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.


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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