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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.

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.

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.

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.

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. $C(\mathbf{D})$ where $\mathbf{D}$ is the unit disk.
4. $C({0,1})=\mathbf{C}^2$.

In each case, the generator is the identity function, just as in your $C(\mathbf{T})$ 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 $C^*(\mathbf{Z}_2\ast \mathbf{Z}_2)=C^*(\mathbf{Z}\rtimes\mathbf{Z}_2)$ and can be realized concretely as the subalgebra of $C([0,1],M_2(\mathbf{C}))$ 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.

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.