[proofplan] The construction is the standard "smooth out the corner" perturbation from the variational point of view. Take a non-trivial Jacobi field $J$ along $\gamma$ with $J(0) = J(t_0) = 0$ and (without loss of generality) $J \perp \dot\gamma$. Define a normal piecewise-smooth field $V$ along $\gamma$ that agrees with $J$ on $[0, t_0]$ and is zero on $[t_0, 1]$; this $V$ has a corner at $t_0$, where its left- and right-derivatives differ by $J'(t_0) \ne 0$. Add to $V$ a small parallel correction $\theta Z_1$ supported near $t_0$, where $Z_1(t_0) = -J'(t_0)$, to get a new field $V_\theta = V + \theta Z_1 \cdot \eta$ for a bump $\eta$ supported near $t_0$. Compute the index form $I(V_\theta, V_\theta)$ — the second variation of length on normal fields. The Jacobi-field property of $J$ on $[0, t_0]$ and the bookkeeping for parallel $Z_1$ on $[t_0, 1]$ produce $I(V_\theta, V_\theta) = -2\theta |J'(t_0)|^2 + O(\theta^2)$. Choosing $\theta$ small and positive gives $I(V_\theta, V_\theta) < 0$, so the variation $H(t, s) := \exp_{\gamma(t)}(s V_\theta(t))$ — fixing $\gamma(0)$ and $\gamma(1)$ and respecting the breakpoint smoothing — has strictly decreasing length to second order. Combined with vanishing first variation along normal fields (since $\gamma$ is a geodesic), this gives $\ell(H(\cdot, s)) < \ell(\gamma)$ for small $s \ne 0$. [/proofplan] [step:Reduce to a normal Jacobi field with a non-zero derivative at the conjugate point] By hypothesis, $\gamma : [0, 1] \to M$ is a geodesic with $\gamma(0) = p$, $\gamma(1) = q$, and $\gamma(t_0)$ is conjugate to $p$ for some $t_0 \in (0, 1)$. By definition of conjugacy, there is a non-trivial Jacobi field $\tilde J$ along $\gamma$ with $\tilde J(0) = \tilde J(t_0) = 0$. By Part 2 of [Conjugate Points and the Exponential Map](/theorems/2732), $\tilde J(t) \perp \dot\gamma(t)$ for every $t \in [0, t_0]$. Restrict $\tilde J$ to $[0, t_0]$ and call this restriction $J$. We claim $J'(t_0) := \nabla_t J(t_0) \ne 0$. Suppose otherwise. Then $J$ is a Jacobi field on $[0, t_0]$ with $J(t_0) = 0$ and $\nabla_t J(t_0) = 0$. By the uniqueness statement for second-order linear ODEs applied to the Jacobi equation in a parallel orthonormal frame along $\gamma$ — the same argument used in [Conjugate Points and the Exponential Map](/theorems/2732) — vanishing initial data at $t_0$ forces $J \equiv 0$ on $[0, t_0]$. This contradicts non-triviality of $\tilde J$. Hence $J'(t_0) \ne 0$. Furthermore, $J'(t_0) \perp \dot\gamma(t_0)$. Indeed, differentiating $g(J, \dot\gamma) \equiv 0$ along $\gamma$ and using $\nabla_t \dot\gamma = 0$: \begin{align*} 0 = \frac{d}{dt} g(J, \dot\gamma) = g(\nabla_t J, \dot\gamma). \end{align*} Evaluating at $t = t_0$ gives $g(J'(t_0), \dot\gamma(t_0)) = 0$. We may therefore assume from here on: - $J$ is a non-trivial Jacobi field on $[0, t_0]$, - $J(0) = J(t_0) = 0$, - $J(t) \perp \dot\gamma(t)$ for all $t \in [0, t_0]$, - $J'(t_0) \ne 0$ and $J'(t_0) \perp \dot\gamma(t_0)$. [/step] [step:Define the broken normal field $V$ and the parallel correction $Z_1$] Define a piecewise-smooth normal vector field $V$ along $\gamma$ by \begin{align*} V : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ V(t) &:= \begin{cases} J(t), & t \in [0, t_0], \\ 0, & t \in [t_0, 1]. \end{cases} \end{align*} The field $V$ is continuous on $[0, 1]$ (both pieces match the value $0$ at $t_0$), smooth on $[0, t_0]$ and on $[t_0, 1]$ separately, normal to $\dot\gamma$ everywhere (the zero field on $[t_0, 1]$ has zero pairing with $\dot\gamma$; $J$ is normal on $[0, t_0]$ by Step 1), and vanishes at the endpoints $V(0) = V(1) = 0$. The covariant derivative $\nabla_t V$ has a jump at $t_0$: \begin{align*} \nabla_t V(t_0^-) = J'(t_0), \quad \nabla_t V(t_0^+) = 0, \end{align*} so the jump is $-J'(t_0)$, a non-zero vector orthogonal to $\dot\gamma(t_0)$. Define $Z_1$ to be the parallel transport of $-J'(t_0)$ along $\gamma$: \begin{align*} Z_1 : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ Z_1(t) &:= P_\gamma^{t_0, t}(-J'(t_0)), \end{align*} so $Z_1$ is smooth, $\nabla_t Z_1 = 0$, $Z_1(t_0) = -J'(t_0) \ne 0$. Since parallel transport along a geodesic preserves orthogonality to $\dot\gamma$ (because $\dot\gamma$ is also parallel), $Z_1(t) \perp \dot\gamma(t)$ for all $t$. Pick $\eta \in C_c^\infty(0, 1)$ with $0 \le \eta \le 1$, $\eta(t_0) = 1$, $\operatorname{supp} \eta \subseteq [t_0 - \delta_0, t_0 + \delta_0]$ for some $\delta_0 \in (0, \min(t_0, 1 - t_0))$. The bump function $\eta$ vanishes at the endpoints of $[0, 1]$. For $\theta \in \mathbb{R}$ small, define the corrected field \begin{align*} V_\theta : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)} \\ V_\theta(t) &:= V(t) + \theta\, \eta(t)\, Z_1(t). \end{align*} The field $V_\theta$ is piecewise smooth (smooth on $[0, t_0]$ and $[t_0, 1]$ separately) with the same corner at $t_0$ as $V$, vanishes at endpoints, and is everywhere normal to $\dot\gamma$ (sum of two normal fields). [guided] We need a piecewise-smooth normal variation field along $\gamma$ that vanishes at the endpoints, has the right structure to interact with the conjugate-point geometry, and produces a strictly negative second variation. The naive first attempt is to extend the Jacobi field $J$ by zero past $t_0$ — Step 1 gave us $J$ on $[0, t_0]$ vanishing at both endpoints, so extending by zero is continuous and respects $V(0) = V(1) = 0$. Concretely, define a piecewise-smooth normal vector field $V$ along $\gamma$ by \begin{align*} V : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ V(t) &:= \begin{cases} J(t), & t \in [0, t_0], \\ 0, & t \in [t_0, 1]. \end{cases} \end{align*} Why is this piecewise smooth and not smooth? Because while the values match at $t_0$ (both pieces are $0$), the covariant derivative does not: from the left, $\nabla_t V(t_0^-) = J'(t_0)$, which Step 1 showed is non-zero; from the right, $\nabla_t V(t_0^+) = 0$. Recording this jump explicitly: \begin{align*} \nabla_t V(t_0^-) = J'(t_0), \quad \nabla_t V(t_0^+) = 0, \end{align*} so the jump is $-J'(t_0)$, a non-zero vector orthogonal to $\dot\gamma(t_0)$ (orthogonality from Step 1). The field $V$ is smooth on $[0, t_0]$ and on $[t_0, 1]$ separately, normal to $\dot\gamma$ everywhere ($J$ normal on $[0, t_0]$ by Step 1; the zero field is trivially normal on $[t_0, 1]$), and vanishes at the endpoints. Will $V$ alone do the job? Computing the index form will show $I(V, V) = 0$ — the Jacobi-equation cancellation forces zero, not negativity. So the conjugate-point hypothesis only buys us "non-strict" inside the index form. We need a small correction whose cross-term with $V$ is strictly negative. What is the right correction? It should be smooth (so it does not introduce new corners), supported near $t_0$ (so its interaction with $V$ localizes at the corner), and have a value at $t_0$ that pairs against $J'(t_0)$ to produce a non-zero scalar. The choice that makes the bookkeeping work is the parallel transport of $-J'(t_0)$. Define \begin{align*} Z_1 : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)}, \\ Z_1(t) &:= P_\gamma^{t_0, t}(-J'(t_0)), \end{align*} so $Z_1$ is smooth, $\nabla_t Z_1 = 0$, and $Z_1(t_0) = -J'(t_0) \ne 0$. Why parallel? Because $\nabla_t Z_1 = 0$ kills $|\nabla_t Z_1|^2$ in the self-interaction $I(Z_1, Z_1)$, so the only $\theta^2$ contribution comes from the curvature term, which stays $O(\theta^2)$. Why $-J'(t_0)$? Because the cross-term boundary contribution involves $g(J'(t_0), Z_1(t_0))$; with $Z_1(t_0) = -J'(t_0)$, this becomes $-|J'(t_0)|^2$, the sign we need. Parallel transport along the geodesic $\gamma$ preserves orthogonality to $\dot\gamma$ (since $\dot\gamma$ is itself parallel), so $Z_1(t) \perp \dot\gamma(t)$ for all $t$. Next we localize the correction near $t_0$. Pick $\eta \in C_c^\infty(0, 1)$ with $0 \le \eta \le 1$, $\eta(t_0) = 1$, and $\operatorname{supp} \eta \subseteq [t_0 - \delta_0, t_0 + \delta_0]$ for some $\delta_0 \in (0, \min(t_0, 1 - t_0))$. The strict containment $\delta_0 < t_0$ and $\delta_0 < 1 - t_0$ guarantees $\eta(0) = \eta(1) = 0$, which kills boundary terms at the global endpoints in the integration by parts. The condition $\eta(t_0) = 1$ is what extracts the full jump $-|J'(t_0)|^2$ at the corner. For $\theta \in \mathbb{R}$ small, define the corrected field \begin{align*} V_\theta : [0, 1] &\to TM \quad \text{(section of $\gamma^* TM$)} \\ V_\theta(t) &:= V(t) + \theta\, \eta(t)\, Z_1(t). \end{align*} The field $V_\theta$ inherits the corner of $V$ at $t_0$ (the correction $\theta\eta Z_1$ is smooth across $t_0$, so it does not affect the jump structure), is smooth on each segment, vanishes at the global endpoints $V_\theta(0) = V_\theta(1) = 0$, and is everywhere normal to $\dot\gamma$ (sum of two normal fields). The hypotheses consumed so far: existence of a non-trivial Jacobi field vanishing at $0$ and $t_0$ (definition of conjugacy, given by Step 1), and $J'(t_0) \ne 0$ (uniqueness for the second-order Jacobi ODE — vanishing both position and velocity at $t_0$ would force $J \equiv 0$, contradicting non-triviality). [/guided] [/step] [step:Compute the index form of $V$ and the cross-term with $Z_1$] Define the index form on piecewise-smooth normal fields $W$ along $\gamma$ vanishing at endpoints: \begin{align*} I(W, W) := \int_0^1 \big[|W'(t)|_g^2 - R(W, \dot\gamma, W, \dot\gamma)\big] d\mathcal{L}^1(t), \end{align*} where $W'(t) := \nabla_t W(t)$ on each smooth segment. By Part 2 of the [Second Variation Formula](/theorems/2729) applied to a variation with normal field $W$ vanishing at $\{0, 1\}$ and respecting the closed-endpoint constraint, the second variation of length is \begin{align*} \frac{d^2}{ds^2} \ell(H(\cdot, s))\Big|_{s = 0} = I(W, W) + \text{(jump terms at corners)}, \end{align*} which we now make precise on each segment of $V_\theta$. We integrate by parts on each smooth segment of $V_\theta$, namely $[0, t_0]$ and $[t_0, 1]$. For a smooth normal field $W$ on a smooth segment $[t', t'']$ vanishing at the segment endpoints (in our case the endpoint vanishing is the global one at $t' = 0$ or $t'' = 1$): \begin{align*} \int_{t'}^{t''} g(W', W')\, d\mathcal{L}^1(t) &= g(W', W)\Big|_{t'}^{t''} - \int_{t'}^{t''} g(W'', W) \, d\mathcal{L}^1(t), \end{align*} where $W''(t) := \nabla_t \nabla_t W(t)$ on the segment. [claim:Index-form expansion of $I(V_\theta, V_\theta)$] [proof] Expand \begin{align*} I(V_\theta, V_\theta) = I(V, V) + 2\theta \cdot I(V, \eta Z_1) + \theta^2 \cdot I(\eta Z_1, \eta Z_1). \end{align*} (a) **Term $I(V, V)$.** On $[0, t_0]$, $V = J$ and $\nabla_t V = J'$. The Jacobi equation gives $J'' = -R(J, \dot\gamma)\dot\gamma$, so \begin{align*} g(J', J') &= \frac{d}{dt} g(J', J) - g(J'', J) = \frac{d}{dt} g(J', J) + g(R(J, \dot\gamma)\dot\gamma, J) \\ &= \frac{d}{dt} g(J', J) + R(J, \dot\gamma, J, \dot\gamma). \end{align*} Integrating from $0$ to $t_0$ with $J(0) = J(t_0) = 0$: \begin{align*} \int_0^{t_0} g(J', J') \, d\mathcal{L}^1(t) = g(J', J)\Big|_0^{t_0} + \int_0^{t_0} R(J, \dot\gamma, J, \dot\gamma)\, d\mathcal{L}^1(t) = \int_0^{t_0} R(J, \dot\gamma, J, \dot\gamma)\, d\mathcal{L}^1(t). \end{align*} Hence \begin{align*} \int_0^{t_0} \big[|J'|_g^2 - R(J, \dot\gamma, J, \dot\gamma)\big] d\mathcal{L}^1(t) = 0. \end{align*} On $[t_0, 1]$, $V \equiv 0$, so the integrand is zero. Therefore \begin{align*} I(V, V) = 0. \end{align*} (b) **Term $I(V, \eta Z_1) = \int_0^1 [g(V', (\eta Z_1)') - R(V, \dot\gamma, \eta Z_1, \dot\gamma)] d\mathcal{L}^1(t)$.** On $[t_0, 1]$, $V = 0$, so the integrand vanishes. On $[0, t_0]$, $V = J$, and recall $\eta(0) = 0$ in our setup (since $\operatorname{supp}\eta \subseteq [t_0 - \delta_0, t_0 + \delta_0]$, so $\eta = 0$ at $0$ if $\delta_0 < t_0$, which we have arranged). Compute on $[0, t_0]$: \begin{align*} g(J', (\eta Z_1)') &= g(J', \eta' Z_1 + \eta \nabla_t Z_1) = \eta' \cdot g(J', Z_1) + 0, \end{align*} since $\nabla_t Z_1 = 0$. Integrate by parts (smooth segment): \begin{align*} \int_0^{t_0} \eta'(t) \cdot g(J'(t), Z_1(t))\, d\mathcal{L}^1(t) = \eta(t) \cdot g(J', Z_1)\Big|_0^{t_0} - \int_0^{t_0} \eta(t) \cdot \frac{d}{dt} g(J'(t), Z_1(t))\, d\mathcal{L}^1(t). \end{align*} The boundary term: $\eta(0) = 0$ (by our choice of supp $\eta$), and $\eta(t_0) \cdot g(J'(t_0), Z_1(t_0)) = 1 \cdot g(J'(t_0), -J'(t_0)) = -|J'(t_0)|_g^2$. The interior: \begin{align*} \frac{d}{dt} g(J'(t), Z_1(t)) = g(\nabla_t J', Z_1) + g(J', \nabla_t Z_1) = g(J'', Z_1). \end{align*} Using $J'' = -R(J, \dot\gamma)\dot\gamma$: \begin{align*} g(J''(t), Z_1(t)) = -R(J, \dot\gamma, Z_1, \dot\gamma)(t). \end{align*} Combining, the contribution to $I(V, \eta Z_1)$ from $[0, t_0]$ is \begin{align*} -|J'(t_0)|_g^2 + \int_0^{t_0} \eta(t) \cdot R(J, \dot\gamma, Z_1, \dot\gamma)\, d\mathcal{L}^1(t) - \int_0^{t_0} \eta(t) \cdot R(J, \dot\gamma, Z_1, \dot\gamma)\, d\mathcal{L}^1(t) = -|J'(t_0)|_g^2. \end{align*} The two interior curvature integrals have opposite signs in the index-form definition: the first emerges from the integration by parts of the $|W'|^2$ piece (with sign $+$ from $\eta \cdot R$), and the second is the explicit curvature term in $I$ (with sign $-R$). They cancel exactly. So \begin{align*} I(V, \eta Z_1) = -|J'(t_0)|_g^2. \end{align*} (c) **Term $I(\eta Z_1, \eta Z_1)$.** This is a smooth bounded quantity depending only on $\eta$, $Z_1$, and the curvature along $\gamma|_{\operatorname{supp}\eta}$. It is independent of $\theta$, finite, and we denote it $C := I(\eta Z_1, \eta Z_1) \in \mathbb{R}$. Combining (a), (b), (c): \begin{align*} I(V_\theta, V_\theta) = 0 - 2\theta\, |J'(t_0)|_g^2 + \theta^2 \cdot C. \end{align*} [/proof] [/claim] For $\theta > 0$ small enough that $\theta^2 |C| < \theta\, |J'(t_0)|_g^2$ — concretely, $0 < \theta < |J'(t_0)|_g^2 / (1 + |C|)$ — we obtain \begin{align*} I(V_\theta, V_\theta) < -\theta\, |J'(t_0)|_g^2 < 0. \end{align*} [/step] [step:Account for the corner of $V_\theta$ at $t_0$ in the second variation] The field $V_\theta$ has a corner at $t_0$: $\nabla_t V_\theta(t_0^-) - \nabla_t V_\theta(t_0^+) = \nabla_t V(t_0^-) - \nabla_t V(t_0^+) = J'(t_0)$ (the corrections $\theta\eta Z_1$ and its derivative are smooth across $t_0$, so the jump comes only from $V$). We claim that for the variation $H(t, s) := \exp_{\gamma(t)}(s V_\theta(t))$, the second variation of length is exactly $I(V_\theta, V_\theta)$ as computed above, with the corner of $V_\theta$ contributing nothing extra. More precisely, the standard piecewise-smooth second variation formula in the form (using Part 2 of [Second Variation Formula](/theorems/2729) on each smooth segment and summing) is \begin{align*} \frac{d^2}{ds^2} \ell(H(\cdot, s))\Big|_{s = 0} = I(V_\theta, V_\theta), \end{align*} because $\gamma$ is itself smooth (no velocity jumps in $\dot\gamma$), so the breakpoint correction term $g(V_\theta(t_i), \dot\gamma(t_i^+) - \dot\gamma(t_i^-))$ from the piecewise integration vanishes at every breakpoint. The corner is in $V_\theta$, not in $\dot\gamma$, and the index-form computation in the previous step already incorporates it correctly through the segmented integration by parts. The first variation of length at $s = 0$ vanishes: by the [First Variation Formula](/theorems/2728), with $\nabla_t \dot\gamma = 0$ (geodesic) and $V_\theta(0) = V_\theta(1) = 0$ (endpoint vanishing), the integral and boundary terms both vanish, giving $\frac{d}{ds}\ell(H(\cdot, s))|_{s=0} = 0$. Therefore the function $s \mapsto \ell(H(\cdot, s))$ has zero first derivative and strictly negative second derivative at $s = 0$: \begin{align*} \ell(H(\cdot, s)) = \ell(\gamma) + \tfrac{1}{2} I(V_\theta, V_\theta) \cdot s^2 + o(s^2) = \ell(\gamma) - \tfrac{1}{2} \theta |J'(t_0)|_g^2 \cdot s^2 + o(s^2) \end{align*} for some implied constants. Hence there exists $s_0 > 0$ such that \begin{align*} \ell(H(\cdot, s)) < \ell(\gamma) \quad \text{for all } s \in (-s_0, s_0) \setminus \{0\}. \end{align*} The variation $H$ fixes the endpoints $\gamma(0) = p$ and $\gamma(1) = q$ for every $s$, since $V_\theta(0) = V_\theta(1) = 0$. Therefore there is a piecewise-smooth variation $H(t, s)$ of $\gamma$ fixing $p$ and $q$ such that $\ell(H(\cdot, s)) < \ell(\gamma)$ for all sufficiently small $s \ne 0$. [guided] With the index-form computation $I(V_\theta, V_\theta) < 0$ in hand, we must turn this into a statement about lengths of nearby paths. The variation we use is geodesic in the fibre direction: \begin{align*} H : [0, 1] \times (-\varepsilon, \varepsilon) &\to M, \\ (t, s) &\mapsto \exp_{\gamma(t)}(s V_\theta(t)). \end{align*} At $s = 0$, $H(\cdot, 0) = \gamma$. At each $s$, the curve $H(\cdot, s)$ is a piecewise-smooth path from $p$ to $q$ since $V_\theta(0) = V_\theta(1) = 0$ implies $H(0, s) = \exp_p(0) = p$ and $H(1, s) = \exp_q(0) = q$ for every $s$. There is a bookkeeping question to settle before we can quote the second-variation formula: the field $V_\theta$ has a corner at $t_0$, with \begin{align*} \nabla_t V_\theta(t_0^-) - \nabla_t V_\theta(t_0^+) = \nabla_t V(t_0^-) - \nabla_t V(t_0^+) = J'(t_0) \end{align*} (the corrections $\theta\eta Z_1$ and $\nabla_t(\theta\eta Z_1) = \theta \eta' Z_1$ are continuous across $t_0$, so the jump in $\nabla_t V_\theta$ comes only from $V$). Does this corner inject extra terms into the second variation? We must check. Apply Part 2 of the [Second Variation Formula](/theorems/2729) on each smooth segment $[0, t_0]$ and $[t_0, 1]$ separately and sum. The breakpoint correction term from the formula has the form $g(V_\theta(t_i), \dot\gamma(t_i^+) - \dot\gamma(t_i^-))$, i.e., it pairs the variation field at the breakpoint against the jump in $\dot\gamma$. In our setting $\gamma$ is smooth, so $\dot\gamma$ has no jumps anywhere — the breakpoint correction vanishes at every interior breakpoint. The corner is in $V_\theta$, not in $\dot\gamma$, and the corner of $V_\theta$ has already been incorporated into the index-form computation: the boundary contribution at $t_0$ from the segmented integration by parts is exactly what produced the $-|J'(t_0)|^2$ in part (b) of the previous step. Therefore \begin{align*} \frac{d^2}{ds^2} \ell(H(\cdot, s))\Big|_{s = 0} = I(V_\theta, V_\theta) < 0, \end{align*} strict negativity by the previous step. We also need the first variation to vanish, so that $s = 0$ is a critical point and the second-derivative test applies. By the [First Variation Formula](/theorems/2728), the first variation along a normal field $V_\theta$ is \begin{align*} \frac{d}{ds}\ell(H(\cdot, s))\Big|_{s = 0} = -\int_0^1 g(V_\theta, \nabla_t \dot\gamma)\, d\mathcal{L}^1(t) + g(V_\theta, \dot\gamma)\Big|_0^1. \end{align*} Both terms vanish: the integral by $\nabla_t \dot\gamma = 0$ ($\gamma$ is a geodesic), and the boundary by $V_\theta(0) = V_\theta(1) = 0$. Now we Taylor-expand $s \mapsto \ell(H(\cdot, s))$ around $s = 0$: \begin{align*} \ell(H(\cdot, s)) = \ell(\gamma) + 0 \cdot s + \tfrac{1}{2} I(V_\theta, V_\theta) \cdot s^2 + o(s^2), \end{align*} and substituting the bound $I(V_\theta, V_\theta) < -\theta\, |J'(t_0)|_g^2$ from the previous step, \begin{align*} \ell(H(\cdot, s)) < \ell(\gamma) - \tfrac{\theta}{2} |J'(t_0)|_g^2 \cdot s^2 + o(s^2). \end{align*} Since the leading correction is strictly negative and quadratic, there exists $s_0 > 0$ such that the $o(s^2)$ remainder cannot dominate it, giving \begin{align*} \ell(H(\cdot, s)) < \ell(\gamma) \quad \text{for all } s \in (-s_0, s_0) \setminus \{0\}. \end{align*} The variation respects the endpoint constraint for every $s$: $H(0, s) = p$ and $H(1, s) = q$ by $V_\theta(0) = V_\theta(1) = 0$. The geometric content: the conjugate point at $t_0$ means that on $[0, t_0]$ the Jacobi field $J$ has index form exactly zero — not strictly positive. So $\gamma$ already fails to be a strict minimum on $[0, t_0]$ along the Jacobi direction. The smoothing correction $\theta \eta Z_1$ exploits this near-degeneracy: it interacts with the corner of $V$ to tip the sign of the second variation from zero to strictly negative. The result: $\gamma$ is a strict local maximum, not a minimum, of length along this two-parameter family for small $s$ — there are genuinely shorter paths from $p$ to $q$ arbitrarily close to $\gamma$. [/guided] [/step]