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Homework Help: Tricky abstract algebra problem!

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that [itex]SL_{2}(ℝ)[/itex] is generated by the set:

    [1 a], [1 0]
    [0 1], [b 1], [itex]a,b \in ℝ[/itex]

    2. Relevant equations

    1. GCD (Greatest common divisor)
    2. The property of special linear group
    3. Some basic linear algebra, like determinant

    3. The attempt at a solution

    [itex]SL_{2}(ℝ)[/itex] is the group consisting of invertible matrices with the determinant 1. Let [itex]A[/itex] be this matrix:

    [a b]
    [c d]

    Then, [itex]det(A) = ad - bc = 1[/itex]

    In order for the determinant to be 1, [itex]gcd(a,b) = gcd(a,c) = gcd(d,b) = gcd(d,c) = 1[/itex].

    I'm not sure if my reasoning is right. I treat ad - bc = 1 as the linear combination. If 1 divides a and 1 divides b, then 1 divides (ad - bc). But -1 also divides a, b and (ad - bc). Thus a = ±1.

    If [itex]a = 1[/itex], then [itex]det(A) = d - bc = 1[/itex]. Here, we need to split into some cases:

    If [itex]c = 0[/itex], then [itex]d = 1[/itex], so we are done, having this matrix:

    [1 b]
    [0 1]

    Similar argument shows that if [itex]a = 1[/itex] and [itex]b = 0[/itex], then d = 1, so we are done, leaving off this matrix:

    [1 0]
    [c 1]

    If [itex]a = -1[/itex], then [itex]d = -1[/itex]. Similar arguments show that if either [itex]b = 0[/itex] or [itex]c = 0[/itex], then we obtain these similar matrices that belong to the special linear group:

    [-1 0]
    [c -1]

    [-1 b]
    [0 -1]

    So these matrices form the set as given.

    What if [itex]c ≠ 0[/itex]? Would I need to use Euclidean Algorithm to work out this problem?
  2. jcsd
  3. Mar 9, 2013 #2

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    Hi NasuSama! :smile:

    a,b,c, and d do not have to be whole numbers do they?
    I think gcd's and the Euclidean algorithm do not apply here.

    What you need to show, is that any allowed matrix can be constructed from the generators.
    Each allowed matrix has 3 numbers that you can pick freely, after which the 4th number is determined.
    So to construct any allowed matrix you need at least (and perhaps exactly) 3 matrices from your generator set.
  4. Mar 9, 2013 #3

    So you mean I can select any matrix that belongs to the given set? I don't quite understand what you are saying here. Sorry about this. Just making sure I'm getting the details clear!
  5. Mar 9, 2013 #4

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    Pick an arbitrary matrix ##\begin{bmatrix}a &b\\c&d \end{bmatrix}## from ##SL_2(\mathbb R)##.
    Then try to find a composition of matrices of the forms ##\begin{bmatrix}1 &x\\0&1 \end{bmatrix}## and ##\begin{bmatrix}1 &0\\x&1 \end{bmatrix}## that construct it.
  6. Mar 10, 2013 #5
    ##\begin{bmatrix} 1&1\\0&1\end{bmatrix}## for the first matrix.
    ##\begin{bmatrix} 1&0\\1&1\end{bmatrix}## for the second matrix.

    Multiply each of the matrix by itself to obtain that form?
  7. Mar 10, 2013 #6

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    We seem to be on a different page.
    That usually means we need to take a look at the related definitions in order to get clear what we're talking about.

    Can you say what a generator is?
    And what it means that a group is generated by a set?
    Can you give an example?
    Last edited: Mar 10, 2013
  8. Mar 10, 2013 #7
    An element is a generator that forms a group. For instance:

    <x> = {1,x,x²,...,xⁿ} with the order n.

    The example is the cyclic group as the one above generated by x, which is the generator.
  9. Mar 10, 2013 #8

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    More specifically from wiki:
    a generating set of a group is a subset such that every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses.

    Now suppose I pick the element ##\begin{bmatrix}1 & \frac 1 2 \\ 1 & \frac 3 2 \end{bmatrix}##.
    Can you express it as the combination of finitely many elements of the (generator) subset and their inverses?
  10. Mar 10, 2013 #9
    The possible combination for that matrix I found from the given generator I have given to you is:

    [1 0][1 ½]
    [1 1][0 1]

    Multiply these matrices altogether to get the one you have given to me.
  11. Mar 10, 2013 #10

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    How about ##\begin{bmatrix} 3&1 \\ -1&0 \end{bmatrix}##?
  12. Mar 10, 2013 #11
    ##\begin{bmatrix} 1&-2 \\ 0&1 \end{bmatrix}## * ##\begin{bmatrix} 1&0 \\ -1&1 \end{bmatrix}## * ##\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}##
  13. Mar 10, 2013 #12
    Too tedious to work out by trials.

    The question is: Is there any fast way to obtain the following matrix, using the multiplication of the matrices in the form ##\begin{bmatrix} 1&a \\ 0&1 \end{bmatrix}## and ##\begin{bmatrix} 1&0 \\ b&1 \end{bmatrix}##?

    ##\begin{bmatrix} a&b \\ c&d \end{bmatrix}##

    Is there a method to work out this problem, or do I need to work out by trials?
  14. Mar 10, 2013 #13

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    Looking good. :)
    So let's go up a notch.

    How about ##\begin{bmatrix} a & b \\ c & d \end{bmatrix}##?
  15. Mar 10, 2013 #14

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    Try ##\begin{bmatrix} 1&0 \\ x&1 \end{bmatrix} \cdot \begin{bmatrix} 1&y \\ 0&1 \end{bmatrix} \cdot \begin{bmatrix} 1&0 \\ z&1 \end{bmatrix}##

    You need 3 matrices since your target matrix contains 3 unknowns (why 3?), while each of your generator matrices can only introduce 1 unknown at a time.
  16. Mar 10, 2013 #15
    I quite understand this now. When I set up with 3 matrices for the second example, I am able to obtain the 0 for the element in 2nd row and 2nd column. If we have 3 matrices, then we can set up the product of the matrices forming the certain matrix. That is enough to obtain the matrix.

    Let me work on the product of the matrices.
  17. Mar 10, 2013 #16

    Actually, there are more than one solution - one by following your way and another... I could have also worked out using the matrices I've previously used for the second problem.

    Here is the product of the matrices.

    \begin{bmatrix} 1&0 \\ (-a + bc + 1)/(ab)&1 \end{bmatrix} * \begin{bmatrix} 1&b \\ 0&1 \end{bmatrix} * \begin{bmatrix} 1&0 \\ (a - 1)/b&1 \end{bmatrix}

    Checked by Wolfram, but Wolfram doesn't give the same results as I computed. Wonder if this is correct.
    Last edited: Mar 10, 2013
  18. Mar 10, 2013 #17

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    I didn't check, but it looks good. ;)
    So now you can construct almost any matrix with determinant 1.
    That almost completes the proof.
  19. Mar 10, 2013 #18
    What would be another part of the proof? I tested out the general product of the matrices.
  20. Mar 10, 2013 #19

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    I just checked and your matrices work out.
    What is missing is that you cannot construct every matrix yet, since you require that a and b are both non-zero.
    It would be a pity if you couldn't construct generic matrices with for instance a zero at the right top.
  21. Mar 10, 2013 #20
    But if you multiply any of the matrices in that form with the identity matrix, then you obtain the value in the 1st row and 1st column and also in 2nd row and 2nd column. That is what I need to be aware of.
  22. Mar 11, 2013 #21

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    I don't understand.
    If you multiply any matrix with the identity matrix, you get the same matrix.

    Btw, if you revise your first matrix to be ##\begin{bmatrix} 1&0 \\ \frac{d-1}{b} &1 \end{bmatrix}##, you cover all matrices with ##b \ne 0##.
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