I've heard this result from my differential manifold class, and I don't know how to prove it.
Does anyone know how to construct such diffeomorphism? Please tell me, thanks a lot.
Any comments are welcome.
I've heard this result from my differential manifold class, and I don't know how to prove it. Does anyone know how to construct such diffeomorphism? Please tell me, thanks a lot. Any comments are welcome. 


Here is a sketch: $SL(2,\mathbb R)/SL(2,\mathbb Z)$ can be identified with the space of lattices of covolume 1 in $\mathbb R^2$. Indeed, $SL(2, \mathbb R)$ can be identified with a couple of vectors, and multiplying this couple by an element of $SL(2, \mathbb Z)$ does not alter the lattice they generate. Let $\Lambda$ be a lattice, $v$ be its shorter vector, $v'$ be the shortest vector non colinear with $v$ so that the basis $(v,v')$ is direct. Denote by $\theta$ the angle (mod $\pi$) made by $v$ with the horizontal direction. Then the couple $(v'/v, \theta)$ determines the lattice. It is an exercise to see that $v'/v$ belongs to the domain $\mathcal D:=\{z\in\mathbb C\big\vert 1/2\le\Re z\le 1/2, \vert z\vert\ge 1\}$. In order two have an isomorphism, it should be noted that the only repetitions in the code concern the points $(z,\theta)$ and $(\bar z, \theta)$ for $\vert z\vert=1$ which determine the same rhombical lattice, as well as the points $(z, \theta)$ and $(z+1, \theta)$ for $\Re z = 1/2$. Let $\mathcal D/\tilde{}$ denote the corresponding quotient of $\mathcal D$. All in all this shows that $SL(2,\mathbb R)/SL(2,\mathbb Z)$ is a Seifert circle bundle over $\mathcal D/\tilde{}$, with one singular fiber of order 2 above the point $i$ (which corresponds to square lattices) and one singular fiber of order 3 above the point $j$ (which corresponds to hexagonal lattices). Now cut $\mathcal D/\tilde{}$ along the two lines $\Re z=1/4$ and $\Re z=1/4$. You obtain two disks $D_i$ and $D_j$, where $D_i$ contains a point of index 2, and $D_j$ contains a point of index 3, and both of them have one point removed on the boundary. Now there is only one Seifert fibered space over a disk with a singular pointof order $p$, namely a solid torus where each fiber except the central one crosses $p$ times every meridian disk. Thus our Seifert fibered space is made of two solid tori glued along their boundary, where we have to identify the fibers on the boundary. What we obtain is a Lens space with one fiber removed. It turns out that in this $(2,3)$case, the obtained Lens space is $S^3$, and the regular fibers are trefoils. 


There is a proof without words in Etienne Ghys' ICM address. and a short proof with words by Rubinstein and Gardiner. (Compositio, 1979) 


A different proof (due to Quillen) is given in Milnor's book Introduction to algebraic KTheory, page 84. 

