The Problem. Compute the conjugate locus of $\mathrm{SL}(2,\mathbb{R})$ endowed with the 'natural' left-invariant Riemannian metric as described in my answer to the MathOverflow Question 108280

The geodesic formula: As explained there, the formula for the geodesic leaving $I_2\in\mathrm{SL}(2,\mathbb{R})$ with velocity $$ v = \begin{pmatrix} v_1 & v_2+v_3\\ v_2-v_3 & -v_1\end{pmatrix} \in {\frak{sl}}(2,\mathbb{R}) $$ 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 obvious 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 locus 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 $\det(v)<0$ 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 $\mathrm{SO}(2)\subset \mathrm{SL}(2,\mathbb{R})$.