These are notes for a presentation I had to give for my Riemannian Geometry class. They follow Chapters 11-17 of Milnor's book Morse Theory very closely.

## Morse Functions

Let $M$ be a manifold and $f:M \to \R$ be a smooth function. A helpful example to keep in mind is the height function of a surface immersed in $\R^3$, the restriction of $f(x,y,z) = z$ to $M \subseteq \R^3$.

A point $p \in M$ is a critical point of $f$ if $df_p:T_pM \to T_{f(p)}\R$ is the zero map. Equivalently, $p$ is a critical point if $(f \circ c)'(0) = 0$ for any curve $c:(-\epsilon, \epsilon) \to M$ such that $c(0) = p$.

At a critical point, $f$ has a well-defined Hessian (second derivative)

In general, the second derivative of a function $f$ is not well defined. We can differentiate twice in any given chart, but the answer in general depends on the chart you use.

We can give a nice coordinate-free expression for the Hessian as follows.

Given $p \in M$ a critical point of $f:M \to \R$, we define the Hessian of $f$ at $p$ to be the bilinear map $H: T_pM \times T_pM \to \R$ given by $H(X,Y) = X \cdot (\tilde Y \cdot f)(p)$ where $\tilde Y$ is a local vector field extending $Y$ to a neighborhood of $p$.

It's not obvious that this is bilinear, or even that it is well-defined (a priori, it could depend on the extension of $Y$). But we can show both of these facts with one neat computation. $X \cdot (\tilde Y \cdot f)(p) - Y \cdot (\tilde X \cdot f)(p) = [\tilde X, \tilde Y] \cdot f (p) = (df)_p([\tilde X, \tilde Y]) = 0$ Therefore, $H(X,Y) = H(Y,X)$. So the Hessian is symmetric. It is clearly linear in the first argument. So the fact that it is symmetric shows that it is well-defined and bilinear.

A critical point is nondegenerate if the Hessian is nondegenerate. A function is Morse if all of its critical points are nondegenerate.
The index of a nondegenerate critical point is the maximum dimension of a subspace of $T_pM$ on which $H$ is negative definite.
If $f$ is a Morse function on $M$ such that $f^{-1}((-\infty, a])$ is compact for each $a$, then $M$ is homotopy equivalent to a $CW$ complex with one cell of dimension $n$ for each critical point of index $n$.

For example, consider the height function on the torus $T^2$. There are 4 critical points: one minimum, two saddle points, and one maximum. These correspond to the CW structure on the torus with one 0-cell, one 1-cells, and one 2-cell.

## The Calculus of Variations

We want to treat the space of piecewise smooth paths on a manifold like an infinite-dimensional manifold. I won't make this idea fully formal (mostly because I don't know the full formalism), but we can use analogies with finite-dimensional manifolds to motivate some useful definitions related to this space of paths. By analogy to the finite-dimensional case, we will define tangent vectors to the space of paths.

## The Path Space of a Smooth Manifold

Let $M$ be a smooth manifold and let $p$ and $q$ be two points of $M$. $p$ and $q$ are allowed to be the same point. We will denote the set of all piecewise-smooth paths from $p$ to $q$ in $M$ by $\Omega(M;p,q)$. If $p$ and $q$ are clear from context, we will just write $\Omega$. Later, we will topologize this space, but we don't need to worry about that yet.

Before defining the tangent space of $\Omega$ and critical points, we will review their definitions for finite-dimensional manifolds. Given a finite-dimensional manifold $M$, we can think of tangent vectors as the velocities of curves. Concretely, given a tangent vector $v \in T_pM$, we can always find a curve $c:(-\epsilon, \epsilon) \to M$ such that $c(0) = p$ and $c'(0) = v$. We can push forward tangent vectors along a map $\phi:M \to N$ by defining $d\phi_p (v) = (\phi \circ c)'(0)$. We say that $p$ is a critical point of $\phi$ if $d\phi_p = 0$, which is to say that $(\phi \circ c)'(0) = 0$ for all curves $c$ through $p$.

Now, we give analogous definitions on $\Omega$. To define tangent vectors on $\Omega$, we have to generalize the idea of a curve on a manifold $M$. We do so with the idea of a variation.

A variation of $\omega$ (keeping endpoints fixed) is a function $\bar \alpha : (-\epsilon, \epsilon) \to \Omega(M;p,q)$ such that
1. $\bar \alpha(0) = \omega$
2. There is a subdivision $0 = t_0 < t_1 < \cdots < t_k = 1$ of $[0,1]$ such that the map $\alpha : (-\epsilon, \epsilon) \times [0,1] \to M$ defined by $\alpha(u,t) = \bar \alpha(u)(t)$ is smooth on each strip $(-\epsilon, \epsilon) \times [t_{i-1}, t_i]$.
More generally, if $(-\epsilon, \epsilon)$ is replaced by a neighborhood of the origin in $\R^n$, we call $\bar \alpha$ an $n$-parameter variation of $\omega$.

We can think of $\bar \alpha$ as a "smooth path" in $\Omega$. Its "velocity vector", $\dd {\bar \alpha} {u} (0)$ is the vector field $W$ along $\omega$ given by $W(t) = \left.\ddo u \right|_{u=0} \bar \alpha(u)(t) = \left.\pd{\alpha(u,t)} u\right|_{u=0}$ Inspired by this, we define the tangent space to $\Omega$ at a path $\omega$ as follows.

A tangent vector to $\Omega(M;p,q)$ at a path $\omega$ from $p$ to $q$ is a piecewise-smooth vector field $W$ along $\omega$ such that $W(0) = W(1) = 0$. We will denote the space of all such vector fields along $\omega$ by $T\Omega_\omega$.

We note that $\dd{\bar\alpha}{u}(0)$ is such a vector field. And given any such vector field, we can find an associated variation by setting $\bar \alpha(u)(t) := \exp_{\omega(t)}(u W(t))$ Now that we have a definition of tangent vectors, we can define critical points.

Given a function $F:\Omega \to \R$ a critical point or critical path of $F$ is a path $\omega \in \Omega$ such that $\left.\dd{F (\bar \alpha (u))}{u}\right|_{u=0}$ is zero for every variation $\bar \alpha$ of $u$.

## The Energy of a Path

On Riemannian manifolds, we often want to talk about the lengths of paths. However, the length functional is kind of annoying because there's a square root involved. To get around this, we define an energy functional which is similar to length, but better behaved.

Given a path $\omega \in \Omega$, the energy of $\omega$ from $a$ to $b$ (for $0 \leq a < b \leq 1$) is $E_a^b(\omega) := \int_a^b \left\|\dd \omega t\right\|^2\;dt$ Frequently, we will write $E$ for $E_0^\ell$.

Recall that the arc-length of a curve from $a$ to $b$ is given by $L_a^b(\omega) := \int_a^b \left\|\dd \omega t \right\|\;dt$ Using the Cauchy-Schwarz inequality, we can relate the length and energy of a curve. $(L_a^b)^2 = \left(\int_a^b \left\|\dd \omega t\right\| \cdot 1\;dt\right)^2 \leq \left(\int_a^b \left\|\dd \omega t \right\|^2\;dt\right)\left(\int_a^b 1^2\;dt\right) = (b-a)E_a^b$ Suppose that $\gamma$ is a minimal geodesic with $\gamma(0) = p$ and $\gamma(1) = q$, and $\omega$ is any other path. Then (using the fact that $\|\dot \gamma\|$ is constant) $E_0^1(\gamma) = \int_0^1 \|\dot\gamma\|^2\;dt = \|\dot\gamma\|^2 = L_0^1(\gamma)^2 \leq L_0^1(\omega)^2 \leq E_0^1(\omega)$ We can only have $L(\gamma)^2 = L(\omega)^2$ if $\omega$ is a reparameterization of a minimal geodesic from $p$ to $q$. And we can only have $L(\omega)^2 = E(\omega)$ if $\omega$ is parameterized proportional to arclength. Thus, we conclude that $E(\gamma) \leq E(\omega)$ with equality iff $\omega$ is a minimal geodesic. That means that the minima of the energy functional are the minimal geodesics from $p$ to $q$.

Now that we understand the minima of the energy functional, we turn to the critical points.

(First Variation Formula) Let $\omega \in \Omega$ be a path with $\omega(0) = p$ and $\omega(1) = q$. Let $V_t = \dot \omega$ be the velocity field, $A_t = \Ddt \dd \omega t$ be the acceleration field. Let $\Delta_t V$ be the discontinuty of the velocity field at time $t$. Then the derivative of energy along a variation $\overline \alpha$ with associated vector field $W_t$ is given by $\frac 12 \left.\dd {E(\bar\alpha(u))}{u}\right|_{u=0} = -\sum_t \inrp {W_t}{\Delta_t V} - \int_0^1 \inrp {W_t}{A_t}$
The proof is the same as Lee's proof of the first variation formula for the length functional.
The path $\omega$ is a critical point for the energy functional iff $\omega$ is a geodesic.
This is also pretty much the same as Lee's proof of the corresponding statement for the length functional

## The Hessian of the Energy Functional at a Critical Path

To do Morse Theory, we need to talk about the Hessian of this functional. The Hessian will be a bilinear functional $E_{**}:T\Omega_\gamma \times T\Omega_\gamma \to \R$ Note that we only define the Hessian at critical points of $E$ (that is, geodesics).

Given vector fields $W_1, W_2 \in T\Omega_\gamma$ we define the Hessian $E_{**}(W_1, W_2)$ as follows:

Pick a two parameter variation $a:U \times [0,1] \to M$ where $U$ is a neighborhood of the origin in $\R^2$, so that $\alpha(0,0,t) = \gamma(t), \; \pd \alpha {u_1} (0,0,t) = W_1(t),\;\pd \alpha {u_2} (0,0,t) = W_2(t)$ Then $E_{**}(W_1, W_2) := \left.\frac{\partial^2 E(\bar \alpha(u_1, u_2))}{\partial u_1 \partial u_2}\right|_{(0,0)}$

It's not obvious from this definition that this is actually well defined (i.e. that it depends only on $W_1$ and $W_2$, and not on the particular variation $\bar \alpha$ that you pick). It turns out that it is well-defined. We can see this using the second variation formula.

(Second variation formula) $\frac 12 \left.\frac{\partial^2 E(\bar \alpha(u_1, u_2))}{\partial u_1 \partial u_2}\right|_{(0,0)} = -\sum_t \inrp {W_2(t)} {\Delta_t \frac{DW_1}{dt}} - \int_0^1 \inrp {W_2}{\frac{D^2 W_1}{dt^2} + R(V,W)V}\;dt$

I won't prove this here, but it's a pretty straightforward computation given the first variation formula, and some other identities, which can be found in Milnor or Lee.

The expression $E_{**}(W_1, W_2) = \frac{\partial^2 E}{\partial u_1 \partial u_2}$ is well-defined, symmetric and bilinear.
The fact that the expression is bilinear and depends only on variation fields $W_1$ and $W_2$ follows from the second variation formula. The symmetry follows from the fact that mixed partial derivatives commute.
It turns out that when $\gamma$ is a minimal geodesic, $E_{**}(W,W) \geq 0$ for all $W$. To see this, we note that $E_{**}(W)$ can be computed in terms of a 1-parameter variation of $\gamma$. Let $\alpha$ be a one-parameter variation corresponding to $W$. We can define a two-parameter variation using $\alpha$ by $\bar \beta(u_1, u_2) = \bar \alpha(u_1 + u_2)$. Then $\pd {\bar \beta}{u_1} = \pd {\bar \beta}{u_2} = \dd {\bar \alpha} u$. Thus, $E_{**}(W,W) = \frac{\partial^2 E\circ\bar\beta}{\partial u_1 \partial u_2} = \frac{d^2 E \circ \bar \alpha}{du^2}$ Since $\gamma$ is a minimal geodesic, $E(\bar\alpha(u)) \geq E(\gamma) = E(\bar \alpha(0))$. Thus, $\displaystyle\frac{d^2E\circ \bar\alpha}{du^2} \geq 0$, so we conclude that $E_{**}(W,W) \geq 0$.

## Jacobi Fields and the Null Space of $E_{**}$

A Jacobi field is a vector field $J$ along a geodesic $\gamma$ which satisfies the Jacobi equation $\frac{D^2 J}{dt^2} + R(\dot\gamma, J)\dot\gamma = 0$

Recall that a Jacobi field $J$ is determined by its initial conditions $J(0), \frac{DJ}{dt}(0) \in TM_{\gamma(0)}$

Two points $p,q \in M$ are conjugate along $\gamma$ if there exists a nonzero Jacobi field $J$ along $\gamma$ which vanishes at $p$ and $q$. The multiplicity of $p$ and $q$ as a conjugate pair is the dimension of the vector space of such Jacobi fields.
The null space of the Hessian $E_{**}$ is the vector space of $W_1 \in T\Omega_\gamma$ such that $E_{**}(W_1, W_2) = 0$ for all $W_2 \in T\Omega_\gamma$. The nullity $\nu$ of $E_**$ is the dimension of the null space. $E_{**}$ is degenerate if $\nu > 0$.
A vector field belongs to the null space of $E_{**}$ iff it is a Jacobi field. The nullity of $E_{**}$ equals the multiplicity of the conjugate pair $p,q$

The proof is a fairly straightforward computation using the second variation formula.

## The Morse Index Theorem

The index $\lambda$ of the Hessian $E_{**}$ is the maximum dimension of a subspace on which $E_{**}$ is negative definite.
(Morse Index Theorem) The index $\lambda$ of $E_{**}$ is equal to the number of points $\gamma(t)$ with $0 < t < 1$ such that $\gamma(t)$ is conjugate to $\gamma(0)$ along $\gamma$, where we count conjugate points with multiplicity. $\lambda$ is always finite.

The proof is pretty involved, so we split it up into steps.

We can split up $T\Omega_\gamma$ into $E_{**}$-orthogonal subspaces so that $E_{**}$ is positive-definite on a subspace of finite codimension.
We know that each point along $\gamma$ is contained in a uniformly normal neighborhood. Since $\gamma([0,1])$ is compact, we can pick a finite cover of $\gamma$ by normal neighborhoods. Thus, we can pick a partition $0 = t_0 < t_1 < \cdots < t_k = 1$ of the unit interval such that $\gamma([t_i, t_{i+1}])$ lies inside a normal neighborhood. Note that this implies that the restriction of $\gamma$ to $[t_i, t_{i+1}]$ is minimal.

Let $T\Omega_\gamma(t_0, t_1, \ldots, t_k) \subseteq T\Omega_\gamma$ be the subspace of vector fields $W$ along $\gamma$ such that

1. $W$ restricted to each $[t_i, t_{i+1}]$ is a Jacobi field
2. $W(0) = W(1) = 0$.
Note that $T\Omega(t_0, \ldots, t_k)$ is finite-dimensional. Let $T' \subseteq T\Omega_\gamma$ be the subspace of vector fields $W$ such that $W(t_i) = 0$ for all $i$. Now, we will show that $T\Omega_\gamma = T\Omega_\gamma(t_0, \ldots, t_k) \oplus T'$, that these subspaces are $E_{**}$-orthogonal, and that $E_{**}$ is positive-definite on $T'$.

Let $W \in T\Omega_\gamma$. Since a Jacobi field along a geodesic contained in a uniformly normal neighborhood is determined by its values on the endpoints, there is unique broken Jacobi field in $W_1 \in T\Omega_\gamma(t_0, \ldots, t_k)$ defined by the property that $W_1(t_i) = W(t_i)$ for each $i$. And $W - W_1 \in T'$. Clearly $T\Omega_\gamma(t_0, \ldots, t_k) \cap T' = 0$. So we conclude that $T\Omega_\gamma = T\Omega_\gamma(t_0, \ldots, t_k) \oplus T'$.

Now, we will show that these subspaces are $E_{**}$-orthogonal. Let $W_1 \in T\Omega_\gamma(t_0, \ldots, t_k)$ and $W_2 \in T'$. Applying the second variation formula, we see $E_{**}(W_1, W_2) = -\sum_t \inrp {W_2(t), \Delta_t \frac{DW_1}{dt}} - \int_0^1 \inrp {W_2} 0 \;dt = 0$

Finally, we note that $E_{**}$ is positive definite on $T'$. The fact that $E_{**}(W,W) \geq 0$ for $W \in T'$ follows from the fact that $E_{**}(V,V) \geq 0$ on minimal geodesics. Since $W$ vanishes at each $t_i$, and $\gamma$ restricted to $[t_i, t_{i+1}]$ is a minimal geodesic, one can show that $E_{**}(W,W) \geq 0$.

Now, we will show that $E_{**}(W,W) = 0$ only if $W = 0$. Suppose $E_{**}(W,W) = 0$. We will show that $W$ must lie in the null space of $E_{**}$. We know that $E_{**}(W, W') = 0$ for $W' \in T\Omega_\gamma(t_0, \ldots, t_k)$. Now, suppose $W_2 \in T'$. By bilinearity of $E_{**}$, we see that $0 \leq E_{**}(W + cW_2, W + cW_2) = 2cE_{**}(V_2, W) + c^2E_{**}(W_2,W_2)$ Since this is true for all $c$ (in particular for all negative $c$), we see that $E_{**}(W_2, W) = 0$. Therefore, $W$ is in the null space of $E_{**}$, which means that it is a Jacobi field. Since the only Jacobi field in $T'$ is 0, we conclude that $W = 0$. So $E_{**}$ is positive definite on $T'$.

Thus, the index of $E_{**}$ equals the index of $E_{**}$ restricted to $T\Omega_\gamma(t_0, \ldots, t_k)$. This shows our claim that the index is finite, since $T\Omega_\gamma(t_0, \ldots, t_k)$ is finite-dimensional.

Now, we will prove the formula for the index. Let $\gamma_\tau$ be the restriction of $\gamma$ to $[0,\tau]$, and let $\lambda(\tau)$ be the index of the associated Hessian $(E_0^\tau)_{**}$. We are going to show a formula for $\lambda(1)$.

$\lambda(\tau)$ is monotone nondecreasing in $\tau$.
Let $\tau < \tau'$. We have a $\lambda(\tau)$-dimensional space of broken Jacobi fields which are zero at $\gamma(0)$ and $\gamma(\tau)$ on which the Hessian is negative definite. Since the vector fields must vanish at $\gamma(\tau)$, we can extend them to vector fields on $\gamma([0, \tau'])$ by making it zero for $t > \tau$. Thus, we obtain a $\lambda(\tau)$-dimensional space of vector fields on which $(E_0^{\tau'})_{**}$ is negative definite. So $\lambda(\tau) \leq \lambda(\tau')$.
$\lambda(\tau) = 0$ for sufficiently small $\tau$
For small $\tau$, $\gamma_\tau$ is a minimal geodesic. We saw that $E_{**}$ is positive definite on minimal geodesics, so $\lambda(\tau) = 0$.
For sufficiently small $\epsilon > 0$, $\lambda(\tau - \epsilon) = \lambda(\tau)$.
This time, assume that $\tau \neq t_i$ for any $i$. Let $H_\tau$ denote the restriction of $E_{**}$ to the finite-dimensional subspace $\Sigma := T\Omega_\gamma(t_0, \ldots, t_s)$ where $0 = t_0 < \cdots < t_s = \tau$ is a partition of $[0, \tau]$. Since this subspace is independent of $\tau$, the quadratic form $H_\tau$ varies continuously with $\tau$ on this subspace (as long as the variation is sufficiently small). Thus, if $H_\tau$ is negative definite on a subspace $V \subseteq \Sigma$, $H_{\tau'}$ will also be negative definite for $\tau$ sufficiently close to $\tau'$. Thus, $\lambda(\tau') \geq \lambda(\tau)$. If $\tau' = \tau - \epsilon$, then the fact that $\lambda$ is monotone nondecreasing tells us that $\lambda(\tau') = \lambda(\tau)$. Thus, $\lambda(\tau-\epsilon) = \lambda(\tau)$ for sufficiently small $\epsilon$.
Why does this argument not show that $\lambda$ is locally constant? It seems to say that $\lambda(\tau') \geq \lambda(\tau)$, and to be symmetric in $\tau$ and $\tau'$?

It's not actually symmetric in $\tau$ and $\tau'$. We may conclude that $\lambda(\tau') \geq \lambda(\tau)$, and that for $\tau''$ sufficiently close to $\tau$, we have $\lambda(\tau'') \geq \lambda(\tau')$. But there's no guarantee that $\tau$ is close enough to $\tau'$ to ensure that $\lambda(\tau) \geq \lambda(\tau')$, which we would need to be the case for $\lambda$ to be locally constant.

Let $\nu$ be the nullity of the Hessian $(E_0^\tau)_{**}$. Then for sufficiently small $\epsilon > 0$, we have $\lambda(\tau + \epsilon) = \lambda(\tau) + \nu$
First, we will show that $\lambda(\tau + \epsilon) \leq \lambda(\tau) + \nu$. We will keep the notation $H_\tau, \Sigma$ from the last lemma. We see that $\dim \Sigma = ns$. Since $H_\tau$ has a null space of dimension $\nu$, we see that $H_\tau$ is positive definite on a subspace $V \subseteq \Sigma$ of dimension $ns - \lambda(\tau) - \nu$. For $\tau'$ sufficiently close to $\tau$, $H_{\tau'}$ is also positive definite on $V$. So $\lambda(\tau') \leq \dim \Sigma - \dim V \leq \lambda(\tau) + \nu$

Next, we will show that $\lambda(\tau + \epsilon) \geq \lambda(\tau) + \nu$. Let $W_1, \ldots, W_{\lambda(\tau)}$ be a basis for the negative-definite subspace of $H_\tau$. Let $J_1, \ldots, J_\nu$ be a basis for the null space of $H_\tau$. Note that the vectors $\frac{DJ_i}{dt}(\tau)\in TM_{\gamma(\tau)}$ must be linearly independent (since the Jacobi fields are all zero there). Thus, we can choose $\nu$ vector fields $X_1, \ldots, X_\nu$ along $\gamma_{\tau + \epsilon}$ so that the matrix $\left(\inrp{\frac{DJ_k}{dt}(\tau)}{X_k(\tau)}\right)$ is equal to the $\nu \times \nu$ identity matrix. (Just invert the matrix $(\frac{DJ_k}{dt}(\tau))$ and extend the vectors to vector fields along $\gamma$). Now, extend the vector fields $W_i$ and $J_k$ to $\gamma_{\tau + \epsilon}$ by setting them to 0 for $\tau \leq t \leq \tau + \epsilon$. Using the second variation formula, we see that $(E_0^{\tau + \epsilon})_{**}(J_h, W_i) = 0$ $(E_0^{\tau + \epsilon})_{**}(J_h, X_k) = 2\delta_{hk}$

Now, let $c$ be small and consider the $\lambda(\tau) + \nu$ vector fields $W_1, \ldots, W_{\lambda(\tau)}, c^{-1}J_1 - cX_1, \ldots, c^{-1}J_\nu - cX_\nu$ along $\gamma_{\tau + \epsilon}$.

Let $A$ be the matrix of $(E_0^{\tau + \epsilon})_{**}$ on $(W_i, X_k)$ and $B$ be the matrix of $(E_0^{\tau + \epsilon})_{**}$ on $(X_h, X_k)$. Then, the matrix of $(E_0^{\tau + \epsilon})_{**}$ is $\begin{pmatrix} (E_0^\tau)_{**}(W_i,W_j) & cA \\ cA^t & -4 \mathbb{I} + c^2 B\end{pmatrix}$ Clearly this is negative definite for small $c$.

This finishes our proof of the Morse Index Theorem.

## A Finite-Dimensional Approxmination to $\Omega^C$

Finally, we will put a topology on $\Omega$. Let $\rho$ denote $M$'s topological metric which is induced by its Riemannian metric.

We define a topological metric on $\Omega(M;p,q)$ as follows. Let $\omega, \omega' \in \Omega(M;p,q)$ with arc-length functions $s(t), s'(t)$ respectively. We define the distance between them to be $d(\omega, \omega') := \max_{0 \leq t \leq 1} \rho(\omega(t), \omega'(t)) + \left[\int_0^1 \left(\dd s t - \dd {s'} t\right)^2\right]^{1/2}$
The last term is present so that the energy functional $E_a^b(\omega) = \int_a^b \left(\dd s t\right)^2\;dt$ is continuous.
Let $c > 0$. We define the closed subset $\Omega^c := E^{-1}([0,c]) \subseteq \Omega$, and the open subset $\Int \Omega^c := E^{-1}([0,c))$

Fix a partition $0 = t_0 < t_1 < \cdots < t_k = 0$ of the unit interval, and define $\Omega(t_0, \ldots, t_k)$ to be the set of piecewise geodesics with vertices at these times. Then define $\Omega(t_0, \ldots, t_k)^c := \Omega^c \cap \Omega(t_0, \ldots, t_k)$ and $\Int \Omega(t_0, \ldots, t_k)^c := (\Int \Omega^c) \cap \Omega(t_0, \ldots, t_k)$.

Let $M$ be a complete Riemannian manifold and $c > 0$ such that $\Omega^c \neq \emptyset$. Then for all sufficiently fine partitions $t_i$, the set $\Int \Omega(\{t_i\})^c$ can be given a finite-dimensional smooth structure.
Let $S$ be the ball centered at $x$ of radius $\sqrt c$. Note that the image of every path in $\Omega^c$ is contained in $S$, since $L^2 \leq E \leq c$. Since $M$ is complete, $S$ is compact. Thus, sufficiently close points are always contained in a common normal neighborhood, so sufficiently close points are connected by a unique geodesic which depends smoothly on the points. Fix $\epsilon > 0$ such that if $\rho(x,y) < \epsilon$, then there is a unique geodesic from $x$ to $y$ of length $< \epsilon$.

Let $\{t_i\}$ be a partition fine enough that $t_i - t_{i-1} \leq \epsilon^2/c$. Then for any broken geodesic $\omega \in \Omega(t0, \ldots, t_k)^c$, we have $(L_{t_{i-1}}^{t_i} \omega)^2 = (t_i - t_{i-1})(E_{t_{i-1}}^{t_i}\omega) \leq (t_i - t_{i-1})(E \omega) \leq \epsilon^2$ So $\omega$ is determined by its values at its vertices. Thus, we can identify $\Omega(t_0, \ldots, t_k)^c$ with a subset of $M^{\times k}$. We can pull back the smooth product structure to get a smooth structure on $\Int \Omega(t_0, \ldots, t_k)^c$.

For convenience, we will write the manifold of broken geodesics $\Int \Omega(t_0, \ldots, t_k)^c$ as $B$. Let $E':B \to \R$ be the restriction of the energy functional to $B$.

The restricted energy functional $E':B \to \R$ is smooth. And for each $a < c$, the set $B^a = (E')^{-1}[0, a]$ is compact, and a deformation retract of the set $\Omega^a$. The critical points of $E'$ are the same as the critical points of $E$ in $\Int \Omega^C$ (i.e. the unbroken geodesics from $p$ to $q$ of length less than $\sqrt c$). The index of the Hessian $E'_{**}$ at each critical point $\gamma$ is equal to the index of the unrestricted Hessian $E_{**}$ at $\gamma$.
Since our broken geodesics depend smoothly on their vertices, the restriction of $E$ to $B$ is clearly smooth. Viewing $B^a$ as a set of $k-1$-tuples $(p_1, \ldots, p_{k-1}) \in S \times \cdots \times S$ subject to a closed condition on length. Thus, $B^a$ is a closed subset of a compact set, so it must be compact.

We will define an explicit retraction $r:\Int \Omega^c \to B$. We start with $\omega \in \Int \Omega^c$. Let $r(\omega)$ be the broken geodesic in $B$ that agrees with $\omega$ at its vertices. Now, we will show that this is a deformation retraction. Let $r_u:\Int \Omega^c \to \Int \Omega^c$ be defined as follows. For $t_{i-1} \leq u \leq t_i$, let

$\begin{cases} r_u(\omega)|_{[0, t_{i-1}]} = r(\omega)|_{[0, t_{i-1}]}\\ r_u(\omega)|_{[t_{i-1}, u]} = \text{minimal geodesic from}\;\omega(t_{i-1})\;\text{to}\;\omega(u)\\ r_u(\omega)|_{[u,1]} = \omega|_{[u,1]} \end{cases}$

Clearly $r_0$ is the identity, $r_1 = r$, and $r$ is smooth. So $B$ is a deformation retract of $\Int \Omega^c$. It's clear that the critical points of $E'$ lie in $B$, since geodesics are broken geodesics, and the first variation formula tells us that these are still the only critical points. And we saw earlier that restricting to broken Jacobi fields does not change the index of $E_{**}$.

Let $M$ be a complete Riemannian Manifold, and let $p,q$ be non-conjugate points along a geodesic of length at most $\sqrt a$. Then $\Omega^a$ is homotopy equivalent to a finite CW complex with one cell of dimension $n$ for each geodesic in $\Omega^a$ where $E_{**}$ has index $n$.
We saw above that $\Omega^a$ is homotopy equivalent to $B$. $B$ is a finite-dimensional manifold equipped with a Morse function $E$, whose critical points are the geodesics where $E_{**}$ has nonzero index. The result follows from elementary Morse theory.

## The Topology of the Full Path Space

The topology we put on $\Omega$ is kind of weird. A more natural topology for this space is the so-called "compact open topology", in which a sequence of functions converges whenever it converges uniformly on every compact subset of the domain. An equivalent description of this topology is that it is induced by the metric $d^*(\omega, \omega') = \max_t \rho(\omega(t), \omega'(t))$

The natural map $(\Omega, d) \to (\Omega, d^*)$ is a homotopy equivalence.
$(\Omega, d^*)$ is homotopy equivalent to a CW complex
(Fundamental Theorem of Morse Theory) Let $M$ be a complete Riemannian manifold and let $p,q$ be a pair of non-conjugate points. Then $\Omega(M;p,q)$ is homotopy equivalent to a countable CW complex which is made of one cell of dimension $n$ for each geodesic from $p$ to $q$ of index $n$.

I will describe the proof here. Let $a_0 < a_1 < \cdots$ be a sequence of real numbers which are not critical values of the energy functional $E$. Pick the numbers so that each interval $(a_i, a_{i+1})$ contains exactly one critical value. Now, consider the sequence $\Omega^{a_0} \subset \Omega^{a_1} \subset \Omega^{a_2} \subset \cdots$

We see that each $\Omega^{a_{i+1}}$ is homotopic to $\Omega^{a_i}$ with a finite number of cells attached, corresponding to the finitely many geodesics in $E^{-1}((a_i, a_{i+1}))$. So we can construct a sequence of $CW$ complexes $K_0 \subset K_1 \subset K_2 \subset \cdots$ such that for each $i$, we have a homotopy equivalence $\Omega^{a_i} \to K_i$. We can take a direct limit to get a map $f:\Omega \to K$. Clearly $f$ induces isomorphisms of homotopy groups in every dimension. Since $\Omega$ is homotopy equivalent to a CW complex, it follows by Whitehead's theorem that $f$ is a homotopy equivalence.