# When are two elements conjugated in GL(2), but not in SL(2)

Let $F$ be an arbitrary field. How can we describe the set of elements in $SL(2,F)$ which are conjugated in $GL(2,F)$ but not in $SL(2,F)$?

I would be happy already with a partial solution as given in the comments.

• Suppose $PAP^{-1}=B$ for $A,B \in \mathrm{SL}(2,\mathbb{F})$ and $P \in \mathrm{GL}(2,\mathbb{F})$. Say $\mathrm{det}(P)=p$, then $\sqrt{p^{-1}}P \in \mathrm{SL}(2,\mathbb{F})$. Notice that $(\sqrt{p^{-1}}P)^{-1}=\sqrt{p}P^{-1}$ and $(\sqrt{p^{-1}}P)A(\sqrt{p^{-1}}P)^{-1}=\sqrt{p^{-1}}\sqrt{p}PAP^{-1}=B$. Partial Answer: Conjugation by an element of $\mathrm{GL}_2$ can always be replaced by conjugation by an element of $\mathrm{SL}_2$ as long as the field is closed under square roots. – Bill Cook Mar 19 '12 at 14:18
• @BillCook: If p is negative then we might come across a problem. Right? So if we are working in $M_n(\mathbb{C})$ I think there will be no problem. Moreover, I think there in $M_n(\mathbb{C})$ there exists a bijection between $GL_n(\mathbb{C})$ and $M_n(\mathbb{C})$ by defining $P\in GL_n(\mathbb{C}) \mapsto (p)^\frac{1}{n} P$ – Saikat Oct 9 '20 at 4:10
• That's why I have the caveat "as long as the field is closed under square roots". This does not apply to the real numbers - only non-negative reals (not the whole field) are closed under square roots. – Bill Cook Oct 9 '20 at 10:49

There is a very general result on this kind of things, but I will have to dig it up when I get back to my office tomorrow. ADDED Here's the reference I promised. It's the beautiful article by G.E. Wall, Conjugacy classes in projective and special linear groups. Bull. Austral. Math. Soc. 22 (1980), no. 3, 339–364.

Anyway, in this particular case, I believe the following should be correct.

All of the conjugates of $A \in \operatorname{SL}(2,F)$ under $\operatorname{GL}(2,F)$ are also conjugate under $\operatorname{SL}(2,F)$ if and only if for each $G \in \operatorname{GL}(2,F)$ there is $S \in\operatorname{SL}(2,F)$ such that $$S^{-1} G^{-1} A G S = A$$ that is, for each $G \in \operatorname{GL}(2,F)$ there is $S \in\operatorname{SL}(2,F)$ such that $$GS \in C_{\operatorname{GL}(2,F)}(A),$$ that is, in every coset of the centralizer $C_{\operatorname{GL}(2,F)}(A)$ there is an element of $\operatorname{SL}(2,F)$, that is $$\operatorname{GL}(2,F) = \operatorname{SL}(2,F) C_{\operatorname{GL}(2,F)}(A),$$ that is, the centralizer $C_{\operatorname{GL}(2,F)}(A)$ contains elements of arbitrary determinant.

In @BillCook's case, every centralizer contains the scalar matrices, which have arbitrary determinant, as all element are squares in $F$.

In @Ludolila's first case, the centralizer of $A$ is made of the elements of the form $$c I + d A = \begin{bmatrix}c&d\\-d&c\end{bmatrix}$$ of determinant $c^2 + d^2 \ge 0$. So if you conjugate (as noted by Ludolila) $A$ by a matrix of negative determinant, you'll never be able to get back conjugating with an element in $\operatorname{SL}(2,F)$. In the second case, the centralizer contains all scalar matrices, so the determinant is arbitrary here.

• Note that the centralizer of $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$, with determinant $a^2$. So your criterion implies that the squaring map is surjective. Bill Cook shows that this is also a sufficient condition. PS: Nice answer! – David E Speyer Feb 12 '13 at 22:11

Great question! Here are some thoughts: First, if the field is closed under square roots (like $\mathbb{C}$), Bill already gave a great answer.

What about $\mathbb{R}$? For example, $A= \left[ \begin{array}{ c c } 0 & 1 \\ -1 & 0 \end{array} \right]$ and $(-A)$ are conjugated in $GL_2(\mathbb{R})$ but not in $SL_2(\mathbb{R})$. Indeed, a simple check shows that if $PAP^{-1}=-A$ then $\det P<0$.

Another observation: even if for two matrices $A,B\in SL_2(\mathbb{R})$ we have $P\in GL_2(\mathbb{R})$ such that $PAP^{-1}=B$ and $\det P<0$, that still doesn't mean that they are not conjugated in $SL_2(\mathbb{R})$ by a different matrix. For example: $A= \left[ \begin{array}{ c c } 2 & 0 \\ 0 & \frac{1}{2} \end{array} \right]$, $B= \left[ \begin{array}{ c c } \frac{1}{2} & 0 \\ 0 & 2 \end{array} \right]$, and $P= \left[ \begin{array}{ c c } 0 & 3 \\ 4 & 0 \end{array} \right]$ . We can still find $P' \in SL_2(\mathbb{R})$ that conjugates $A$ and $B$.

So, what went wrong in the first example? Maybe it has something to do with the fact that $A= \left[ \begin{array}{ c c } 0 & 1 \\ -1 & 0 \end{array} \right]$ has a characteristic polynomial which is irreducible over $\mathbb{R}$?