-2

Given the special unitary group, we can define

$$ M = -iX = \begin{bmatrix} b_{11} & -ia_{12}+b_{12} & ... & -a_{1n} + b_{1n} \\ ia_{12}+b_{12} & b_{22} & ... & ... \\ ... & ... & ... & ... \\ ia_{1n}+b_{1n} & ... & ... & b_nn = -\sum_{k=1}^{n-1} b_{kk} \end{bmatrix}$$

where all elements $a,b$ are Real.

How would one find the dimension of $su(n)$ from this? I know the answer is supposed to be $n^2 -1$ but I don't know how to prove it.

Also, one can show that the Pauli and Gen-Mann matrices span $su(2)$ and $su(3)$ respectively. How can we generalize this to an arbitrary n to find a set of basis martrices that are orthogonal under the inner product?

Also, what are the normalizations required to ensure that: $2tr(t^at^b) = \delta^{ab}$?

Qmechanic
  • 220,844
Anna
  • 115

1 Answers1

0

Too many questions for one post. I'll answer the first and the last since the second is a little bit involved and you can find the solution looking to the book Lie Algebras in particle physics by Georgi.

Take the $SU(N)$ Lie algebra. Being a complex Lie group you'll need at least $2N^2$ parameters, $N^2$ for the real part and $N^2$ for the imaginary. If you see your matrix $U$ in in $SU(N)$ on the diagonal you have no imaginary part, so $N$ parameters go away. Moreover the imaginary part is antisymmetric and so half of the parameters are fixed: not counting the diagonal there are $$\frac{N^2-N}{2}$$ fixed imaginary parameters. For the real ones they are symmetric and so another $\frac{N^2-N}{2}$ fixed parameters. What remains is the last diagonal element which is fixed by all the others and so another $1$ fixed parameters.

The number of free parameters is then $$2N^2-\frac{N^2-N}{2}-\frac{N^2-N}{2}-N-1 = 2N^2 - N^2 + N-N-1 = N^2-1$$

The normalisation stays the same $$\text{Tr } t^a t^b = \frac{1}{2}\delta^{ab}$$

Davide Morgante
  • 4,160