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Homework Help: Group theory question- very basic

  1. Sep 13, 2005 #1
    A little intro: This is for a whirlwind intro to Group Theory as part of another class (QFT) in which we are not proving anything, simply introducing definitions and theorems. We are not using a text book, simply some notes the professor has written up for this intro and the notes are very incomplete. They also do not give any proof or explaination. So here is the problem:

    Show that the number of group parameters for [tex] O(N) [/tex] and [tex] SO(N) [/tex] are [tex] \frac{N(N-1)}{2} [/tex]

    Does this make sense as the solution:

    We have [tex] n^2 [/tex] parameters. For [tex] O(N) [/tex] we have the condition that [tex] O^T O = 1[/tex]. This condition places [tex] \frac{n(n+1)}{2} [/tex] constraints on the parameters. This leaves [tex] n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2} [/tex]

    Since [tex] O(N) [/tex] has [tex] detO(N) = \pm 1[/tex] and [tex] SO(N) [/tex] has [tex] detSO(N)=+1[/tex], simply picking the sign of the determinent doesn't force any additional constraint on the parameters.

    I can only show this statement: "This condition places [tex] \frac{n(n+1)}{2} [/tex] constraints on the parameters." for specific examples, like N=2 and 3. I can see how it is extended to higher orders, but I cannot prove it in the general case for N. Any ideas? Also, any good books on Group Theory and Lie Algebras for Physicists?
  2. jcsd
  3. Sep 14, 2005 #2

    George Jones

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    The matrix equation O^T O = 1 gives n^2 scalar equations - one for each element of the identity matrix. By symmetry, the equations for the elements below the main diagonal are the same as the equations above the main diagonal, so the number of constraint equations is the the number of elements on the main diagonal, n, plus the number of elements above the main diagonal, say x. x equals the total number of elements in the matrix minus the number of elements on the diagonal, all divided by 2, i.e., x = (n^2 - n)/2.

    Cosequently, the number of contraint equations is

    n + x = n(n + 1)/2.

    Another way to calculate this is using an arithmetic series. The number of constraint equation for the top row of the identity matrix is n. The number of constraint equations for the second row of the identity matrix is n - 1, the number of elements in this row starting with the main diagonal and moving right, etc. So the total number of constraint equations is

    n + (n -1) + (n - 2) + ... + 2 + 1 = n(n + 1)/2.

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