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A smooth nondegenerate component of the fixed-point set of an isometry is a totally geodesic submanifold. If a geodesic starts tangent to the fixed component, its image under the isometry has identical position and tangent. Uniqueness of the geodesic equation makes the two geodesics coincide throughout their common domain, so the original geodesic stays in the fixed set.

For a Riemannian submersion with totally geodesic submanifolds as fibers, the positive Laplace-Beltrami operator commutes with pullback of functions. Horizontal terms in the Riemannian Hessian are pulled back from the base, and vertical terms vanish. Minimal fibers suffice because only the trace of the vertical second fundamental form enters.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

Consider the discrete mapIt is an isometry: $\cos \theta$ and $d\varphi$ both change sign, leaving $\cos \theta d\varphi$ unchanged, while $d{\theta }^2$ and ${\sin }^2\theta d{\varphi }^2$ are also unchanged. A local component of its fixed-point set is $\theta =\pi /2$, $\varphi ={\varphi }_0$.

If a geodesic starts tangent to that component, the isometry fixes both its initial position and its initial tangent. Applying the isometry therefore gives a geodesic with the same initial data. Uniqueness of the geodesic equation makes it the same curve, so it remains in the fixed component. This proves that the selected submanifold is a totally geodesic submanifold; it has dimension three, not two. The proof continues through a horizon when expressed in regular coordinates, so coordinate singularities of the original chart do not invalidate it.

The geodesic conserved quantities from Killing vectors associated with ${\partial }_{\psi }$ and ${\partial }_{\varphi }$ areDots denote derivatives with respect to an affine parameter. On the chosen submanifold $\dot{\varphi }=0$ and $\cos \theta =0$, so $L_{\varphi }=0$. Requiring also $L_{\psi }=0$ givesMore generally, away from the polar-coordinate axes, both displayed angular charges being zero imply $\dot{\varphi }=0$ and the same relation. In particular, zero conserved angular momentum does not imply constant $\psi$: the motion follows the local dragging of the angular coordinate by the rotating geometry.

For a nondegenerate induced metric tensor, a totally geodesic submanifold has vanishing second fundamental form; equivalently, every ambient geodesic initially tangent to it remains in it. This extends the notion of a totally geodesic hypersurface to arbitrary codimension.