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".

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 subset of $C$.

My question

Is the space $C'$ of nondegenerate classical paths connected?

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