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John Jiang
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Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let \begin{equation} X = \left(\begin{array}{cc} J & 0 \\ 0 & \alpha J \end{array}\right) \end{equation}$$ X = \begin{pmatrix} J & 0 \\ 0 & \alpha J \end{pmatrix} $$

be in block diagonal form, where $J = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$$J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed.

Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let \begin{equation} X = \left(\begin{array}{cc} J & 0 \\ 0 & \alpha J \end{array}\right) \end{equation}

be in block diagonal form, where $J = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed.

Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let $$ X = \begin{pmatrix} J & 0 \\ 0 & \alpha J \end{pmatrix} $$

be in block diagonal form, where $J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed.

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John Jiang
  • 4.5k
  • 25
  • 47

Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let \begin{equation} X = \left(\begin{array}{cc} J & 0 \\ 0 & \alpha J \end{array}\right) \end{equation}

be in block diagonal form, where $J = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed.