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The essence of Stone's theorem

  1. Jun 14, 2013 #1

    mma

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    Could someone shortly summarise the essence of Stone's theorem ? What is the difference between Stone's theorem and the statement "the Lie-algebra of the group of orthogonal matrices consists of skew-symmetric matrices"? How Stone's theorem is related to the general notion of the exponential map between Lie-algebras and Lie-groups? What is the essential difference between Stone's theorem and its corresponding version for the finite dimensional orthogonal group? What is the significance of the strongly continuity of the one-parameter unitary subgroup? What can we say about the one-parameter subgroups that are not strongly continuous?

    I would greatly appreciate if somebody could enlighten me.

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    Last edited: Jun 14, 2013
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  3. Jun 16, 2013 #2

    mma

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    Perhaps I can formulate my question more specifically.

    What is wrong in the following? How can it be made precise, and what fails in it, if the Hilbert space is infinite-dimensional?

    Any one-parameter subgroup of the isometry-group of a finite or infinite dimensional, real or complex Hilbert space is a curve running in the group across the unit element. The tangent vector [itex]v[/itex] of this curve at the unit element is a skew-symmetric transformation of the Hilbert-space, and [itex]t \mapsto \exp(tv)[/itex] is the one-parameter subgroup itself. We say that [itex]v[/itex] is the infinitesimal generator of the one-parameter subgroup [itex]t \mapsto \exp(tv)[/itex]. So every one-parameter subgroup determines a skew-symmetric transformation as its infinitesimal generator. Conversely, for every skew adjoint vector [itex]v[/itex], [itex]t \mapsto \exp(tv)[/itex] is the one-parameter subgroup of the isometry-group.

    How comes here the strongly continuity?
     
    Last edited: Jun 16, 2013
  4. Jun 16, 2013 #3

    dextercioby

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    Stone's theorem enters the picture in connecting the strongly continuous unitary (irreducible) representations of a Lie group on a (complex separable) Hilbert space to the strongly continuous representations of the Lie algebra of the group on the same Hilbert space.
     
  5. Jun 16, 2013 #4

    mma

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    What do you mean exactly?
     
  6. Jun 16, 2013 #5

    dextercioby

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    Since the exponential mapping sends vectors in the Lie algebra into group elements in a neighborhood of identity, one uses this fact when trying to represent group and algebra elements as operators on a Hilbert space. See the theorem of Nelson as formulated in B. Thaller's book <The Dirac equation>.
     
  7. Jun 16, 2013 #6

    mma

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    Thanks, I'll try to rake something from this book.
     
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