I'm trying to understand the proof of the Morse Index Theorem for geodesics given in do Carmo's book on Riemannian Geometry, specifically the following excerpt from Chapter 11, page 246:
It's the final sentence that I can't quite figure out. Here's the situation:
We have a geodesic $\gamma \colon [0, a] \to M$, and for any $t \in (0, a)$, we consider the index form $I_t$, which is a quadratic form defined on the space $\mathcal V(0,t)$, the space of piecewise smooth vector fields along $\gamma|_{[0,t]}$ that vanish at the end points. The definition of $I_t$ is $$ I_t (V, V) = -\int_0^t \langle V'' + R(V, \gamma') \gamma', V \rangle - \sum_{i=1}^{k-1} \langle V'(t_i^+) - V'(t_i^-), V(t_i) \rangle, \qquad (\star) $$ where $V'$ denotes the covariant derivative of $V$ along $\gamma$, and similarly $V''$ is the covariant derivative of $V'$ along $\gamma$. Here $0 = t_0 < t_1 < \cdots < t_{k-1} < t_k = t$ are such that $V|_{[t_{i-1}, t_i]}$ is smooth for $i=1, \ldots, k$. Thus $V'$ can have a finite number of jump discontinuities, so $V''$ may fail to be defined at the $t_i$, but this does not affect the definition of $I_t (V, V)$ above. [Note that I have taken the liberty of replacing do Carmo's non-standard convention for the Riemann curvature tensor with the standard one. That is, $V'' + R(V, \gamma') \gamma' = 0$ is the Jacobi equation.]
One of the confusing things about Section 11.2 of the book is that do Carmo has two different subdivisions of $[0, a]$, which are given the same notation. The first one $0 = t_0 < \cdots < t_k = t$ in the previous paragraph, which depends on $V \in \mathcal V(0,t)$, gives the expression for $I_t (V,V)$ for a piecewise smooth vector field $V$ along $\gamma|_{[0,t]}$ that has discontinuities in $V'$ at the $t_j$. Then, in order to study the index of the quadratic form $I_t$, do Carmo introduces a new subdivision $0 = t_0 < \cdots < t_k = a$ (depending only on the geodesic $\gamma$, and not on any particular $V$) such that each $\gamma_{[t_{i-1}, t_i]}$ is minimizing on $[t_{i-1}, t_i]$ with no conjugate points. This is fine (I might have used different notation to make it clear that this is something different, although they will coincide a bit further below.) Let us fix this decomposition once and for all.
Given this subdvision, earlier in Section 11.2 of the book, do Carmo proves that $\mathcal V(0,t) = \mathcal V^+(0,t) \oplus \mathcal V^-(0,t)$, where $V^-(0,t)$ is the subspace consisting of those elements of $\mathcal V(0,t)$ such that $V|_{[t_{i-1}, t_i]}$ is a Jacobi field, for all $i = 1, \ldots, k$. [In fact, do Carmo write $V|_{(t_{i-1} t_i)}$ is a Jacobi field for all $i$, but I think he must mean the closed interval $[t_{i-1}, t_i]$ in order to make sense of the isomorphism $F_t \colon \mathcal V^-(0,t) \to S_j$ defined in the next paragraph. Please let me know if I am wrong.] The space $\mathcal V^-(0,t)$ is clearly finite-dimensional. The space $\mathcal V^+(0,t)$ is the subspace consisting of those elements of $\mathcal V(0,t)$ that vanish at the $t_i$. I understand the proof that $\mathcal V(0,t) = \mathcal V^+(0,t) \oplus \mathcal V^-(0,t)$.
We're almost at the place where I get confused. [Thanks for your patience up to now!] Fix a $t \in (0,a)$. We want to study the quadratic form $I_t$. We can without loss of generality choose the (second) decomposition above (the one that depends only on $\gamma$) so that $t_{j-1} < t < t_j$ for some $j$. Note that since he now restricts $I_t$ to $\mathcal V^-(0,t)$, whose elements are smooth except at the $t_i$, the first $j$ elements of the two decompositions $0 = t_0 < \cdots < t_{j-1} < t$ described above are now in fact the same.
Since $\gamma|_{[t_{i-1}, t_i]}$ has no conjugate points, any Jacobi field along $\gamma|_{[t_{i-1}, t_i]}$ is uniquely determined by its boundary values. Thus, for $t \in (t_{j-1}, t_j)$, the map $$F_t \colon \mathcal V^-(0,t) \to T_{\gamma(t_1)} M \oplus \cdots \oplus T_{\gamma(t_{j-1})} M =: S_j$$ given by $$F_t (V) = (V(t_1), \ldots, V(t_{j-1}))$$ is a linear isomorphism, so for $t \in (t_{j-1}, t_j)$ we can regard $I_t$ as a quadratic form on the fixed vector space $S_j$, which has dimension $n(j-1)$, where $n=\dim M$. This is all fine.
Now do Carmo says: "since the elements of $V^-(0,t)$ are broken Jacobi fields, it follows from (2) that $I_t$ depends continuously on $t \in (t_{j-1}, t_j)$." (Equation (2) is what I call $(\star)$ above.) I am trying to parse this statement and justify it. [I certainly believe it.] First, we are regarding all the $I_t$ for $t \in (t_{j-1}, t_j)$ as quadratic forms on $S_j$. Then, I believe, continuity of $I_t$ in $t$ should be the same as $t \mapsto I_t(v,v)$ is continuous in $t$ for each fixed $v \in S_j$. By the definition of $\mathcal V^-(0,t)$, we know that $V = F_t^{-1} (v)$ is a Jacobi field on each $[t_{i-1}, t_i]$, so $V'' + R(V, \gamma') \gamma' = 0$, and the first term in $(\star)$ vanishes, so we have only the second term in $(\star)$. For $v \in S_j$ fixed, if $t \in (t_{j-1}, t_j)$ is varied, then $F_t^{-1} (v)$ can only possibly change on $[t_{j-1}, t]$. This means that $V(t_j)$ and $V'(t_j^-)$ are also unchanged, and only $V'(t_j^+)$ can change.
So my question is this: Why does $V'(t_j^+)$ vary continuously with $t \in (t_{j-1}, t_j)$, where $V|_{[t_{j-1}, t_j]}$ is the unique Jacobi field along $\gamma|_{[t_{j-1}, t_j]}$ satisfying: $V(t_j)$ is a fixed given vector, and $V(t) = 0$. It seems that I need continuous dependence of $V'$ on the boundary values of $V$, but the "boundary point" $t \in (t_{j-1}, t_j)$ is moving.
Possibly this is all completely trivial and I am missing something obvious. In fact, I hope that is the case. Thanks for reading through all of this!
