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Special Linear Group

  1. Jul 6, 2005 #1

    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 R² 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.
     
  2. jcsd
  3. Jul 7, 2005 #2

    matt grime

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    what do you know about mobius maps? they send circles and straight lines to circles and straight lines (and that means they can send circles to straightlines and stright liens to circles. they are also composed of simpler maps: translation, inversion, that is you can write any of them as a composite of maps z to 1/z and w to w+k. i';m sure if you google for them you'll find out lots of information.

    it may help (though i don;t know for sure) to show that the orbit is a graph and then that it has no closed loops, and perhaps generators will help here.
     
    Last edited: Jul 7, 2005
  4. Jul 7, 2005 #3

    AKG

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    It's not in my book, but I saw that stuff yesterday on Mathworld. Mathworld says that:

    A linear fractional transformation is a composition of translations, rotations, magnifications, and inversions.

    but you only mention translations and inversion. However, that is for the entire Mobius Group, as my book calls it, consisting of the matrices of [itex]GL_2(\mathbb{C})[/itex]. I'm looking at a very small subset of that set, [itex]SL_2(\mathbb{Z})[/itex]. No element of this set can correspond to a magnification. That doesn't mean that |g(z)| = |z| for every g in the set, it just means that there is no fixed constant a such that for all z, g(z) = az. Also, as far as I can tell, the only translations that will occur will occur in the real direction (to the right).
    I'm not sure how to show that it's a graph. Moreover, it would not only have to be a graph, but a connected one. As it stands, I have very little idea as to what [itex]SL_2(\mathbb{Z})(\alpha )[/itex] will look like, it could be a bunch of arcs floating around the plane without all being connected.

    I'll try to think more about the maps sending lines and circles to lines or circles. Maybe there are some magnifications, I'll have to look again.
     
  5. Jul 7, 2005 #4

    matt grime

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    sorry, i omitted multipliation too z to az.

    but any mobius transformation is also determined by where it sends 3 points, not sure if that helps at all. not much use here am i?
     
  6. Jul 7, 2005 #5

    matt grime

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    had a though or two.z to z+k translates parallel to the real axis. what about the other maps? w=(az+b)/(cz+d) it suffices to assume that z is on that arc, and now consider the inverse map z=M(w) for some w, say z=(ew+f)/(gw+h) then since z is on the unit circle |ew+f|=|gw+h| so you can work out all the inverse images of the unit circle, but its a group so the inverse images are also all the images.
     
  7. Jul 7, 2005 #6

    AKG

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    I'm pretty sure that there are no magnifications. The matrix:
    Code (Text):
    ( cb 0 )
    ( 0  b )
    would be a magnification by c, for some non-zero real c, and non-zero complex b. But the determinant of this matrix is (cb)(b), so if the determinant must be 1 and if cb and b must be integers, then c = 1, and so we have a magnifcation by 1, which is no magnification at all. Now I would like to say that all magnifications take this form. That is, a magnification by c corresponds to a unique matrix, but obviously, it corresponds to any matrix of the above form for any value of b. So it might even be possible that a matrix with a less nice form might also correspond to a magnification, one that may actually be both a magnification and an element of SL2(Z). However, if it is true that only multiples of the above matrix are magnifications by c, then there are no magnifications corresponding to matrices of SL2(Z).

    Assuming something similar for translations, knowing that
    ( a w )
    ( 0 a )

    is a translation through w, means that the only translations corresponding to matrices in SL2(Z) are those that move the arc some rational distance in the horizontal direction only (left or right, but no upwards or downwards component). Actually, the condition that the determinant be 1 implies that only integral translations to the left or right are possible.

    The matrix
    ( 0 a )
    ( a 0 )

    gives inversion, but since the determinant here is -a², and since the square of any integer is positive, the determinant can never be 1, so no matrix corresponds to inversion. Mathworld mentions rotation as well, and that will be tougher to deal with since it's hard to think of a Mobius transformation off the top of my head that will rotate a complex number. Since we can make any translation we want if we consider all of GL2(C), we can treat rotations as simply rotations about the origin. Then rotations must preserve the norm. I wonder what conditions are placed on the entries of a matrix if it must correspond to an element of SL2(Z) and also must preserve the norm. I suspect that there are no rotations, since the question says that these matrices act on the upper half plane, but rotations will send [itex]\alpha[/itex] eventually to the lower half plane.

    However, I know I've done something wrong. I'm not getting any lines, I've only found that translations are possible, but this won't make a tree. I suspect that my assumption that two matrices correspond to the same transformation iff one is a complex multiple of the other is wrong, but then it becomes very difficult to figure out what SL2(Z) can do to [itex]\alpha[/itex]. Any suggestions?
     
  8. Jul 7, 2005 #7

    shmoe

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    Do you know a fundamental region of [itex]SL_2(\mathbb{Z})[/itex]?

    Do you know a set of generators of [itex]SL_2(\mathbb{Z})[/itex]?
     
  9. Jul 7, 2005 #8

    AKG

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    How would I use this? It seems to me that there are too many possibilities for e, f, g, h.
     
  10. Jul 7, 2005 #9

    AKG

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    I don't know what a fundamental region is, and I don't know that it matters if I know a set of generators of [itex]SL_2(\mathbb{Z})[/itex], since I don't think [itex]SL_2(\mathbb{Z}[/itex] is homomorphic to the Mobius group. Matrix multiplication and Mobious transformation composition are quite different. And actually, no, I can't think of a set of generators for [itex]SL_2(\mathbb{Z})[/itex] off the top of my head. In fact, just as there is no non-trivial set of generators for the multiplicative group of positive real numbers, I can't see how there could be one for these matrices either.
     
  11. Jul 7, 2005 #10

    matt grime

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    really? you might want to recheck that.
     
  12. Jul 7, 2005 #11

    matt grime

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    read the question again - it asks you to sketch some of the images, so do it for a few choices of small e,f,g and h and try and spot a pattern. it's onyl by doing examples that you bein to understand the general.
     
  13. Jul 7, 2005 #12

    shmoe

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    If f and g are two mobius transformations with correponding matrices F and G, what is the matrix of f composed with g?

    see http://mathworld.wolfram.com/ModularGroupGamma.html for generators.

    A fundamental region is essentially a little patch that will cover the upper half plane under the group action such that no two points in this region are equivalent under this group action. e.g. if your group action on the real line is given by the integers (via translation) a fundamental region will be [0,1). See http://64.233.167.104/search?q=cach...tm+modular+group+gamma&hl=en&client=firefox-a for a pretty picture.
     
  14. Jul 7, 2005 #13

    AKG

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    Oops, well I did a question on the Mobius group a long time ago and I guess I forgot, but yes the multiplications are similar. In fact looking at my notes, I see that after what appears to have been painstaking efforts, I found that the inverse of a transformation induced by M is induced by M-1, and that the Mobius Group was isomorphic to GL2(C)/Z(GL2(C)) = GL2(C)/{kI : k in C}.

    So it would appear then that my assumption was right, that two matrices correspond to the same transformation iff they are complex multiples of each other. As far as I can tell, the stuff I said about magnifications, inversion, and translations is correct. Since there are only translations to the right and left by integral distances, not all rotations can be treated as rotations about the origin, so maybe there are rotations about other centers which, together with the translations, make [itex]SL_2(\mathbb{Z})(\alpha )[/itex] a tree. I'll try some e, f, g, h that don't make a translation, and see if they make a rotation, or what.
     
  15. Jul 7, 2005 #14

    AKG

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    I see, I guess you meant what my book calls "fundamental tree."
     
  16. Jul 7, 2005 #15

    AKG

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    Thanks for that link to the generators. S will send z to -1/z = -1z*/|z|² which is a reflection in the imaginary axis, then a magnification by |z|-2. T is the translation to the right by 1.
     
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