Thread: Hilbert A-module

Suppose X is a Hilbert A-module. I would like to show that for each there is a unique such that

The plan is to use adjoinable operators for each



where we take to be a right Hilbert A-module with the inner product defined by

I have shown that the operators are the adjoint of each other

Originally Posted by Mauritzvdworm

Suppose X is a Hilbert A-module. I would like to show that for each there is a unique such that

The plan is to use adjoinable operators for each



where we take to be a right Hilbert A-module with the inner product defined by

I have shown that the operators are the adjoint of each other

It's true that the operators are adjoints of each other, but I don't see how to use that to find y, given x.

To me this looks a bit similar to the polar decomposition of an element of a von Neumann algebra, as done in Gert Pedersen's book C*-algebras and their automorphism groups. In fact, if then . Of course, that construction only works if is invertible. But you can avoid that snag as follows. First note that we may assume that A is unital (if not, then adjoin an identity e to it and extend the action of A on X to an action of the unitised algebra on X). Now define a sequence of elements and prove, (as in Pedersen's Proposition 2.2.9) that is Cauchy.

I don't see how to prove the uniqueness part of the problem. If then , from which you get and hence . But that's as far as I can get.

I think I have found a way to handle to problem, but it is somewhat involved. Here I will give the idea

Let us define the following operators


and let
We can now use these to define an operator on the direct sum Hilbert A-module in the following way


This operator clearly is adjoinable (just take the complex conjugate en transpose of the matrix). We can now also show that the operator defined above is in

This can be done using embeddings of the form


and similarly for the others.

now for each is selfs adjoint element in which anticommutes with

We now exploit the functional calculus together with the function and find that and anticommutes with

We can also show that every self adjoint element in is of the form
(The proof of this is also somewhat involved and for practical reasons I will omit it here)

Now we have
for some

Now we just multiply out the matrices and solve for the bottom left corner which will yield
then we have


so finally we get

That's a very nice argument. I haven't checked every detail of it, but it looks convincing. Certainly the cube root function has to lie at the heart of the proof.