Realizing universal C*-algebras as concrete C*-algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:24:34Z http://mathoverflow.net/feeds/question/120128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120128/realizing-universal-c-algebras-as-concrete-c-algebras Realizing universal C*-algebras as concrete C*-algebras SiOn 2013-01-28T17:01:03Z 2013-02-01T23:54:08Z <p>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:</p> <ol> <li>universal C*-algebra generated by single self-adjoint element whose norm is 1.</li> <li>universal C*-algebra generated by single positive element whose norm is 1.</li> <li>universal C*-algebra generated by single normal element whose norm is 1.</li> <li>universal C*-algebra generated by single projection.</li> </ol> http://mathoverflow.net/questions/120128/realizing-universal-c-algebras-as-concrete-c-algebras/120130#120130 Answer by Tobias Fritz for Realizing universal C*-algebras as concrete C*-algebras Tobias Fritz 2013-01-28T17:42:45Z 2013-01-28T19:15:37Z <p><strong>Edit</strong>: as pointed out in the comments, the following answers the question for <em>unital</em> C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra.</p> <p>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.</p> <p>But in your four examples, the universal C*-algebras are all commutative, and simple answers are possible:</p> <ol> <li>$C([-1,1])$.</li> <li>$C([0,1])$.</li> <li>$C(\mathbb{D})$ where $\mathbb{D}$ is the unit disk.</li> <li><code>$C(\{0,1\})=\mathbb{C}^2$</code>.</li> </ol> <p>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.</p> <p>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 <a href="http://www.mscand.dk/article.php?id=1231" rel="nofollow">this paper</a> of Raeburn and Sinclair.</p> <p>So why do I think that a general solution is impossible? Consider the word problem for groups: <a href="http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.ijm/1256044631&amp;page=record" rel="nofollow">there are</a> 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.</p> http://mathoverflow.net/questions/120128/realizing-universal-c-algebras-as-concrete-c-algebras/120195#120195 Answer by Martin Brandenburg for Realizing universal C*-algebras as concrete C*-algebras Martin Brandenburg 2013-01-29T10:23:01Z 2013-01-29T16:23:24Z <p>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 <em>terminal</em> 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 $&lt;1$ doesn't work). In a similar way one gets the other answers mentioned by Tobias Fritz.</p> http://mathoverflow.net/questions/120128/realizing-universal-c-algebras-as-concrete-c-algebras/120557#120557 Answer by Aaron Tikuisis for Realizing universal C*-algebras as concrete C*-algebras Aaron Tikuisis 2013-02-01T23:14:58Z 2013-02-01T23:14:58Z <p>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".)</p> <p>First, check that $A$ really is generated by a set of elements satisfying the given relations.</p> <p>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.</p> <p>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 $t$.</p>