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Homework Help: How do i show the lie algebra of the group

  1. Mar 18, 2010 #1
    how do i show the lie algebra of the group [itex]SO(p,q)[/itex] is

    [itex] \mathfrak{so}(p,q) = \Biggl\{ \left( \begin{array}{cc} X & Z \\ Z^t & Y \end{array} \right) \vline X \in M_p ( \mathbb{R} ), Y \in M_q ( \mathbb{R} ), Z \in M_{p \times q} ( \mathbb{R} ), X^t=-X, Y^t=-Y \Biggr\}[/itex]

    i'm not really getting anywhere so any help is appreciated. thanks.
     
  2. jcsd
  3. Mar 21, 2010 #2
    Re: Groups

    nobody? aargh.
     
  4. Mar 21, 2010 #3

    Dick

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    Re: Groups

    A matrix g is in the Lie group if g^(T)*eta*g=eta, where eta is a unit diagonal matrix with signature (p,q), right? So if G is in the Lie algebra then exp(t*G^T)*eta*exp(t*G)=eta. Do the usual thing and differentiate and set t=0 to get the condition on G. Now split G up into block form and see what that condition is on each block. There's no special tricks here.
     
    Last edited: Mar 22, 2010
  5. Mar 22, 2010 #4
    Re: Groups

    yeah. this is pretty much what i've been trying.
    (disclaimer: my notation differs a bit from yours but it should be understandable!)

    using the above defn, if we write A=exp(tau X) then
    [itex]A^t \eta A = \eta \Rightarrow X=-\eta^{-1} X^t \eta=-\eta X^t \eta[/itex] since [itex]\eta^{-1}=\eta[/itex] (since it's a diagonal matrix)

    thus if we write X as a block matrix [itex]X= \left( \begin{array}{cc} B & C \\ D & E \end{array} \right)[/itex] and if we write [itex]\eta = \left( \begin{array}{cc} \mathbb{I}_{p} & 0 \\ 0 & \mathbb{I}_{q} \end{array} \right)[/itex]

    then if we follow through the couple of lines of matrix multiplication, our condition on X simplifies to

    [itex] \left( \begin{array}{cc} B & C \\ D & E \end{array} \right) = \left( \begin{array}{cc} -B & 0 \\ 0 & - E \end{array} \right)[/itex]

    this doesn't look to promising at this point. also i haven't even made use of the determinant condition yet and i've already established that the upper right and lower left blocks are zero, which according to the answer, they shouldn't be.
     
  6. Mar 22, 2010 #5

    Dick

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    Re: Groups

    For one thing eta=diag(-I_p,I_q) or diag(I_p,-I_q) depending on where you like to put the sign, right? For another X^(T)=[[B^T,D^T],[C^T,E^T]]. I think you are forgetting to transpose the block matrices.
     
  7. Mar 22, 2010 #6
    Re: Groups

    ok. i've managed to get to the answer without using the condition of unit determinant though. surely i've done something wrong?
    or perhaps this condition is used on the next part of the question, which is to establish that the lie algebra has dimesnion [itex]\frac{n(n-1)}{2}[/itex] where [itex]n=p+q[/itex].

    do you have any advice on how to go about provingthe dimension? thanks.
    so far i' ve said that because of the condition on B it will have dimension p/2 and similarly E has dim q/2 but don't know how to deal with Z???

    p.s. i also have to do the same for the group [itex]SU(p,q)[/itex]. i don't actually know the definition of SU(p,q) though? i can define [itex]SU(n)= \{ a \in M_n ( \mathbb{C} ) | \bar{a}^t a = \mathbb{I} , \text{det } a = 1 \}[/itex] alright but the p and q bit messes it up. unlike the SO case where we had a eta matrix to give the p,q signature to.
     
    Last edited: Mar 22, 2010
  8. Mar 22, 2010 #7

    Dick

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    Re: Groups

    Just add up the number of free variables in your matrix blocks. The unit determinant choice only selects components of the Lie group, it doesn't affect the dimension of the Lie algebra.
     
  9. Mar 22, 2010 #8
    Re: Groups


    (i)so the unit determinant must affect the lie algebra in some way otherwise O(p,q) and SO(p,q) would have identical lie algebras, no?

    (ii)okay. as for the dimension:
    there are p free variables in B and q free variables in E. C has (pq)^2 free variables. i think i've gone wrong since this is looking nothing like what i want!

    (iii) what is the definition of SU(p,q)?

    thanks again.
     
  10. Mar 22, 2010 #9

    Dick

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    Re: Groups

    O(p,q) and SO(p,q) do have the same Lie algebra don't they? SO(p,q) is the component of O(p,q) containing the identity. They have the same dimension. You'd better try counting those variables again. What are the dimensions of B, C, D and E? SU(p,q) is the simlar to SO(p,q) except the matrices can be complex and transpose is replaced by conjugate transpose.
     
  11. Mar 22, 2010 #10
    Re: Groups

    ok. i agree on the first point that O(p,q) and SO(p,q) have the same lie algebra. the det 1 condition neatly falls in to place from the fact that matrices in the lie algebra have 0's all down the diagonal.

    right next bit:

    B has dim p^2. we only want to count the upper triangular components (not including the diagonal). this give (p^2-p)/2. and E has dimension q^2. again count just that upper triangle. this gives (q^2-q)/2.

    add these together we get

    (p^2+q^2 - (p+q))/2 = (n^2-n)/2 which is the answer.

    the only problem i have with this is that we haven't counted any components of C, have we? i don't understand where this piece has gone? yet it seems to work. for example consider a 4x4 matrix with 2x2 blocks then a basis for B would be [0,1][-1,0] and a basis for E would be [0,1][-1,0] and for C it would be {[1,0][0,0],[0,1][0,0],[0,0][1,0],[0,0][0,1]} this is a 6 dimensional basis and (4^2-4)/2=6 as required.
    why is this working when we didn't count any of the basis elements for C????


    ok.

    should i work with the definition that

    [itex]SU(p,q) = \{ A \in M_n ( \mathbb{C} ) | \bar{A}^t \eta A = \eta \}[/itex]
     
  12. Mar 22, 2010 #11

    Dick

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    Re: Groups

    p^2+q^2 isn't equal to (p+q)^2. Come on, that's a rookie mistake. Your definition is U(p,q). For SU(p,q) you need to add the determinant condition. Unlike O(p,q) the determinant does affect the Lie algebra.
     
  13. Mar 22, 2010 #12
    Re: Groups

    ok. yeah i missed the det condition when i wrote it out but put it in in the calculation and got the right answer for the SU(p,q) one.

    okay so back to this dimension business.
    we have the p^2 +q^2 already counted. C has entirely free components. that means it contributes pq=2pq/2. and then that all combines to give us the p^2+2pq+q^2=(p+q)^2.

    finally i have to work out the lie algebra of S0*(n). i have been working with the defn

    [itex]SO*(n) = \{ A \in M_n ( \mathbb{H} ) | \bar{A}^t j \mathbb{I} A = j \mathbb{I} \quad \text{ & det } A = +1 \}[/itex] is this correct?
     
    Last edited: Mar 22, 2010
  14. Mar 22, 2010 #13

    Dick

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    Re: Groups

    Stop being so sloppy in counting. You had (p^2+q^2)/2 and you added 2pq and got (p^2+2pq+q^2)/2. You tell me what went wrong.
     
  15. Mar 22, 2010 #14
    Re: Groups

    sorry i corrected this almost as soon as i posted it! see my amended post.
     
  16. Mar 22, 2010 #15
    Re: Groups

    also, i have in the next part that

    [itex]\mathfrak{su}(p,q) = \Biggl\{ \left( \begin{array}{cc} X & Z \\ \bar{Z}^t & Y \end{array} \right) \vline X \in M_p ( \mathbb{C} ) , Y \in M_q ( \mathbb{C} ) , Z \in M_{p \times q} ( \mathbb{C} ) , \bar{X}^t = - X , \bar{Y}^t = - Y , \text{tr}X + \text{tr}Y=0 \Biggr\}[/itex]

    now i need to show the dimension of this is n^2 - 1.

    i have that this time since we're working in C and not R, we must also count diagonal elements of X and Y. so we get a contribution of (p^2+p)/2 and (q^2+q)/2. now Z will contribute pq=2pq/2 as before.
    therefore, we have at the moment [(p+q)^2 + (p+q)]/2=(n^2+n)/2

    then the constraint of vanishing trace forces us to subtract 1 from this giving

    (n^2+n)/2-1

    i can't seem to get that first term to go to n^2 regardless of what i try. any hints? thanks.
     
  17. Mar 22, 2010 #16

    Dick

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    Re: Groups

    Hint: you are counting the real dimension of the Lie algebra. Z doesn't contribute pq anymore, it contributes 2pq. Each entry has an independent real and imaginary part.
     
  18. Mar 22, 2010 #17
    Re: Groups

    ahhh. i see. so dim X = p^2 + p, dim Y = q^2+q, dim Z=2pq

    adding together we get n^2 + n

    so to get the right answer we need to subtract n+1 as a result of the constraint tr X + tr Y = 0. why is this? surely it's just one constraint so we should only have to subtract 1?
     
  19. Mar 22, 2010 #18

    Dick

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    Re: Groups

    How did you get dim(X)=p^2+p? I think you want to count that again.
     
  20. Mar 22, 2010 #19
    Re: Groups

    well initially i had it as (p^2+p)/2 but this just gives us the number of independent components. surely each of these will need to be counted twice for its' real and imaginary part?
     
  21. Mar 22, 2010 #20

    Dick

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    Re: Groups

    Careless counting again. Take it from the beginning. Think carefully this time.
     
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