TITLE: Realizing universal $C^*$-algebras as concrete $C^*$-algebras
QUESTION [18 upvotes]: 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:

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

REPLY [9 votes]: 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 $t$.