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Nuclear operators and trace

  1. Jan 23, 2010 #1

    I came across a line (http://www.springerlink.com/content/t523l30514754578/) about how the trace of a linear operator is not, in general, independent of the choice of orthonormal basis. The link states that such an operator may have a trace that converges for one basis but not another.

    This sounds really neat but I haven't been able to find much about it, other than to learn that if you want your trace to exist and you want it independent of an orthonormal basis then you should restrict to nuclear operators.

    Ultimately I'm looking for a counterexample, where the trace converges for one basis and not the other. But I'll take anything at this point.


  2. jcsd
  3. Jan 25, 2010 #2
    Hmm, this must be a good one if noone seems to know. Is anyone else interested in finding out? Maybe we could split a list of functional analysis texts and slowly start combing? I've emailed a professor I know who has some experience in functional analysis, but I don't know any personally. Though I'm on the verge of emailing random analysts.


  4. Jan 25, 2010 #3


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    Well, it seems the obvious thing to do is to construct a non-nuclear operator and play with it.

    (I'm using wikipedia's Nuclear Operator page)

    What's a good infinite-dimensional space? How about all square-summable sequences. Also, I'll pick the simplest orthonormal basis and the simplest sequence that converges to zero, but whose sum diverges.

    Let dn be the sequence with a 1 in the n-th position, and 0 elsewhere.

    Then, define
    [tex]L(s) = \sum_{n = 1}^{+\infty} \frac{1}{n} s(n) d_n[/tex]​
    (where s(n) is the n-th term of s)

    (This is well-defined, right? We should check)

    If we consider the orthonormal basis <dm>, the trace would be infinite:
    [tex]\sum_m \langle d_m, L(d_m) \rangle = \sum_{m,n} \frac{1}{n} d_m(n) \langle d_m, d_n\rangle = \sum_n \frac{1}{n} = +\infty[/tex]​

    If we instead consider the basis [itex]e_m = i^m d_m[/itex], we get
    [tex]\sum_m \langle e_m, L(e_m) \rangle = \sum_{m,n} \frac{1}{n} e_m(n) \langle e_m, d_n\rangle =
    \sum_n \frac{1}{n} (-1)^n = -\log 2[/tex]​

    And now I think the problem seems clear -- if we slip in some other set of constants, we get yet another sum. We could probably reorder the basis to reorder the sum to change it's value (its conditionally convergent, not absolutely). And so forth. I haven't even tried mixing the basis elements yet (i.e. with linear combinations)!
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