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Levi Theorem

  1. Nov 14, 2008 #1

    I've got big problems with understanding abstract algebra, the way we deal with it in the seminar on Lie algebras. In just four weeks we progressed up to Levi and Malcev theorems, which are actually the culmination, the say, of classical Lie algebras theory. I didn't think, that the material would become so dense and abstract in so small amount of time. So it's my fault in the end. But I need help, because I have to make a presentation of these two theorems.

    I'd like to note that I've got absolutely no background in Lie group theory. They said that some knowledge of linear algebra would suffice...

    I attached a pdf with the statement of Levi's theorem and the treatment of the first case when kernel is not a minimal ideal.

    What I don't understand is:
    1. why the formula for the dimensions holds.
    2. why [itex]\beta_1(s)[/itex] isomorphic to s.
    3. why [itex]dim(ker\alpha_2)=dim(n_1)[/itex]

    and why is there beta with tilde present. Is it a typo?
    Last edited: Nov 11, 2009
  2. jcsd
  3. Nov 19, 2008 #2


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    By the first isomorphism theorem,
    [tex]\dim \ker \alpha_1 = \dim \mathfrak{g}/\mathfrak{n}_1 - \dim \mathfrak{s} = \dim \mathfrak{g}/\mathfrak{n}_1 - \dim \mathfrak{g}/\mathfrak{n} = \dim \mathfrak{n} - \dim \mathfrak{n}_1.[/tex]
    The last equality follows from the fact that dim(v/w)=dimv-dimw.
    [itex]\beta_1[/itex] has a left inverse, and hence is injective.
    By definition, [itex]\ker\alpha_2 = \{x : x + \mathfrak{n}_1 = \mathfrak{n}_1\} = \mathfrak{n}_1[/itex].
    The tilde is there probably because the domain [itex]\beta_2[/itex] is being adjusted to all of [itex]\mathfrak{g}/\mathfrak{n}_1[/itex]. But don't worry about it - it's mostly irrelevant.
    Last edited: Nov 19, 2008
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