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The trace $\operatorname{tr}(A)$ of a matrix $A$ is the sum of its diagonal entries. Apparently if $A\in \operatorname{SL}(2,\mathbb{R})$ and $|\operatorname{tr}(A)|<2$, then $A$ is conjugate in $\operatorname{SL}(2,\mathbb{R})$ to a matrix of the form

$$\left(\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right).$$

Why is this? I seem to have forgotten my linear algebra.

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Neat question! I don't ever remember learning such a thing in linear algebra. Is there any indication that it generalizes, or does it seem to be a special quirk of SL(2,R)? –  rschwieb May 15 '12 at 12:04
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Well, I came across this in the context of Fuchsian groups, which are subgroups of $\operatorname{PSL}(2,\mathbb{R})$. I don't know of a generalisation. –  Tara B May 15 '12 at 12:17
    
There's probably some material on this in the more technical books on modular forms: Miyake, Rankin, etc. –  Dylan Moreland May 15 '12 at 12:23
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2 Answers

up vote 6 down vote accepted

Check theorem 3.1 from this document: Decomposing $SL_2(R)$

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Thank you, that's a very helpful reference. –  Tara B May 15 '12 at 12:08
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When we assume that $|tr(A)|<2$ for some $A\in SL(2,\mathbb{R})$, then the roots of the characteristic polynomial $$ 0=\det(A-tI_2)=t^2-tr(A) t+\det A=t^2-tr(A) t+1 $$ are complex conjugates of each other, and hence on the unit circle, so of the form $e^{\pm i\theta}$. This already implies that $A$ would be conjugate to that rotation matrix in $SL(2,\mathbb{C})$. To show that they are conjugate also in $SL(2,\mathbb{R})$ requires a bit more.

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It is pointless to attempt to top a link to Keith Conrad's notes:-). I'm afraid I couldn't easily find the correct basis of $\mathbb{R}^2$. Must come from the eigenvectors somehow, but I don't see it now. –  Jyrki Lahtonen May 15 '12 at 12:04
    
It turned out that the first idea:1) pick an eigenvector $V=(z_1,z_2)\in \mathbb{C}^2$,2) write $V^*=(z_1^*,z_2^*)$, 3) take their "real combinations" $(V+V^*)$ and $i(V-V^*)$, and 4) write the matrix of $A$ w.r.t. this basis, does work after all. I just fumbled the calculation on my first attempt somehow. –  Jyrki Lahtonen May 15 '12 at 12:33
    
Right, that's the approach in Keith Conrad's notes. It's annoying when you think something doesn't work because of a calculating error! –  Tara B May 15 '12 at 12:35
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