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The Continuous Functional Calculus

  1. May 16, 2006 #1
    Although this problem is meant to be easy I cant quite work it out.

    Let [itex]U(A)[/itex] denote the set of unitary elements of a C*-algebra A. Ive 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, apparantly 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. Im assuming I will have to use the continuous functional calculus theorem somewhere.
     
    Last edited: May 16, 2006
  2. jcsd
  3. May 16, 2006 #2

    AKG

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    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.
     
  4. May 16, 2006 #3
    You could be right actually, it may be a typo.
     
  5. May 17, 2006 #4
    Yeah, I think the spectral radius of a unitary element equals the norm of the element:

    [tex]\|u\| = \sigma(u)[/tex]
     
  6. May 17, 2006 #5

    Hurkyl

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    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: May 17, 2006
  7. May 17, 2006 #6

    matt grime

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    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: May 17, 2006
  8. May 17, 2006 #7
    Matt, when you write the angled brackets: <iota> and <u> do you mean the algebras "generated" by iota and u respectively?
     
  9. May 17, 2006 #8

    matt grime

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    Quoting myself:

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

    of course it should read 'between' not 'from'
     
  10. May 17, 2006 #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 doesnt make sense, even after what you guys wrote. I dont see how phi can be defined, or why it should be defined!
     
    Last edited: May 17, 2006
  11. May 17, 2006 #10

    matt grime

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    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: May 17, 2006
  12. May 17, 2006 #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 definately C(T) -> A.
     
    Last edited: May 17, 2006
  13. May 17, 2006 #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 im 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: May 17, 2006
  14. May 17, 2006 #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: May 17, 2006
  15. May 17, 2006 #14

    matt grime

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    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: May 17, 2006
  16. May 17, 2006 #15

    matt grime

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    What is B? Where did it come from?
     
  17. May 17, 2006 #16
    uu* = 1 if u is unitary. So phi is a map from C(T) to the subalgebra defined by my unitary element. I dont 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.
     
  18. May 17, 2006 #17

    matt grime

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    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.
     
  19. May 17, 2006 #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 Im not sure?
     
  20. May 18, 2006 #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 Im not sure?
     
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