Let $G$ be a linear, semisimple Lie group with no compact factors. The unipotent elements of $G$ are those that have only eigenvalue 1.

I've seen it asserted that $G$ is generated by its unipotent elements: see Exercise #2 $\S 4.5$ in Dave Witte Morris' book on Arithmetic Groups. The hint in the book is that you consider the simple factors. But even considering $SL(2, \mathbb{R})$, it is unclear to me why it is generated by its unipotent elements.

I am aware that any matrix in $G$ can be written as a product of commuting hyperbolic, elliptic, and unipotent element, but I am unsure of how you might generally express the hyperbolic and elliptic elements as unipotent elements.


2 Answers 2


I am reading the same book, and I come up with one way to prove this (hope I'm correct)

By the Appendix (A5.3) we know that for simple noncompact Lie group, it contains the nontrivial unipotent elements. Then consider the subgroup generated $H$ by the unipotent element of $G_i$, since conjugate of unipotent are still unipotent so this subgroup is a normal subgroup and hence $H= G_i$. So the noncompact factor of $G$ is generated by the unipotent elements. `


I think I did the case $SL_2(\mathbb{R}):$

Let $$A=\begin{pmatrix} a&b\\c&d \end{pmatrix} \in SL_2(\mathbb{R}).$$

Notice that $$A\cdot\begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} -b&a\\-d&c \end{pmatrix}, $$


$$\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}\cdot A = \begin{pmatrix} c&d\\-a&-b \end{pmatrix},$$

and also

$$\begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} 1&0\\-1&1 \end{pmatrix}\begin{pmatrix} 1&1\\0&1 \end{pmatrix}\begin{pmatrix} 1&0\\-1&1 \end{pmatrix}.$$

So, using these facts, you'll be able to reduce your problem to show that

$$ A = \begin{pmatrix} a&0\\0&1/a \end{pmatrix}$$ is generated by unipotent elements. But

$$ \begin{pmatrix} a&0\\0&1/a \end{pmatrix} = \begin{pmatrix} 1&0\\1/a-1&1 \end{pmatrix} \begin{pmatrix} 1&1\\0&1 \end{pmatrix} \begin{pmatrix} 1&0\\a-1&1 \end{pmatrix} \begin{pmatrix} 1&-1/a\\0&1 \end{pmatrix}.$$


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