[proofplan] We compare the index form with the Jacobi equation along $\gamma$. First we isolate the non-negative part of the index form on intervals whose right endpoint is not conjugate to $\gamma(a)$, using the [index lemma](/theorems/5349). Then, near each conjugate time, we construct exactly as many negative directions as the multiplicity of the conjugate point. Finally, we choose a partition separating the conjugate times, decompose vector fields by their endpoint values on the partition, and add the local contributions. [/proofplan]

[step:Introduce the Jacobi operator and conjugate multiplicities] For a piecewise smooth vector field \begin{align*} V:[a,b]\to TM,\qquad t\mapsto V(t)\in T_{\gamma(t)}M, \end{align*} define the Jacobi operator along $\gamma$ by \begin{align*} \mathcal J V = \nabla_t\nabla_t V+R(V,\dot\gamma)\dot\gamma. \end{align*} A smooth vector field $J$ along $\gamma$ is a Jacobi field if $\mathcal J J=0$. For each $c\in(a,b]$, define the evaluation map \begin{align*} A_c:T_{\gamma(a)}M&\to T_{\gamma(c)}M,\\ v&\mapsto J_v(c), \end{align*} where $J_v$ is the unique Jacobi field along $\gamma|_{[a,c]}$ satisfying \begin{align*} J_v(a)=0,\qquad \nabla_tJ_v(a)=v. \end{align*} The point $\gamma(c)$ is conjugate to $\gamma(a)$ along $\gamma$ precisely when $\ker A_c\neq\{0\}$, and its multiplicity is \begin{align*} m(c)=\dim\ker A_c. \end{align*} We use the [Conjugate-Instant Isolation Theorem for Jacobi Fields](/page/Conjugate-Instant%20Isolation%20Theorem%20for%20Jacobi%20Fields) in the following form: for every $d\in(a,b]$, there is a number $\rho_d>0$ such that either $d$ is not conjugate and no $t\in(a,b]$ with $0<|t-d|<\rho_d$ is conjugate, or $d$ is conjugate and no $t\in(a,b]$ with $0<|t-d|<\rho_d$ and $t\neq d$ is conjugate. This theorem applies because, after parallel-trivialising $TM$ along $\gamma$ by the parallel transport maps \begin{align*} P_a^t:T_{\gamma(a)}M&\to T_{\gamma(t)}M, \end{align*} the Jacobi equation becomes a smooth linear second-order ordinary differential equation on the fixed [vector space](/page/Vector%20Space) $T_{\gamma(a)}M$ with smooth coefficients. The first-order system has a smooth fundamental solution, and the conjugate instants are precisely the instants at which the Jacobi tensor represented by $A_t$ is singular. The isolation theorem rules out interior accumulation: if a sequence of distinct conjugate instants in $(a,b]$ converged to $d\in(a,b]$, then every neighbourhood of $d$ would contain a conjugate instant different from $d$, contradicting the displayed isolation property at $d$. The endpoint hypothesis says $\ker A_b=\{0\}$, so the isolation theorem gives a number $\rho_b>0$ such that no $t\in(b-\rho_b,b]$ is conjugate to $a$. At the initial endpoint, Taylor expansion of the Jacobi field $J_v$ gives \begin{align*} A_c(v)=(c-a)P_a^c v+O((c-a)^2|v|) \end{align*} as $c\downarrow a$, where $P_a^c:T_{\gamma(a)}M\to T_{\gamma(c)}M$ is parallel transport along $\gamma$. Since $P_a^c$ is a linear isomorphism, $A_c$ is invertible for all sufficiently small $c>a$. Hence the conjugate times in $(a,b)$ form a discrete subset contained in a compact subinterval of $(a,b)$. By compactness, this discrete subset is finite. List the conjugate times in $(a,b)$ as \begin{align*} a<c_1<\cdots<c_N<b. \end{align*} [/step]

[step:Prove nonnegativity before the first conjugate point]Let $d\in(a,b]$ be such that no point $\gamma(t)$ with $t\in(a,d]$ is conjugate to $\gamma(a)$ along $\gamma|_{[a,d]}$. We claim that the index form on \begin{align*} \mathcal V_0(\gamma|_{[a,d]}) = \{V \text{ piecewise smooth along }\gamma|_{[a,d]}:V(a)=V(d)=0\} \end{align*} is non-negative. Let $V\in\mathcal V_0(\gamma|_{[a,d]})$. Because $A_t$ is invertible for every $t\in(a,d]$ when there is no conjugate point in $(a,d]$, there exists a unique smooth vector field \begin{align*} X:[a,d]\to TM,\qquad X(t)\in T_{\gamma(t)}M, \end{align*} such that $X$ is a Jacobi field, $X(a)=0$, and $X(d)=V(d)=0$. Thus $X=0$. The [Index Lemma for Geodesics](/page/Index%20Lemma%20for%20Geodesics) in this endpoint-zero case gives \begin{align*} I_{\gamma|_{[a,d]}}(V,V)\ge I_{\gamma|_{[a,d]}}(X,X)=0. \end{align*}[/step]

[guided]We need a precise reason why negative directions cannot appear on an interval without conjugate points. Let \begin{align*} V:[a,d]\to TM,\qquad t\mapsto V(t)\in T_{\gamma(t)}M, \end{align*} be an arbitrary element of \begin{align*} \mathcal V_0(\gamma|_{[a,d]}) = \{V \text{ piecewise smooth along }\gamma|_{[a,d]}:V(a)=V(d)=0\}. \end{align*} The no-conjugate hypothesis says that $A_t:T_{\gamma(a)}M\to T_{\gamma(t)}M$ is invertible for every $t\in(a,d]$. In particular the Jacobi boundary value problem \begin{align*} \mathcal J X=0,\qquad X(a)=0,\qquad X(d)=0 \end{align*} has only the zero solution, because any such $X$ equals $J_v$ with $v=\nabla_tX(a)$ and satisfies $A_dv=X(d)=0$. Therefore the Jacobi field with the same endpoint values as $V$ is the zero map \begin{align*} X:[a,d]\to TM,\qquad t\mapsto 0\in T_{\gamma(t)}M. \end{align*} The [Index Lemma for Geodesics](/page/Index%20Lemma%20for%20Geodesics) applies because there is no conjugate point to $\gamma(a)$ in $(a,d]$; it states that, among piecewise smooth vector fields with prescribed endpoint values, the Jacobi field with those endpoint values minimizes the index form. Applying it to $V$ and $X$ gives \begin{align*} I_{\gamma|_{[a,d]}}(V,V)\ge I_{\gamma|_{[a,d]}}(X,X)=0. \end{align*} Thus every element of $\mathcal V_0(\gamma|_{[a,d]})$ has non-negative index form. This is exactly the mechanism that prevents non-conjugate intervals from contributing to the Morse index.[/guided]

[step:Compute the Dirichlet index jump at each conjugate instant]Fix a conjugate time $c_i$. Choose numbers $r_i,s_i$ such that \begin{align*} a<r_i<c_i<s_i<b \end{align*} and such that no conjugate time lies in $(r_i,s_i]$ except $c_i$. Define \begin{align*} K_i = \{J:\,[a,c_i]\to TM \text{ Jacobi field along }\gamma|_{[a,c_i]}:J(a)=0,\ J(c_i)=0\}. \end{align*} Then $\dim K_i=m(c_i)$. For every nonconjugate $t\in(r_i,s_i]\setminus\{c_i\}$, define the Dirichlet variation space \begin{align*} \mathcal V_0(t) = \{V \text{ piecewise smooth along }\gamma|_{[a,t]}:V(a)=V(t)=0\} \end{align*} and let \begin{align*} \iota(t)=\operatorname{ind}\bigl(I_{\gamma|_{[a,t]}}\big|_{\mathcal V_0(t)}\bigr). \end{align*} We now state precisely the local jump result being used. After parallel-trivialising $TM$ along $\gamma$ and rescaling each interval $[a,t]$ linearly to $[a,c_i]$, the Dirichlet index forms define a $C^1$ family of bounded symmetric Fredholm quadratic forms \begin{align*} Q_t:H^1_0([a,c_i];T_{\gamma(a)}M)&\to\mathbb R \end{align*} for $t\in(r_i,s_i]$ near $c_i$, where $Q_t$ represents $I_{\gamma|_{[a,t]}}$ under this transport. The [Dirichlet Index-Jump Lemma](/page/Dirichlet%20Index-Jump%20Lemma) says that if $Q_t$ is non-degenerate for $t\neq c_i$, has kernel $K_i$ at $t=c_i$, and has non-degenerate crossing form \begin{align*} \Gamma_i(J)=\left.\frac{d}{dt}\right|_{t=c_i}Q_t(\tilde J_t,\tilde J_t) \end{align*} for any $C^1$ extension $t\mapsto \tilde J_t$ with $\tilde J_{c_i}=J$, then \begin{align*} \iota(s_i)-\iota(r_i)=-\operatorname{sgn}(\Gamma_i), \end{align*} where $\operatorname{sgn}(\Gamma_i)$ is the number of positive eigenvalues minus the number of negative eigenvalues of $\Gamma_i$ on $K_i$. Its hypotheses are satisfied on $(r_i,s_i]$: the Jacobi operator is a smooth formally self-adjoint second-order operator along $\gamma$, $r_i$ and $s_i$ are nonconjugate to $a$, the only degeneracy of the Dirichlet problem in $(r_i,s_i]$ occurs at $c_i$, and the degeneracy space is $K_i$. Green's identity for the Jacobi operator gives the crossing form \begin{align*} \Gamma_i:K_i&\to\mathbb R,\\ J&\mapsto -g_{\gamma(c_i)}(\nabla_tJ(c_i),\nabla_tJ(c_i)). \end{align*} If $J\in K_i$ and $\Gamma_i(J)=0$, then $\nabla_tJ(c_i)=0$. Together with $J(c_i)=0$, uniqueness for the Jacobi equation with terminal data gives $J=0$. Hence $\Gamma_i$ is negative definite on $K_i$, so $\operatorname{sgn}(\Gamma_i)=-\dim K_i$. The jump formula therefore gives \begin{align*} \iota(s_i)-\iota(r_i)=\dim K_i=m(c_i). \end{align*} Equivalently, as the right endpoint passes through $c_i$, the endpoint-zero index form gains exactly $m(c_i)$ negative directions and no other directions change sign.[/step]

[guided]We now compute the jump for the correct quadratic forms: the index forms on vector fields that vanish at both endpoints. Fix the conjugate time $c_i$ and choose $r_i,s_i$ with \begin{align*} a<r_i<c_i<s_i<b \end{align*} such that $(r_i,s_i]$ contains no conjugate time except $c_i$. The kernel appearing at the crossing is \begin{align*} K_i = \{J:\,[a,c_i]\to TM \text{ Jacobi field along }\gamma|_{[a,c_i]}:J(a)=0,\ J(c_i)=0\}, \end{align*} and by definition $\dim K_i=m(c_i)$. For each nonconjugate endpoint $t\in(r_i,s_i]\setminus\{c_i\}$, set \begin{align*} \mathcal V_0(t) = \{V \text{ piecewise smooth along }\gamma|_{[a,t]}:V(a)=V(t)=0\} \end{align*} and define \begin{align*} \iota(t)=\operatorname{ind}\bigl(I_{\gamma|_{[a,t]}}\big|_{\mathcal V_0(t)}\bigr). \end{align*} These are the forms whose index appears in the theorem. We do not use the endpoint quadratic form on Jacobi fields with arbitrary endpoint value, because degeneracy of that form is not the same as conjugacy. We apply the [Dirichlet Index-Jump Lemma](/page/Dirichlet%20Index-Jump%20Lemma) for the Jacobi index form. To make the sign and the space precise, parallel-trivialise $TM$ along $\gamma$ and rescale each interval $[a,t]$ linearly to $[a,c_i]$. This transports the Dirichlet index form to a $C^1$ family of bounded symmetric Fredholm quadratic forms \begin{align*} Q_t:H^1_0([a,c_i];T_{\gamma(a)}M)&\to\mathbb R. \end{align*} The lemma says that if $Q_t$ is non-degenerate for $t\neq c_i$, has kernel $K_i$ at $t=c_i$, and has non-degenerate crossing form \begin{align*} \Gamma_i(J)=\left.\frac{d}{dt}\right|_{t=c_i}Q_t(\tilde J_t,\tilde J_t) \end{align*} for any $C^1$ extension $t\mapsto \tilde J_t$ with $\tilde J_{c_i}=J$, then \begin{align*} \iota(s_i)-\iota(r_i)=-\operatorname{sgn}(\Gamma_i), \end{align*} where $\operatorname{sgn}(\Gamma_i)$ is the number of positive eigenvalues minus the number of negative eigenvalues of $\Gamma_i$ on $K_i$. The hypotheses are satisfied here. The Jacobi operator \begin{align*} \mathcal J V=\nabla_t\nabla_tV+R(V,\dot\gamma)\dot\gamma \end{align*} is formally self-adjoint with respect to the Riemannian metric and integration over the interval, because Green's identity for the index form leaves only endpoint terms. The choices of $r_i$ and $s_i$ make the endpoints nonconjugate, and the isolation of conjugate instants ensures that $c_i$ is the only degeneracy in $(r_i,s_i]$. The degeneracy space is exactly $K_i$. It remains to identify the crossing form. The same rescaling and Green-identity computation gives \begin{align*} \Gamma_i:K_i&\to\mathbb R,\\ J&\mapsto -g_{\gamma(c_i)}(\nabla_tJ(c_i),\nabla_tJ(c_i)). \end{align*} This form is negative definite. If $\Gamma_i(J)=0$, then $\nabla_tJ(c_i)=0$; since also $J(c_i)=0$, uniqueness for the Jacobi equation with terminal data forces $J=0$. Hence $\operatorname{sgn}(\Gamma_i)=-\dim K_i$, and the jump formula gives \begin{align*} \iota(s_i)-\iota(r_i)=\dim K_i=m(c_i). \end{align*} Thus the local contribution of the conjugate instant is precisely its multiplicity.[/guided]

[step:Choose nonconjugate endpoints and add the Dirichlet jumps] Choose numbers \begin{align*} a<\tau_0<c_1<\tau_1<c_2<\cdots<c_N<\tau_N=b \end{align*} such that each $\tau_j$ is not conjugate to $\gamma(a)$ and each interval $(\tau_{i-1},\tau_i]$ contains exactly one conjugate time, namely $c_i$, for $1\le i\le N$. If $N=0$, choose any $\tau_0=b$. This is possible because the conjugate times are finite, the endpoint $b$ is not conjugate to $\gamma(a)$, and there is no conjugate time sufficiently close to $a$. Let $\iota(t)$ denote the index of $I_{\gamma|_{[a,t]}}$ on \begin{align*} \mathcal V_0(\gamma|_{[a,t]}) = \{V \text{ piecewise smooth along }\gamma|_{[a,t]}:V(a)=V(t)=0\} \end{align*} whenever $t$ is not conjugate to $\gamma(a)$. Since there is no conjugate point in $(a,\tau_0]$, the nonconjugate-interval argument gives \begin{align*} \iota(\tau_0)=0. \end{align*} The Dirichlet index-jump computation from the previous step applies on each interval $(\tau_{i-1},\tau_i]$, whose only conjugate time is $c_i$, and gives \begin{align*} \iota(\tau_i)-\iota(\tau_{i-1})=m(c_i),\qquad 1\le i\le N. \end{align*} Summing these equalities yields \begin{align*} \operatorname{ind}(I_\gamma)=\iota(b)=\sum_{i=1}^N m(c_i). \end{align*} The additivity is entirely at the level of the endpoint-zero Dirichlet index, so no comparison with an endpoint form on Jacobi fields with arbitrary terminal value is used. [/step]

[step:Pass from the Sobolev Dirichlet index to the stated piecewise smooth index] The jump argument computed the index after completing the endpoint-zero piecewise smooth fields in the [Sobolev space](/page/Sobolev%20Space) $H^1_0(\gamma^*TM)$. We verify that this is the same index as the one in the theorem. Let \begin{align*} \mathcal P_0(\gamma) = \{V \text{ piecewise smooth along }\gamma:V(a)=V(b)=0\} \end{align*} and let $H^1_0(\gamma^*TM)$ denote the Sobolev completion of smooth compactly supported vector fields along $\gamma$ with zero endpoint trace. The inclusion \begin{align*} \mathcal P_0(\gamma)&\subset H^1_0(\gamma^*TM) \end{align*} has dense image in the $H^1$ norm, because parallel trivialisation identifies the statement with density of piecewise smooth compactly supported functions in $H^1_0([a,b];\mathbb R^n)$. The index form extends continuously to $H^1_0(\gamma^*TM)$, since along the compact interval $[a,b]$ the curvature term $R(\cdot,\dot\gamma)\dot\gamma$ is bounded and the derivative term is the $L^2$ [inner product](/page/Inner%20Product) of covariant derivatives. Therefore every finite-dimensional negative subspace of $H^1_0(\gamma^*TM)$ can be approximated, basis vector by basis vector, by elements of $\mathcal P_0(\gamma)$ so that the Gram matrix of the index form remains negative definite. Conversely, every negative subspace of $\mathcal P_0(\gamma)$ is a negative subspace of $H^1_0(\gamma^*TM)$. Thus \begin{align*} \operatorname{ind}_{\mathcal P_0(\gamma)}(I_\gamma) = \operatorname{ind}_{H^1_0(\gamma^*TM)}(I_\gamma). \end{align*} The equality computed by the Dirichlet jump formula is therefore the Morse index on the precise piecewise smooth space stated in the theorem. [/step]

[step:Record why the jump computation gives both bounds] The [Dirichlet Index-Jump Lemma](/page/Dirichlet%20Index-Jump%20Lemma) is an equality statement for the Morse index of the endpoint-zero index form, not only a construction of local negative directions. On every interval of endpoints with no conjugate time, the transported family \begin{align*} t\mapsto Q_t:H^1_0([a,c];T_{\gamma(a)}M)\to\mathbb R \end{align*} passes through non-degenerate symmetric Fredholm forms, so spectral continuity keeps its index constant. At a conjugate instant $c_i$, the same lemma identifies the entire change in index with $-\operatorname{sgn}(\Gamma_i)$ on \begin{align*} K_i = \{J:\,[a,c_i]\to TM \text{ Jacobi field along }\gamma|_{[a,c_i]}:J(a)=0,\ J(c_i)=0\}. \end{align*} Since that crossing form is negative definite, the whole change is \begin{align*} \dim K_i=m(c_i). \end{align*} Thus the lower bound and upper bound are obtained simultaneously: no additional negative directions can appear away from conjugate instants, and each conjugate instant contributes exactly its multiplicity. [/step]

[step:Conclude equality and identify the endpoint hypothesis] The Dirichlet jump computation, together with the equality between the Sobolev and piecewise smooth indices, gives \begin{align*} \operatorname{ind}(I_\gamma) = \sum_{i=1}^N m(c_i) = \sum_{c\in(a,b)} m(c). \end{align*} The assumption that $\gamma(b)$ is not conjugate to $\gamma(a)$ ensures that the endpoint evaluation map at $b$ is non-singular, so no null Jacobi field with $J(a)=J(b)=0$ lies on the boundary of the interval. Thus the index counts precisely the conjugate points in the open interval $(a,b)$, with their multiplicities. This proves the Morse Index Theorem for geodesic segments. [/step]