I need to the following formula: let $g,h\in SL_2\mathbb{R}$, clearly $[g,h]\in SL_2\mathbb{R}$ since its determinant is one. Reading a publication I found the following equality:
$$tr[g,h]=tr^2(g)+tr^2(h)+tr^2(gh)-tr(g)tr(h)tr(gh)-2. $$
I wasn't able to prove it, maybe I don't know some useful identity about the trace. Thanks you so much.
On
The statement above is true in general for matrices $g,h\in M_2(\mathbb{C})$. If we denote by $\tilde{g}, \tilde{h}$ the adjugate of the matrices $g,h$, it's not difficult to see that $$tr(gh\tilde{g})=det(g)tr(h)$$ and $$tr(gh)+tr(\tilde{g}h)=tr(g)tr(h).$$ Now: to prove the equality above, we apply that relation repeatedly:
$$tr(gh\tilde{g}\tilde{h})=tr((\widetilde{hg})gh)=tr^2(gh)-tr(hg^2h)=tr^2(gh)-tr(h)tr(g^2h)+tr(\tilde{h}g^2h)=$$
$$=tr^2(gh)-tr(g)tr(h)tr(gh)+tr^2(h)det(g)+tr(g^2)det(h).$$
Since $tr(g^2)=tr^2(g)-2det(g)$ we have:
$$tr(gh\tilde{g}\tilde{h})=tr^2(gh)-tr(g)tr(h)tr(gh)+tr^2(h)det(g)+tr^2(g)det(h)-2det(g)det(h).$$
In particular: if $g,h\in SL_2(\mathbb{R})$, then $det(g)=det(h)=1$ and $tr(gh\tilde{g}\tilde{h})=tr([g,h])$, and the equality is proved.
We may assume $g=\begin{pmatrix}0&-1\\1&u\end{pmatrix}$ (otherwise, $g$ is a scalar matrix and the result is obviously true). Then (Maple) $tr(g.h.adj(g).adj(h))-tr(g)^2-tr(h)^2-tr(gh)^2+tr(g)tr(h)tr(gh)+2=(u^2-2)(\det(h)-1)$.