which the pure states are the extreme points. There is a nice characterization of the state spaces of C*-algebras in E. Alfsen, H. Hanche-Olsen and F.W. Shultz: State Spaces of C∗-Algebras, Acta Math. 144 (1980) 267–305, which came up at this other question. However, there is yet another option which I learned a little about from Pedersen's book, where the quasi-state space Q is used. A quasi-state is a positive functional of norm at most 1. (Other than 0, the extreme points of Q are also the pure states.) A C*-algebra can be studied as affine functions on Q; see section 3.10 of Pedersen for details.

  • $\begingroup$ Thanks, your answer and the referrence are very helpful. $\endgroup$ – Roger Liu Feb 14 '10 at 6:45

@Meyer has provided a very good answer, so I just want to add something from my notes from Professor Rieffel's $C^\ast$-algebra class about the primitive ideal space, which is a canonical topological space associated to a $C^\ast$-algebra (in fact, it is associated to any normed $*$-algebra, but let's keep it "easy"...).

Let $A$ be a $C^\ast$-algebra.

Definition: An ideal $I\subset A$ is called primitive if it is the kernel of an irreducible representation. Denote the set of all primitive ideals by $Prim(A)$.

Note that $Prim(A)$ is always a set as it is a subset of $P(A)$, the power set of $A$. An important theorem:

If $A$ is GCR, then the map $\widehat{A}\to Prim(A)$ by $(\pi, H)\mapsto \ker(\pi)$ is a bijection (where $\widehat{A}$ is the "set" of equivalence classes of irreducible representations).

This is far from the case if $A$ is NCR (there is some set theory here, such as the term "unclassifiable," that I don't want to get into as I am not a set theorist), but $Prim(A)$ is still a set, so we still get a topological space using the following fact:

All primitive ideals are prime.

The space of prime ideals of $A$ comes with the Jacobson, or "hull-kernel" topology, so we get the relative topology on $Prim(A)$.

A few facts:

  • In general, $Prim(A)$ is not Hausdorff, but it is $T_0$ (see @Meyer's comment for a counterexample).
  • $Prim(A)$ is locally compact.
  • If $A$ is separable, $Prim(A)$ has the Baire Category Property.

Once again, this is all from a course I took from Professor Rieffel. I hope it helps!

  • $\begingroup$ Nice answer, and thanks for the compliment. I wasn't careful with set theoretic considerations when I wrote "space of equivalence classes of irreducible representations" (which of course can be made precise). $\endgroup$ – Jonas Meyer Jan 26 '10 at 9:09
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    $\begingroup$ I do have one small quibble (which I am posting as a new comment so that it may later be deleted): The primitive ideal space of the compacts is {{0}}, so it is Hausdorff. An example that comes to mind is the C*-algebra generated by the unilateral shift, whose quotient by the compacts is isomorphic to the continuous functions on the circle. The primitive ideal space "is" the circle unioned with {{0}}, and the closure of {{0}} is the whole space. $\endgroup$ – Jonas Meyer Jan 26 '10 at 9:22
  • $\begingroup$ Doesn't the primitive ideal space of $B_0(H)$ consist of two points $\\{\\{0\\},B_0(H)\\}$, as the trivial representation is irreducible as well? $\endgroup$ – Dave Penneys Jan 26 '10 at 23:36
  • $\begingroup$ By convention, zero representations don't count as irreducible, or equivalently primitive ideals are assumed to be proper. This is analogous to prime ideals being defined to be proper in ring theory, and for similar reasons. Including the algebra as a primitive ideal causes a problem in the definition of the topology; the empty set wouldn't be closed. (And including it would provide no information.) Not counting the zero rep as irreducible is often only implicitly assumed, which can be annoying, but for example one wants the primitive ideal space of C([0,1]) to just be [0,1]. $\endgroup$ – Jonas Meyer Jan 27 '10 at 0:00
  • $\begingroup$ thanks! i will edit the answer to refer to your comment as a counterexample. $\endgroup$ – Dave Penneys Jan 27 '10 at 0:18

Concerning your last question, I would say you should view your C*-algebra itself as the (coordinate ring on the) "non-commutative topological space." The spaces you suggest are commutative. Your C*-algebra might be considered as a non-commutative substitute of the coordinate ring on one of them.


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