[proofplan] We compare two ways of encoding first-order variations of the geodesic $\gamma(t)=\exp_p(tv)$. Varying the initial velocity from $v$ to $v+sw$ produces a geodesic variation, hence a Jacobi field, and evaluating this field at $t=1$ is exactly applying $d(\exp_p)_v$ to the tangent vector represented by $w$. The kernel condition is therefore equivalent to producing a Jacobi field vanishing at both endpoints. Conversely, every Jacobi field along $\gamma$ with $J(0)=0$ is determined by its initial covariant derivative, so it is realised by a unique initial-velocity variation. [/proofplan] [step:Identify tangent vectors to $T_pM$ by translation] Let $E:=T_pM$, regarded as a finite-dimensional real [vector space](/page/Vector%20Space). Define the translation map \begin{align*} \tau_v:E &\to E \\ u &\mapsto v+u. \end{align*} Its differential at $0 \in E$ is the linear isomorphism \begin{align*} d(\tau_v)_0:E \to T_vE. \end{align*} Let $\mathcal{D}_p \subset T_pM$ denote the open domain of the exponential map \begin{align*} \exp_p:\mathcal{D}_p &\to M. \end{align*} Throughout the proof, when $w \in T_pM$ is used as a tangent vector at $v \in T_pM$, it means the vector $d(\tau_v)_0(w) \in T_v(T_pM)$. [/step] [step:Choose a uniform interval for all initial velocity variations] For every $w \in T_pM$, there exists $\varepsilon>0$ such that \begin{align*} t(v+sw) \in \mathcal{D}_p \end{align*} for all $t \in [0,1]$ and all $s \in (-\varepsilon,\varepsilon)$. [claim:The segment $\{tv:0\leq t\leq 1\}$ has a uniform tubular margin inside $\mathcal{D}_p$ in the direction $tw$] For each fixed $w \in T_pM$, there exists $\varepsilon>0$ such that $t(v+sw) \in \mathcal{D}_p$ for all $t \in [0,1]$ and $|s|<\varepsilon$. [/claim] [proof] Define the compact segment map \begin{align*} \rho:[0,1] &\to E \\ t &\mapsto tv. \end{align*} The hypothesis that $\gamma(t)=\exp_p(tv)$ is defined for all $t\in[0,1]$ means precisely that $\rho([0,1])\subset \mathcal{D}_p$. Since $\mathcal{D}_p$ is open in the finite-dimensional [normed vector space](/page/Normed%20Vector%20Space) $E$ and $\rho([0,1])$ is compact, there exists $\delta>0$ such that \begin{align*} \{u\in E: \operatorname{dist}(u,\rho([0,1]))<\delta\}\subset \mathcal{D}_p. \end{align*} If $w=0$, choose any $\varepsilon>0$. If $w\ne0$, choose \begin{align*} \varepsilon:=\frac{\delta}{|w|}>0. \end{align*} Then for $t\in[0,1]$ and $|s|<\varepsilon$, \begin{align*} |t(v+sw)-tv|=|tsw|\leq |s|\,|w|<\delta, \end{align*} so $t(v+sw)\in\mathcal{D}_p$ by the choice of $\delta$. [/proof] [/step] [step:Compute the endpoint of the initial velocity variation] Fix $w \in T_pM$. By the previous step, choose $\varepsilon>0$ such that $t(v+sw) \in \mathcal{D}_p$ for all $t \in [0,1]$ and all $s \in (-\varepsilon,\varepsilon)$, and define \begin{align*} \alpha:[0,1]\times(-\varepsilon,\varepsilon) &\to M \\ (t,s) &\mapsto \exp_p(t(v+sw)). \end{align*} For each fixed $s \in (-\varepsilon,\varepsilon)$, the curve \begin{align*} \alpha_s:[0,1] &\to M \\ t &\mapsto \alpha(t,s) \end{align*} is the geodesic with initial point $p$ and initial velocity $v+sw$. Therefore $\alpha$ is a smooth geodesic variation of $\gamma=\alpha_0$. Let \begin{align*} J:[0,1] &\to TM \\ t &\mapsto \frac{\partial \alpha}{\partial s}(t,0) \end{align*} be its variation field. Since $\alpha$ is a geodesic variation, $J$ is a Jacobi field along $\gamma$. At $t=0$, \begin{align*} J(0)=\frac{\partial}{\partial s}\bigg|_{s=0}\exp_p(0)=\frac{\partial}{\partial s}\bigg|_{s=0}p=0. \end{align*} At $t=1$, the curve \begin{align*} \beta:(-\varepsilon,\varepsilon) &\to M \\ s &\mapsto \alpha(1,s)=\exp_p(v+sw) \end{align*} satisfies $\beta=\exp_p\circ\tau_v\circ \sigma$, where \begin{align*} \sigma:(-\varepsilon,\varepsilon) &\to E \\ s &\mapsto sw. \end{align*} By the chain rule, \begin{align*} J(1) &=\beta'(0) \\ &=d(\exp_p)_v\bigl(d(\tau_v)_0(\sigma'(0))\bigr) \\ &=d(\exp_p)_v\bigl(d(\tau_v)_0(w)\bigr). \end{align*} [guided] The point of this step is to connect the geometric variation with the [linear map](/page/Linear%20Map) in the statement. We start with a vector $w \in T_pM$ and vary the initial velocity $v$ in the affine direction $w$. The previous step guarantees that there is $\varepsilon>0$ for which $t(v+sw)\in\mathcal{D}_p$ whenever $t\in[0,1]$ and $s\in(-\varepsilon,\varepsilon)$. Hence the map \begin{align*} \alpha:[0,1]\times(-\varepsilon,\varepsilon) &\to M \\ (t,s) &\mapsto \exp_p(t(v+sw)) \end{align*} is defined. For fixed $s$, the curve $t \mapsto \exp_p(t(v+sw))$ is exactly the geodesic starting at $p$ with initial velocity $v+sw$. Hence $\alpha$ is a geodesic variation of $\gamma$. The variation field \begin{align*} J:[0,1] &\to TM \\ t &\mapsto \frac{\partial \alpha}{\partial s}(t,0) \end{align*} is therefore a Jacobi field along $\gamma$. Now evaluate this field at the endpoints. At $t=0$, every curve in the variation starts at $p$, since $\exp_p(0)=p$. Thus \begin{align*} J(0)=\frac{\partial}{\partial s}\bigg|_{s=0}\alpha(0,s) =\frac{\partial}{\partial s}\bigg|_{s=0}p =0. \end{align*} At $t=1$, the variation curve in the $s$-direction is \begin{align*} \beta:(-\varepsilon,\varepsilon) &\to M \\ s &\mapsto \exp_p(v+sw). \end{align*} Writing $s \mapsto sw$ as a curve in $T_pM$ and then translating by $v$, this curve is $\exp_p\circ\tau_v\circ\sigma$, where \begin{align*} \sigma:(-\varepsilon,\varepsilon) &\to T_pM \\ s &\mapsto sw. \end{align*} Since $\sigma'(0)=w$, the chain rule gives \begin{align*} J(1) =\beta'(0) =d(\exp_p)_v\bigl(d(\tau_v)_0(w)\bigr). \end{align*} Thus the endpoint value of the Jacobi field is precisely the image of the tangent vector represented by $w$ under $d(\exp_p)_v$. [/guided] [/step] [step:Use a kernel vector to construct a vanishing Jacobi field] Assume $\ker d(\exp_p)_v$ contains a nonzero vector. Under the translation identification from the first step, choose $w \in T_pM\setminus\{0\}$ such that \begin{align*} d(\exp_p)_v\bigl(d(\tau_v)_0(w)\bigr)=0. \end{align*} By the uniform-domain step, choose $\varepsilon>0$ such that $t(v+sw)\in\mathcal{D}_p$ for all $t\in[0,1]$ and $s\in(-\varepsilon,\varepsilon)$. Construct the geodesic variation $\alpha$ and its variation field $J$ as above using this $\varepsilon$. The previous step gives $J(0)=0$ and \begin{align*} J(1)=d(\exp_p)_v\bigl(d(\tau_v)_0(w)\bigr)=0. \end{align*} Let $\nabla_{\dot{\gamma}}J(0)\in T_pM$ denote the covariant derivative of the vector field $J$ along $\gamma$ at $t=0$. For a two-parameter map such as $\alpha$, let $\frac{D}{dt}$ and $\frac{D}{ds}$ denote covariant differentiation along the $t$-curves and $s$-curves of the variation, respectively. The field $J$ is nonzero because its initial covariant derivative is \begin{align*} \nabla_{\dot{\gamma}}J(0) &=\frac{D}{dt}\bigg|_{t=0}\frac{\partial \alpha}{\partial s}(t,0) \\ &=\frac{D}{ds}\bigg|_{s=0}\frac{\partial \alpha}{\partial t}(0,s) \\ &=\frac{D}{ds}\bigg|_{s=0}(v+sw) \\ &=w. \end{align*} The interchange of covariant derivatives uses that the coordinate vector fields $\partial_t$ and $\partial_s$ on $[0,1]\times(-\varepsilon,\varepsilon)$ commute and that the Levi-Civita connection is torsion-free. The identity $\frac{\partial \alpha}{\partial t}(0,s)=v+sw$ holds because every varied geodesic $t\mapsto\alpha(t,s)$ starts at the fixed point $p$, so the initial velocity is an element of the single vector space $T_pM$. Since $w\ne0$, the Jacobi initial data $(J(0),\nabla_{\dot{\gamma}}J(0))=(0,w)$ are not both zero. Hence $J$ is a nonzero Jacobi field along $\gamma$ with $J(0)=J(1)=0$, so $\gamma(1)$ is conjugate to $p$ along $\gamma$. [/step] [step:Realise every Jacobi field vanishing at $0$ by varying the initial velocity] Assume that $\gamma(1)$ is conjugate to $p$ along $\gamma$. By definition, there exists a nonzero Jacobi field \begin{align*} J:[0,1] \to TM \end{align*} along $\gamma$ such that $J(0)=0$ and $J(1)=0$. Define \begin{align*} w:=\nabla_{\dot{\gamma}}J(0)\in T_pM. \end{align*} Since the Jacobi equation is a second-order linear ordinary differential equation along $\gamma$, a Jacobi field is uniquely determined by its initial data $(J(0),\nabla_{\dot{\gamma}}J(0))$. Because $J$ is nonzero and $J(0)=0$, we must have $w\ne0$. By the uniform-domain step, choose $\varepsilon>0$ such that $t(v+sw)\in\mathcal{D}_p$ for all $t\in[0,1]$ and $s\in(-\varepsilon,\varepsilon)$. Let $\widetilde{J}$ be the variation field obtained from \begin{align*} \alpha:[0,1]\times(-\varepsilon,\varepsilon) &\to M \\ (t,s) &\mapsto \exp_p(t(v+sw)). \end{align*} As computed above using this defined variation, $\widetilde{J}(0)=0$ and $\nabla_{\dot{\gamma}}\widetilde{J}(0)=w$. Therefore the uniqueness theorem for the Jacobi equation gives $\widetilde{J}=J$ on $[0,1]$. Evaluating at $t=1$ yields \begin{align*} 0=J(1)=\widetilde{J}(1) =d(\exp_p)_v\bigl(d(\tau_v)_0(w)\bigr). \end{align*} Since $w\ne0$ and $d(\tau_v)_0$ is an isomorphism, $d(\tau_v)_0(w)\ne0$. Thus $d(\exp_p)_v$ has a nonzero kernel vector. [/step] [step:Conclude the equivalence] The previous two steps prove both implications. A nonzero kernel vector of $d(\exp_p)_v$ produces a nonzero Jacobi field along $\gamma$ vanishing at $0$ and $1$, and every nonzero Jacobi field along $\gamma$ vanishing at $0$ and $1$ arises from such a kernel vector under the translation identification \begin{align*} d(\tau_v)_0:T_pM \to T_v(T_pM). \end{align*} Therefore $\gamma(1)$ is conjugate to $p$ along $\gamma$ if and only if $d(\exp_p)_v$ has nonzero kernel. [/step]