#### AKG

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*The elements of [itex]SL_2(\mathbb{Z})[/itex], the group of 2 x 2 matrices which have integer entries and determinant +1, act on the upper half of the complex plane as Möbius transformations. Let [itex]\alpha[/itex] denote the arc of the unite circle which joins [itex]\exp (i\pi /3)[/itex] to [itex]\exp (i\pi /2)[/itex] in the upper half plane and define [itex]\Gamma[/itex] to be the union of all segments [itex]g(\alpha )[/itex] where [itex]g \in SL_2 (\mathbb{Z})[/itex]. Draw a picture of [itex]\Gamma[/itex] and check that it represents a tree. Show that the action of [itex]SL_2(\mathbb{Z})[/itex] on this tree is not a free action.*

The Möbius transformation induced by the 2 x 2 matrix with entries a, b, c, and d will send the complex number z to

(az + b)/(cz + d) in

**C**U {∞}.

Is there any easy way to see what the orbit is? If we treat the plane as

**R**² in the natural way, the point (x,y) is sent to:

(a + c + bd + (ad + bc)x, y)/(c² + 2cdx + d²)

= (a + c + bd + (ad + bc)x, y)/(c² + 2cdx + d²)

This doesn't seem to help too much. Does the fact that the determinant is 1 tell me anything special, in terms of preserving the shape in some way? Anyways, assuming that we can tell what the image of the original arc will be if we can determine the image of it's endpoints, we have two endpoints, (0,1) and (sqrt(3)/2, 0.5). In the first case, we get:

(a + c + bd, 1)/(c² + d²)

Knowing that ad - bc = 1, does that place any nice restrictions on a + c + bd? I don't even want to touch the other case right now.