Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adjoint representation

  1. Sep 16, 2007 #1
    Why has the adjoint representation a higher dimension than the basis matrices it acts on? for example here

    Why is e_1 two dim and ad(e_1) four dim?

    Isn't ad(X) Y a simple matrix multiplication here? But then multiplying 4x4 with 2x2 matrices, what does it mean?

  2. jcsd
  3. Sep 16, 2007 #2
    Think carefully. The elements of a Lie algebra are vectors. The given example happens to be gl(2), which allows those vectors to be interpreted as matrices, but the dimensionality of gl(2) is actually 4.
  4. Sep 16, 2007 #3

    thanks genneth
  5. Sep 17, 2007 #4
    Wait, still don't get it.

    what the heck is, for example ad(e_1) x e_2?
    How can we multiply 4 x 4 with 2 x 2 matrices?

    And ae_1 + be_2 + ce_3 + de_4 is a 2x2 matrix right?
  6. Sep 17, 2007 #5
    You need to hold two views of the algebra in your head -- and it's only for this example. gl(n) are nxn matrices. But they also form a vector space, of dimension n^2. Linear operators on that vector space are NOT matrices of nxn -- they would be (n^2)x(n^2). Thus, the adjoint representation are matrices of (n^2)x(n^2), acting on a n^2 dimensional vector space.
  7. Sep 17, 2007 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    This is the problem when people think too much in terms of bases. You are not mutlitpying a 4x4 matrix by a 2x2. You are letting a lie algebra acts as endomorphisms on a vector space. Think of reps, as as maps from gxV to V, in this case we send the pair (g,v) to [gv]. Things in the left slot are elements of g, and things in the right slot are elements of some vector space, which just happens to be g again in this case.
  8. Sep 17, 2007 #7
    thanks again

    but, how do I compute the adjoint representation in some basis (like basis given in example)?

    Why is my vec space n^2?
  9. Sep 17, 2007 #8
    Like I said, the elements of gl(n) are nxn matrices, which form a vector space. How many independent vectors can you find in it?
  10. Sep 17, 2007 #9

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Since you didn't state what your basis was, we can't tell you. But the very definition of the adjoint rep is that


    Now, what is there left to discuss?
  11. Sep 17, 2007 #10
    Here they give a basis for gl(2) from which they compute an adjoint representation. I can't see how they do it, how they go from 2x2 to 4x4 matrices.
  12. Sep 17, 2007 #11
    You seem very fixated on the 2x2 matrices. Stop it! They stop being matrices as soon as you start thinking of them as a vector space, and then they're a 4 dimensional vector space. Imagine unfolding a matrix, like this:

    (a b) ___> (a b c d)
    (c d)

    You can pick *any* 4 independent vectors for the basis, but the usual (1000), (0100), (0010) and (0001) are pretty good ones, as they're trivially independent, by inspection. If you run that map above backwards, you'll find them corresponding to the basis matrices given. Now, you're given the action of the adjoint representation in the original matrix view of things; construct the actual adjoint 4x4 representation by inspection on what the basis do to each other.
  13. Sep 18, 2007 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Even more explicitly, what is [e_1,e_i] for i=1..4? Now you know what ad(e_1) does to the 4 basis vectors, so you can write down the matrix for ad(e_1). It would help if you remembere to think of ad(e_1) as an endomorphism of a 4-d space.
  14. Sep 18, 2007 #13
    many thanks again, got it now

    by the way, I just learnt here about structure constants, why has no one brought them up yet?
    Last edited: Sep 18, 2007
  15. Sep 18, 2007 #14

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Because they are unnecessary for the discussion.
  16. Sep 18, 2007 #15
    Why are they unnecessary? Don't they give you the matrix of adjoint representation once you have a basis?

    I just found out I'm not alone being confused. Not long ago there was this thread on PF.

    Especially here, that's exactly my problem (post 16)!!!

    and what Matt Grime writes in post 18 drives it home for me!

    Got it! Finally!

    (concrete examples is how most people learn, I'm no exception here...wish that view would be shared by at least one author of those trillion books on lie groups and lie algebras out there...)
    Last edited: Sep 18, 2007
  17. Sep 18, 2007 #16

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    How did you propose to get the structure constants? 'Structure constants' is just a fancy name for what you have to write down. Giving them a fancy name doesn't alter what you had to do in the slightest. I didn't mean that concept of structure constants is unimportant, but that there was absolutely no need to mention them, i.e. to give a name to what you were doing already.
  18. Sep 18, 2007 #17
    It's the old issue that people over-rely on basis representations. The fact is that mathematicians don't choose to work in the abstract just because they're masochistic and weird -- there are genuine advantages. It's worthwhile spending time to try and free yourself from the need to have a basis, and therefore able to put numbers to things, and just work with the abstract concepts. After all, adjoint representations aren't even that high up on the hierarchy of increasingly abstract things, and as the abstraction climbs, it may get to a point where a basis is no longer even possible (i.e. you're not in a vector space any more)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook