Relation between generators of $\mathfrak{sl}_2(\mathbb{R})$ and $SL_2(\mathbb{R})$

Question

So $\mathfrak{sl}_2(\mathbb{R})$ is generated by the following three traceless matrices.

\begin{equation} X_1 = \begin{bmatrix} 0&1\\1&0\end{bmatrix},\quad X_2 = \begin{bmatrix} -1&0\\0&1\end{bmatrix}, \quad X_3 = \begin{bmatrix} 0&-1\\1&0\end{bmatrix} \end{equation}

The exponential map goes from the lie algebra to the lie group such that for $X\in \mathfrak{sl}_2(\mathbb{R})$, $e^{tX} = \gamma_X(t)$, i.e the integral curve generated by $X$. So applying this map to the three generators gives;

\begin{equation} e^{tX_1} = \begin{bmatrix}\cosh(t)&\sinh(t)\\ \sinh(t)&\cosh(t)\end{bmatrix}, \quad e^{tX_2} = \begin{bmatrix}e^{-t}&0\\0&e^t\end{bmatrix}, \quad e^{tX_3} = \begin{bmatrix}\cos(t)&-\sin(t)\\ \sin(t) &\cos(t)\end{bmatrix} \end{equation}

So my question is, do these three global flows generate $SL_2(\mathbb{R})$? I don't see how the first matrix does since it is only trivially a shear mapping. I don't really understand what information we can get about the group if we know the algebra.

Can someone explicitly show that these three matrices generate $SL(2,\mathbb{R})$? Because people are saying it does but I really don't see it.

EDIT: So the reason I am having trouble seeing this is because I was thinking about 'generating' $SL(2,\mathbb{R})$ from the Iwasawa decomposition. The Iwasawa decomposition says that $SL(2,\mathbb{R}) = KAN$ such that every $g\in SL(2,\mathbb{R})$ has a unique representation $g = kan$ where $k \in K, a\in A, n \in N$. Where \begin{equation} K=\left\{\left(\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\right\}, \quad A=\left\{\left(\begin{array}{cc} r & 0 \\ 0 & 1 / r \end{array}\right): r>0\right\}, \quad N=\left\{\left(\begin{array}{ll} 1 & x \\ 0 & 1 \end{array}\right)\right\} \end{equation}

So I was having trouble showing that these three matrices generate $SL(2,\mathbb{R})$ since I didn't know how to show that $e^{tX_1}$ and/or some combination of $e^{tX_2}$ and $e^{tX_3}$ gives you $N$.

However, the Cartan decomposition seems to imply that any $g \in SL(2,\mathbb{R})$ can be represented as $g = k_1 a k_2$ for some $k_1,k_2 \in K$ and $a \in A$. Does this mean that we don't even need $e^{tX_1}$ to generate $SL(2,\mathbb{R})$? My understanding is that the Cartan decomposition is not unique for each element. So does the Iwasawa decomposition need $N$ in order to have this uniqueness but it is otherwise not necessary to generate the whole group?

Answer 1

To conclude the discussion: the Cartan decomposition (aka the SVD) implies that already the 1-parameter subgroups of positive diagonal matrices and of rotations generate $SL(2, {\mathbb R})$. In particular, you do not need the subgroup $\exp({\mathbb R} X_1)$, the other two subgroups suffice.

Answer 2

Note that $\mathfrak{sl}(2, \mathbb{R})$ is 3-dimensional, so given that the number of matrices is 3, 'generating' $\mathfrak{sl}(2, \mathbb{R})$ is kind of a low bar here: they just need to span it as a vector space (i.e. be linear independent) and we do not need to use the Lie algebra structure at all. On the group side things are different.

It is true that the three matrices 'generate' the group in the sense that every $g \in SL(2, \mathbb{R})$ can be written as a product $e^{t_1X_1}e^{t_2X_2}e^{t_3X_3}$ (i.e. as a product of elements of the three one-parameter subgroups you describe) but we need the full one-parameter subgroups for that, i.e. all possible values of $t$.

Still... the three one-dimensional curves are of course only a tiny fraction of the huge three-dimensional donut-interior that $SL(2, \mathbb{R})$ is. So while we didn't need the Lie-algebra structure to generate the Lie algebra from these generators we very much need the group structure to generate the group from their respective curves.

It is not entirely clear either that if we would have picked a random different vectorspace basis of $\mathfrak{sl}(2, \mathbb{R})$, the resulting one parameter subgroups would still genarate the group. I think they do, but I forgot the argument; maybe someone can comment on that.

However I do know why the one-parameter subgroups spawning from these matrices generate the group and so there is a strong suspicion that whoever chose this basis of the vectorspace $\mathfrak{sl}(2, \mathbb{R})$ over all the other random sets of three linearly independent matrices was in fact secretly thinking about the group structure.

It is a bit late here/now, I will edit in the 'why is every $g \in SL(2, \mathbb{R})$ writable as $e^{t_1X_1}e^{t_2X_2}e^{t_3X_3}$ for these $X_1, X_2, X_3$'-part tomorrow or later this week.

However I can already point out the meta-remark that seeing why this works does not really answer your question 'what information we can get about the group if we know the algebra?' but rather goes in the opposite direction: 'how does knowing the group point out why some vectorspace bases of the algebra are nicer/more useful/more beautiful than others?'.

[Then again one could argue that we didn't need any group-level argument to show that this basis is more beautiful and nicer than others, the equations $[X_2, X_1] = 2X_3, [X_2, X_3] = 2X_1, [X1, X_3] = 2X_2$ already do that for us! (By which I mean that a random different basis would be more ugly and less useful in the precise sense that the bracket of two basis element would be a real linear combination of all three rather than an integer multiple of just one of them.) However this particular type of beauty doesn't seem to have a clear counterpart on the group level.

So yeah this kind of argues that there is not much point for me to later edit in the missing part of the answer as it only answers the first question (in what sense do $X_1, X_2, X_3$ generate the group?) and not the second, but I still intend to do it because it is a nice and interesting construction.]