I have a fairly specific question. My intuition says the answer is "yes", but there is a natural generalizations in which I take out all the "physics", and then I think the answer is "no".

Edit number 2: the question without all the background

In response to Andrew's comments, here's the question I want to ask without all the infinite-dimensional preamble:

On $\mathbb R^d$ with its usual metric, pick a differential one-form $b$ and a smooth function $c$, and suppose that each has compact support. Consider the following (nondegenerate, nonlinear, second-order) differential equation for a path $\gamma(t)$: $$ \ddot \gamma = db \cdot \dot\gamma + dc $$ This is the Euler-Lagrange equation, and so I will abbreviate it as (EL). In coordinates, it is: $$ \ddot \gamma^i = (\partial\_i b\_j - \partial\_j b\_i) \dot\gamma^j + \partial\_i c $$ Since (EL) is nondegenerate and $b,c$ have compact support, every solution to (EL) extends to have domain all of $\mathbb R$, and the solutions are in bijection with the tangent bundle ${\rm T}\mathbb R^d = \mathbb R^{2d}$ by identifying $\gamma$ with $(\dot\gamma(0),\gamma(0))$.

For each $(v,q) \in {\rm T}\mathbb R^d$, define a second-order linear differential operator $h\_{(v,q)}$, given in coordinates by: $$ h\_{(v,q)}[\eta]^j(t) = \ddot\eta^j(t) + \bigl(\partial\_i b\_j|\_{\gamma(t)} - \partial\_j b\_i|\_{\gamma(t)}\bigr) \dot\eta^i(t) + \bigl( \partial\_i \partial\_k b\_j|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j b\_k|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j c|\_{\gamma(t)}\bigr) \eta^j(t) $$ where $\gamma$ is the solution to (EL) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$.

Let $C = {\rm T}\mathbb R^d \times \mathbb R\_{\>0}$. For $(v,q,T) \in C$, consider the operator $h\_{(v,q)}$ as a map $$ h\_{(v,q,T)} : \bigl\{ \eta: [0,T] \to \mathbb R^d \text{ s.t. } \eta(0) = 0 = \eta(T) \bigr\} \to \bigl\{ \eta: [0,T] \to \mathbb R^d \bigr\}$$ Define $C' \subseteq C$ to be the set $\{ (v,q,T) \in C \text{ s.t. } \ker h\_{(v,q,T)} = 0\}$.

Then I have the following questions:

1. Is $C'$ open (with the topology induced from $C$)? I asserted that it was, because the coefficients of the second-order operator depend smoothly, and I think that kernels can only jump in dimension at closed regions. But I'm not 100% sure.
2. Is $C'$ (path) connected? This was my originally-posed question, and it definitely requires that $b,c$ have compact support.
3. Actually, for my research I need that for each $T > 0$, we have $C' \cap {\rm T}\mathbb R^d \times \{T\}$ is path connected. Since $C'$ includes every $\gamma \in C$ with $\gamma([0,T])$ always outside the support of $b,c$, and since this set is path connected if the supports are compact, 3. implies 2., but perhaps 3. is stronger. Also, perhaps 3. does not require that $b,c$ have compact support?

Bonus question: I used the metric exactly once in (EL) and exactly once in (HJ), to compare the folks with raised indices to the ones with lowered indices. Does anything happen if I change the signature of the metric?

The rest is what I wrote before:

Background and definitions

On $\mathbb R^d$ (with its usual metric), pick a differential one-form $b$ and a smooth function $c$. The tangent bundle $T\mathbb R^d$ is just $\mathbb R^{2d}$; define the Lagrangian $L: T\mathbb R^d \to \mathbb R$ by $L(v,q) = \frac12 |v|^2 + b(q)\cdot v + c(q)$, where $v$ is the fiber coordinate on $T\mathbb R^d$, $q$ is the base coordinate on $\mathbb R^d$, and $\cdot$ is the canonical pairing of a one-form with a vector. A path of length $t$ is a smooth map $\gamma: [0,t] \to \mathbb R^d$; it has a canonical lift $(\dot\gamma,\gamma): [0,t] \to T\mathbb R^d$. The action of a path $\gamma$ of length $t$ is the integral $A[\gamma] = \int_0^t L(\dot\gamma(\tau),\gamma(\tau))d\tau$. By adjusting signs, one can include paths of negative length; a path of length $0$ is a point in $T\mathbb R^d$ and has zero action.

Consider the set $P$ of all paths (of arbitrary length); it is an infinite-dimensional smooth manifold. There are various natural projections from $P$ to finite dimensions. The "initial-value map" $P \to T\mathbb R^d \times \mathbb R$ takes a path $\gamma: [0,t]\to \mathbb R^d$ to the triple $(\dot\gamma(0),\gamma(0),t)$. I will be more interested in the "boundary-value map" $P \to \mathbb R^d \times \mathbb R^d \times \mathbb R$ taking $\gamma \mapsto (\gamma(0),\gamma(t),t)$. The fiber over a point in $\mathbb R^d \times \mathbb R^d \times \mathbb R$ is an affine space modeled on the space of Dirichlet paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.

I like to think of the action $A$ as a Morse function on fibers of the boundary-value map. Let $C \subset P$ be the set of classical paths, i.e. paths $\gamma$ so that $dA|\sb \gamma \cdot \xi = 0$" /> if $\xi$ is Dirichlet ($dA|\sb \gamma$" /> is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing). Equivalently, $\gamma \in C$ if $\gamma$ satisfies the Euler-Lagrange equations $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$. Since the Euler-Lagrange equations are second-order nondegenerate, the initial-value map restricts to a diffeomorphism of $C$ to an open subset of $T\mathbb R^d \times \mathbb R$ containing $T\mathbb R^d \times \{0\}$.

If I really want to think of $A$ as a Morse function, I should require that its critical points (the classical paths) be nondegenerate. Let $\gamma$ be a (classical) path of length $t$, and $V$ the vector space of Dirichlet paths of length $t$. Then the second derivative or Hessian of $A$ is well-defined as a map $H : V \to V^*$. In fact, the Hessian makes sense as a second-order linear differential operator on the space of all paths of length $t$. Let's say that a classical path is nondegenerate if $\ker H = 0$ (or, rather, does not intersect the space $V$ of Dirichlet paths). The set $C'$ of nondegenerate classical paths is an open (I'm pretty sure) subset of $C$.

My question

Is the space $C'$ (path) connected?

Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?

Edit

The answer to my original question is "no". Let $d = 1$, $b = 0$, and $c(q) = \frac12 q^2$. Then a classical path of length $t$ is nondegenerate if and only if $t$ is not an integer multiple of $\pi$. This is a very nongeneric Lagrangian (it is the harmonic oscillator, and is exactly solvable). Also, I think with my definitions, paths of length $0$ are always degenerate.

So let me ask a more restricted question. Let's suppose that $b$ and $c$ are only supported in a compact neighborhood. Then classical paths that do not enter this neighborhood are precisely the straight lines, and they are all generic (provided $t \neq 0$). Is it true that the space of classical nondegenerate paths with positive length is connected with the restriction that $b,c$ have compact support?