# Finding the basis of $\mathfrak{so}(2,2)$ (Lie-Algebra of $SO(2,2)$)

I want to calculate the Basis of the Lie-Algebra $\mathfrak{so}(2,2)$. My idea was, to use a similar Argument as in this Question. The $SO(2,2)$ is defined by:
$$SO(2,2) := \left\{ X \in Mat_4(\mathbb R): X^t\eta X = \eta,\; \det(X) = 1 \right\}$$
(With $\eta = diag(1,1,-1,-1)$)

With the argument from the link, i get the following equation:
$\forall X \in \mathfrak{so(2,2)}$:
$$X^t\eta + \eta X = 0.$$
My idea was to use the block decomposition:
$$X = \left(\begin{matrix} A & B\\ C & D \end{matrix} \right), \; \eta = \left(\begin{matrix} \mathbb I & 0\\ 0 & -\mathbb I\\ \end{matrix}\right).$$
I get the following equation:
$$\left(\begin{matrix} A^t & -B^t\\ C^t & -D^t\\ \end{matrix}\right) + \left(\begin{matrix} A & B\\ C & D \end{matrix}\right) = 0.$$
Is this correct? I also don’t really know, what to do with the $det(X) = 1$ condition.

#### Solutions Collecting From Web of "Finding the basis of $\mathfrak{so}(2,2)$ (Lie-Algebra of $SO(2,2)$)"

The determinant condition implies that the trace of an element of the Lie algebra must be zero (see the Jacobi formula).

In your computations, you forgot to transpose $X$ (and you messed up the second matrix multiplication). You should get
$$\pmatrix{A^t&-C^t\\B^t&-D^t} + \pmatrix{A&B\\-C&-D} = 0.$$
You get that $A$ and $D$ must be antisymmetric (so the trace condition is automatically satisfied), and that $B^t = C$, so
$$X=\pmatrix{A&B\\B^t&D},$$
with $A,D\in\mathfrak{so}(2)$ and $B\in\mathfrak{gl_2} = Mat_2(\mathbb{R})$.

It follows that
$$\begin{split} \mathfrak{so(2,2)} = span\left\{ \left(\begin{matrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ \end{matrix}\right), \\ \left(\begin{matrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ \end{matrix}\right), \right\} \end{split}$$
Therefore $\dim_{\mathbb R}(\mathfrak{so(2,2)}) = 6$?