The Continuous Functional Calculus

[Oxymoron]

Although this problem is meant to be easy I can't quite work it out.

Let $U(A)$ denote the set of unitary elements of a C*-algebra A. I've already shown that if u is unitary in A then the spectrum of u:

$$\sigma(u) \subset \mathbb{T} = \{z\in\mathbb{C}\,:\,|z|=1\}$$

which was easy.

Now, apparently I can deduce that there exists a *-homomorphism $\phi\,:\,C(\mathbb{T})\rightarrow A$ from the compact space of continuous operators on the spectrum to the C*-algebra, such that $\phi(\iota)\mathbb{C} = u$. Where $\iota\,:\,\mathbb{T}\rightarrow\mathbb{C}$ is the function defined by $\iota(z) := z$.

Now, i figured that the obvious choice for this *-homomorphism is the exponential function $e^{it}$ - but I could be wrong. However, if I am right, how should I go about proving that this is the correct deduction according to the question asked. I am assuming I will have to use the continuous functional calculus theorem somewhere.

 

[AKG]

I don't understand most of what you're saying, since I haven't studied it, but wouldn't you mean $\phi (\iota ) = u$, not $\phi (\iota )\mathbb{C} = u$? This is something of a stab in the dark, since I don't know what most of this stuff means, but are unitary elements in some way uniquely determined by their spectra? Perhaps you can map an element of f of C(T) to the unitary element whose spectrum is f(X), where X is some appropriate subset of T? Like if u is some nxn unitary matrix, then its spectrum is its eigenvalues (?) and if you let X be the n complex roots of unity, then f gets mapped to the unitary matrix with spectrum f(X)?

I don't know if this helps at all.

 

[Oxymoron]

You could be right actually, it may be a typo.

 

[Oxymoron]

but are unitary elements in some way uniquely determined by their spectra?

Yeah, I think the spectral radius of a unitary element equals the norm of the element:

$$\|u\| = \sigma(u)$$

 

[Hurkyl]

Suppose A consists of 2x2 matrices. Then,

$$\left(<br /> \begin{array}{cc}<br /> i & 0 \\<br /> 0 & -i<br /> \end{array}<br /> \right)$$

and

$$\left(<br /> \begin{array}{cc}<br /> -i & 0 \\<br /> 0 & i<br /> \end{array}<br /> \right)$$

are both unitary, and have the same spectrum.

 

[matt grime]

I think the point is that you just define phi(iota)=u. This defines a function from <iota> to <u> ie a map from the C* algebras defined by these elements. Now, you need to verify that

1. it is a * map (preserves norms)
2. it is a homomorphism between the subalgebras (which is trivial)
3. extends to a map of the whole of C(T) probably by some theorem that I can't remember the name of.

iota spans the polynomial sub algebra, by the way, which is dense in the space of continuous functions if we throw in the constants (I'm assuming A is unital might help here too).

 

[Oxymoron]

Matt, when you write the angled brackets: <iota> and <u> do you mean the algebras "generated" by iota and u respectively?

 

[matt grime]

Quoting myself:

"a map from the C* algebras defined by these elements"

of course it should read 'between' not 'from'

 

[Oxymoron]

So $\phi$ is a map from the continuous operators on the spectrum of $u$ to the C*-algebra such that $\phi$ acting on the identity function: $\iota(z):=z$ returns $u$. Does this mean that $\phi(\iota(z)) = u$ implies $\phi(z) = u$?

I mean, all I know is
1. $A$ is a C^*-algebra
2. $u$ is unitary
3. $\sigma(u) \subset \mathbb{T}$
and from this I am (somehow) supposed to be able to see that there must be a * preserving homomorphism from $C(\mathbb{T})$ to A.

It doesn't make sense, even after what you guys wrote. I don't see how phi can be defined, or why it should be defined!

 

[matt grime]

There are several points of concern right now.

1. Does the book state that phi is a map from C(sigma(u)) to A or from C(T) to A? You keep swapping between the two. the spectrum of u is merely contained in T it is not equal to it.

2. phi(z) does not make sense. phi acts on functions it sends iota to u where iota is the map iota(z)=z a map from T to the complex numbers.

 

[Oxymoron]

Ok, Ill do as you say and just take it for granted that there IS such a phi from C(T) -> A. So now all I have to do is prove that it IS a *-homomorphism?

QUESTION
If u is unitary in A (which is a C*-algebra with 1) then $\sigma(u) \subset \mathbb{T} = \{z\in\mathbb{C}\,:\,|z|=1\}$. Deduce that there is a *-homomorphism $\phi\,:\,C(\mathbb{T}) \rightarrow A$ such that $\phi(\iota) = u$ where $\iota\,:\,\mathbb{T} \rightarrow\mathbb{C}$ is the function defined by $\iota(z):=z$.

1. It is definitely C(T) -> A.

 

[Oxymoron]

2. But phi acting on the identity function equals u doesn't it?

So if I put any old function f into phi, say phi(f), then this does not necessarily mean that phi(f) = u. But if I act on the identity function, i, then phi(i) does equal u. Is this right?

This is all confusing me a little. Since phi is a function from a space of continuous functions to a C*-algebra and I am also told that phi acts on functions but only when it acts on certain functions (i.e. the identity function) does it return my unitary element u.

 

[Oxymoron]

The next part of the question says that if $u \in U(A)$ where U(A) is the set of all unitary elements of A, and if $\|u-1\| < 2$ then there is an element $b \in A$ such that $b^* = b$ and $u = exp(ib)$.

Just in case you wanted to know where the question was heading.

EDIT: B has changed to b thanks to Matt's eagle eyes.

 

[matt grime]

Yes, phi(i)=u, and therefore phi(i^2)=u^2, note that i^2 is the map sending z to z^2, ie i^2(z)= i(z)i(z) in this sense and it is NOT the composition of functions. So you see phi extends to a map on all polynomial functions on T, and thus to all functions since the polynomials are dense so it is clearly a map from C(T) to <u> a sub algebra of A, it is also clearly a homomorphism of algebras, so it really is only the case of deciding if it is a *-map. It has been some years since I looked at these things so I'm very rusty on exactly what the definitions are.

 

[matt grime]

The next part of the question says that if $u \in U(A)$ where U(A) is the set of all unitary elements of A, and if $\|u-1\| < 2$ then there is an element $b \in A$ such that $b^* = B$ and $u = exp(ib)$.

Just in case you wanted to know where the question was heading.

What is B? Where did it come from?

 

[Oxymoron]

Posted by Matt Grime

So you see phi extends to a map on all polynomial functions on T, and thus to all functions since the polynomials are dense so it is clearly a map from C(T) to <u> a sub algebra of A, it is also clearly a homomorphism of algebras, so it really is only the case of deciding if it is a *-map, but what is i(z)i*(z)? It is zz*=1 since z is in T, and what is uu* if u is unitary?

uu* = 1 if u is unitary. So phi is a map from C(T) to the subalgebra defined by my unitary element. I don't quite see how phi extends to a map on all "polynomial" functions on T. Where did polynomials come from? I mean, once you bring polynomials into the picture you can start talking about dense and thus homomorphisms.

 

[matt grime]

iota is a polynomial, it is a function, call it f if it helps and f(z)=z so it's a poly, 2f(z)=2z, f(z)f(z)=z^2 iota generates all the polynomials. You're thinking of composition of functions but the operations on C(T) are pointwise multiplication and addition.

 

[Oxymoron]

What does $\|u-1\|<2$ have to do with $u = \exp(ib)$ where b is self-adjoint. I think I can see that the exponential function might have something to do with the unit circle, but I am not sure?

 

[Oxymoron]

What does $\|u-1\|<2$ have to do with $u = \exp(ib)$ where b is self-adjoint. I think I can see that the exponential function might have something to do with the unit circle, but I am not sure?