Most probable path of diffusion process Suppose we have an Ito diffusion $X_{t}$ on $\mathbb{R}$ given by
\begin{align*}
dX_{t} = A(X_{t})dt + B(X_{t}) dW_{t} \qquad (1)
\end{align*}
where $W_{t}$ is a standard Brownian motion. If $B = 1$, it is well-known that for a twice continuously differentiable curve $u:[0,T]\rightarrow \mathbb{R}$ it is true that
\begin{align*}
P\bigl( \sup_{  0 \leq t \leq T } | X_{t} - u(t) | 
< \epsilon \bigr) \underset{\epsilon \rightarrow 0^{+}}{\sim} 
e^{-\frac{1}{2} \int_{0}^{T} \mathcal{L}(u(t),u'(t)) dt }
\end{align*}
where
\begin{align*}
\mathcal{L}(u,u')  = \bigl( A( u ) - u' \bigr)^{2} + A'(u)
\end{align*}
is the Onsager-Machlup function. Thus, minimization of $\mathcal{L}(u(t),u'(t))$ using the Euler-Lagrange equation will yield the most probable path.
Now, if $B = B(X_{t})$ is state-dependent, is has been shown by Dürr and Bach (Commun. Math. Phys. 60: 153–170, 1978) that, loosely speaking, the Onsager-Machlup function cannot be defined as a Lagrangian for the most probable path. On the other hand, Y. Takahashi and S. Watanabe (Springer Lecture Notes in Math. 851: 432–463, 1980) have proven that, if $M$ is a Riemannian manifold, $X_{t}$ is a diffusion process with generator $\frac{1}{2} \Delta + f$ (where $\Delta$ is the Laplace–Beltrami operator and $f$ is a vector field) and $u:[0,T] \rightarrow M$ is a smooth curve then
\begin{align*}
P\bigl( \rho( X_{t} , u(t) )  < \epsilon \text{ for all } t \in [0,T] \bigr)\underset{\epsilon \rightarrow 0^{+}}{\sim} e^{-\frac{1}{2} \int_{0}^{T} \mathcal{L}(u(t),u'(t)) dt }
\end{align*}
where $\rho$ is the Riemannian distance and $\mathcal{L}$ is a function on the tangent bundle $TM$ given by
\begin{align*}
\mathcal{L}(u,u') = \lVert f(u) - u' \rVert^{2} + \text{div } f(u) - \frac{1}{6} R(u)
\end{align*}
Here $\lVert \cdot \rVert$ is the Riemannian norm on the tangent space $T_{u}(M)$ and $R(u)$ is the scalar curvature.
But how does this general result relate to (1)? That is, what are $\rho$ and $\mathcal{L}$ (and in particular $\lVert \cdot \rVert$, $f$, $\text{div }f$, and $R$) in this case? As far as I understand (and I am not very familiar with manifolds or diffusions on them) $\rho$ would depend on $B$.
 A: This answer is a little late, but thank you for this question; it is very interesting (I hadn't known about the Onsager-Machlup formulation for most probable paths)!
In any case, if the generator is $\mathcal{A} = \Delta_g/2 + f$, we can write this as (see here):
$$ \mathcal{A} = \frac{1}{2}g^{jk}\partial_j\partial_k - \frac{1}{2}g^{jk}\Gamma_{jk}^i \partial_i + f^\ell\partial_\ell $$
using the Einstein summation convention. Notice that this is not for general Ito diffusions on the manifold, but rather for a Brownian motion with additional drift on the manifold.
This means the resulting diffusion process in local coordinates has an SDE given by:
$$
dX_t = b(X_t)\,dt + \sigma(X_t)\,dW_t
$$
where the coefficients are:
\begin{align}
b^i &= \left( - \frac{1}{2}g^{jk}\Gamma_{jk}^i + f^i \right)\\
\sigma &= \sqrt{g^{-1}} 
\end{align}
using the formula here, where the square root is a matrix root (well-defined and unique, because $g$ is symmetric positive definite).
Ok, so the actual question is what happens when the manifold is $M=\mathbb{R}$?
Well, $g=g^{-1}=I$ in coordinates and so $\Gamma_{jk}^i=0$. Thus, $R(x)=0$ and $\text{div}_g Q = \partial_x Q = Q'(x)$. Further, the Riemannian norm becomes the Euclidean norm (which is just $||\cdot||=|\cdot|$ in 1D) since $g=I$.
So:
$$ dX_t = A(X_t)dt + B(X_t)dW_t = f(X_t) dt + dW_t $$
Let's substitute this into Takahashi and Watanabe's result:
\begin{align}
\mathcal{L}(u,u') &= || f(u) - u' ||_g^2 + \text{div}_g f(u) - \frac{1}{6}R(u) \\
&= | f(u) - u' |^2 + f'(u)
\end{align}
which is exactly the case for $\mathbb{R}$ you had, with $f=A$ and $B=1$.
Also, the Riemannian (geodesic) distance in uncurved $\mathbb{R}$ is simply:
$$ \rho(X_t,u(t)) = |X_t - u(t)| $$
as expected.
So, indeed, it is a direct generalization.
