Spin Chronicles Part 38: Inside a representation

 Suppose ρ is a *-representation of a *-algebra A (with unit 1) on a Hilbert space H. We will assume that both A and H are finite-dimensional. Then one of two things can happen: either ρ is irreducible, or ρ is reducible. Let us examine the case of irreducible ρ first.

ρ like rose. Inside a representation

The case of ρ irreducible: every non-zero vector is cyclic.

Suppose ρ is irreducible. That means H has no nontrivial invariant subspaces. Choose any non-zero vector x in H. Consider ρ(A)x:

ρ(A)x = {ρ(a)x: a∈A}.

Then ρ(A)x is a linear subspace of H (Why?). 

Moreover ρ(A)x is an invariant subspace (Why?). 

By definition it must be trivial. Since it is different from {0} (Why?) - it must be the whole H. 

We say that the vector x is cyclic for ρ(A). 

A vector is cyclic for an algebra, if acting with the algebra elements on this vector you get the whole space (in general a dense subspace, but we are in finite dimensions, so every dense subspace is the whole space).

The case of ρ reducible.

Now suppose ρ is reducible. That means there is a nontrivial invariant subspace F. 

But then F^⟂ - the orthogonal complement of F in H, is also invariant (Why?). 

Now we have two representations of A, one on F, and one on F^⟂. 

 We choose a non-zero vector in each of them and repeat the reasoning. Since we assume the dimension of H is finite, this procedure must end after a finite number of steps with all irreducible representations.  

So, we have a decomposition

H = F₁⊕F₂⊕...⊕F_k,

and vectors x₁,x₁,...,x_k, with x_i being a cyclic vector for ρ(A) acting on F_i: ρ(A)x_i = F_i.

Vectors and states.

Let us go back to a general case. We have A, H, and ρ, with H being different from {0}. Let us choose a nonzero x vector in H, and let us suppose that x is normalized, wit ||x||=1. 

Then x defines a state functional f_x on A by

f_x(a) = (x,ρ(a)x).

It is easy to check (check it!) that f_x(a*)=f_x(a)*, f_x(a*a) ≥ 0 for all a in A, and f_x(1) = 1.

Notice that vectors x and ix, define the same state. In fact, vectors x and cx, where c i a complex number with |c|=1, define the same f. 

So there is some redundancy when replacing states (positive normalized functionals on A) by norm 1 vectors in H. 

Perhaps the source of this redundancy is already contained in A itself? In quantum theory A is assumed to be an algebra over complex numbers. What is the physical meaning of the product ca, where c is a complex number and a is an "observable" in A? Nobody knows it in a general case. 

For our Clifford algebra the meaning is geometric, but what it has to do with quantum theory? Good question.

In the next post we will finally see the GNS construction: starting from a state f, and constructing H, ρ, and a cyclic vector x_f. This representation sometimes, for pure states f,  will be irreducible. 

We will see that this construction gives us a particular vector x_f representing f, not just a class {cx: |c|=1} of physically "equivalent" vectors. There is some mystery here.

P.S, 18-01-25 9:42 I have read Kassandrov's paper Biquaternion Electrodynamics and Weyl-Cartan Geometry of Space-Time

I do not understand his math. I am having problems with his definition of analytic functions, I am having problems with his use of it. Will have to check papers written by others on this subject. He provides a couple of references. I will have to see if it is done in a better way in his 1992 monograph. I like his general idea, but I am not able to grasp his math. I prefer the definition-lemma-theorem method than talk-talk-talk. I know, I am doing talk-talk-talk here, on my blog. But this is an interactive forum, not a scientific publication.