Understanding Group C*-Algebras: Intuitive Expl.

[Monocles]

I am having trouble understanding the definition of a group C*-algebra, i.e. take a group algebra of a group G and then complete it with respect to the norm

$$\| a \| = \sup \{ | \pi_u(a) | \}$$

where $$\pi_u(a)$$ is a unitary representation of G. I am looking for a way to think about this intuitively - what does it mean to complete with respect to such a strange looking norm? Is it necessary to figure what every unitary possible unitary representation of G is in order to determine what the group C*-algebra of G is? Or is this typically a non-constructive definition?

I know that the group C*-algebra of an abelian group G is the algebra of continuous functions on the Pontryagin dual of G (I forgot what norm this is with respect to, though, but I assume uniform convergence?). Can anyone give an intuitive explanation as to why?

 

[Hurkyl]

I can't answer your question, but I hope I can convince you the norm is a reasonable one -- even the "obviously" right one.

C*-algebras were distilled from the notion of an algebra of operators acting on Hilbert spaces. I take from your post that you're already content with the idea of the norm of an operator being a reasonable notion.

So, the notion of a norm in a C*-algebra should, at least, have some relationship with the notion of the norm of an operator. Among other things, the C*-norm of an element should be at least as large as the norm of the corresponding operator in some representation of that algebra. i.e.

$$|| x || \geq \sup |\pi_u(x)|$$​

The neat thing, IIRC, is that for any particular algebra element there is always a representation in which element and its corresponding operator have the same norm. So it's actually an equality.


And since we already have a notion that we like groups to act on Hilbert spaces as unitary operators...

 

[Monocles]

That actually helps a lot! I will probably have more questions to ask, but I'll give this at least a day to mull over before then.

 

[Monocles]

Ok, I have a better idea of what is going on now that I have learned about the continuous functional calculus (I had not when I started this topic the other day).

I have been trying to understand group C*-algebras from the point of view of universal C*-algebras. The universal C*-algebra C*(u) generated by by a single unitary operator u is precisely the C*-algebra of continuous complex valued functions on the unit circle. I think I understand partially why this is:

1. The spectrum of a unitary always lies within the unit circle.
2. By the continuous functional calculus, polynomials in u and u* in C*(u) correspond to polynomials in z and z* (z* is complex conjugate of z) on the spectrum of u.
3. By the Stone-Weierstrass theorem, continuous functions can be approximated in norm by polynomials. Since C*-algebras are closed, we close the polynomials with respect to the uniform norm to obtain all continuous functions on the spectrum of u.

My question is then, why does the spectrum of C*(u) consist of the entire unit circle? By the continuous functional calculus, I would think that it should just consist of continuous functions on the spectrum of u, but of course since u is just an abstract unitary operator we can only say that the spectrum of u lies within the unit circle. So is the fact that the spectrum of u must lie within the unit circle automatically force us to consider the entire unit circle to be the spectrum of u?

Other examples do this as well, for example the universal unital C*-algebra generated by a self-adjoint operator of norm less than or equal to 1 is precisely the C*-algebra of continuous complex-valued functions on [-1,1].

Thanks.

EDIT: I had a thought - they are called universal because they satisfy some sort of universal property, right? I am new to universal properties, though, so I'm not quite sure what it would be.

 

[Monocles]

Another thought that I'm having trouble put into precise wording:

Is the reason that the spectrum 'expands' to all possible values of the spectrum when considering the universal C*-algebra precisely due to the supremum over all possible representations in the norm? Meaning, if we have some concrete unitary operator u and represent it on a Hilbert space, it will have a specific spectrum. But other representations will give it different spectrums. So, in order to take into account all possible spectra, the universal C*-algebra generated by a unitary has the entire unit circle as its spectrum.

Does this sound correct?

 

[Hurkyl]

EDIT: I had a thought - they are called universal because they satisfy some sort of universal property, right? I am new to universal properties, though, so I'm not quite sure what it would be.

If I was forced to guess, I would say that

If $X$ is a unitary element in a C*-algebra A, then there is a unique C*-algebra homomorphism f from C*(u) to A such that $f(u) = X$​

I imagine you can answer your questions by showing that the continuous functions on the unit circle have this property.


But anyways, a more elementary thing to note: what can we say if $X - \lambda$ was not invertible in A?

 

[Monocles]

OK, I am slowly understanding. I understand now that the group C*-algebra of G is universal with respect to unitary representations of G.

I also think I have figured out why the universal C*-algebra generated by a single unitary is a specifically the closure of polynomials on the unit circle. The universal C*-algebra generated by a single unitary u consists of polynomials in u and u*. It would just be polynomials on the entire complex plane then, if it weren't for the requirement that uu*=u*u = 1, which in terms of the continuous functional calculus means that zz*=z*z = 1, which means that |z|=|z*|=1. Thus, they are polynomials on the unit circle specifically. From this point of view, there is no need to consider the algebra as being 'continuous functions on the spectrum of u', since the relations themselves already dictate that these are polynomials on the unit circle.

I'm almost feeling confident enough to move onto crossed products now - thanks again for your help.

 

[Hurkyl]

It would just be polynomials on the entire complex plane then,
...
Thus, they are polynomials on the unit circle specifically.

Alas, those are the same ring. I think there's some argument like this, but I'm having trouble pinning down the details precisely.

 

[Monocles]

Ah, you are right. Hmmm. Just when I thought I had it all figured out!

 

[Monocles]

OK, howabout this. While they are isomorphic as rings, I suspect that they are not isomorphic as C*-algebras, because that would imply that the unit circle and the complex plane are homeomorphic. Does that sound right?

 

[Hurkyl]

Well, they aren't even C*-algebras yet, or even have a norm equipped! The intended norm is presumably the sup norm, so z has different norm in the two rings (1 in 1, infinite in the other).

I will agree there is clearly an injective ring homomorphism from the ring of polynomials in z and 1/z into C*(u). (that sends z to u and 1/z to u*)

 

[Monocles]

Ahhh, I had not considered an explicit example like the norm of f(z) = z in the complex plane vs. the unit circle. Then it is very clear how the two are distinguished from one another! I think my problem was that I was only thinking about norm in terms of what sequences converge and not whether the 'same' element would have a different norm in different C*-algebras. You have been a great help.