# Lie groups -- Real parameters

• A
• LagrangeEuler

#### LagrangeEuler

When we have a Lie group, we want to obtain number of real parameters. In case of orthogonal matrices we have equation
$$R^{\text{T}}R=I$$,
that could be written in form
$$\sum_i R_{i,j}R_{i,k}=\delta_{j,k}$$.
For this real algebra ##SO(N)## there is ##n=\frac{N(N-1)}{2}## real parameters. Why this is the case when unitary matrix is not symmetric?

Why this is the case when unitary matrix is not symmetric?
Why is what the case? ##\dim U_n(\mathbb{C}) =n^2## and with the restriction ##\det =1## we get ##\dim SU_n(\mathbb{C})= n^2-1##

Orthogonal matrices are not necessarily symmetric, but since RTR is symmetric, we get at most n(n+1)/2 constraints in n^2 variables. Thus we are left n(n-1)/2 degrees of freedom, with some hand-waving involved.

• WWGD
One approach is to note that the number of real parameters is the same as the dimensions of the Lie algebra. We consider a group element ##R## and linear-ize to first order to produce an element ##R' =R+\epsilon K ## which must obey the group properties up to first order.
\begin{align*} &\Big(R^T + \epsilon K^T\Big)\Big(R+\epsilon K \Big) =R^T R & + \epsilon(K^T R + R^T K) \\ &\text{at }R=I \text{ we must have } &K^T + K = 0\end{align*}
Thus we must have zeros along the diagonal. The upper triangular matrix part of the matrix is just the negative transpose of the lower triangular matrix, that is ##K_{ij}=-K_{ji}##. So the number of real parameters for K is the same as the number of upper triangular components which is ##n(n-1)/2##. As the dimension of the Lie algebra and the group are the same, the number of parameters for the group is also ##n(n-1)/2##.