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Lie algebra, ideal and isomorphism

  1. Sep 1, 2007 #1
    Suppose [itex]A\subset\mathfrak{g}[/itex] and [itex]I\subset\mathfrak{g}[/itex] are subalgebras of some Lie algebra, and I is an ideal. Is there something wrong with an isomorphism [itex](A+I)/I \simeq A/I[/itex], [itex]a+i+I=a+I\mapsto a+I[/itex], for [itex]a\in A[/itex] and [itex]i\in I[/itex]? I cannot see what could be wrong, but all texts always give a theorem [itex](I+J)/J\simeq I/(I\cap J)[/itex] instead.
     
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  3. Sep 1, 2007 #2

    Chris Hillman

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    Review your sources?

    "All texts always give a theorem"?

    It is never a bad idea to cite your sources, and usually a good idea. In fact, if you had checked your sources before posting, you might have answered your own question.

    I suggest that you review:

    1. the definition of ideal in Lie theory versus ring theory,
    2. the definiton of a module in ring theory,
    3. the precise statement of Noether's formula [itex](A+B)/B \simeq A/(A \cap B)[/itex].

    If this doesn't help, ask again.
     
  4. Sep 1, 2007 #3
    It could be my post was more confusing than I intended, because I don't know this "Noether's formula" if it exists also elsewhere than in theory of Lie groups. I have two sources: http://www.amazon.com/Lie-Groups-Introduction-Anthony-Knapp/dp/0817642595 and lecture notes of course in University of Helsinki. They both present this theorem, and Knapp puts it like this:

    I thought I understood this, until I noticed that to me it seems that also the other claim, I presented in OP, is correct. And it seems strange if they both were correct, so now I think that there is something wrong with the claim, but I cannot see what.

    (Now I used the same notation as Knapp does, which is different from the notation in the OP. Sorry for the confusion)
     
    Last edited: Sep 1, 2007
  5. Sep 1, 2007 #4
    Or is there anything wrong with the claim? Could the theorem continued to be

    [tex]
    \cdots\simeq\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}) \simeq \mathfrak{b}/\mathfrak{a} ?
    [/tex]

    If it can, then why is it written in the intersection form, when this one is simpler?
     
  6. Sep 1, 2007 #5

    Chris Hillman

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    I'm not trying to be difficult...

    ... but you'll learn more if you figure it out yourself.

    I really think that reviewing the definitions and so on would be a very good idea! However, ask again if this really doesn't help.

    Incidently, Prof. Knapp was my analysis teacher (when I was an undergraduate at Cornell; Knapp later moved to SUNY Stony Brook).
     
  7. Sep 2, 2007 #6

    matt grime

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    Chris's idea of reviewing things carefully is a good one, but I think it doesn't emphasise this point enough, and might make you try to look too hard for something that is obvious

    What is A/I if A is just some subalgebra of g and I is some ideal of g?
     
    Last edited: Sep 2, 2007
  8. Sep 2, 2007 #7
    Basic stuff:

    For an element of a algebra, X, and a subalgebra [itex]\mathfrak{a}[/itex], the coset is defined [itex]X+\mathfrak{a}=\{X+A\;|\;A\in\mathfrak{a}\}[/itex]. An important property is

    [tex]
    X+\mathfrak{a}=X'+\mathfrak{a} \quad\Leftrightarrow\quad X-X'\in\mathfrak{a}
    [/tex]

    If [itex]\mathfrak{g}[/itex] is a Lie algebra, and [itex]\mathfrak{a}[/itex] is its subalgebra, a quotient set is defined [itex]\mathfrak{g}/\mathfrak{a}=\{X+\mathfrak{a}\;|\;X\in\mathfrak{g}\}[/itex]. For this to be a Lie algebra, [itex]\mathfrak{a}[/itex] should be an ideal. Otherwise the bracket operation is not necessarily well defined.

    Comments on possible mistakes here are welcome.

    Doubts:

    The claim

    [tex]
    \mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b})\simeq \mathfrak{b}/\mathfrak{a}
    [/tex]

    doesn't make fully sense, because the right side is a Lie algebra if [itex]\mathfrak{a}[/itex] is an ideal, but for the left side both [itex]\mathfrak{a}[/itex] and [itex]\mathfrak{b}[/itex] should be ideals. Am I on the right track? But if we assume them both to be ideals, I still cannot see what's wrong with the proof.

    I don't see any problems arising from a possible case where we have strictly [itex]\mathfrak{b}\subset\mathfrak{a}[/itex]. [itex]\mathfrak{b}/\mathfrak{a}[/itex] merely becomes a trivial group, but it's still well defined.

    Proof:

    Let [itex]\mathfrak{a},\mathfrak{b}\subset\mathfrak{g}[/itex] be ideals of some Lie algebra. Then

    [tex]
    f:\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b})\to\mathfrak{b}/\mathfrak{a},\quad\quad f(B+\mathfrak{a}\cap\mathfrak{b})=B+\mathfrak{a}
    [/tex]

    is an isomorphism.

    The map is well defined because

    [tex]
    B+\mathfrak{a}\cap\mathfrak{b} = B' + \mathfrak{a}\cap\mathfrak{b} \quad\implies\quad B-B'\in\mathfrak{a}\cap\mathfrak{b} \quad\implies\quad B-B'\in\mathfrak{a} \quad\implies\quad B+\mathfrak{a} = B'+\mathfrak{a}
    [/tex]

    Similarly it is also injective because

    [tex]
    B-B'\notin \mathfrak{a}\cap\mathfrak{b} \quad\big(\textrm{and}\; B-B'\in\mathfrak{b}\big) \quad\implies\quad B-B'\notin\mathfrak{a}
    [/tex]

    The map is surjective because for arbitrary [itex]B+\mathfrak{a}[/itex] we can choose [itex]B+\mathfrak{a}\cap\mathfrak{b}[/itex].

    The map is Lie algebra homomorphism because

    [tex]
    f([B+\mathfrak{a}\cap\mathfrak{b},\;B'+\mathfrak{a}\cap\mathfrak{b}]) = f([B,B']+\mathfrak{a}\cap\mathfrak{b}) = [B,B']+\mathfrak{a} = [B+\mathfrak{a},\;B'+\mathfrak{a}] = [f(B+\mathfrak{a}\cap\mathfrak{b}),\; f(B'+\mathfrak{a}\cap\mathfrak{b}])
    [/tex]

    I still don't know if I'm supposed to figure out that my claim is wrong or that it is right.

    Very cool.
     
  9. Sep 2, 2007 #8

    matt grime

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    Nope, you're looking too hard. If I have A a subalgebra of g, and I an ideal of g, how are you defining the quotient space A/I? I know how to define the quotient space g/I, for example, because I is an ideal of g, but is I an ideal of A?
     
  10. Sep 2, 2007 #9
    Ahaa... so it doesn't make sense to write A/I according to the standard definitions, because I is not necessarily an ideal of A. But if I still define it like this

    [tex]
    A/I=\{X+I\;|\;X\in A\}
    [/tex]

    It seems to be isomorphic to [itex]A/(A\cap I)[/itex].
     
  11. Sep 2, 2007 #10

    matt grime

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    But is A\cap I an ideal? I.e. does it make sense to write A/(A\cap I)? As a vector space, you're certainly correct, but that isn't necessarily a Lie algebra homomorphism.
     
  12. Sep 2, 2007 #11
    In fact I think I proved this! I thought it was something that the book assumes, that reader must be able to figure out on his own.

    So [itex]A\subset\mathfrak{g}[/itex] and [itex]I\subset\mathfrak{g}[/itex] are subalgebras of some Lie algebra, and I is an ideal. Now I want to prove that [itex]A\cap I[/itex] is an ideal in A (oh and yeah, not in [itex]\mathfrak{g}[/itex]...)

    Let [itex]x\in A[/itex] and [itex]i\in A\cap I[/itex] be arbitrary.

    [itex]x,i\in A[/itex], so because A is a subalgebra, we have [itex][x,i]\in A[/itex].

    [itex]x\in A[/itex] and [itex]i\in I[/itex], so because I is an ideal, we have [itex][x,i]\in I[/itex].

    Therefore [itex][x,i]\in A\cap I[/itex], and [itex][A, A\cap I]\subset A\cap I[/itex].
     
    Last edited: Sep 2, 2007
  13. Sep 3, 2007 #12

    Chris Hillman

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    Back to square one, with benefit of more experience

    Jostpuur, I recommend that you carefully reread my first reply above and see if it makes more sense now. Then review all the definitions again (and the statement of Noether's formula--- make sure you know all the hypotheses). I really think you'll learn more if you figure this out yourself since your question is based upon misconceptions which are apparently due to not thinking about the issues I raised.

    (I hope everyone realizes that I am trying not to give jostpuur a fish, but to teach him how to fish, as the proverb goes.)
     
  14. Sep 13, 2007 #13
    This is the only context where I have seen this formula. But I can guess from Chris' comments, that there is some other theorem, not related to lie algebras, where this same formula is also present. I don't have this theorem at any sources now, and it seems difficult to find it by google and keywords "noether's formula" (OMG, the 9th google hit is my own thread! :surprised, googled 2007-09-13)

    I thought I already understood this. The equation

    [tex]
    \mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}) \simeq \mathfrak{b}/\mathfrak{a}
    [/tex]

    was correct in the sense, that if you write down both sides with expression [itex]A/B:=\{a+B\;|\;a\in A\}[/itex], they are isomorphic, but it is bad notation because in the definition of a quotient group we assume [itex]B\subset A[/itex], and hence the left side is preferred.

    Now if that's wrong, then please explain. I'm believing in this now.
     
  15. Sep 13, 2007 #14

    matt grime

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    ahem. assume? No, it is part of the definition.

    the left side of what?
     
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