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.
 A: 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 $B\in\mathfrak{gl_2} = Mat_2(\mathbb{R})$.
A: Your equation can be rewritten:
$$X^T\eta + \eta X = 0 \iff X = -(\eta X\eta)^T$$
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 $\eta^1_1 = -\eta^3_3$).
This line of reasoning arrives at the same resulting block matrix form as the other answer.
A: 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 $\dim_{\mathbb R}(\mathfrak{so(2,2)}) = 6$?
