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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 4th 2010
    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeSep 4th 2010

    I supplemented this with simple Lie algebra.

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 4th 2010

    Couldn’t we whittle down the last two conditions on simple Lie algebra to one by saying there is exactly one proper ideal (the zero ideal)? Because the trivial Lie algebra has no proper ideals.

    Also, while the definition in simple Lie algebra seems to be standard, I wouldn’t mind seeing discussion of why the 1-dimensional Lie algebra shouldn’t be considered simple. I don’t really understand this myself.

    • CommentRowNumber4.
    • CommentAuthorTobyBartels
    • CommentTimeSep 4th 2010

    Couldn’t we whittle down the last two conditions

    Yes, certainly. I was just spelling everything out in detail. But the last two conditions simply say that the Lie algebra is a simple object.

    On second thought, the non-abelian condition covers the non-trivial condition too. So there really is no point in having three separate conditions. I’ll edit it.

    I don’t really understand this myself.

    Me neither. Could it be that we are really working in the category of Lie algebras whose largest abelian subalgebra is trivial? But then why that?

    It could be a historical problem, that the too simple to be simple condition, which should have been simply that the algebra is nontrivial, was made too strong, requiring that the algebra is nonabelian.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeSep 6th 2010
    • (edited Sep 6th 2010)

    I started LieAlg.

    Also I added remarks at simple Lie algebra and semisimple Lie algebra that these correspond to simple and semisimple Lie groups, respectively.

    Concerning terminology: I suppose for Lie algebras and -groups it well predates any attempts to isolate the abstract meaning of “simple object” in category theory? Possibly the case of abelian Lie algebras, notably in the classification results, was felt to be too trivial to deal with and discarded from the start. Otherwise each table of the A,B,C,DA,B,C,D- and E,F,GE,F,G-classification of simple Lie groups would contain two more “exceptional” entries, \mathbb{R} and U(1)U(1), and maybe that was felt to be too boring to be included.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2010

    My immediate reaction would be to say that U(1)U(1) would correspond to the empty Coxeter diagram, so the rejection of this group might be consistent with historical tendencies to reject empty things.

    But I’m not so sure of this position. There are various nice ways of characterizing semisimple Lie algebras that fail for abelian Lie algebras, e.g.,

    • Any solvable ideal is the zero ideal

    • Any abelian ideal is the zero ideal

    • The adjoint representation is faithful

    • The Killing form is nondegenerate

    and it appears to me that much of the general theory (as traditionally presented) devolves on nice facts like these. Maybe the general theory could be rewritten so that abelian Lie algebras are naturally part of the story, but it’s far from clear that the story would be any nicer.

    This would be a nice MO question, were it not for the fact that most mathematicians there (and elsewhere) exhibit a very impatient attitude toward “trivial” things, not realizing that in the long run it usually pays to think about them, as theory often becomes cleaner and easier to manage if care is taken with trivial cases. For example, it is better to allow covering spaces to have empty fibers over some points, since then the category of covering spaces over XX becomes a topos (of representations of the fundamental groupoid) and is much, much cleaner to deal with.

    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeSep 6th 2010

    My feeling is that in common usage a semisimple Lie algebra is by definition a finite sum of simple Lie algebras, isn’t it ? Are there common exceptions to this usage ?

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2010

    Sure, that’s arguably the definition, especially if we want “semisimple” to match other uses of the word. But in the theory of finite-dimensional Lie algebras, there are lots of other starting points, and the ones I mentioned, which lead up to nondegeneracy of the Killing form, are the ones given in a book I happen to have close to hand (Fulton-Harris), and they give a nice approach to complete reducibility which leads to your definition. Another avenue is by Weyl’s unitary trick.

    The question still remains: conventions aside, why shouldn’t the 1-dimensional Lie algebra also be considered simple?

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeSep 6th 2010

    created a stub for orthogonal Lie algebra (with precisely no non-tautological information so far, but got to run now…)

    • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeSep 6th 2010

    The question still remains: conventions aside, why shouldn’t the 1-dimensional Lie algebra also be considered simple?

    Many of the common theorems about semisimple Lie algebras do not extend to the case when you allow such factors.

    • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeSep 6th 2010

    Am I crazy, or is the list of semisimple Lie alebras at semisimple Lie algebra actually a list of simple Lie algebras? To be sure, every simple Lie algebra is semisimple, but it seems to me that the list would do better at simple Lie algebra. In fact, I think that I'll move things around so that both articles show a complete classification (at least over an algebraically closed field of characteristic $0$), not just a list of examples.

    • CommentRowNumber12.
    • CommentAuthorzskoda
    • CommentTimeSep 6th 2010

    Toby 11: Is it true that the Cartan classification is the same for all algebraically closed fields of characteristics zero. I know that in logic a big class of statements is valid iff it is valid for complex numbers, but is this classification in that logical class ? Is the Cartan classification really the same for p-adics ?

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeSep 6th 2010

    Am I crazy,

    No, it’s me. :-)

    • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2010

    @Zoran #10: yes, and in fact I mentioned a few such at #6. I’m not convinced this is end of story, but for now I’m not going to worry about it.

    @Zoran #12: the algebraic closure of the pp-adics is isomorphic to \mathbb{C}. It’s an early victory of model theory that algebraically closed fields are completely characterized up to isomorphism by their characteristic and transcendence degree over their prime field. I’m pretty sure the classification holds for any transcendence degree over \mathbb{Q}.

    • CommentRowNumber15.
    • CommentAuthorTobyBartels
    • CommentTimeSep 6th 2010

    I’ve verified that I’m not crazy, created simple Lie group, and put a classification at simple Lie algebra.

    @ Todd #8

    The empty Dynkin diagram corresponds to the trivial Lie algebra, not the line. In general, any diagram whose components are Dynkin diagrams corresponds to a semisimple Lie algebra given by a direct sum of the components. In this way, we get theorems such as 𝔰𝔬 4=𝔰𝔩 2𝔰𝔩 2\mathfrak{so}_4 = \mathfrak{sl}_2 \oplus \mathfrak{sl}_2 (that is, 𝔡 2=𝔞 1𝔞 1\mathfrak{d}_2 = \mathfrak{a}_1 \oplus \mathfrak{a}_1, if you define the diagram D 2D_2 by continuing to remove dots down the main row).

    @ Zoran #12

    I’m pretty sure this is true in finite dimensions, since the classification of simple Lie algebras of a fixed finite dimension is a finite first-order statement, so we simply prove the statement for each dimension in turn. Thus, it remains only to prove that a simple Lie algebra must have finite dimension. However, the pp-adic fields are not themselves algebraically closed, so classifying Lie algebras over them (rather than over their algebraic closures) should be at least as difficult as classifying real Lie algebras.

    • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2010

    Okay, I see my mistake. Thanks.

    • CommentRowNumber17.
    • CommentAuthorTobyBartels
    • CommentTimeSep 7th 2010

    @ Urs #13

    That’s what happens when you copy from Wikipedia!

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