dimension of $SO(2,1)$

Question

I am looking for the dimension of the indefinite special orthogonal group $SO(2,1)$.

I computed its Lie algebra $$\mathfrak{so}(2,1)=\{M\in M_3(\Bbb R):^t\!\!MJ=-JM,\ \mathrm{Tr}(M)=0 \},$$ where $J=\begin{pmatrix} 1 & 0 &0\\ 0 & 1 &0 \\ 0&0&-1 \end{pmatrix}$.

Let $M=\begin{pmatrix} a & b &c\\ d & e &f \\ g&h&i \end{pmatrix}\in \mathfrak{so}(2,1)$.

Simple calculations using definition of $\mathfrak{so}(2,1)$ show that $a=b=d=e=i=0$, so $\dim(\mathfrak{so}(2,1))=4$.

But this doesn't make sense since $\dim(O(2,1))=3<4$.

Thank you for your help in finding my mistake.

Answer 1

The calculations were wrong as I forgot the transpose. The final solution is 3 and M looks like:

$M=\begin{pmatrix} 0 & b &c\\ -b & 0 &f \\ c&f&0 \end{pmatrix}$