Because of Iwasawa Decomposition(link), I know that $SL(2,\mathbb{R})$ is also decomposable by $K,A,$ and $N$, where $K$ is $SO(2)$, $A$ is the set of diagonal matrices with $(x,\frac{1}{x})$, and $N$ is the group of matrices $((1,x),(0,x))$.

However, when I read "Ergodic theory, with a view towards number theory," which is written by Manfred Einsiedler, it states that on the page 284,

Now the subgroup U = $\left\{\begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix}: b \in \mathbb{R}\right\} \leq SL_{2}{(\mathbb{R})}$ together with

$w = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ generates $SL_{2}{(\mathbb{R})}$, since $wUw^{-1} = \left\{\begin{pmatrix}1 & 0 \\ -b & 1 \end{pmatrix}: b \in \mathbb{R} \right\}$.

But, I don't know how to this $w$ and $U$ generates the whole $SL_{2}(\mathbb{R})$. How can I show that they are generator of $SL_{2}(\mathbb{R})$?

  • 2
    $\begingroup$ Perhaps it's a bit much, but it suffices to know that an element of $SL_2$ admits an $LU$ decomposition iff its upper left entry is non-zero, and a $UL$ decomposition iff the bottom left entry is non-zero. $\endgroup$ Sep 11, 2016 at 12:26
  • $\begingroup$ @Omnomnomnom Wow.. I forgot it. Yes, LU-decomposition may works. I'll try it. Thank you! $\endgroup$
    – user124697
    Sep 11, 2016 at 12:38
  • $\begingroup$ That should be the bottom right entry in the second. And your welcome. $\endgroup$ Sep 11, 2016 at 12:39
  • $\begingroup$ Your $N$ matrices don't have determinant $=1$. $\endgroup$
    – mr_e_man
    Sep 16, 2018 at 19:21
  • $\begingroup$ It is possible to write a matrix in $SL(2,R)$ as a product of lower and upper triangular matrices,all of whose diagonal entries are $1$. $\endgroup$ Aug 26, 2019 at 10:59

1 Answer 1


This is 4 years late, but here's an answer for others. All lower and upper triangular matrices mentioned below are to have 1s on the diagonal, they are taken from the given set of generators.

First, we look at the product $$\begin{bmatrix}1&0\\\alpha&1\end{bmatrix} \begin{bmatrix}x&0\\0&1/x\end{bmatrix} \begin{bmatrix}1&\beta\\0&1\end{bmatrix} = \begin{bmatrix} x&\beta x\\\alpha x& \alpha\beta x+1/x\end{bmatrix}$$

Given $\begin{bmatrix}a&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$, when $a\neq 0$, we can solve for $\alpha, x, \beta$. Since we have lower and upper triangular matrices, it then suffices to show that the generators in question also generate the diagonal matrices. We will come to the $a = 0$ case later.

Now we look at a product of the form LULU: $$\begin{bmatrix} 1&0\\\alpha&1\end{bmatrix} \begin{bmatrix} 1&\beta\\0&1\end{bmatrix} \begin{bmatrix} 1&0\\\gamma&1\end{bmatrix} \begin{bmatrix} 1&\delta\\0&1\end{bmatrix}= \begin{bmatrix} 1+\beta\gamma&\delta(1+\beta\gamma)+\beta\\ \alpha(1+\beta\gamma)+\gamma&\alpha\delta(1+\beta\gamma)+\alpha\beta+\gamma\delta+1 \end{bmatrix}$$

Assuming $1+\beta\gamma\neq 0$, to get the anti-diagonal terms to be $0$, we take $$\alpha = \frac{-\gamma}{1+\beta\gamma}, \delta = \frac{-\beta}{1+\beta\gamma}$$ then the bottom right term becomes $$\frac{\beta\gamma}{1+\beta\gamma}+\frac{-2\beta\gamma}{1+\beta\gamma}+1 = \frac{1}{1+\beta\gamma}$$ as required.

Therefore, a matrix of the form $\begin{bmatrix}x&0\\0&1/x\end{bmatrix}$ is generated by lower and upper diagonal matrices with ones on the diagonal for suitable choice of $\beta, \gamma$.

When $1+\beta\gamma = 0$, the above product becomes $$\begin{bmatrix} 0&\beta\\ \gamma& 1+\alpha\beta+\gamma\delta \end{bmatrix}$$

We can take $\delta = 0$ and make suitable choices so that the matrix becomes $\begin{bmatrix} 0&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$.

So, any matrix in $SL_2(\mathbb{R})$ is of the form LUL or LULU where the matrices have 1s on the diagonal. The same argument holds good over other fields as well.

  • $\begingroup$ Honestly, I didn't even remember why I question this; but thank you very much for your clear explanation! $\endgroup$
    – user124697
    Dec 19, 2020 at 4:09

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