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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.

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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 $\dim_{\mathbb R}(\mathfrak{so(2,2)}) = 6$?

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