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Exceptional Lie algebras

  1. Jul 4, 2010 #1

    tom.stoer

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    I have some questions regarding the exceptional Lie algebras e(n), n=6,7,8.

    Can anybody explain to me what prevents us from constructing e(9) from e(8)? What goes wrong? One can use the e(8) lattice vectors and try to construct an e(9) vector; one could go even further and try e(10) etc. I know that the Cartan Matrix becomes zero (or negative for 10, ...) in that case (which is forbidden), but what does that mean if one would try to write down the generators for e(9)? What's wrong with them as Lie algebra generators?

    Another question I have is related to E(n) as symmetry groups. For the A, B, C and D series one can understand the (fundamental or defining representation of) Lie groups acting on a certain vector space and leaving a certain scalar product invariant. For SO(n) it's

    [tex]\vec{x}^t \vec{y}[/tex]

    with [tex]\vec{x}, \vec{y} \in R^n[/tex], for SU(n) it's

    [tex](\vec{x}^\ast)^t \vec{y}[/tex]

    with [tex]\vec{x}, \vec{y} \in C^n[/tex]. What about E(n)? Is there a similar scalar product which is invariant? Are there other invariants?
     
  2. jcsd
  3. Jul 4, 2010 #2
    Hi Tim,

    I don't really have a good understanding of it myself, but a good place to start would be
    http://math.ucr.edu/home/baez/octonions/node13.html
    which has good references that give more detailed discussions.

    A quick summary is the 3 infinite sequences, SO, SU and SP of classical groups relate to the three associative division algebras; R, C and H resp. The exceptional groups are related to the largest (and nonassociative) division algebra, the octionians - there's only a finite number of them because it's hard to get group associativity out of a nonassociative division algebra.

    The standard/modern approach to semisimple Lie algebra classification is via Cartan matrices and Dynkin diagrams, but that isn't very enlightening about why the exceptional algebras exist. The alternate/older approach is via the study of invariants (which give you the standard definitions of the classical Lie groups). Modern texts using this approach are rarer - an example that I know of is the "birdtracks" book by Cvitanovic
    http://www.nbi.dk/GroupTheory/
    In this book (using nonstandard diagrammatic methods) he does what you asked and finds and classifies the possible invariants and relates them to the Lie groups.
    He also has a good discussion of the Dynkin diagram vs invariants approach in the intro.

    I hope that helps,

    Simon
     
  4. Jul 5, 2010 #3

    tom.stoer

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    Thanks a lot!

    Regarding the invariants I asked John one or two years ago; he was not aware of any such construction for E(n). And unfortunately I never really understood his idea about the octo-octonionic projective plane.

    I hope the web book will help; looks very promising

    Tom
     
  5. Jul 12, 2010 #4
    A question: Lie algebras verify Jacobi identity:

    [[a,b],c] + [[b,c],a] + [[c,a],b] = 0

    This is another identity with four summands:

    [[[a,b],c],d] + [[[b,c],d],a] + [[[c,d],a],b] + [[[d,a],b],c] = 0

    I don't know the name. I have checked that some Lie algebras verify it. I have also checked that i can find brackets that verify the last identity and not the Jacobi identity.
    ... Are they algebras? Name?
     
  6. Jul 27, 2010 #5

    tom.stoer

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    Hello Simon, hello all,

    unfortunately the book does not explain (or I do not understand :-( what I was looking for; so I'll try it again: any answers regarding my questions in post #1

    Thanks a lot
     
  7. Jul 31, 2010 #6
  8. Jul 31, 2010 #7

    tom.stoer

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    thanks a lot
     
  9. Jul 31, 2010 #8

    tom.stoer

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    They always come to the same conclusion.

    Exceptional algebras are constructed over the octonions which explains why the series terminates and why there are no simple bilinear symmetry structures. The bist hint is always Baez http://math.ucr.edu/home/baez/octonions/node16.html. Others say that the series does not terminate but that one can construct Kac-Moody algebras (but I don't see how to get the infinitely many generators).

    So it seems that it's up to me to understand better all this exactly means ...
     
  10. Aug 25, 2010 #9

    tom.stoer

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    ... push ...
     
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