How do I in general realize a universal C*-algebra generated by some generators and relation as concrete C*-algebras? For example, I know that universal C*-algebra generated by a single unitary is $C(\mathbb{T})$ by functional calculas. I am looking at the following examples to work on:

  1. universal C*-algebra generated by single self-adjoint element whose norm is 1.
  2. universal C*-algebra generated by single positive element whose norm is 1.
  3. universal C*-algebra generated by single normal element whose norm is 1.
  4. universal C*-algebra generated by single projection.

Edit: as pointed out in the comments, the following answers the question for unital C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra.

It may depend on what exactly you mean by "concrete", but I highly doubt that there is a general solution to this; finding a concrete realization of a universal C*-algebra requires classifying all the representations of the given generators and relations on Hilbert space, and this is an extremely difficult problem in general. For a more rigorous argument, see below.

But in your four examples, the universal C*-algebras are all commutative, and simple answers are possible:

  1. $C([-1,1])$.
  2. $C([0,1])$.
  3. $C(\mathbb{D})$ where $\mathbb{D}$ is the unit disk.
  4. $C(\{0,1\})=\mathbb{C}^2$.

In each case, the generator is the identity function, just as in your $C(\mathbb{T})$ example. It is a good exercise to verify the required universal property in each of these cases.

Another good example is the C*-algebra freely generated by two projections. This turns out to be the group C*-algebra $C^*(\mathbb{Z}_2\ast \mathbb{Z}_2)=C^*(\mathbb{Z}\rtimes\mathbb{Z}_2)$ and can be realized concretely as the subalgebra of $C([0,1],M_2(\mathbb{C}))$ containing those matrix-valued functions which are diagonal on the endpoints $0$ and $1$. See this paper of Raeburn and Sinclair.

So why do I think that a general solution is impossible? Consider the word problem for groups: there are groups given in terms of generators and relations for which there is no algorithm that can decide whether a given word in the generators represents the unit element. Now we can look at the maximal group C*-algebra of such a group. This C*-algebra is itself given by the same generators and relations together with additional relations requiring the generators to be unitary. If your intended meaning of a "concrete representation" comprises the existence of an algorithm that decides whether a given formal combination of generators represents $0$, then it follows that such a concrete representation cannot exist.

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    $\begingroup$ Looks like you're taking "C*-algebra" to mean "unital C*-algebra". The answers would be slightly different if you don't assume there are units. $\endgroup$ – Nik Weaver Jan 28 '13 at 18:38
  • $\begingroup$ @Nik: thanks, I just clarified this. I haven't thought about the non-unital case. $\endgroup$ – Tobias Fritz Jan 28 '13 at 18:46
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    $\begingroup$ The undecidability just means you cannot have an algorithmically effective description. I suppose that it is possible that the universal and reduced $C^*$-algebras of a group with undecidable word problem are the same and in some sense the reduced $C^*$-algebra is concrete in the sense that it is given by operators on a Hilbert space. $\endgroup$ – Benjamin Steinberg Jan 28 '13 at 19:21
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    $\begingroup$ @Benjamin: the reduced C*-algebra of a discrete group coincides with the universal one if and only if the group is amenable. And yes, there are amenable groups with undecidable word problem! (Kharlampovich constructed a solvable fp group with undecidable wp.) Hence my question about what the OP exactly means by "concrete": if it entails decidability, then the answer to the OP's main question is negative. On the other hand, if e.g. the left regular representation of a group still counts as concrete, then why not regard the universal representation as concrete as well? $\endgroup$ – Tobias Fritz Jan 28 '13 at 19:37
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    $\begingroup$ Thanks, Benjamin and Tobias, for this interesting discussion! Let me contribute an example illustrating the subtlety behind "concreteness". Consider the free group on 2 generators $a,b$; in the group ring, consider the element $M=a+a^{-1}+b+b^{-1}$. In the universal $C^*$-algebra, the spectrum of $M$ is $[-4,4]$, while in the reduced $C^*$-algebra, it is $[-2\sqrt{3},2\sqrt{3}]$. So, using functional calculus, we may construct "explicit" elements in the kernel of the canonical homomorphism from the universal to the reduced $C^*$-algebra. Would you describe these elements as "concrete"? $\endgroup$ – Alain Valette Jan 30 '13 at 0:33

Just a supplement to the answer by Tobias Fritz: All your examples are obviously commutative, since there is only one generator which is normal. Thus the question is really about finding certain terminal compact Hausdorff spaces. For example 1. comes from the terminal compact Hausdorff space $X$ equipped with a continuous function $X \to \mathbb{C}$ which is self-adjoint and norm $1$, i.e. whose image equals $[-1,1]$. This is obviously $[-1,1]$, equipped with the identity. You get the same answer when the norm is supposed to be $\leq 1$ (but $<1$ doesn't work). In a similar way one gets the other answers mentioned by Tobias Fritz.


Here is a further supplement: a tip for how to check if a $C^\ast$-algebra $A$ is the universal $C^\ast$-algebra for a given presentation. (I probably learned this from Terry Loring's book "Lifting solutions to perturbing problems in $C^\ast$-algebras".)

First, check that $A$ really is generated by a set of elements satisfying the given relations.

Second, check that every irreducible representation of the universal $C^\ast$-algebra is a representation of $A$. Say your generators are $x_1,\dots,x_n$. Then an irreducible representation would be generated by elements $X_1,\dots,X_n$. Since the centre of an irreducible representation is trivial, anything built out of the $X_i$'s that $*$-commutes with all the $X_i$'s is a scalar - so this approach works well if your relations entail a certain amount of commutativity, since commuting elements.

For example, to show that the universal $C^*$-algebra on a self-adjoint element of norm at most $1$ is $C_0([-1,1] \setminus \{0\})$, the second part above would go as follows. Let $X$ be the generator in an irreducible representation. Then $X$ is a scalar, which is self-adjoint (i.e. real) and has norm at most $1$. So $X = t \in [-1,1]$, and this representation corresponds to evaluating at this point $t$.

  • $\begingroup$ Am I correct in thinking that this is a some sort of version of the Yoneda lemma? It seems to me that, essentially, you are checking that your $C^\ast$-algebra $A$ satisfies the same universal mapping property as the universal one with those generators and relations, so it must be the universal one, by abstract nonsense. $\endgroup$ – MTS Feb 2 '13 at 1:38
  • $\begingroup$ Perhaps this is a way to view it, although I am unfamiliar with the Yoneda Lemma (and I'd be grateful if you elaborated). However, note that what's useful about the method I mentioned is that one needn't check that for any $C^\ast$-algebra $B$ with generators satisfying the given relations, there is a $\ast$-homomorphism $A \to B$ sending generators to respective generators. Rather, one only needs to check this when $B$ is irreducible. This is a consequence of the fact that, for any $C^\ast$-algebra, the kernels of all irreducible representations have trivial intersection. $\endgroup$ – Aaron Tikuisis Feb 3 '13 at 12:57
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    $\begingroup$ The clarify Arron's remark. The only representations you need to check are where $B = B(\mathcal H)$ for some Hilbert space, and only when the representation is irreducible. This means the commutant of the set of operators satisfying the relation is just the scalars. $\endgroup$ – Terry Loring Sep 15 '13 at 13:38
  • $\begingroup$ A nice example is the relations $a^2-a$ and $\|A\|\leq C$, for which I am seeking referneces: mathoverflow.net/questions/152007. $\endgroup$ – Terry Loring Dec 22 '13 at 20:47

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