Given the symplectic group $\mathrm{Sp}(2,\mathbb{R})$, represented as real $2\times 2$ matrices, I would like to compute the geodesic from the identity matrix $1\!\!1$ to the group element \begin{align} \left( \begin{array}{cc} \omega & 0\\ 0 & \frac{1}{\omega} \end{array}\right)\,, \end{align} where I define my right-invariant metric on the Lie algebra, such that the following the following basis of the Lie algebra is orthonormal: \begin{align} K_1=\left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\,,\quad K_2=\left( \begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right)\,,\quad K_3=\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\,. \end{align} I believe that the geodesic should be given by just \begin{align} \gamma(t)=\exp\left(t\ln{(\omega)}K_1\right)\,, \end{align} with $t\in[0,1]$.
Is there a direct way to see from the metric that this is a geodesic? Also, due to the right-invariance of the metric, we should have $3$ Killing vector fields that may help to find the geodesic?
