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I'm trying to understand the proof of the Oppenheim conjecture using Ratner's theorem, and I don't immediately see why $SO(2,1)$ is generated by unipotents. Why is $SO(2,1)$ generated by unipotents? More generally, are there equivalent conditions to being generated by unipotents that are easier to check?

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You might see if this helps. It is a second post about unipotents in Lorentz groups.… –  Jack Schmidt Apr 24 '10 at 22:18

2 Answers 2

up vote 6 down vote accepted

Groups like SO$(2,1)$ have been studied in several frameworks: geometry and generation of classical groups over various ground fields, where unipotent elements tend to appear as transvections; real Lie groups, where the structure theory of groups like this is well developed (as in Helgason's old book, for example, republished by AMS); real points of (almost) simple algebraic groups, where papers by Borel and Tits have worked out the abstract group structure and generation in considerable generality. From the last point of view, I guess the crucial point about your question is that the group is isotropic over $\mathbb{R}$ and thus noncompact as a Lie group unlike SO(3). In particular, it has nontrivial unipotent elements (here of infinite order as group elements); the subgroup they generate in the algebraic group setting is closed and normal as well as defined over $\mathbb{R}$.

Your example is old and references can be quoted, but you need to keep in mind the different approaches possible. What is simplest here depends a lot on your viewpoint.

P.S. The blog entry by Terry Tao suggested by Jack Schmidt gives a nice way to visualize the 3-dimensional situation, but the general perspectives I've sketched are useful in higher dimensions.

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Why does non-compact mean that it must have non-trivial unipotents? And does the subgroup they generate being closed and normal imply it has to be all of $So(2,1)$? I don't know much about the general structure of Lie groups-everything I know comes from Fulton & Harris. I know even less about algebraic groups. –  Sam Ruth Apr 25 '10 at 19:37
The structure of arbitrary real semisimple Lie groups is complicated to develop (while "unipotent" elements are more visible for linear algebraic groups), so your first question takes work to sort out. For compact Lie groups the special feature is that every element is conjugate to one in a fixed maximal torus, hence semisimple. In the noncompact case, generation by unipotents appears concretely for classical groups like SO$(n,1)$; but in general Borel-Tits structure theory for algebraic groups is probably most helpful and covers all fields of definition: noncompact becomes "isotropic". –  Jim Humphreys Apr 25 '10 at 22:30
Thanks Jim. Is there a good book on algebraic groups and Borel-Tits structure theory? I'm ultimately interested in doing ergodic theory, if that helps narrow it down. –  Sam Ruth Apr 26 '10 at 14:33
Short answer: no. Books by Borel, Springer, me have limited coverage of structure over arbitrary fields. Try AMS online surveys by Tits (1965) PSPUM/9, Springer (1979) PSPUM/33.1: (and note Springer's 3.4 for your original question). Extensive details are in papers by Borel-Tits (IHES papers online at Groupes reductifs (Publ. Math. IHES 27, 1965), etc. Lie groups are just an example there. But books by Margulis, Zimmer focus more on ergodic theory, less on background. Good luck! –  Jim Humphreys Apr 26 '10 at 18:35

For $SO(2,1)$ the problem is easy, since the group is isomorphic to $SL(2,\RR)$. To see this consider the action by conjugation of $SL(2,R)$ on the vector space of $2 \times 2$ matrices with real coefficients and trace zero. This action preserves determinant, which is a quadratic form of signature $(2,1)$. This gives you an isomorphism. Then all you need to do is to track the obvious unipotents in $SL(2)$ and see their image in $SO(2,1)$.

In the context of Oppenheim conjecture, you can also refer to any one of the early papers of Margulis. He explicitly described the required unipotents.

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But in the action you describe, −I acts as the identity. So it's really an action of $PSL(2,R)$ right? Wikipedia says this map gives you an isomorphism between $PSL(2,R)$ and $SO^+(2,1)$. Does this map lift somehow to an isomorphism between $SL(2,R)$ and $SO(2,1)$? –  Sam Ruth Apr 26 '10 at 14:32

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