# example of group action

Let $a,b,c$ be integers and let $[a,b,c]$ denote the mapping

 $[a,b,c]\colon{\mathbb{Z}}\times{\mathbb{Z}}\to{\mathbb{Z}},(x,y)\mapsto ax^{2}% +bxy+cy^{2}.$

Let $G$ be the group of $2\times 2$ matrices such that $\det{A}=\pm 1\;\forall\;A\in G$. The substitution

 $\left(\begin{array}[]{l}x\\ y\end{array}\right)\mapsto A\left(\begin{array}[]{l}x\\ y\end{array}\right)$

 $[a,b,c](a_{11}x+a_{12}y,a_{21}x+a_{22}y)=a^{{}^{\prime}}x^{2}+b^{{}^{\prime}}% xy+c^{{}^{\prime}}y^{2},$

where

 $\displaystyle a^{{}^{\prime}}$ $\displaystyle=$ $\displaystyle aa_{11}^{2}+ba_{11}a_{21}+ca_{21}^{2}$ (1) $\displaystyle b^{{}^{\prime}}$ $\displaystyle=$ $\displaystyle 2aa_{11}a_{12}+2ca_{21}a_{22}+b(a_{11}a_{22}+a_{12}a_{21})$ $\displaystyle c^{{}^{\prime}}$ $\displaystyle=$ $\displaystyle aa_{12}^{2}+ba_{12}a_{22}+ca_{22}^{2}$

So we define

 $[a,b,c]\ast A:=[a^{{}^{\prime}},b^{{}^{\prime}},c^{{}^{\prime}}]$

to be the binary quadratic form with coefficients $a^{{}^{\prime}},b^{{}^{\prime}},c^{{}^{\prime}}$ of $x^{2},xy,y^{2}$, respectively as in (1). Putting in $A=\begin{matrix}1&0\\ 0&1\end{matrix}$ we have $[a,b,c]\ast A=[a,b,c]$ for any binary quadratic form $[a,b,c]$. Now let $B$ be another matrix in $G$. We must show that

 $[a,b,c]\ast(AB)=([a,b,c]\ast A)\ast B.$

Set $[a,b,c]\ast(AB):=[a^{{}^{\prime\prime}},b^{{}^{\prime\prime}},c^{{}^{\prime% \prime}}]$. So we have

 $\displaystyle a^{{}^{\prime\prime}}$ $\displaystyle=$ $\displaystyle a\left(a_{11}b_{11}+a_{12}b_{21}\right)^{2}+c\left(a_{21}b_{11}+% a_{22}b_{21}\right)^{2}+b\left(a_{11}b_{11}+a_{12}b_{21}\right)(a_{21}b_{11}+a% _{22}b_{21})$ (2) $\displaystyle=$ $\displaystyle a^{{}^{\prime}}b_{11}^{2}+c^{{}^{\prime}}b_{21}^{2}+\left(2a% \cdot a_{11}a_{12}+2c\cdot a_{21}a_{22}+b\left(a_{11}a_{22}+a_{12}a_{21}\right% )\right)b_{11}b_{21}$ $\displaystyle c^{{}^{\prime\prime}}$ $\displaystyle=$ $\displaystyle a\left(a_{11}b_{12}+a_{12}b_{22}\right)^{2}+c\cdot\left(a_{21}b_% {12}+a_{22}b_{22}\right)^{2}+b\cdot\left(a_{11}b_{12}+a_{12}b_{22}\right)\left% (a_{21}b_{12}+a_{22}b_{22}\right)$ (3) $\displaystyle=$ $\displaystyle a^{{}^{\prime}}b_{12}^{2}+c^{{}^{\prime}}b_{22}^{2}+\left(2a% \cdot a_{11}a_{12}+2ca_{21}a_{22}+b\left(a_{11}a_{22}+a_{12}a_{21}\right)% \right)b_{12}b_{22}$

as desired. For the coefficient $b^{{}^{\prime\prime}}$ we get

 $\displaystyle b^{{}^{\prime\prime}}$ $\displaystyle=$ $\displaystyle 2a\left(a_{11}b_{11}+a_{12}b_{21}\right)\left(a_{11}b_{12}+a_{12% }b_{22}\right)$ $\displaystyle+$ $\displaystyle 2c\cdot\left(a_{21}b_{11}+a_{22}b_{21}\right)\left(a_{21}b_{12}+% a_{22}b_{22}\right)$ $\displaystyle+$ $\displaystyle b\left(\left(a_{11}b_{11}+a_{12}b_{21}\right)\left(a_{21}b_{12}+% a_{22}b_{22}\right)+\left(a_{11}b_{12}+a_{12}b_{22}\right)\left(a_{21}b_{11}+a% _{22}b_{21}\right)\right)$

and by evaluating the factors of $b_{11}b_{12},b_{21}b_{22}$, and $b_{11}b_{22}+b_{21}b_{12}$, it can be checked that

 $b^{{}^{\prime\prime}}=2a^{{}^{\prime}}b_{11}b_{12}+2c^{{}^{\prime}}b_{21}b_{22% }\\ +\left(b_{11}b_{22}+b_{21}b_{12}\right)\left(2aa_{11}a_{12}+2c\cdot a_{21}a_{2% 2}+b\left(a_{11}a_{22}+a_{12}a_{21}\right)\right).$

This shows that

 $[a^{{}^{\prime\prime}},b^{{}^{\prime\prime}},c^{{}^{\prime\prime}}]=[a^{{}^{% \prime}},b^{{}^{\prime}},c^{{}^{\prime}}]\ast B$ (4)

and therefore $[a,b,c]\ast(AB)=([a,b,c]\ast A)\ast B$. Thus, (1) defines an action of $G$ on the set of (integer) binary quadratic forms. Furthermore, the discriminant    of each quadratic form  in the orbit of $[a,b,c]$ under $G$ is $b^{2}-4ac$.

Title example of group action ExampleOfGroupAction 2013-03-22 13:50:00 2013-03-22 13:50:00 Thomas Heye (1234) Thomas Heye (1234) 11 Thomas Heye (1234) Example msc 11E16 msc 16W22 msc 20M30