Conjugate points along a geodesic and their multiplicity

The smooth fields along $\gamma$ form a real vector space under pointwise operations: in a pulled-back frame, addition and scalar multiplication preserve smooth coefficient functions by [F1], and the vector-space axioms hold in each tangent fiber by [F2].

Suppose $\gamma$ is constant at $p$. If $n=0$, [F1] and [F2] imply the only field along $\gamma$ is zero. Otherwise, take a finite basis $e_1,\dots ,e_n$ of $T_{p}M$ using [F2] and [F10], and extend each vector to a parallel section $E_i$ by [F14]. These sections span every fiber: any vector at any time has a parallel extension by [F14], and its value at $a$ is a linear combination of the $e_i$, whose parallel extension is that same combination of the $E_i$ by uniqueness. They are independent at every time: a linear combination vanishing at one time is a parallel section with zero value, so uniqueness in [F14] makes it identically zero, and its initial coefficients vanish by basis independence. Write an arbitrary Jacobi field as $J(t)={\sum }_i{x}_i(t)E_i(t)$, with smooth coefficients by [F1]. Since $T=0$, curvature multilinearity [F5] and [F3] give $D_t^{2}J=0$. Using the product rule [F4] twice and $D_t{E}_i=0$ from [F15], this equation becomes ${\sum }_i{x}_i^{\mathrm{\prime }\mathrm{\prime }}{E}_i=0$, so every $x_i^{\mathrm{\prime }\mathrm{\prime }}=0$. Applying [F16] to $x_i^{\mathrm{\prime }}$ and then to $x_i-c_{i}t$ gives $x_i(t)=c_{i}t+d_i$. From $x_i(a)=x_i(b)=0$ and $a<b$, we get $c_i=d_i=0$. Therefore ${\mathcal{K}}_{\gamma }(a,b)=\{0\}$ and the constant-geodesic endpoints are not conjugate.

Put $T=\dot{\gamma }$ and define $L(J)=D_t^{2}J+R(J,T)T$ on the smooth-field space of step 1.1. By [F4] and [F5], $L$ is real-linear. Thus its kernel, the Jacobi fields by [F3], is a linear subspace; the endpoint conditions $J(a)=J(b)=0$ are preserved under the same pointwise operations. Hence ${\mathcal{K}}_{\gamma }(a,b)$ is a real vector space.

Define $\mathrm{\Phi }:{\mathcal{K}}_{\gamma }(a,b)\to T_{\gamma (a)}M$ by $\mathrm{\Phi }(J)=D_{t}J(a)$. By [F4] and [F7], $\mathrm{\Phi }$ is linear. If $\mathrm{\Phi }(J)=0$, then $J(a)=0$ and $D_{t}J(a)=0$, so [F8] gives $J=0$; hence $\mathrm{\Phi }$ is injective. Let $U=\mathrm{\Phi }[{\mathcal{K}}_{\gamma }(a,b)]$.

Since $\mathrm{\Phi }$ is linear, $U$ contains zero and is closed under addition and scalar multiplication, so it is a linear subspace of $T_{\gamma (a)}M$ by [F6]. By [F9], it is finite-dimensional with $\dim U\le n$. Take a basis $u_1,\dots ,u_d$ of $U$ by [F10]. Each $u_i$ has a unique preimage $J_i\in {\mathcal{K}}_{\gamma }(a,b)$ because $U$ is the image and $\mathrm{\Phi }$ is injective. The fields $J_1,\dots ,J_d$ span ${\mathcal{K}}_{\gamma }(a,b)$: for any $J$, express $\mathrm{\Phi }(J)$ in the basis $u_i$ and use injectivity of $\mathrm{\Phi }$ to recover the same linear combination of the $J_i$. They are independent because applying $\mathrm{\Phi }$ to a vanishing linear combination gives a vanishing linear combination of the $u_i$. Thus ${\mathcal{K}}_{\gamma }(a,b)$ is finite-dimensional of dimension $d=\dim U\le n$, so the stated multiplicity is well-defined.

Suppose $\gamma$ is nonconstant. By [F12], its speed is constant. If it vanished at any time, then $T=0$ throughout; in local charts the coordinate functions of $\gamma$ would have zero derivative and hence be locally constant by [F16], making $\gamma$ constant on the connected interval. Thus $T(a)$ and $T(b)$ are nonzero; in particular $n>0$. If $T(a)\in U$, there is $J\in {\mathcal{K}}_{\gamma }(a,b)$ with $D_{t}J(a)=T(a)$. The Jacobi field $K(t)=(t-a)T(t)$ from [F13] has the same initial data, because [F11] gives $D_{t}K(a)=T(a)$. Uniqueness [F8] gives $J=K$, but $K(b)=(b-a)T(b)\mathrm{\ne }0$, contradicting $J(b)=0$. Therefore $U$ is a proper subspace. By [F9], $\dim U\mathrm{\ne }n$ and $\dim U\le n$; [F17] gives $\dim {\mathcal{K}}_{\gamma }(a,b)=\dim U\le n-1$.

The interval is nondegenerate because $a<b$. The zero field belongs to $\mathcal{K}$ but does not witness conjugacy, which requires a nonzero field. For $n=1$, step 5.1 gives dimension zero for the endpoint-vanishing space of a nonconstant geodesic. If $n=0$, every tangent fiber is zero by [F2], so $\dot{\gamma }=0$ and local coordinate functions are locally constant by [F16]; connectedness makes $\gamma$ constant, and step 1.2 applies. A supplied geodesic on $[a,b]$ rules out the empty-manifold case. All endpoint derivatives are one-sided by [F3], [F8], [F11], and [F14]. The proof uses one finite basis of a single tangent space and uniquely determined Jacobi and parallel fields; [F9] also uses no choice principle, so neither AC nor ${\mathrm{A}\mathrm{C}}_{\omega }$ is assumed or used. By the stated definition, a nonzero $J\in {\mathcal{K}}_{\gamma }(a,b)$ gives conjugate endpoints; conversely, conjugate endpoints provide such a nonzero $J$.