Are a $2\times 2$ matrix in $\text{SL}(2,\mathbb{Z})$ and its transpose conjugate in $\text{GL}(2,\mathbb{Z})$?

Question

I've been studying some math by myself this summer, and have recently been doing some reading about the groups $\text{GL}(2,\mathbb{Z})$, $\text{SL}(2,\mathbb{Z})$, etc. I've been trying to get a better grasp of conjugation in these groups, but have been unable to figure out something relatively simple:

Are the following two matrices conjugate in $\text{SL}(2,\mathbb{Z})$? If not maybe in $\text{GL}(2,\mathbb{Z})$?

$$\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \begin{bmatrix} a & c\\ b & d \end{bmatrix}\in\text{SL}(\mathbb{Z}[t,t^{-1}]) $$

If so is there a nice conjugating element? Thanks in advance! I don't have the greatest linear algebra background, but am loving group theory, which might be why I am not getting this.

Answer 1

Two such matrices are not always conjugate.

If the matrices $$ A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad A^T= \begin{pmatrix} a & c \\ b & d \end{pmatrix} $$ are conjugate in $GL_2(\mathbb{C})$, then $XA=A^TX$ for some matrix $$ X= \begin{pmatrix} x & y \\ u & v \end{pmatrix} \in GL_2(\mathbb{C}). $$ Let $b\neq0$ or $c\neq0$. We have $u=y$ and $$ xb+y(d-a)-vc=0. $$

Let $$ A= \begin{pmatrix} t^2+t+1 & t \\ 1+t & 1 \end{pmatrix}. $$ The equality $xt-y(t^2+t)-v(1+t)=0$ must be fulfilled for some $x,y,v\in\mathbb{C}$ simultaneously not equal to $0$. If $t$ is an independent variable, this is not possible.

Correction. I replaced the matrix $A$.

Answer 2

I am not sure what adjoining $[t,t^{-1}]$ accomplishes, so I will ignore that.

Set $G=\text{GL}(2,\mathbb{Z})$ and $H=\text{SL}(2,\mathbb{Z})$.
Call $M\in \text{Mat}(2,\mathbb{C})$ "transpose-conjugate by $S\subset\text{Mat}(2,\mathbb{C})$" ("tc by $S$" for short) if and only if $\exists T\in S: T^{-1}MT=M^T$.
Call $M\in \text{Mat}(2,\mathbb{C})$ "transpose-conjugate" ("tc" for short), if and only if $M$ is transpose-conjugate by $G$.

The answer has three parts:

1. $M_1:=\begin{pmatrix}1&s\\0&1\end{pmatrix}\in H$ for any $s\in\mathbb{Z}\setminus\{0\}$ is not tc by $H$, but is tc by $G$.
2. $M_2:=\begin{pmatrix}2&5\\-9&-22\end{pmatrix}\in H$ is not tc.
3. Further remarks on the property tc.

Part 1. Let $M:=M_1:=\begin{pmatrix}1&s\\0&1\end{pmatrix}\in H$ for any $s\in\mathbb{Z}\setminus\{0\}$. Take a general $T:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in H$. The determinant is $D:=\text{det}(T)=ad-bc=1$. As kabenyuk said, we need to test: $$ N:=MT-TM^T\stackrel{\text{?}}{=}0. $$ We get $N:=\begin{pmatrix}-bs+cs&sd\\-sd&0\end{pmatrix}$. Since $s\neq 0$ we have $d=0$. Furthermore $-b+c=0$. Since $d=0$ we have $1\stackrel{\text{!}}{=}\det(T)=a\cdot 0-b\cdot c=-bc$. The system $[-b+c=0,1=-bc]$ has no solutions in $\mathbb{Z}$ (but $(b,c)\in\{(i,i),(-i,-i)\}$ over $\mathbb{C}$ are solutions). Thus $M\in H$ cannot be conjugated to $M^T$ via matrices in $H$, i.e. $M$ is not tc by $H$.
However $M$ and $M^T$ are conjugate via matrices in $G\setminus H$, e.g. $T=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ does the job for all $s$.

Part 2. Now look at $M:=M_2:=\begin{pmatrix}2&5\\-9&-22\end{pmatrix}\in H$. Take a general $T:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in G$. The determinant is $D:=\text{det}(T)=ad-bc\in\{\pm 1\}$. Test again: $$ N:=MT-TM^T\stackrel{\text{?}}{=}0. $$ We get $N:=\begin{pmatrix}-5b+5c&9a+24b+5d\\-9a-24c-5d&-9b+9c\end{pmatrix}$. The diagonal entries boil down to $b=c$. Substituting this in the off-diagonal terms shows, that both yield essentially the same equations $9a+24c+5d=0$. We distinguish two cases according to $D\in\{\pm 1\}$ and use Maple to solve the resulting systems:
case A. $D=1$: Maple produces two (families of) solutions, for both $d$ is a free parameter. Set $\sigma=\sqrt{11d^2-9}$, then $(a,b,c,d)=$ $$\left( 3d\pm\frac{8}{9}\sigma,-\frac{4}{3}d\mp\frac{1}{3}\sigma,-\frac{4}{3}d\mp\frac{1}{3}\sigma,d \right). $$ As kabenyuk pointed out, we need to do some computations in quadratic extensions of $\mathbb{Q}$: We need to have $q(d):=\sigma^2=11d^2-9$ a square (in $\mathbb{Z}$). If $q(d)$ is a square it is so $\bmod 11$, thus we need $-9\equiv 2\bmod 11$ to be a square, but the squares $\bmod 11$ are $\{0,1,3,4,5,9\}$. Thus $q(d)$ is never a square and we get no integral (not even rational) transformation matrices. This settles A.
case B. $D=-1$: Except for changing $\sigma=\sqrt{11d^2+9}$ the solutions provided by Maple have the same form as before: $(a,b,c,d)=$ $$\left( 3d\pm\frac{8}{9}\sigma,-\frac{4}{3}d\mp\frac{1}{3}\sigma,-\frac{4}{3}d\mp\frac{1}{3}\sigma,d \right). $$ We need $q(d):=\sigma^2=11d^2+9$ to be a square (in $\mathbb{Z}$). It will turn out that for all $d$ with $q(d)\in\mathbb{Z}$ all possible $a$ are proper fractions with denominator $3$. Here is a list indexed by $j$ (the meaning of $j$ becomes clearer after discussing Pell's equation) with the smallest (in magnitude) $d$ with $q(d)\in\mathbb{Z}$, associated pairs of $a$ and the remainder of the numerator $\tilde{a}\bmod 3$ of these $a$. $$\begin{bmatrix} j & d & a & \tilde{a}\bmod 3 \\ -3 & -3591 & \left\{-\frac{559}{3},-\frac{64079}{3}\right\}& \{1,2\}\\ -2 & -180 & \left\{-\frac{28}{3},-\frac{3212}{3}\right\} & \{1,2\}\\ -1 & -9 & \left\{-\frac{1}{3},-\frac{161}{3}\right\} & \{1,2\}\\ 0 & 0 & \left\{\pm\frac{8}{3}\right\} & \{1,2\}\\ 1 & 9 & \left\{\frac{1}{3},\frac{161}{3}\right\} & \{1,2\}\\ 2 & 180 & \left\{\frac{28}{3},\frac{3212}{3}\right\} & \{1,2\}\\ 3 & 3591 & \left\{\frac{559}{3},\frac{64079}{3}\right\} & \{1,2\} \end{bmatrix}$$ A generalized Pell equation is of the form $X^2-MY^2=N$. See the wikipedia page or lecture notes by Keith Conrad. Here and here and there are some resources to compute solutions for (non-generalized) Pell equations. If we set $M=11, N=9$ any solution $(X,Y)$ will produce a suitable $d=Y$ and any $d$ can be realized this way.
From the general theory we know that there is a bounded set $F=\{(X_{F,i},Y_{F,i})\}_{i=1}^m$ of fundamental solutions to the generalized equation $X^2-MY^2=N$ and a generating solution $(X',Y')$ to the normalized Pell equation $X'^2-MY'^2=1$ giving rise to $u=X'+Y'\sqrt{11}$ such that every solution to the generalized Pell equation $X^2-MY^2=N$ satisfies $X+Y\sqrt{11}=(X_{F,i}+Y_{F,i}\sqrt{11})(X'+Y'\sqrt{11})^{j}$ for some $1\leq i\leq m$ and $j\in\mathbb{Z}$. Doing some computer calculations we get in our case: $$ F:=\{(\pm3,0)\}, u=10+3\sqrt{11} \quad\Rightarrow\quad X+Y\sqrt{11}=\pm 3(10+3\sqrt{11})^{j} $$ Up to sign we can index the solutions by the exponent $j\in\mathbb{Z}$.
The $Y$-part of these solutions gives $d$. Because $a=3d\pm\frac{8}{9}\sigma=3d\pm\frac{8}{9}\sqrt{11d^2+9}$ we need to show, that $11d^2+9=11Y^2+9\not\equiv 0\bmod 27$ in order to prove that $3a\tilde{a}\not\equiv 0\mod 3$, which implies $a\in\mathbb{Q}\setminus\mathbb{Z}$. We show by induction the following

Lemma: If $(X,Y)$ is a solution to $X^2-11Y^2=9$ then $X\equiv \pm 3\bmod 27$ and $Y\in\{0,\pm 9\}\bmod 27$.
proof of Lemma: for $j=0$ we have the solutions $(\pm 3,0)$ which obviously satisfy the lemma.
For $j+1>0$ we assume that a solution $(X_j,Y_j)$ with index $j$ satisfies the lemma. The solution $X_{j+1},Y_{j+1}$ satisfies $X_{j+1}+Y_{j+1}\sqrt{11}=\pm(X_j+Y_j\sqrt{11})(10+3\sqrt{11})$, i.e. $X_{j+1}=\pm(10X_j+33Y_j),Y_{j+1}=\pm(3X_j+10Y_j)$. Computing $\bmod 27$ together with the induction hypothesis on $(X_j,Y_j)$ shows that the property of the lemma also holds for $j+1$.
For $j-1<0$ we have $X_{j-1}+Y_{j-1}\sqrt{11}=\pm(X_j+Y_j\sqrt{11})(10+3\sqrt{11})^{-1}=\pm(X_j+Y_j\sqrt{11})(10-3\sqrt{11})$, i.e. $X_{j+1}=\pm(10X_j-33Y_j),Y_{j+1}=\pm(-3X_j+10Y_j)$. Computing $\bmod 27$ again induction goes through.

By the lemma we have $Y\in\{0,\pm 9\}\bmod 27$ which implies $11d^2+9=11Y^2+9\equiv 9\not\equiv 0\bmod 27$. This finishes case B. $\text{det}(T)=-1$.

In summary this proves, that $M_2=\begin{pmatrix}2&5\\-9&-22\end{pmatrix}\in H$ is not tc by $G$.

Part 3.

property 1: Symmetric matrices are trivially tc, they can be ignored when studying tc.

property 2: TFAE:

1. $M=\begin{pmatrix}e&f\\g&h\end{pmatrix}$ is tc,
2. $M^T$ is tc,
3. $-M$ is tc,
4. any conjugate $M'=U^{-1}MU$ is tc, in particular $\begin{pmatrix}h&g\\f&e\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}^{-1}M\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is tc,
5. $M':=\begin{pmatrix}e&-f\\-g&h\end{pmatrix}$ is tc.

The equivalences 1. $\Leftrightarrow$ 2. $\Leftrightarrow$ 3. are easy. 1. $\Leftrightarrow$ 4. follows from $$ T^{-1}MT=M^T \Leftrightarrow T'^{-1}M'T'=M'^T, \text{ where } T'=U^{-1}TU^{-T} $$ and 1. $\Leftrightarrow$ 5. from $$ \begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}e&f\\g&h\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}E&F\\G&H\end{pmatrix}\Rightarrow $$ $$ \begin{pmatrix}a&-b\\-c&d\end{pmatrix}^{-1}\begin{pmatrix}e&-f\\-g&h\end{pmatrix}\begin{pmatrix}a&-b\\-c&d\end{pmatrix}=\begin{pmatrix}E&-F\\-G&H\end{pmatrix} $$

property 3: If any entry of $M=\begin{pmatrix}e&f\\g&h\end{pmatrix}\in H$ vanishes then $M$ is tc. This can be seen from the following list of $M\in H=\text{SL}(2,\mathbb{Z})$ and $T\in G=\text{GL}(2,\mathbb{Z})$ which satisfy $T^{-1}MT=M^T$ (Note that part 1. is part of this list): $$ M:=\begin{pmatrix}0&\pm 1\\\mp 1&h\end{pmatrix}, T:=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad M:=\begin{pmatrix}\pm 1&0\\g&\pm 1\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&0\end{pmatrix}, $$ $$ M:=\begin{pmatrix}\pm 1&f\\0&\pm 1\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad M:=\begin{pmatrix}e&\pm 1\\\mp 1&0\end{pmatrix}, T:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ This remains true even for $M\in G\setminus H$ as implied by these matrices where $2\mid g_e\in\mathbb{Z}$ and $2\nmid g_o\in\mathbb{Z}$ together with some easy to check further cases: $$ M:=\begin{pmatrix}1&0\\g_e&-1\end{pmatrix}, T:=\begin{pmatrix}1&\frac{g_e}{2}\\\frac{g_e}{2}&\left(\frac{g_e}{2}\right)^2-1\end{pmatrix},\quad M:=\begin{pmatrix}1&0\\g_o&-1\end{pmatrix}, T:=\begin{pmatrix}2&g_o\\g_o&\frac{g_o^2-1}{2}\end{pmatrix}. $$

These properties imply: If we want to study tc for $M=\begin{pmatrix}e&f\\g&h\end{pmatrix}\in H$ we may assume wlog: $$0<e\neq h,|e|\leq|h|, 0<f\neq g, |f|\leq |g|.$$ ($e\neq h$ is implied by one of the example families below.)

One can use these properties to search for $M\in H$ with "small entries" that might not be tc. For $\max\{|e|,|f|,|g|,|h|\}\leq 25$ and $|eh|\leq 50$ a heuristic search only left three candidates for not-tc up to the symmetries of property 2: $$ M_2=\begin{pmatrix}2&5\\-9&-22\end{pmatrix},\quad \begin{pmatrix}2&5\\7&18\end{pmatrix},\quad \begin{pmatrix}2&5\\9&23\end{pmatrix}. $$ For $\max\{|e|,|f|,|g|,|h|\}\leq 15$ without restrictions on the diagonal the heuristic search leaves us with two candidates for not-tc: $$ \begin{pmatrix}7&9\\10&13\end{pmatrix},\quad \begin{pmatrix}9&7\\14&11\end{pmatrix}. $$

Magma verifies quickly that all of these candidates are also not tc:

G := GeneralLinearGroup(2,Integers());
ML := [elt<G|2,5,-9,-22>,elt<G|2,5,7,18>,elt<G|2,5,9,23>,elt<G|7,9,10,13>,elt<G|9,7,14,11>,elt<G|4,3,13,10>];
M := ML[2];
MT := Transpose(M);
b,T := AreGLConjugate(M, MT);
M;
MT;
b;

Finally here is a list of tc $M\in G$ together with some $T\in G$ as witnesses (here $x,y,z\in\mathbb{Z}$ are any integers, such that $\det(M)\in\{-1,1\}$): $$ M:=\begin{pmatrix}1&1\\x&x+1\end{pmatrix}, T:=\begin{pmatrix}-1&1\\1&0\end{pmatrix},\quad M:=\begin{pmatrix}1&1\\x&x-1\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&x-2\end{pmatrix}, $$ $$ M:=\begin{pmatrix}1&x\\x^2&x^3+1\end{pmatrix}, T:=\begin{pmatrix}x&-1\\-1&0\end{pmatrix},\quad M:=\begin{pmatrix}x&y\\z&x\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&0\end{pmatrix}, $$ $$ M:=\begin{pmatrix}3&2\\4&3\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad M:=\begin{pmatrix}3&1\\8&3\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&0\end{pmatrix}, $$ $$ M:=\begin{pmatrix}2&3\\9&14\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&4\end{pmatrix},\quad M:=\begin{pmatrix}4&3\\9&7\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&1\end{pmatrix}, $$ $$ M:=\begin{pmatrix}7&3\\9&4\end{pmatrix}, T:=\begin{pmatrix}0&1\\1&-1\end{pmatrix},\quad M:=\begin{pmatrix}14&3\\9&2\end{pmatrix}, T:=\begin{pmatrix}3&2\\2&1\end{pmatrix}, $$ $$ M:=\begin{pmatrix}28&3\\9&1\end{pmatrix}, T:=\begin{pmatrix}3&1\\1&0\end{pmatrix}, $$

Remark: This answer went through a lot of editing to make it clearer and to reflect simplifying insights.