# 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.

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 $B\in\mathfrak{gl_2} = Mat_2(\mathbb{R})$.

$$X^T\eta + \eta X = 0 \iff X = -(\eta X\eta)^T$$
where we have used $$\eta^2 = I$$. This tells us a few things about $$X$$:
• The diagonal entries must be zero, since e.g. for the $$a$$th diagonal entry, $$X^a_a = -(\eta^a_a)^2 X^a_a = -X^a_a$$. (automatically satisfies trace condition)
• If an off-diagonal entry $$X^a_b$$ is nonzero, the opposite (as in transposed) entry must have a relative sign of $$-\eta^a_a\cdot\eta^b_b$$ by a similar argument.
Thus, if $$\eta$$ is definite (like for SO(4)), the $$X$$ are anti-symmetric, while if $$\eta$$ is indefinite (like for SO(2,2)), then some $$X$$ are anti-symmetric, (e.g. if $$a = 1$$ and $$b = 2$$, so $$\eta^1_1 = \eta^2_2$$) and some are symmetric (e.g. if $$a=1$$ and $$b=3$$, so $$\eta^1_1 = -\eta^3_3$$).
It follows that $$\begin{split} \mathfrak{so(2,2)} = span\left\{ \left(\begin{matrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ \end{matrix}\right), \\ \left(\begin{matrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix}\right), \left(\begin{matrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ \end{matrix}\right), \right\} \end{split}$$ Therefore $$\dim_{\mathbb R}(\mathfrak{so(2,2)}) = 6$$?