Clean energy critical manifolds in Clifford-centriole polygon spaces

Put $x_0=o$, $x_n=q$, and let $Q_n$ be the open set of $(x_1,\dots ,x_{n-1})\in P^{n-1}$ with $\mathrm{dist}(x_{j-1},x_j)<r$. Define $E_n=n{\sum }_j\mathrm{dist}(x_{j-1},x_j{)}^2$. On $E_n\le c$ every edge length is at most $\sqrt{c\mathrm{/}n}<r$. Thus this sublevel is closed in the compact product and lies inside $Q_n$. A fixed path component is closed here: sufficiently close vertex families yield pointwise short-geodesic homotopies. First variation at a vertex is twice the incoming velocity minus twice the outgoing velocity. Hence the critical polygons are exactly sampled smooth geodesics.

For $l\ge 1$ write $\mathfrak{h}=\{Z\in \mathfrak{s}\mathfrak{o}(p):[Z,J_i]=0(i<l)\}$, $\mathfrak{k}=\{Z\in \mathfrak{h}:[Z,J_l]=0\}$ and $\mathfrak{m}=\{Z\in \mathfrak{h}:Z{J}_l=-J_{l}Z\}$. Then $\mathfrak{h}=\mathfrak{k}\oplus \mathfrak{m}$, $[\mathfrak{m},\mathfrak{m}]\subseteq \mathfrak{k}$, and $T_{J_l}{P}_l=\mathfrak{m}{J}_l$. Along $\gamma (t)=\exp (tX)J_l$, $X\in \mathfrak{m}$, the parallel curvature operator is $-{\mathrm{ad}}_X^2\mathrm{/}4$ on $\mathfrak{m}$. It is nonnegative since ${\mathrm{ad}}_X$ is skew-adjoint for the trace metric. In a parallel eigenframe, a Jacobi field vanishing at zero is $tV$ on the zero eigenspace and $\sin (\sqrt{\kappa }t)V\mathrm{/}\sqrt{\kappa }$ on an eigenspace of eigenvalue $\kappa >0$. The terminal value vanishes precisely for the latter directions with $\sqrt{\kappa }\in \pi \mathbb{N}$.

At a critical geodesic decompose any endpoint-zero $H^1$ field as $W=J+Z$, where $J$ is the unique edgewise Jacobi field with the same vertex values and $Z$ vanishes at every vertex. Short convex edges have no conjugate points, so those Jacobi interpolants exist uniquely. Integration by parts on every edge gives $I(J,Z)=0$. On the edgewise zero-vertex space the index form is positive definite: if $Y$ is the Jacobi fundamental matrix with $Y(0)=0$ and $Y^{\mathrm{\prime }}(0)=I$, then before the first conjugate point $S=Y^{\mathrm{\prime }}{Y}^{-1}$ is symmetric and satisfies $S^{\mathrm{\prime }}+S^2+R=0$. Expanding $\mathrm{\mid }{Z}^{\mathrm{\prime }}-SZ{\mathrm{\mid }}^2$ and integrating gives $I(Z,Z)=\int \mathrm{\mid }{Z}^{\mathrm{\prime }}-SZ{\mathrm{\mid }}^2$; the boundary terms vanish, first for smooth zero-endpoint fields and then by $H^1$ approximation. Equality forces $Z^{\mathrm{\prime }}=SZ$ and the zero terminal value forces $Z=0$. Thus the polygon Hessian is twice $I$ on the interpolant space and has exactly the full Hessian's index and kernel.

For such an eigenvector $Y\in \mathfrak{m}$ put $B=[X,Y]\mathrm{/}(4\kappa )\in \mathfrak{k}$. The identity ${\mathrm{ad}}_X^{2}Y=-4\kappa Y$ gives $[B,X]=Y$. Conjugation of $\gamma$ by $\exp (sB)$ is a geodesic variation fixing both $J_l$ and $-J_l$, with initial derivative $Y{J}_l$. By uniqueness it is the corresponding Jacobi field. Therefore all endpoint-zero Jacobi fields are tangent to the orbit of the compact centralizer of $J_1,\dots ,J_l$. For $l=0$ use conjugation in $\mathrm{S}\mathrm{O}(p)$, which fixes $I$ and $-I$, with the same bracket calculation. Its orbit is compact and embedded; its tangent is exactly the Hessian kernel by step 2.1.

Choose a normal slice to one such orbit. The Hessian on that slice is invertible. Apply the inverse theorem to the normal component of the differential of $E_n$ in slice coordinates: its only zero near the zero section is the zero section. Hence the critical locus near that orbit is exactly the orbit, proving the Morse–Bott condition. The critical locus in a compact energy sublevel is compact, so finitely many such neighborhoods cover it. Each contains just one critical component; consequently there are finitely many components and critical values. No classification of all symmetric spaces or centriole chains is needed.