 Solution

Parametrize the nonconstant closed geodesic at unit speed on $[0,L]$. Its parallel transport around the loop fixes the tangent $\dot{\gamma }(0)$. Since the manifold is orientable, it preserves orientation on the full tangent space, and hence has determinant one on the normal space. This normal space has odd dimension because the manifold has even dimension.

An odd-dimensional special orthogonal transformation has a fixed vector. Indeed, nonreal eigenvalues of a real orthogonal transformation occur in conjugate pairs with product one; the real eigenvalues are $\pm 1$, and determinant one makes the number of $-1$ eigenvalues even. Odd dimension therefore requires a $+1$ eigenvalue. Transport a unit fixed normal vector around the geodesic to obtain a periodic parallel field $V$, with $D_{t}V=0$ and $V\perp \dot{\gamma }$.

The variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$ is defined uniformly for small $s$, because the image of the original loop is compact. Periodicity gives closed curves and a homotopy through loops. With the standard positive-sectional-curvature Riemannian index form, the second variation of geodesic energy isThe first variation vanishes for a closed geodesic, since the endpoint terms cancel. Therefore $E({\gamma }_s)<E({\gamma }_0)=L/2$ for sufficiently small nonzero $s$. The Cauchy-Schwarz inequality givesThus the closed geodesic is homotopic to a strictly shorter closed curve. This proves the instability of a closed geodesic in positive even-dimensional curvature; completeness of the manifold is unnecessary for this compact-loop variation. Constant loops are excluded by the usual nonconstant meaning of a closed geodesic in this assertion.