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Dimension of a Lie algebra

  1. Jan 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Show the (real) dimension of su(n) is ##n^2-1##.

    2. Relevant equations


    3. The attempt at a solution
    ##su(n) = \{ A \in M_n(\mathbb{C}) | A+A^T = 0, tr(A) =0 \}##
    Maybe the solution is obvious, because I can't find a thing online about how to do this. But I can't see how to do it! I imagine since I'm after a real dimension, the first thing to do is somehow consider real and complex parts separately. But I'm not sure exactly how to do that, or how I'm meant to then find the dimension of this Lie algebra. Could someone briefly outline the method? I really appreciate any help, thanks! :)
     
    Last edited: Jan 25, 2017
  2. jcsd
  3. Jan 25, 2017 #2

    fresh_42

    Staff: Mentor

    How many real entries do the matrices ##A \in \mathfrak{su}(n)## have? And how many conditions on this entries exist? Are these conditions linear? If so, what do you know about linear equation systems?
     
  4. Jan 25, 2017 #3
    There are n^2 real entries, because if the entries are complex you could split A into the sum of a matrix of real components with a matrix of the imaginary components. There are two conditions on the entries, but I don't know what would make a condition linear - the conditions don't involve powers so in that regard they're linear.

    A system of linear equations in 3D, for example, describes a set of planes.
     
  5. Jan 25, 2017 #4

    fresh_42

    Staff: Mentor

    There are ##2n^2## real numbers: ##A_{ij}=x_{ij}+iy_{ij}##. Now which conditions for these numbers do you get from ##A+A^T=0##, and what does the ##"T"## mean?
     
  6. Jan 25, 2017 #5
    T is the hermitian conjugate, couldn't make the dagger symbol work in latex though. Transpose and take complex conjugate. So that would mean that ##x_{ij}+iy_{ij} + x_{ji}-iy_{ji} = 0## for all i, j I think. That would mean ##y_{ij}=y_{ji}##, and ##x_{ij} = -x_{ji}##. If that's right.
     
  7. Jan 25, 2017 #6
    The trace leads to the condition ##x_{ii} + iy_{ii} = 0 ## so ##x_{ii} =- iy_{ii}##. Doesn't mean that actually! But apparently I can't delete posts.

    ##\Sigma x_{ii} + iy_{ii} = 0##.
     
  8. Jan 25, 2017 #7

    fresh_42

    Staff: Mentor

    Yes. So how many in total? And how many conditions do we get from ##trace(A)=0\,?##
     
  9. Jan 25, 2017 #8

    fresh_42

    Staff: Mentor

    So we have ##2n^2## variables and a certain number of linear (independent) conditions (equations) for them.
    Now we only need the formula ##dim (V) = rank(\varphi) + dim(ker(\varphi))## for a linear mapping ##\varphi: V \rightarrow V## which is only another way to write the degree of freedom (dimension of the solutions of) a system of linear equations. (The equations describe ##\varphi##, the variables span ##V## and the kernel of ##\varphi## is given by the linear conditions.)
     
  10. Jan 25, 2017 #9
    Wow ok, never seen anything like that before. Is ##2n^2## the rank of ##\phi##? And how do I find the kernel from the linear conditions? As to how many in total, I think the condition ##A + A^T = 0## reduces the number of unique entries because the entries ##y_{ij}## are duplicated, I'm just trying to work out by how much they would reduce.
     
  11. Jan 25, 2017 #10

    fresh_42

    Staff: Mentor

    No, ##rank(\varphi)## is the number we are looking for. We have ##2n^2## real variables, and equations like ##x_{ij}=-x_{ji}\; , \;y_{ij}=y_{ji}##. The only issue here is not to forget, that this also implies some conditions for the diagonal. These conditions all can be written as ##\ldots =0##, i.e. they describe the kernel of ##\varphi##.
    Yes. ##2## conditions on each ##A_{ij}## with ##i\neq j##, ##n## conditions on the diagonal (which and why not ##2n\,##?) plus on condition for the trace.
     
  12. Jan 25, 2017 #11
    Well, ##y_{ij} = y_{ji} ## implies for the diagonal that ##y_{ii}=y_{ii}##. But for ##x##, the implication is that ##x_{ii}=-x_{ii}##. That would mean ##x_{ii} =0##. Is that ##n## conditions? It reduces the number of parameters by n, and means that ##\Sigma iy_{ii} = 0##, which is the condition on the trace I'm guessing. How do these conditions relate to the kernel? Reading about it, the kernel is the set of solutions to the =0 equations. Is the dimension then equal to the number of solutions -1? Just like if you have linear equations in 3 variables it describes planes which are 2D (in 3D space of course)?
     
  13. Jan 25, 2017 #12

    fresh_42

    Staff: Mentor

    We have ##dim(V) = rank(\varphi) + dim(ker(\varphi)) = 2n^2\,,## because there are as many variables, or coordinates in ##V=\mathbb{R}^{2n^2}\,.## The kernel of the linear mapping ##\varphi## is ##M_\varphi \, \vec{x}=0## where ##M_\varphi## is the matrix of ##\varphi## according to our coordinates ##\vec{x}##, which are the ##2n^2## possible real values of ##A \in \mathfrak{su}(n)##.
    The rows of ##M_\varphi## are the coefficients of our equations, resp. conditions. They are basically from ##\{-1,0,1\}##. But they are all linearly independent, so we have as many dimensions of ##ker(\varphi)## as we have conditions on ##\vec{x}=(x_{ij},y_{ij})\,.##
    So all that's left to do, is counting the conditions (= equations = non-zero rows of ##M_\varphi##) and subtract it from ##2n^2##.
     
  14. Jan 25, 2017 #13
    If the condition ##x_{ii} = 0## counts as ##n## equations, and the condition on the trace is another equation, and ##x_{ij}=-x_{ji}## is a condition on everything that's not on the diagonal so that's ##n^2 - n##, and ##n^2-n## for the ##y## equation then:

    ##2n^2 = n^2 - n + n^2 - n + n + 1 + rank(\varphi)## so after all that do you end up with ##n-1##?
     
  15. Jan 25, 2017 #14

    fresh_42

    Staff: Mentor

    I get ##dim_\mathbb{R}\, \mathfrak{su}(n)= rank(\varphi)= \underbrace{2n^2}_{dim(V)}-\underbrace{2\cdot \frac{n^2-n}{2}}_{non-diagonal\, x_{ij}\, and\, y_{ij}}-\underbrace{n}_{diagonal \, x_{ii}}-\underbrace{1}_{trace}=n^2-1\,##
    where we have only ##x_{ii}## on the diagonal, therefore ##n## conditions.
     
  16. Jan 25, 2017 #15
    I have one extra ##n^2 - n##. I was thinking ##n^2 - n## conditions from ##x_{ij}## and ##n^2 - n## for ##y_{ij}##, because all ##n^2## elements except those ##n## on the diagonal have the condition imposed on them, why is it divided by 2?
     
  17. Jan 25, 2017 #16
    Ah. Because one equation deals with 2 components? So you actually get half the number of equations as the number of elements they're imposed on?
     
  18. Jan 25, 2017 #17

    fresh_42

    Staff: Mentor

    You double counted the conditions: ##y_{ij}=y_{ji}## is only one, so you can only count one half, e.g. all upper triangular elements.
     
  19. Jan 25, 2017 #18
    Brilliant, got it! Thank you for being so incredibly patient, and for your help, I feel like I understand this so much better! :)
     
  20. Jan 25, 2017 #19
    Well, I didn't understand it at all before!
     
  21. Jan 25, 2017 #20

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If ##A = X + iY##, where ##X,Y## are real ##n \times n## matrices, we have ##A^{\dagger}= X^T - iY^T##, so ##X^T = -X## and ##Y^T = Y##. How many independent elements does ##X## have? Ditto for ##Y##. How does the condition ##\text{tr}(A) = i \text{tr}(Y) = 0## further restrict the entries?

    BTW: to get ##A^{\dagger}##, just use A ^ {\dagger} (no spaces, etc.)
     
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