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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.