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It is well known that the Lie algebra of $SO(4,1)$ is given by the expression $$A^tB+BA=0,\quad (\ast)$$ where $B=\text{diag}(1,1,1,1,-1)$, that is, $$\mathfrak{so}(4,1)=\{A\in\text{Mat}(5,\mathbb{C}):A^tB+BA=0\}.$$ Take an element $A\in\mathfrak{so}(4,1)$ and write it as $$A = \begin{pmatrix} W & x \\ y^t & z \end{pmatrix},$$ where $W\in\text{Mat}(4,\mathbb{C})$, $x,y\in\mathbb{C}^2$, $z\in\mathbb{C}$. In this block decomposition, $B = \begin{pmatrix} \mathbb{I}_4 & 0 \\ 0 & -1\end{pmatrix}$ and $(\ast)$ becomes $$\begin{pmatrix} W^t & y \\ x^t & z\end{pmatrix} \begin{pmatrix} \mathbb{I}_4 & 0 \\ 0 & -1\end{pmatrix} + \begin{pmatrix} \mathbb{I}_4 & 0 \\ 0 & -1\end{pmatrix} \begin{pmatrix} W & x \\ y^t & z\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}.$$

Writing out separately the equation for each block we get the conditions $$W^t = -W, \qquad y = x, \qquad z = 0,$$ so, \begin{align} \mathfrak{so}(4,1) &= \left\{ \begin{pmatrix} W & x \\ x^t & 0\end{pmatrix} : W^t = -W \right\} \\ &= \left\{ \begin{pmatrix} 0 & -w_1 & -w_2 & -w_3 & x_1 \\ w_1 & 0 & -w_4 & -w_5 & x_2 \\ w_2 & w_4 & 0 & -w_6 & x_3 \\ w_3 & w_5 & w_6 & 0 & x_4 \\ x_1 & x_2 & x_3 & x_4 & 0 \end{pmatrix} \right\}. \end{align} It is also well known that every $A\in\mathfrak{so}(4,1)$ can be written uniquely as $$\begin{pmatrix} W & x \\ x^t & 0\end{pmatrix}=\begin{pmatrix} W & 0 \\ 0 & 0\end{pmatrix}+\begin{pmatrix} 0 & x \\ x^t & 0\end{pmatrix},$$ which is the Cartan decomposition $\mathfrak{so}(4,1)=\mathfrak{l}\oplus\mathfrak{p}$.


Now, I also have seen that $SO(4,1)$ can be realized as $$SO(4,1)=\{A\in\text{Mat}(5,\mathbb{C}):A^tBA=B, \det(A)=1\},$$ where $$B=\text{diag}(-1,1,1,1,1),$$ (an even with some other stranger matrices like $B=\begin{pmatrix} 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 \end{pmatrix}$). Using the same idea as before, and writting now the element $A\in\mathfrak{so}(4,1)$ as $A = \begin{pmatrix} z & x^t \\ y & W \end{pmatrix}$, one can get the following representation by blocks for $\mathfrak{so}(4,1)$: \begin{align} \mathfrak{so}(4,1) &=\left\{\begin{pmatrix} 0 & x^t \\ x & W\end{pmatrix} : W^t = -W \right\} \\ &=\left\{\begin{pmatrix} 0 & x_1 & x_2 & x_3 & x_4 \\ x_1 & 0 & -w_1 & -w_2 & -w_3 \\ x_2 & w_1 & 0 & -w_4 & -w_5 \\ x_3 & w_2 & w_4 & 0 & -w_6 \\ x_4 & w_3 & w_5 & w_6 & 0 \end{pmatrix} \right\}. \end{align} With this representation it is claimed in a paper that the Cartan decomposition $\mathfrak{so}(4,1)=\mathfrak{l}\oplus\mathfrak{p}$ is given by $$\mathfrak{l}=\left\{\begin{pmatrix} A_1 & 0 \\ 0 & A_2\end{pmatrix} : A_1^tI_{1,1}+I_{1,1}A_1 = 0, A_2+A_2^t = 0 \right\},$$ and $$\mathfrak{p}=\left\{\begin{pmatrix} 0 & B_1 \\ -B_1^tI_{1,1} & 0\end{pmatrix}\right\},$$ where $I_{1,1}=\text{diag}(-1,1)$.


My questions

  1. I checked the conditions for the following blocks $A_1, A_2$ and $B_1$ and they hold, are these the right blocks for the decomposition?

$A_1=\begin{pmatrix} 0 & x_1 \\ x_1 & 0\end{pmatrix},\quad$ $A_2=\begin{pmatrix} 0 & -w_4 & -w_5 \\ w_4 & 0 & -w_6 \\ w_5 & w_6 & 0 \end{pmatrix},\quad$ $B_1=\begin{pmatrix} x_2 & x_3 & x_4 \\ -w_1 & -w_2 & -w_3 \end{pmatrix}$.

  1. Now, if all the previous is correct and I am given with a matrix for which I want to know whether it belongs to $\mathfrak{l}$ or $\mathfrak{p}$, how should I proceed? Since there seems to be different representations for the Lie algebra and also for the Cartan decomposition, how can I know if a given matrix belongs to any of these subspaces of $\mathfrak{so}(4,1)$.

Any help is really appreciated.

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The Cartan decomposition is not unique, but only unique up to conjugacy. In the case of $\mathfrak g:=\mathfrak{so}(4,1)$ you can describe it in a way which is independent of the relazation is follows. Your algebra by definition is the algebra of all endomorphisms of a five-dimensional space $V$ endowed with a Lorentzian inner product. The main input for the Cartan decomposition is given by choosing a subspace $W\subset V$ of dimension $4$ on which the Lorentzian inner product is positive definite. Then the stabilizer of $W$ in $\mathfrak g$ is isomorphic to $\mathfrak{so}(4)\times\mathfrak{so}(1)=\mathfrak{so}(4)$ and this is the Lie algebra $\mathfrak k$ of a maxiaml compact subgroup. Via the restriciton of the adjoint action $\mathfrak k$ acts on $\mathfrak g$ leaving the subspace $\mathfrak k$ invariant. By general results, it follows that there is a $\mathfrak k$-invariant complementary subspace $\mathfrak p\subset\mathfrak g$ to that invariant subspace (which in this situation is uniquely determined). The Cartan decomposition then is $\mathfrak g=\mathfrak k\oplus\mathfrak p$.

In particular, the blocks $A_1$, $A_2$ and $B_1$ has the wrong sizes, you need $B_1$ to be a $2\times 2$-block. (The decomposition you describe also points out - via the two $A$-blocks - the fact that $\mathfrak{so}(4)$ is not simple but only semisimple which is specific to dimension $4$.) But the matrices $A_1$ and $A_2$ do not form blocks in the strict sense of the word.

So the question whether a given matrix lies in $\mathfrak k$ or in $\mathfrak p$ strictly speaking does not make sense unless you fix one Cartan decomposition. Also then "most" matrices do not lie in either of the two factors. A reasonable related notion may be to look at the trace of $A^2$ (which always is negative on $\mathfrak k$ and positive on $\mathfrak p$).

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  • $\begingroup$ Is the notion of the trace of $A^2$ independent of the choice of the Cartan decomposition? $\endgroup$ – Edu Apr 24 '17 at 9:39
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    $\begingroup$ I am sorry, maybe my formulation was a bit misleading. What I meant is that whenever you have a realization of $\mathfrak{so}(4,1)$, then you can look at matrices in there and take the trace of the square of such a matrix. (By general results, this coincides up to a non-zero multiple with the Killing form on the algebra.) This defines a quadratic form on $\mathfrak{so}(4,1)$ and for any choice of Cartan decomposition, this quadratic form is negative definite on $\mathfrak k$ and positive definite on $\mathfrak p$. (This comes from the basic definition of Cartan involutions.) $\endgroup$ – Andreas Cap Apr 24 '17 at 10:10
  • $\begingroup$ Could you explain why the blocks $A_1$, $A_2$ and $B_1$ are wrong? By the representation of $\mathfrak{so}(4,1)$ that I get using the matrix $B=\text{diag}(-1,1,1,1,1)$ (the one that the authors use in this paper) are the only blocks that make sense to me... $\endgroup$ – Edu Apr 24 '17 at 11:02
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    $\begingroup$ Yes, that's the right decomposition.I did not understand originally that the claim about the blocks refers to the diagonal $B$. With the suggested blocks, things cannot work out dimension-wise. If $A_1$ can be multiplied by $I_{1,1}$ it must be of size $2$, but then the space of $A_1$'s is one-dimensional. Likewise the $A_2$ block can only contribute 3 more dimensions, but $\mathfrak k$ has to have dimension 6. Your analysis of the block sizes looks reasonable, so there has to be an error elsewhere. $\endgroup$ – Andreas Cap Apr 25 '17 at 18:19
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    $\begingroup$ The paper you refer to deals with a symmetric decomposition $\mathfrak{so}(1,4)=\mathfrak k\oplus\mathfrak p$ with $\mathfrak k=\mathfrak{so}(1,1)\oplus\mathfrak{so}(3)$ and $\mathfrak p$ of dimension $6$. This exactly corresponds to the block decomposition you have indicated (with $B$ providing $6$ dimensions), and everything is fine. The name "Cartan decomposition" usually is reserved for the case that $\mathfrak k$ is a maximal compact subalgebra (which is also very important from the point of view of symmetric spaces). So it is a problem of wording only. $\endgroup$ – Andreas Cap Apr 26 '17 at 9:44

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