Show that whenever $X$ is a trace free $2 \times2$ matrix, then $A(s)=\exp(sX)$

Question

In this problem we try to find the critical points (ideally minima) of the energy functional on curves in the hyperbolic plane. Generally, for any Riemannian manifold (U, g) the energy functional is given by $$E(\gamma) = \frac{1}{2} \int_I g_{\gamma (t)} (\gamma '(t), \gamma '(t))dt$$

for curves $\gamma : I → U.$

Recall that $SL(2,\mathbb{R})$ acts by isometries on $\mathbb{H}^2$ (the upper half hyperbolic plane) via: $$ Az = \frac {az + b}{cz + d}$$

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Let $V = \gamma^\prime$ where the variation $\gamma_s = A(s) \cdot \gamma$ is given by a curve $s \mapsto A(s)$ in $SL(2, \mathbb{R})$ with $A(0) = I_2$. In other words, we choose as a variation the curve $\gamma$ moved by a $1-$parameter family of isometries. Show that whenever $X$ is a trace free $2 \times 2$ matrix, then $A(s) = \exp(sX) \in SL(2,\mathbb{R})$ and calculate $V$ for the cases where $X$ is of the following three cases
$$\text{Case 1:}~X = \left(\begin{array}{cc} 1&0\\ 0&-1 \end{array} \right),\quad \text{Case 2:} ~ X = \left(\begin{array}{cc} 0&1\\ 1&0 \end{array} \right), \quad \text{Case 3:}~ X = \left(\begin{array}{cc} 0&1\\ -1&0 \end{array} \right) $$

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So for this, it seems, there are two parts:

I. Showing that whenever X is trace free and a 2 $\times$ 2 matrix then it must take the form above.

• For this I'm not so sure why exactly A(s) would take that form. Is it possibly because A $\in$ SL(2, $\mathbb{R}$)?

• I know the general properties of the special linear group are that they have determinant one and that conceptually they are the group of linear transformations of the real plane that preserve oriented area.

• The eigenvalues of any element T $\in$ SL(2, $\mathbb{R}$) are given by the characteristic polynomial $$\lambda^{2} + tr(T)\lambda + 1 = 0$$ and this leads to determinant: $$ tr(A)^{2} - 4$$ for which we may classify our Matrix A as elliptic and thus conjugate to a rotation (since |tr(A)| = 0 < 2) and since rotations typically have a form similar to exp$^{sX}$ that leads me to believe this is the an okay path to go down? Any suggestions, help?

II. Here I can calculate A(s) easily when X is defined but I'm unsure how to calculate V = $\gamma$' from the determination of A(s).

• Perhaps is it since $\gamma$' is equal to V should I solve for $\gamma$: $\gamma$ = $\frac{\gamma_s}{A(s)}$ and then differentiate (w.r.t s I'm guessing?) Then we're given A(s) and given $\gamma_s$ in terms of A(s) and $\gamma$ and so we would have our solution for V, in each case, in terms of s and $\gamma$? Doesn't feel right..

Any help on either component is much appreciated!

Answer 1

$SL(2,\mathbb{R})$ is a Lie group with lie algebra $sl(2,\mathbb{R})$. Thus the one parameter curves are given by $e^{sX}$.

One can also see this directly, on that if $f(s)\in SL(2,\mathbb{R})$ is continuous with $$f(s+t)=f(s)f(t)$$ then we see that $$f^{\prime}(s)=f^{\prime}(0)f(s)$$ and thus we see from this differential equation that $$f(s)=e^{sX}$$ (where $X=f^{\prime}(0)$). Finally the equation $$\det e^X=e^{tr X}$$ gives the desired properties of $X$.

Answer 2

The following argument shows that for every trace free $X\in M_n(\mathbb{R})$ and every $t$, we have $e^{tX}\in SL_n(\mathbb{R}).$

The most important fact is that "trace is the derivative of the determinant". Explicitly, if $A_t$ is a one parameter family of invertible matrices, we have$$\frac{d}{dt}\det A_t=\det A_t\cdot\mathrm{tr}\left(A_t^{-1}\dot{A}_t\right),$$where $\dot{A}_t:=\frac{d}{dt}A_t$. Applying this to the one parameter family $A_t=e^{tX}$, $$\frac{d}{dt}\det e^{tX}=\det e^{tX}\cdot\mathrm{tr}\left(e^{-tX}e^{tX}X\right)=\det e^{tX}\cdot\mathrm{tr}X=0.$$Clearly, when $t=0$ we have $\det e^{tX}=1$, and this completes the proof.