# A question about $O(3,1)$

Recall that $O(3,1)$ is the collection of matrices $A\in M_4(\mathbb R)$ such that $$A\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}A^T=\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}.$$ Let $\psi\in \mathfrak o(3,1)$ be an element in its Lie algebra. Can we find a $Q\in O(3,1)$ such that $$Q\psi Q^{-1}=\begin{pmatrix} 0&a&&\\-a&0&&\\&&0&b\\&&b&0\end{pmatrix}$$ for some $a,b\in \mathbb R$?

Formally, for any $\psi\in \mathfrak o(3,1)$ can we find an $\hat \psi \in \mathfrak o(2)\oplus \mathfrak o(1,1)$, such that $\psi, \hat \psi$ belong to the same $O(3,1)$-cojugate class?

No. E.g. $\psi=\begin{pmatrix}0&0&-1&1\\0&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}$ is nilpotent: $\psi^3=0$. If your $\phi=\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&b&0\end{pmatrix}$ was $Q\psi Q^{-1}$, we would have $\phi^3=\begin{pmatrix}0&-a^3&0&0\\a^3&0&0&0\\0&0&0&b^3\\0&0&b^3&0\end{pmatrix}=0$, whence $a=b=\phi=\psi=0$, contradiction.

Edit (Re: your amended question below): If all (or even some) eigenvalues of $\psi$ are nonzero then yes, $\psi$ is conjugate to one of your $\phi$'s. It is equivalent but easier to classify adjoint orbits of the covering group $\mathrm{SL}(2,\mathbf C)$, whose Lie algebra consists of all $$Z=J(\mathbf z) := \frac1{2i}\begin{pmatrix}z_3&z_1-iz_2\\z_1+iz_2&-z_3\end{pmatrix}, \qquad \mathbf z = \mathbf a + \mathbf bi\in\mathbf C^3.$$ The Lie algebra isomorphism maps $Z$ to $\psi=\begin{pmatrix}j(\mathbf a)&\mathbf b\\{}^t\mathbf b&0\end{pmatrix}$ where $j(\mathbf a)=\left(\begin{smallmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{smallmatrix}\right)$.

Now $\det(\lambda\mathbf1-2iZ)=\lambda^2 - (\mathbf z,\mathbf z)=(\lambda+\zeta)(\lambda-\zeta)$ where $\zeta$ is a square root of $(\mathbf z,\mathbf z):=$ $z_1^2+z_2^2+z_3^2$ $=\|\mathbf a\|^2-\|\mathbf b\|^2+2i(\mathbf a,\mathbf b)$. So there are two cases:

• if $(\mathbf z,\mathbf z)= 0$ then $Z$ is nilpotent, hence either 0 or (Jordan) conjugate to $\smash{\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)}$ $= J(i\mathbf e_1-\mathbf e_2)$ which maps to my nilpotent $\psi$ above (eigenvalues: 0).

• if $(\mathbf z,\mathbf z)\ne 0$ then $Z$ is diagonalizable and hence conjugate to $\smash{\frac1{2i}\left(\begin{smallmatrix}\zeta&0\\0&-\zeta\end{smallmatrix}\right)}=J(\zeta\mathbf e_3)$ which maps to your $\phi$ if $\zeta = a + bi$ (eigenvalues: $\pm ai,\pm b$).

• Thanks! You are correct. I see this case now, but how about the case of all eigenvalues do not equal to zero? – A.T.Saaki Aug 11 '15 at 16:08
• Or in a coordinate-free language, the $ab$-matrix is diagonalizable (semisimple), with eigenvalues $\pm ai, \pm b$, whereas the Lie algebra $\frak{o}(3,1)$ (being semisimple and noncompact) has non-zero nilpotent elements. For a Lie algebra with a higher rank simple component, there are also elements that are neither nilpotent nor semisimple (cf Jordan decomposition). – Victor Protsak Aug 12 '15 at 4:43
• @VictorProtsak Yes, please email me, too. – Dietrich Burde Aug 14 '15 at 14:12
• I feel quite sorry! Your answer helps me a lot and I did accept your answer. But I think the left bottom of my mouse get something wrong (sometimes the one click becomes double click)! Moreover, as Protsak noted, does this phenomenon appear in higher dimensional $\mathfrak{o}(m,n)$? I am not familiar with this topic, would you offer some literatures about it? Thanks again! This comment maybe too late for you. – A.T.Saaki Aug 16 '15 at 14:54
• @A.T.Saaki No problem! Yes, the situation is similar (though more complicated, as Victor notes) for all semisimple groups. For the big picture I'd recommend §3 of Vogan's ICM paper. For more details see Collingwood-McGovern which despite the title, also has a chapter on semisimple orbits. – Francois Ziegler Aug 16 '15 at 16:58