In the last post, SU(1,1) parametrization, we have concluded that the group SU(1,1) has the shape of a doughnut. Its elements can be uniquely parametrized by an angle and a complex number inside of the unit disk:
Thus SU(1,1) can be parametrized using the Cartesian product of the circle and the (interior of) the disk. But parametrization is one thing, and the actual construction is a different thing. Today we will see how every SU(1,1) matrix can be uniquely decomposed into a product of disk parametrized positive part, and circle parametrized unitary part.
We already know that every SU(1,1) matrix is of the form
We also know (see Eqs. (11,14,15) from SU(1,1) parametrization) that it determines and so that
But then, as you can easily check, can be written as a product (so called polar decomposition)
The matrix is evidently Hermitian: It is also positive. Hermitian matrix is positive when all its eigenvalues are positive. In two dimensions there are two eigenvalues. Their product is the determinant, and so it is equal to 1. Thus either both eigenvalues are positive or both are negative. But the trace of , which is the sum of eigenvalues, is positive is positive, therefore both eigenvalues must be positive, and so is positive.
Exercise: calculate the two eigenvalues of
In fact every positive matrix from SU(1,1) is of the form for some Since suppose is in SU(1,1) and is positive. We write it in the form as in Eq. (6). Positive matrix must be Hermitian, therefore must be real. This happens only for or But so for we would have which is impossible for a positive matrix. Therefore and
The matrix is evidently unitary, In fact, every unitary matrix from SU(1,1) must be of the type Why? Suppose in SU(1,1) is unitary. Write it in the form Suppose Then But is unitary and is Hermitian, so Thus squares of the eigenvalues of are both equal to and since is positive, both must be equal to 1. Therefore and so
Now, being familiar with the polar decomposition, we can fly, like these polar birds
and we will.