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I need help with the following abstract algebra problem. It is not homework, but I need a solution.

Let $$SL_2(\mathbb{R}) = \left\{ \big( \begin{smallmatrix} a & b\\ c & d\\\end{smallmatrix} \big)\;\middle\vert\;a,b,c,d \in \mathbb{R}, ad - bc = 1 \right\},$$

$$SO_2(\mathbb{R}) = \left\{ \big( \begin{smallmatrix} a & b\\ -b & a\\\end{smallmatrix} \big)\;\middle\vert\;a,b,c,d \in \mathbb{R}, a^2 + b^2 = 1 \right\}.$$

Prove that $ SL_2(\mathbb{R})$ is a group, that $SO_2(\mathbb{R}) \leq SL_2(\mathbb{R})$, and that $SL_2(\mathbb{R})$ can be represented as a union of nonintersecting left classes with respect to $SO_2(\mathbb{R})$ in the form:

$$SL_2(\mathbb{R}) = \bigcup_{\substack{r,p \in \mathbb{R}\\ r>0}} \begin{pmatrix} r & 0 \\ r^{-1}p & r^{-1} \\ \end{pmatrix} \cdot SO_2(\mathbb{R})$$

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$SL_2(R)$ is a group since $det(A)*det(B)=det(AB)$ and the other group axioms are easily satisfied. $SO_2(R)$ is a subgroup for $SL_2(R)$ by definition also, by setting $c = -b$ and $d = -a$. –  rckrd Jun 6 '12 at 19:34
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R is the set of all real numbers. –  Adam Jun 6 '12 at 19:39
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A group is always the union of the non-intersecting left cosets with respect to one of his subgroups. You only need to prove that the cosets have that form. –  Vittorio Patriarca Jun 6 '12 at 19:40
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@Adam: In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, people are also much happier to help those who demonstrate that they've tried the problem themselves first. –  Zev Chonoles Jun 6 '12 at 19:45
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Note that it doesn't matter at all that there is a number of the form $r^{-1}p$ in the last matrix: this can be any real number since $p$ is arbitrary. –  Egbert Jun 6 '12 at 20:28

3 Answers 3

up vote 2 down vote accepted

$\def\SL\{\mathrm{SL}}\def\SO{\mathrm{SO}}$

  1. $\SL_2(\mathbb{R})$ is a group.

    Note that multiplication of matrices is associative (it corresponds to composition of linear transformations); the identity matrix is the neutral element, and has determinant $1$, so it is in $\SL_2(\mathbb{R})$. The determinant of a product is the product of the determinants, so if $A,B\in\SL_2(\mathbb{R})$, then so is $AB$, since $\det(AB) = \det(A)\det(B)=1\times 1 = 1$. Finally, if $A$ has nonzero determinant, then it is nonsingular so it has a multiplicative inverse; and since $1=\det(I) = \det(AA^{-1})=\det(A)\det(A^{-1})$, it follows that $\det(A^{-1}) = \frac{1}{\det(A)}$; so if $A\in\SL_2(\mathbb{R})$, then it is invertible and $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{1}=1$, so $A^{-1}\in\SL_2(\mathbb{R})$ as well. Thus, $\SL_2(\mathbb{R})$ is a group.

  2. $\SO_2(\mathbb{R})$ is a subgroup of $\SL_2(\mathbb{R})$.

    From the definition, it is clear that $\SO_2(\mathbb{R})$ is a subset of $\SL_2(\mathbb{R})$, since the determinant of a matrix of the form $\left(\begin{array}{rr}a&b\\-b&a\end{array}\right)$ is $a^2+b^2$, which by assumption will equal $1$. Also, the identity is an element of $\SO_2(\mathbb{R})$.

    To show it is a subgroup, you need to show that it is closed under products and inverses. For products, assume that $a^2+b^2=c^2+d^2=1$. Then: $$\left(\begin{array}{rr} a&b\\-b&a \end{array}\right) \left(\begin{array}{rr}c&d\\-d&c \end{array}\right) = \left(\begin{array}{cc} ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{array}\right),$$ which is of the desired $\left(\begin{array}{rr}x&y\\-y&x\end{array}\right)$, with $x=ac-bd$, $y=ad+bc$. And the determinant is $1$, again because the product of two matrices with determinant $1$ is of determinant $1$, so $(ac-bd)^2 + (ad+bc)^2 = 1$. Or, if you just expand, we get: $$\begin{align*} (ac-bd)^2 + (ad+bc)^2 &= a^2c^2 - 2abcd + b^2d^2 + a^2d^2+2abcd+b^2c^2 \\ &= a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\ &= (a^2+b^2)c^2 + (b^2+a^2)d^2 \\ &= c^2+d^2=1.\end{align*}$$ So if $A,B\in \SO_2(\mathbb{R})$, then $AB\in\SO_2(\mathbb{R})$. That is, $\SO_2(\mathbb{R})$ is a submonoid of $\SL_2(\mathbb{R})$.

    For the inverse, note that the inverse of $\left(\begin{array}{rr}a&b\\-b&a\end{array}\right)$ with $a^2+b^2=1$ is $\left(\begin{array}{rr}a&(-b)\\-(-b)&a\end{array}\right)$, so if $A\in\SO_2(\mathbb{R})$, then $A^{-1}\in \SO_2(\mathbb{R})$. Thus, $\SO_2(\mathbb{R})$ is a subgroup of $\SL_2(\mathbb{R})$.

  3. If $r,p,s,q\in\mathbb{R}$, $r,s\gt 0$, and $$\left(\begin{array}{rr} r & 0\\ r^{-1}p & r^{-1}\end{array}\right)^{-1}\left(\begin{array}{rr} s & 0\\ s^{-1}q & s^{-1}\end{array}\right)\in SO_2(\mathbb{R})$$ then $r=s$ and $p=q$. Indeed, the product is equal to $$\left(\begin{array}{rr} r^{-1} & 0\\ -r^{-1}p & r\end{array}\right)\left(\begin{array}{rr} s & 0\\ s^{-1}q & s^{-1} \end{array}\right) = \left(\begin{array}{cc} \frac{s}{r} & 0\\ -\frac{ps}{r} + \frac{qr}{s} & \frac{r}{s}\end{array}\right).$$ If this is in $\SO_2(\mathbb{R})$, then $\frac{r}{s}=\frac{s}{r}$, so $r^2=s^2$, hence (since both are positive) $r=s$; and then $-\frac{ps}{r}+\frac{qr}{s} = -p+q$ must equal $0$, so $p=q$.

    Therefore, the lateral classes listed are pairwise distinct.

  4. If $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\in\SL_2(\mathbb{R})$, then there exist $r\gt 0$, $p\in\mathbb{R}$, such that $$\left(\begin{array}{cc} a&b\\c&d\end{array}\right)\in\left(\begin{array}{rr}r&0\\r^{-1}p&r^{-1}\end{array}\right)\SO_2(\mathbb{R}).$$ If $b=0$ and $a\gt 0$, then we are done: we can take $r=a$, $p=ac$; note that since $ad-bc=1$, we will necessarily have $d=\frac{1}{a}=r^{-1}$. If $b=0$ and $a\lt 0$, then $$\left(\begin{array}{cc} a&0\\c&d\end{array}\right)=\left(\begin{array}{rr}-a&0\\-c&-d\end{array}\right)\left(\begin{array}{rr}-1&0\\0&-1\end{array}\right)\in \left(\begin{array}{rr}-a&0\\-c&-d\end{array}\right)\SO_2(\mathbb{R}),$$ which has the desired form with $r=-a$, $p=ac$.

    If $a=0$, then note that $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{rr}0 &1\\-1&0\end{array}\right) = \left(\begin{array}{rr}-b & a\\-d & c\end{array}\right)$$ so we are back to the previous case.

    Finally, if $a\neq 0$ and $b\neq 0$, we want to find $x$ and $y$ with $x^2+y^2=1$ such that $$\left(\begin{array}{rr} a&b\\c&d\end{array}\right)\left(\begin{array}{rr}x&y\\-y&x\end{array}\right)$$ has a $0$ in the $(1,2)$ entry. The $(1,2)$ entry is $ay+bx$, so we need $y=-\frac{b}{a}x$. Then plugging that into $x^2+y^2 = 1$ yields that we want $x=\frac{a}{\sqrt{a^2+b^2}}$. Indeed, $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{cc}\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{array}\right),$$ then the right hand factor is in $\SO_2(\mathbb{R})$, and the product has a $0$ on the upper right corner, which is all we need.

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You only have two things to check:

  1. If $M\in SL_2(\mathbb{R})$, then there exists real numbers $r,p$, with $r>0$, and a matrix $A\in SO_2(\mathbb{R})$ such that $$M=\begin{pmatrix} r & 0 \\ r^{-1}p & r^{-1} \\ \end{pmatrix}A.$$ (This will show that the union convers everything.)
  2. If $$\begin{pmatrix} r_1 & 0 \\ r_1^{-1}p_1 & r_1^{-1} \\ \end{pmatrix} \cdot SO_2(\mathbb{R})=\begin{pmatrix} r_2 & 0 \\ r_2^{-1}p & r_2^{-1} \\ \end{pmatrix} \cdot SO_2(\mathbb{R}),$$ then $r_1=r_2$ and $p_1=p_2$. (This will show that the cosets are distinct, i.e. non-intersecting.)
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Hints

To check that $SL_2$ is a group (under matrix multiplication), you need to do the following:

  1. Show that it is closed under multiplication. Hint $\mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B)$.
  2. Find the unit element. (That would be easy, I suppose)
  3. Show that each element has a multiplicative inverse. Hint $\mathrm{det}(A^{-1})=\mathrm{det}(A)^{-1}$.
  4. Show that the multiplication is associative.

To show that $SO_2$ is a subgroup, you need to do the following:

  1. Show that it is non-empty.
  2. Show that for every pair $A,B\in SO_2$, the matrix $AB^{-1}$ is in $SO_2$. Hint: use what you know about the determinant.

Hopefully this outline helps to get you started. For the last problem, suppose that

$$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in SL_2.$$

Multiplying this matrix with

$$\frac{1}{\sqrt{a^2+b^2}}\left(\begin{array}{cc} a & -b \\ b & a\end{array}\right)$$

(is it possible that this matrix is zero?) on the right side should give you something nice (pun intended) :)

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