Bonnet conjugate radius theorem

The test field. [F4, F6, F7, given] Put $b:=\pi \mathrm{/}\sqrt{k}$ and let $e\in N_0$ be a unit normal vector, which exists because $\dim N_0=n-1\ge 1$ by [F6]. Let $E$ be the parallel field along $\gamma {\mathrm{\mid }}_{[0,b]}$ with $E(0)=e$, and put $V(t):={\mathrm{sn}}_k(t)E(t)$ for $t\in [0,b]$. By [F6] $\mathrm{\mid }E\mathrm{\mid }\equiv 1$, and metric compatibility of the Levi-Civita connection makes $g(E,T)$ constant along $\gamma$; its value $g(e,T(0))=0$ at $t=0$ shows $E(t)\perp T(t)$ for every $t$, with $(E(t),T(t))$ orthonormal. Since ${\mathrm{sn}}_k$ is smooth with ${\mathrm{sn}}_k(0)={\mathrm{sn}}_k(b)=0$ by [F4], the field $V$ is smooth and belongs to ${\mathcal{X}}_0(\gamma {\mathrm{\mid }}_{[0,b]})$; and $V\mathrm{\ne }0$, because ${\mathrm{sn}}_k(b\mathrm{/}2)>0$ while $\mathrm{\mid }E(b\mathrm{/}2)\mathrm{\mid }=1$.

The index form of the test field is nonpositive. [F2, F3, F4, F5, given] Because $E$ is parallel, $D_{t}V={\mathrm{sn}}_k^{\mathrm{\prime }}E$, so $g(D_{t}V,D_{t}V)={\mathrm{sn}}_k^{\mathrm{\prime }2}$; and the quadratic term is $g(R(V,T)T,V)={\mathrm{sn}}_k^{2}g(R(E,T)T,E)={\mathrm{sn}}_k^{2}K(\mathrm{span}\{E,T\}),$ where [F3] was used with the orthonormal pair $(E,T)$ and the four-tensor identity of [F2]. By the hypothesis $K\ge k$ of the Given and ${\mathrm{sn}}_k^2\ge 0$, the integrand is pointwise at most ${\mathrm{sn}}_k^{\mathrm{\prime }2}-k{\mathrm{sn}}_k^2$, so $I_{\gamma }(V,V)\le {\int }_0^b({\mathrm{sn}}_k^{\mathrm{\prime }2}-k{\mathrm{sn}}_k^2)dt\mathrm{.}$ By the model equation of [F4] and the product rule in [F5], $({\mathrm{sn}}_k{\mathrm{sn}}_k^{\mathrm{\prime }}{)}^{\mathrm{\prime }}={\mathrm{sn}}_k^{\mathrm{\prime }2}+{\mathrm{sn}}_k{\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}={\mathrm{sn}}_k^{\mathrm{\prime }2}-k{\mathrm{sn}}_k^2$; Newton–Leibniz in [F5] therefore evaluates the integral as $[{\mathrm{sn}}_k{\mathrm{sn}}_k^{\mathrm{\prime }}{]}_0^b={\mathrm{sn}}_k(b){\mathrm{sn}}_k^{\mathrm{\prime }}(b)-{\mathrm{sn}}_k(0){\mathrm{sn}}_k^{\mathrm{\prime }}(0)=0-0=0,$ using ${\mathrm{sn}}_k(0)={\mathrm{sn}}_k(b)=0$ from [F4]. Hence $I_{\gamma }(V,V)\le 0$.

The index lemma with zero endpoint values bounds the same integral from below. [F1, F2, step 1.1] Suppose, toward the conjugacy conclusion, that no $t\in (0,b]$ makes $\gamma (0)$ and $\gamma (t)$ conjugate along $\gamma {\mathrm{\mid }}_{[0,t]}$. Then the hypothesis of [F1] holds on $[0,b]$, and applying [F1] with $u=w=0\in T_{\gamma (0)}M$ (respectively $T_{\gamma (b)}M$) produces a unique Jacobi field $J$ along $\gamma {\mathrm{\mid }}_{[0,b]}$ with $J(0)=J(b)=0$, together with the inequality $I_{\gamma }(J,J)\le I_{\gamma }(V,V)$ for the admissible field $V$ of step 1.1, with equality if and only if $V=J$. By the assumed absence of conjugacy there is no nonzero such $J$, so $J=0$ and $I_{\gamma }(J,J)=I_{\gamma }(0,0)=0$. Thus $0\le I_{\gamma }(V,V)$.

Equality is forced and contradicts the nontriviality of the test field. [step 1.2, step 2.1] Step 1.2 gives $I_{\gamma }(V,V)\le 0$ and step 2.1 gives $0\le I_{\gamma }(V,V)$, so $I_{\gamma }(V,V)=0$ and the inequality of [F1] is an equality. By the equality clause of the index lemma quoted in [F1] this forces $V=J$; but $J=0$ and $V\mathrm{\ne }0$ by step 1.1. This contradiction shows that the assumption "no $t\in (0,b]$ makes $\gamma (0)$ and $\gamma (t)$ conjugate" is false, so there is $t\in (0,b]$ with $\gamma (0)$ and $\gamma (t)$ conjugate along $\gamma {\mathrm{\mid }}_{[0,t]}$, as claimed.

Scope of the hypotheses and boundary cases. [step 3.1, F4, F7] The dimension hypothesis $n\ge 2$ is exactly what [F6] needs to produce one unit normal vector; in dimension $1$ the normal space is zero and no test field exists. This theorem assumes $k>0$ and makes no assertion for $k\le 0$: the model sine then has no positive zero, so the test-field argument in step 1.2 gives no conjugacy conclusion. The case $t=b$ is included, so the bound is "by $\pi \mathrm{/}\sqrt{k}$" and not "before $\pi \mathrm{/}\sqrt{k}$"; the round sphere of constant curvature $k$ realizes conjugacy exactly at $b$, so the bound cannot be improved. Completeness is used only through [F7] to have the geodesic defined on the closed interval $[0,b]$, and no minimizing hypothesis, no cut-locus hypothesis and no upper curvature bound are used. The argument selects one unit normal vector and one parallel field, both explicitly, and spends no choice beyond [A1].