The Continuous Functional Calculus

In summary, the conversation discusses the existence and properties of a *-homomorphism from the space of continuous functions on the unit circle to a C*-algebra A, where the unitary elements of A are shown to have a spectrum contained in the unit circle. This *-homomorphism is shown to map the identity function to a unitary element in A and can be extended to all polynomial functions on the unit circle, making it a homomorphism of algebras. The question of whether it is a *-map is still being debated.
  • #1
Oxymoron
870
0
Although this problem is meant to be easy I can't quite work it out.

Let [itex]U(A)[/itex] 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:

[tex]\sigma(u) \subset \mathbb{T} = \{z\in\mathbb{C}\,:\,|z|=1\}[/tex]

which was easy.

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

Now, i figured that the obvious choice for this *-homomorphism is the exponential function [itex]e^{it}[/itex] - 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.
 
Last edited:
Physics news on Phys.org
  • #2
I don't understand most of what you're saying, since I haven't studied it, but wouldn't you mean [itex]\phi (\iota ) = u[/itex], not [itex]\phi (\iota )\mathbb{C} = u[/itex]? 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.
 
  • #3
You could be right actually, it may be a typo.
 
  • #4
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:

[tex]\|u\| = \sigma(u)[/tex]
 
  • #5
Suppose A consists of 2x2 matrices. Then,

[tex]
\left(
\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}
\right)
[/tex]

and

[tex]
\left(
\begin{array}{cc}
-i & 0 \\
0 & i
\end{array}
\right)
[/tex]

are both unitary, and have the same spectrum.
 
Last edited:
  • #6
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).
 
Last edited:
  • #7
Matt, when you write the angled brackets: <iota> and <u> do you mean the algebras "generated" by iota and u respectively?
 
  • #8
Quoting myself:

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

of course it should read 'between' not 'from'
 
  • #9
So [itex]\phi[/itex] is a map from the continuous operators on the spectrum of [itex]u[/itex] to the C*-algebra such that [itex]\phi[/itex] acting on the identity function: [itex]\iota(z):=z[/itex] returns [itex]u[/itex]. Does this mean that [itex]\phi(\iota(z)) = u[/itex] implies [itex]\phi(z) = u[/itex]?

I mean, all I know is
1. [itex]A[/itex] is a C^*-algebra
2. [itex]u[/itex] is unitary
3. [itex]\sigma(u) \subset \mathbb{T}[/itex]
and from this I am (somehow) supposed to be able to see that there must be a * preserving homomorphism from [itex]C(\mathbb{T})[/itex] 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!
 
Last edited:
  • #10
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.
 
Last edited:
  • #11
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 [itex]\sigma(u) \subset \mathbb{T} = \{z\in\mathbb{C}\,:\,|z|=1\}[/itex]. Deduce that there is a *-homomorphism [itex]\phi\,:\,C(\mathbb{T}) \rightarrow A[/itex] such that [itex]\phi(\iota) = u[/itex] where [itex]\iota\,:\,\mathbb{T} \rightarrow\mathbb{C}[/itex] is the function defined by [itex]\iota(z):=z[/itex].

1. It is definitely C(T) -> A.
 
Last edited:
  • #12
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.
 
Last edited:
  • #13
The next part of the question says that if [itex]u \in U(A)[/itex] where U(A) is the set of all unitary elements of A, and if [itex]\|u-1\| < 2[/itex] then there is an element [itex]b \in A[/itex] such that [itex]b^* = b[/itex] and [itex]u = exp(ib)[/itex].

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

EDIT: B has changed to b thanks to Matt's eagle eyes.
 
Last edited:
  • #14
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.
 
Last edited:
  • #15
Oxymoron said:
The next part of the question says that if [itex]u \in U(A)[/itex] where U(A) is the set of all unitary elements of A, and if [itex]\|u-1\| < 2[/itex] then there is an element [itex]b \in A[/itex] such that [itex]b^* = B[/itex] and [itex]u = exp(ib)[/itex].

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


What is B? Where did it come from?
 
  • #16
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.
 
  • #17
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.
 
  • #18
What does [itex]\|u-1\|<2[/itex] have to do with [itex]u = \exp(ib)[/itex] 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?
 
  • #19
What does [itex]\|u-1\|<2[/itex] have to do with [itex]u = \exp(ib)[/itex] 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?
 

What is the Continuous Functional Calculus?

The Continuous Functional Calculus is a mathematical theory that extends the traditional operator calculus to include continuous functions. It allows for the manipulation of operators and functions in a more general setting, making it a powerful tool for solving problems in areas such as analysis and differential equations.

What is the difference between the Continuous Functional Calculus and the traditional operator calculus?

The main difference between the Continuous Functional Calculus and the traditional operator calculus is that the former deals with continuous functions, while the latter is limited to discrete operators. This means that the Continuous Functional Calculus can be used in a wider range of mathematical problems and applications.

What are some real-world applications of the Continuous Functional Calculus?

The Continuous Functional Calculus has a wide range of applications in various fields such as physics, engineering, and economics. Some specific examples include using it to solve differential equations in physics, analyzing control systems in engineering, and modeling economic systems.

How does the Continuous Functional Calculus handle non-linear operators?

The Continuous Functional Calculus has the ability to handle both linear and non-linear operators. This is because it deals with functions, which can be non-linear, rather than just discrete operators. This makes it a powerful tool for solving problems involving non-linear operators.

What are some important properties of the Continuous Functional Calculus?

Some important properties of the Continuous Functional Calculus include linearity, which allows for the combination of operators and functions, and continuity, which ensures that the results obtained are meaningful and applicable in real-world situations. Other properties include the ability to handle both bounded and unbounded operators, and the existence of an inverse operator for invertible functions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
0
Views
418
  • Calculus and Beyond Homework Help
Replies
3
Views
465
  • Calculus and Beyond Homework Help
Replies
1
Views
449
  • Calculus and Beyond Homework Help
Replies
3
Views
354
  • Calculus and Beyond Homework Help
Replies
27
Views
605
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
960
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
674
Back
Top