Isometry group of $SL(2,\mathbb{R})$ What is the cleanest way to see that the $SL(2,\mathbb{R})$ group manifold has the isometry group $SL(2,\mathbb{R})_{L} \times SL(2,\mathbb{R})_{R}$?
Please be gentle with your answers - I am a physicist by training! My main motivation behind this question is to understand the isometry group of $AdS_{3}$.
 A: OK, here is a sketch of the proof the that the isometry group of $X=SL(2,R)$ equipped with the pseudo-Riemannian metric defined via the Killing form, is locally isomorphic to $SL(2,R)\times SL(2,R)$. (There are few caveats which explain "locally" in this statement: (1)  $SL(2,R)\times SL(2,R)$ acts on $SL(2,R)$ with kernel isomorphic to $Z_2$, which is the diagonal embedding of the center of $SL(2,R)$. (2) The actual isometry group will be a finite index extension of the quotient  $(SL(2,R)\times SL(2,R)/Z_2)$.)
First of all, the action of $SL(2,R)_L$ (by left multiplications) is transitive on $X$. Hence, given any $g\in Isom(X)$, we can compose $g$ with an element $h_1\in SL(2,R)_L$ so that $h_1g(e)=e$ (where $e\in SL(2,R)$ is the identity element). Thus, it suffices to consider only thouse isometries $g\in Isom(X)$ which stabilize $e$. Every such isometry induces a linear isometry $dg_e: T_eX\to T_eX$ of the tangent space $T_eX= sl(2,R)$, the Lie algebra of $SL(2,R)$. Let's understand how this linear isometry is acting. The metric preserved by this isometry is the Killing form, it has signature $(1,2)$. On the other hand, the group $SL(2,R)$ already acts on its own Lie algebra via the adjoint representation (each $h\in SL(2,R)$ acts on $SL(2,R)$ via conjugation $Inn(h): x\mapsto hxh^{-1}$ and the adjoint representation $Ad(h)$ is the corresponding infinitesimal action on the tangent space $T_e SL(2,R)$). The image of the adjoint representation is isomorphic to $SO(1,2)_0$, the identity component of the Lorentz group (i.e. the subgroup of $SO(1,2)$ preserving the future light cone). Therefore, after passing to index 2 subgroup of $Isom(X)$, we may assume that our element $g\in Isom(X)$ (fixing $e$) satisfies $dg_e=Ad(h_2)$ for some $h_2\in SL(2,R)$. I claim that $g=h_2$ in this situation. After replacing $g$ with $Inn(h_2)\circ g$, we may assume that $dg_e$ is the identity map $T_eX\to T_eX$. It is a general fact about connected pseudo-Riemannian manifolds $X$ that if an isometry $g\in Ixom(X)$ fixes a point $x\in X$  and acts as the identity map on the tangent space $TxX$, then $g=id$. To prove it, consider the pseudo-Riemannian exponential map $exp_x: T_xX\to X$.  Restricted to a small neighborhood $B$ of $0\in T_xX$ the exponential map is a diffeomorphism to its image.  Since $g$ maps geodesics to geodesics preserving their parameterization, we obtain that for each $v\in B$, $y=\exp_x(v)$, 
$$
g(y)=\exp_x(dg_x(v))= \exp_x(v)=y. 
$$ 
Thus, $g$ restricts to the identity map on $\exp_x(B)$. Now the argument is easier if the pseudo-Riemannian metric is real-analytic (as it is the case for $SL(2,R)$), which implies that every isometry of $X$ is real-analytic. Hence, $g=id$, since it restricts to the identity map on a nonempty open subset and $X$ is connected. The general case (where $X$ is merely a smooth pseudo-Riemannian manifold) requires a little bit more work which I will skip. 
In conclusion (ignoring the index 2 issue): Starting with an arbitrary $g\in Isom(X)$ we found two elements $h_1\in SL(2,R)_L$ and $h_2\in SL(2,R)$ such that the composition
$$
Inn(h_2^{-1}) \circ h_1\circ g = id. 
$$ 
qed
A: This answer is similar to Moishe's one, with possibly a simpler layout.
Write $X$ the pseudo-Riemannian manifold, and $x_0$ the base-point. The tangent space at 1 is a 3-dimensional space with a quadratic form of signature $(2,1)$, whose isometry group is isomorphic to $\mathrm{PGL}_2(\mathbf{R})$ (realized by its adjoint action). 
Let $s$ be the unique nonzero homomorphism from $\mathrm{PGL}_2(\mathbf{R})$ to a group of 2 elements, and $G=\{(x,y)\in\mathrm{PGL}_2(\mathbf{R})^2:s(x)=s(y)\}$. We can identify $X$ with the quotient of $G$ by the diagonal subgroup $H=\{(x,x):x\in\mathrm{PGL}_2(\mathbf{R})\}\subset G$, the $G$-action on $X=G/H$ preserving the pseudo-Riemannian metric and $H$ being the stabilizer of $x_0$.
We claim that $G$ is the full isometry group $I$. Indeed, if $f$ is an isometry, first (1) after post-composing by an element of $G$, we can write $f=g\circ f'$ with $g\in G$ and $f'(x_0)=x_0$, and then (2) after post-composing by an element of $H$, we can write $f'=h\circ f''$ with $h\in H$ and $f'$ acting trivially on the tangent space at $x_0$. (For (1) we use that $G$ acts transitivity, and for (2) we use that $I_{x_0}$ preserves the quadratic form on the tangent space and that $H$ achieves all isometries of the latter.) Then (as used in Moishe's post) since $f''$ is an isometry fixing a point and the tangent space of this point, it is the identity. So $I=G$.
