Exercise 2.9.2

Question

Determine the representation of
$$U = \begin{pmatrix} 6 & 11 \ 13 & 24 \end{pmatrix}$$
as a power product of $S$ and $T$ as in Exercise 2.9.1.

Richard Ganaye · Accepted Answer

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\begin{proof}
We take a reduced form $f$ whose automorphism group is $\{-I, I\}$, for instance $f = (1,0,5)$ (see Theorem 2.5.10).

Here $$ S = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}, \qquad T =  \begin{pmatrix} 1 &  1\ 0 & 1 \end{pmatrix},$$
and 
$$(a,b,c)\cdot S = (c, -b,a), \qquad (a,b,c) \cdot T^k = (a, b+ 2ak, ak^2 + bk+c).$$

If we apply $U^{-1}= \begin{pmatrix} 24 & -11 \ -13 & 6 \end{pmatrix}$ to $f$, we obtain
$$g = f\cdot U^{-1} = (1421, -1308, 301).$$
Since $g \overset{+}{\sim} f$ , if we reduce the form $g$ as in Chapter 5, we obtain anew the reduced form $f$.
\begin{align*}
&g_1 = g\cdot S = (301, 1308, 1421),\
&g_2 = g_1 \cdot T^{-2} = (301, 104, 9),\
&g_3 = g_2 \cdot S = (9, - 104, 301),\
&g_4 = g_3 \cdot T^6 = (9, 4,1),\
&g_5 = g_4 \cdot S = (1, -4, 9),\
&g_6 = g_5 \cdot T^2 = (1,0,5).
\end{align*}
This gives
$$f=(f\cdot U^{-1}) \cdot  ST^{-2}ST^6S T^2 = f\cdot (U^{-1}ST^{-2}ST^6S T^2),$$
therefore $U^{-1}ST^{-2}ST^6S T^2 \in \mathrm{Aut}^+(f) = \{\pm I\}$. Here $ST^{-2}ST^6S T^2 = -U = S^2 U$, thus
$$U = S^{-1} T^{-2} S T^6 S T^2.$$

Verification :

\begin{align*}
S^{-1}T^{-2} &=  \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 &  -2\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix},\
S^{-1}T^{-2} S &= \begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix} \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 &  0\ 2 & 1 \end{pmatrix},\
S^{-1}T^{-2} ST^6 &= \begin{pmatrix} 1 &  0\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 &  6\ 0 & 1 \end{pmatrix} \phantom{-}\  = \begin{pmatrix} 1 &  6\ 2 & 13 \end{pmatrix},\
S^{-1}T^{-2} ST^6 S &= \begin{pmatrix} 1 &  6\ 2 & 13 \end{pmatrix}\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 6 & -1 \ 13 & -2 \end{pmatrix},\
S^{-1}T^{-2} ST^6 S T^2 &=  \begin{pmatrix} 6 & -1 \13 & -2 \end{pmatrix} \begin{pmatrix} 1 &  2\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 6 & 11 \ 13 & 24 \end{pmatrix}=U.
\end{align*}
\end{proof}

Exercise 2.9.2

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Exercise 2.9.2

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