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Functional Analysis question

  1. Apr 12, 2006 #1
    I have a commutative Banach algebra A with identity 1. If A contains an element [itex]e[/itex] such that [itex]e^2 = e[/itex] and [itex]e[/itex] is neither 0 nor 1 (I think this also means to say that it contains a non-trivial idempotent), then the maximal ideal space of A is disconnected.

    Currently I am trying to show this but Im not getting very far. Here is a summary of what I think I may need to show this:

    Because the question involves the maximal ideal space Im assuming I have to use the Gelfand transform somewhere. In particular it might be interesting to see what the Gelfand transform of the idempotent element e is.
  2. jcsd
  3. Apr 12, 2006 #2
    Question: Does the idempotent e allow us to decompose the Banach algebra A into a direct sum of ideals:

    [tex]A = I_1\oplus I_2[/tex]

    If so, I could write the first ideal as being the set generated by e, ie.

    [tex]I_1 = eA = \{x\in A\,:\,x = ex\}[/tex]


    [tex]I_2 = \{x\in A\,:\,ex = 0\}[/tex]

    although Im not too sure about the relevancy of these two ideals, or if they are at all constructible.

    Although, what I am sure about is that if I can create such ideals then it is clear that they are disjoint (closed) ideals and that for any element of the Banach algebra, x,

    [tex]x = ex+(1-ex)[/tex]

    which is an element of each ideal, and so

    [tex]A = I_1\oplus I_2[/tex]

    The problem remains: Does the idempotent allow me to construct these two disjoint ideals? What is the motivation behind such a decomposition? Does such a decomposition even help me show that the maximal ideals space [itex]\Delta_A[/itex] is disconnected?
  4. Apr 12, 2006 #3

    matt grime

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    Forget functional analysis, it's not important.

    Just let A be a ring, m is a maximal ideal iff A/m is a field or at least an integral domain, right? So what do e and (1-e) quotient out to if m is maximal and e is an idempotent (forget non-trivial too; that's an unnecessry proof by contradiction) and what do integral domains (fields) not have?
  5. Apr 12, 2006 #4
    Matt, thankyou for this new angle for approaching my problem. I must say I understand a little bit more about algebra than I do about functional analysis.

    An integral domain is a commutative ring with unity [itex]1\neq 0[/itex] and does not have any divisors of 0. All fields are integral domains.

    If M is my maximal ideal in A, then since A is commutative with unity, A/M is also a nonzero commutative ring with unity IF M = A.

    Let [itex](a+M)\in A/M[/itex], with [itex]a\neq M[/itex] so that a+M is not the additive identity element of A/M. Suppose that a has no multiplicative inverse in A/M. Then the set

    [tex](A/M)(a+M) = \{(b+M)(a+M)\,|\,(b+M)\in A/M\}[/tex]

    does not contain 1+M. So [itex](A/M)(a+M)[/itex] is an ideal of A/M.

    It is non-trivial because [itex]a\notin M[/itex], and it is proper because it does not contain 1+M. But this contradicts our assumption that M is a maximal ideal, therefore a+M must have a multiplicative inverse in A/M.

    This is how one of my proofs from algebra last year went. Im not sure how I will get to say that the quotient'ing' of the two elements e and e-1 will give me disjoint subsets other than one will not contain elements of the other. I believe I am still lacking the crucial information to make that call. I will keep thinking...
    Last edited: Apr 12, 2006
  6. Apr 19, 2006 #5
    Ive been given a hint on how Im meant to do this problem:

    Consider the case [itex]A = C(X)[/itex] where X is compact. Then apply this case to [itex]\hat{e}[/itex] (the Gelfand transform of e), and you will want to prove that [itex]\hat{e}[/itex] is neither 0 or 1. The spectral radius formula and the observation that [itex](1-e)^2 = 1-e[/itex] should help.

    Im still struggling to show that [itex]\Delta_A[/itex] is not connected. :(
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