Why does the Lie group ##SO(N)## have ##n=\frac{N(N-1)}{2}## real parameters?

[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?

 

[fresh_42]

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 $\dim SU_n(\mathbb{C})= n^2-1$

 

[suremarc]

Orthogonal matrices are not necessarily symmetric, but since R^TR 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.

 

[MisterX]

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 $n(n-1)/2$.