Conjugate points on $\text{SL}(2,\mathbb{R})$ Consider $\text{SL}(2,\mathbb{R})$ with the left-invariant metric obtained by translating the standard Frobenius product at $T_I\text{SL}(2,\mathbb{R})$. (i.e $g_I(A,B)=\operatorname{tr}(A^TB)$ for $A,B \in T_I\text{SL}(2,\mathbb{R})$).
One can show that the geodesics starting at $I$ are of the form of
$$ \gamma_v(t) = e^{tV^T}e^{t(V-V^T)}, \operatorname{tr}(V)=0$$
I am trying to prove the following claims:


*

*If $\det(V) \le 0$, then there are no conjugate points along the geodesic $\gamma_v$.

*If $\det(V)>0$, the first conjugate point is at $t = \pi/\sqrt{\det(V)}$.


(This is an attempt to understand the comments made by Robert Bryant here).
The first question is equivalent to showing that the exponential map $d(exp_I)_V$ is non-singular.
When trying to calculate
$$d(exp_I)_V(W)= \dfrac{d}{dt}\Big|_{t=0}\exp_I(V+tW)=\dfrac{d}{dt}\Big|_{t=0}e^{V^T+tW^T}\cdot e^{V+tW-V^T-tW^T} $$ 
$$=e^{V^T}\dfrac{d}{dt}\Big|_{t=0}e^{V-V^T+t(W-W^T)}+\dfrac{d}{dt}\Big|_{t=0}e^{V^T+tW^T}\cdot e^{V-V^T}$$
We can evaluate the derivatives via the formula
$$ \dfrac{d}{dt}\Big|_{t=0}e^{V+tW}= \int_0^1 e^{\alpha V}We^{(1 - \alpha)V}\,d\alpha $$
However, In am not sure how to proceed. I tried to diagonalize $V$:
$V=\begin{pmatrix} a & b \\  c & -a  \\  \end{pmatrix}$. Then $\det(V) \ge 0 \iff a^2+bc \ge 0$,
and the eigenvalues are $\lambda_i=\pm \sqrt{a^2+bc}$.
I do not see how to proceed (with the proof, not the diagonalization...)
Any ideas how to continue? Perhaps a different approach?
(Remainder: We need to prove $d(exp_I)_V(W)=0 \Rightarrow W=0$).
 A: The geodesic leaving $I_2\in\mathrm{SL}(2,\mathbb{R})$ with velocity
$$
v = \begin{pmatrix} v_1 & v_2+v_3\\ v_2-v_3 & -v_1\end{pmatrix} 
\in {\frak{sl}}(2,\mathbb{R})\simeq\mathbb{R}^3
$$ 
is given by $\gamma_v(t) = e^{t\,v^T}e^{t\,(v{-}v^T)}$.  Thus, the geodesic exponential mapping for this metric is
$$
E(v) = e^{v^T}e^{(v-v^T)}.
$$
(Here, '$v^T$' denotes the transpose of $v$.)
Meanwhile, since $v^2 = -\det(v)\,I_2$, it follows that the formula for the Lie group exponential of $v$ is 
$$
e^v = c\bigl(\det(v)\bigr)\,I_2 + s\bigl(\det(v)\bigr)\,v
$$
where $c$ and $s$ are the entire analytic functions defined on the real line that satisfy $c(t^2) = \cos(t)$ and $s(t^2) = \sin(t)/t$ (and hence satisfy $c(-t^2) = \cosh(t)$ and $s(-t^2) = \sinh(t)/t$). Note that, in particular, these functions satisfy the useful identities
$$
c(y)^2+ys(y)^2 = 1,\qquad c'(y) = -\tfrac12\,s(y),\qquad
\text{and}\qquad\ s'(y) = \bigl(c(y)-s(y)\bigr)/(2y).
$$
Using this, the identity $\det(v) = {v_3}^2-{v_1}^2-{v_2}^2$, and the above formulae, we can compute the pullback under $E$ of the canonical left invariant form on $\mathrm{SL}(2,\mathbb{R})$ as follows.
$$
E^*(g^{-1}\,\mathrm{d}g) 
= E(v)^{-1}\,\mathrm{d}\bigl(E(v)\bigr)
 = e^{-(v-v^T)}\left[e^{-v^T}\,\mathrm{d}(e^{v^T}) 
          + \mathrm{d}(e^{(v-v^T)})\, e^{-(v-v^T)})\right]e^{(v-v^T)}.
$$
Expanding this using the above formula for the Lie group exponential and setting 
$$
E^*(g^{-1}\,\mathrm{d}g)  = \begin{pmatrix} \omega_1 & \omega_2+\omega_3\\ \omega_2-\omega_3 & -\omega_1\end{pmatrix},
$$
we find, after setting $\det(v) = \delta$ for brevity, that
$$
\omega_1\wedge\omega_2\wedge\omega_3
= s(\delta)\left(s(\delta)
  -2({v_1}^2{+}{v_2}^2)\frac{\bigl(c(\delta)-s(\delta)\bigr)}{\delta}\right)
     \,\mathrm{d}v_1\wedge\mathrm{d}v_2\wedge\mathrm{d}v_3\,.
$$
(Note, by the way, that $\frac{c(\delta)-s(\delta)}{\delta}$ is an entire analytic function of $\delta$.)

It follows that the degeneracy locus for the geodesic exponential map 
$E:{\frak{sl}}(2,\mathbb{R})\to \mathrm{SL}(2,\mathbb{R})$ is the union of the loci described by the two equations
$$
s\bigl(\det(v)\bigr) = 0\tag1
$$
and
$$
s\bigl(\det(v)\bigr)
  -2({v_1}^2{+}{v_2}^2)\frac{\bigl(c\bigl(\det(v)\bigr)-s\bigl(\det(v)\bigr)\bigr)}{\det(v)} = 0.\tag2
$$
Now, $s(t)\ge 1$ when $t\le 0$, while $s(t) = 0$ for $t>0$ implies that $t = (k\pi)^2$ for some integer $k>0$.  Thus, the first locus is given by the hyperboloids
$$
\det(v) = {v_3}^2-{v_1}^2-{v_2}^2 = k^2\pi^2,\quad k= 1,2,\ldots
$$ 
Meanwhile, when $t\le 0$, the expression $\frac{c(t)-s(t)}{t}$ is strictly negative, while $s(t)\ge 1$, so it follows that the second locus has no points in the region $\det(v)\le 0$, i.e., no geodesic $\gamma_v$ with $\det(v)\le0$  has any conjugate points.  
Finally, a little elementary analytic geometry
shows that the locus described by (2) is a countable union of surfaces 
$\Sigma_k$ of revolution in ${\frak{sl}}(2,\mathbb{R})$ 
that can be described in the form
$$
{v_3}^2 = ({v_1}^2+{v_2}^2) + \bigl(k + f_k({v_1}^2+{v_2}^2)\bigr)^2\pi^2,
\qquad k = 1,2,\ldots
$$
where $f_k:[0,\infty)\to[0,\tfrac12)$ is a strictly increasing real-analytic
function on $[0,\infty)$ that satisfies $f_k(0)=0$.
In particular, it follows that, for a $v\in\Sigma_k$, 
we have $ k^2\pi^2\le \det(v)< (k+\tfrac12)^2\pi^2$.

Consequently, the first conjugate locus is the image under $E$ of the hyperboloid $\det(v) = \pi^2$.  Note that, by the above formulae, this image
in $\mathrm{SL}(2,\mathbb{R})$ is simply the subgroup $\mathrm{SO}(2)\subset
\mathrm{SL}(2,\mathbb{R})$.
A: ${\rm Tr}\ x=0$ implies that $$ x^2 +({\rm det}\ x) I=0 $$
where $$ x:= \left(
                                                \begin{array}{cc}
                                                  a & b \\
                                                  c & -a \\
                                                \end{array}
                                              \right)  $$
Case 1 - $\omega:=\sqrt{a^2+bc} >0$ Then $$ e^x= \cosh\ \omega I +
\frac{\sinh\ \omega}{\omega} x$$
Case 2 - $\omega:=\sqrt{-(a^2+bc)} >0$ : $$
e^x= \cos\ \omega I + \frac{\sin\ \omega }{
\omega} x $$
Case 3 - $a^2+bc=0$ : $$ e^x=I+x $$
Example : If $$x:= \left(
                                                \begin{array}{cc}
                                                  0 & \pi \\
                                                  -\pi & 0 \\
                                                \end{array}
                                              \right),\ y:= \left(
                                                \begin{array}{cc}
                                                  y_1 & y_2 \\
                                                  y_2 & -y_1 \\
                                                \end{array}
                                              \right)$$ where $
                                              y$ is a
symmetric matrix with ${\rm Tr}\ y=0$, then let
$x_\varepsilon:=x+\varepsilon y$. So
$$ \frac{d}{d
\varepsilon }\bigg|_{ \varepsilon=0}\ e^{x_\varepsilon^T}
=\frac{-1}{\pi} x^T\ \frac{d}{d\varepsilon
}\bigg|_{\varepsilon=0} \sqrt{{\rm det}\ x_\varepsilon }=0$$
Hence ${\rm SL}(2,\mathbb{R})$ has a conjugate point at $e^x$.
