Combining eq. (205) with the relation $p_2(\mathbf{x}, t, \mathbf{y}, \tau) = p(\mathbf{x}, t|\mathbf{y}, \tau)p_{ss}(\mathbf{y})$, we get:

$$
\begin{aligned}
\partial_t \bar{p}(\mathbf{x}, t|\mathbf{y}, \tau) &= \partial_t \bar{p}_2(\mathbf{x}, t, \mathbf{y}, \tau) \bar{p}_{ss}^{-1}(\mathbf{y}) \\
&= \bar{p}_{ss}^{-1}(\mathbf{y}) \partial_t p_2(\mathbf{m}(\mathbf{y}), T - \tau, \mathbf{m}(\mathbf{x}), T - t) \\
&= -\bar{p}_{ss}^{-1}(\mathbf{y}) \frac{\partial}{\partial(T-t)} p(\mathbf{m}(\mathbf{y}), T - \tau|\mathbf{m}(\mathbf{x}), T - t) p_{ss}(\mathbf{m}(\mathbf{x})) \\
&= \bar{p}_{ss}^{-1}(\mathbf{y}) p_{ss}(\mathbf{m}(\mathbf{x})) \mathcal{W}_{\mathbf{m}(\mathbf{x})}^{\dagger} p(\mathbf{m}(\mathbf{y}), T - \tau|\mathbf{m}(\mathbf{x}), T - t) \\
&= \bar{p}_{ss}^{-1}(\mathbf{y}) p_{ss}(\mathbf{m}(\mathbf{x})) \mathcal{W}_{\mathbf{m}(\mathbf{x})}^{\dagger} p_2(\mathbf{m}(\mathbf{y}), T - \tau, \mathbf{m}(\mathbf{x}), T - t) p_{ss}^{-1}(\mathbf{m}(\mathbf{x})) \\
&= \bar{p}_{ss}^{-1}(\mathbf{y}) p_{ss}(\mathbf{m}(\mathbf{x})) \mathcal{W}_{\mathbf{m}(\mathbf{x})}^{\dagger} p_{ss}^{-1}(\mathbf{m}(\mathbf{x})) \bar{p}_2(\mathbf{x}, t, \mathbf{y}, \tau) \\
&= p_{ss}(\mathbf{m}(\mathbf{x})) \mathcal{W}_{\mathbf{m}(\mathbf{x})}^{\dagger} p_{ss}^{-1}(\mathbf{m}(\mathbf{x})) \bar{p}(\mathbf{x}, t|\mathbf{y}, \tau) \\
&= \mathbf{m}^* p_{ss}(\mathbf{x}) \mathcal{W}_{\mathbf{x}}^{\dagger} p_{ss}^{-1}(\mathbf{x}) \mathbf{m}^* \bar{p}(\mathbf{x}, t|\mathbf{y}, \tau).
\end{aligned}
$$

Finally, this last equality, together with eq. (206), leads to the result (204) that we wanted to prove.

Let us now show that, in the $\mathbf{m}$-invariant gauge $\mu = \mu(\mathbf{x})d\mathbf{x}^1 \dots d\mathbf{x}^d$, the Fokker-Planck operator $\bar{W}_\mu$ of the MT-reversed dynamics applied to a $\mu$-density $p^\mu$ reads

$$ \bar{W}_{\mu} p^{\mu} = -\operatorname{div}_{\mu} \left[ \left( \mathbf{m}_{*} \left[ 2\mathbf{D} \cdot d \ln p_{ss}^{\mu} - \mathbf{a}_{\mu} \right] \right) p^{\mu} - \left[ \mathbf{m}_{*} \mathbf{D} \right] \cdot d p^{\mu} \right], \quad (207) $$

where the definitions $\mathbf{m}_*$ (here the pushforward of contravariant tensor fields) and $\mathbf{m}^*$ (the pullback of real-valued functions) are given in section 2.2.1. To this end, let us pick a smooth map $f: \mathcal{M} \to \mathbb{R}$, and compute the effect of $\bar{W}^\dagger$ on it. Taking the $\mathbb{L}^2(\mu)$-adjoint of expression (204) of $\bar{W}$, and using the fact that $(\mathbf{m}^*)^\dagger = \mathbf{m}^*$ since $\mu$ is $\mathbf{m}$-invariant, we get

$$ \bar{W}^{\dagger} f = - \mathbf{m}^{*} p_{ss}^{-1} \mu^{-1} \partial_{i} \mu \left[ a^{i} p_{ss} \mathbf{m}^{*} f - D^{ij} \partial_{j} p_{ss} \mathbf{m}^{*} f \right] \quad (208) $$

$$ = - \mathbf{m}^* p_{ss}^{-1} (\mathbf{m}^* f) \mu^{-1} \partial_i \mu [a^i p_{ss} - D^{ij} \partial_j p_{ss}] \quad (209) $$

$$ - \mathbf{m}^* p_{ss}^{-1} [a^i p_{ss} - D^{ij} \partial_j p_{ss}] \partial_i \mathbf{m}^* f $$

$$ + \mathbf{m}^* p_{ss}^{-1} \mu^{-1} \partial_i \mu [p_{ss} D^{ij} \partial_j \mathbf{m}^* f] , $$

where we used the chain rule to go from the first to the second equality. Now, since $p_{ss}$ is the stationary $\mu$-density, the first line of eq. (209) vanishes and, upon using the chain rule again on the last line, we get

$$ \bar{W}^{\dagger} f = - \mathbf{m}^{*} [a^{i} - 2 D^{ij} \partial_{j} \ln p_{ss}] \partial_{i} \mathbf{m}^{*} f + \mathbf{m}^{*} \mu^{-1} \partial_{i} \mu [D^{ij} \partial_{j} \mathbf{m}^{*} f] . \quad (210) $$

Then, by using the definition of the pullback and pushforward (given in section 2.2.1), the first term on the right-hand side of eq. (210) can be re-written as

$$
\begin{aligned}
\{-\mathbf{m}^*[a^i - 2D^{ij}\partial_j \ln p_{ss}] \partial_i \mathbf{m}^* f\}(\mathbf{x}) &= \mathbf{m}^* \left\{ [2D^{ij}\partial_j \ln p_{ss} - a^i](\mathbf{x}) \frac{\partial \mathbf{m}^k}{\partial x^i}(\mathbf{x}) \frac{\partial f}{\partial x^k}(\mathbf{m}(\mathbf{x})) \right\} \\
&= \frac{\partial \mathbf{m}^k}{\partial x^i}(\mathbf{m}(\mathbf{x})) [2D^{ij}\partial_j \ln p_{ss} - a^i](\mathbf{x}) \frac{\partial f}{\partial x^k}(\mathbf{x}) \\
&= (\mathbf{m}_*[2\mathbf{D} \cdot d \ln p_{ss} - \mathbf{a}]) (\mathbf{x}) \cdot d f(\mathbf{x}).
\end{aligned}
$$

Then, $\mathbf{m}$ being an involution, we have that $\frac{\partial \mathbf{m}^i}{\partial x^i}(\mathbf{m}(\mathbf{x})) \frac{\partial \mathbf{m}^n}{\partial x^\ell}(\mathbf{x}) = \delta^i_\ell$, which implies that the term into the squared brackets in the last term on the right-hand side of eq. (210) reads

$$
\begin{aligned}
D^{ij}(\mathbf{x}) \frac{\partial \mathbf{m}^k}{\partial x^j}(\mathbf{x}) \frac{\partial f}{\partial x^k}(\mathbf{m}(\mathbf{x})) &= D^{\ell j}(\mathbf{x}) \frac{\partial \mathbf{m}^i}{\partial x^n}(\mathbf{m}(\mathbf{x})) \frac{\partial \mathbf{m}^n}{\partial x^\ell}(\mathbf{x}) \frac{\partial \mathbf{m}^k}{\partial x^j}(\mathbf{x}) \frac{\partial f}{\partial x^k}(\mathbf{m}(\mathbf{x})) \\
&= \frac{\partial \mathbf{m}^i}{\partial x^n}(\mathbf{m}(\mathbf{x})) [\mathbf{m}_* \mathbf{D}]^{nk}(\mathbf{m}(\mathbf{x})) \frac{\partial f}{\partial x^k}(\mathbf{m}(\mathbf{x})) \\
&= [\mathbf{m}_* ([\mathbf{m}_* \mathbf{D}] \cdot d f)]^i(\mathbf{x}).
\end{aligned}
$$

The last term of eq. (210) can then be re-written as

$$ \mathbf{m}^* \mu^{-1} \partial_i \mu [D^{ij} \partial_j \mathbf{m}^* f] = \mathbf{m}^* \operatorname{div}_\mu [\mathbf{m}_* ([\mathbf{m}_* \mathbf{D}] \cdot d f)] . \quad (211) $$
Further, we prove in appendix E that the pushforward and the divergence operator commutes (on skew-symmetric tensors), i.e. $\mathrm{div}_{\mu} [m_{*} ([m_{*}D] \cdot df)] = m_{*}\mathrm{div}_{\mu} ([m_{*}D] \cdot df)$. But $\mathrm{div}_{\mu} ([m_{*}D] \cdot df)$ being only a scalar, its pushforward by the involution $m$ coincides with its pullback, which means that

$$m^* \mathrm{div}_{\mu} [m_* ([m_* D] \cdot df)] = m^* m^* \mathrm{div}_{\mu} ([m_* D] \cdot df) = \mathrm{div}_{\mu} ([m_* D] \cdot df), \quad (212)$$

where we used the fact that $m$ is involutive. In turn, the adjoint of $\bar{W}$ applied to $f$ reads

$$\bar{W}^\dagger f = (m_* [2D \cdot d \ln p_{ss} - a]) \cdot df + \mathrm{div}_\mu ([m_* D] \cdot df) . \quad (213)$$

Finally, upon taking the adjoint of this last equation by the $L^2(\mu)$-scalar product (for which $d$ and $\mathrm{div}_{\mu}$ are each other's adjoints), we get the Fokker-Planck operator (207) of the MT-reversed process of (109) in the $m$-invariant gauge $\mu$.

### A.3 MT-dual Fokker-Planck operator: arbitrary gauge

Let us now determine the Fokker-Planck operator $\bar{W}_{\lambda}$ of the MT-reversed process of dynamics (109) in an arbitrary gauge $\lambda$. To this end, we are simply going to start from the invariant gauge $\mu$ and make a gauge transformation $\mu \to \lambda$, as explained in section 4.2. In section 4.2, we show that such a gauge transformation changes the probability density and the drift respectively as $p^{\mu} \to p^{\lambda} = p^{\mu} \mu / \lambda$ and $a_{\mu} \to a_{\lambda} = a_{\mu} + D \cdot d \ln \frac{\mu}{\lambda}$. To get the Fokker-Planck equation of the MT-dual in this new gauge, we just multiply that in the initial gauge $\mu$ by $\mu / \lambda$, the latter being given by eq. (207). This gives

$$\partial_t p_t^{\lambda} = -\frac{\mu}{\lambda} \mu^{-1} \partial_i \mu \left\{ \left[ m_* \left( 2D \cdot d \ln p_{ss}^{\lambda} \frac{\lambda}{\mu} - a_{\lambda} - D \cdot d \ln \frac{\lambda}{\mu} \right) \right]^i p_t^{\lambda} \frac{\lambda}{\mu} - [m_* D]^{ij} \partial_j p_t^{\lambda} \frac{\lambda}{\mu} \right\} . \quad (214)$$

Then factorizing by $\lambda / \mu$ in the curly bracket, using the chain rule and the fundamental property of the logarithm, we get

$$\partial_t p_t^{\lambda} = -\mathrm{div}_{\lambda} \left\{ m_* \left[ 2\mathbf{D} \cdot d \ln p_{ss}^{\lambda} - a_{\lambda} + \mathbf{D} \cdot d \ln \frac{\lambda}{\mu} \right] p_t^{\lambda} - \left( [m_* \mathbf{D}] \cdot d \ln \frac{\lambda}{\mu} \right) p_t^{\lambda} - [m_* \mathbf{D}] \cdot d p_t^{\lambda} \right\} \quad (215)$$

Finally, using the $m$-invariance of $\mu$, together with the fact that $m_*(D \cdot d \ln \frac{m_* \lambda}{m_* \mu}) = [m_* D] \cdot d \ln \frac{\lambda}{\mu}$, we get

$$\partial_t p_t^{\lambda} = -\mathrm{div}_{\lambda} \left\{ m_* \left[ 2\mathbf{D} \cdot d \ln p_{ss}^{\lambda} - a_{\lambda} - \mathbf{D} \cdot d \ln \frac{m_* \lambda}{\lambda} \right] p_t^{\lambda} - [m_* \mathbf{D}] \cdot d p_t^{\lambda} \right\}, \quad (216)$$

In turn, this allows to read out the $\lambda$-drift and diffusion tensor of the MT-dual of dynamics (109), as given in section (4.1).

## B Entropy production rate with respect to an EMT-reversal

In this appendix, we compute the entropy production rate of dynamics (109) with respect to a given $\mathrm{emt}$-reversal. In particular, we assume that the reversibility condition (113) is satisfied, i.e. that

$$m_* \mathcal{E}(D) = D, \quad (217)$$

and that the tensor field $D$ is everywhere invertible. As discussed in the main text, this implies that we can take $D^{-1}$ as a Riemannian metric on $M$. Let us denote by $\nu$ the corresponding Riemannian volume measure.

To the best of our knowledge, there exist two different ways of defining a covariant path probability for dynamics (109) (see [20] and references therein). In both cases, the action can be written as:

$$S[(x_t)_{t \in [0, \tau]}] = \int_0^\tau dt \left\{ \frac{1}{4} [\dot{x}_t - a_\nu] \cdot D^{-1} [\dot{x}_t - a_\nu] + \frac{1}{2} \mathrm{div}_\nu (a_\nu) + f(x_t, [D]) \right\}, \quad (218)$$

where $f(x, [D])$ is a function of the $D^{ij}$ and their derivatives (up to the second order) at $x$ but does not depend on $\dot{x}$ nor $a$. Below we compute the entropy production rate and show that it is
gauge invariant. Determining whether the action (218), which is always written in the gauge $\nu$ in the literature, is itself gauge-invariant is an interesting question that is beyond the scope of this article.

The action of the $\mathfrak{emt}$-dual process reads

$$ \bar{S}[(\mathbf{x}_t)_{t \in [0, T]}] = \int_0^T dt \left\{ \frac{1}{4} [\dot{\mathbf{x}}_t - \bar{\mathbf{a}}_\nu] \cdot \mathbf{D}^{-1} [\dot{\mathbf{x}}_t - \bar{\mathbf{a}}_\nu] + \frac{1}{2} \mathrm{div}_\nu (\bar{\mathbf{a}}_\nu) + f(\mathbf{x}_t, [\mathbf{D}]) \right\}, \quad (219) $$

where $\bar{\mathbf{a}}_\nu$ is the $\nu$-drift of the dual process, whose expression is given by eq. (114) (or alternatively by eq. (116)), and we used that $\bar{\mathbf{D}} = \mathbf{m}_*\mathfrak{e}(\mathbf{D}) = \mathbf{D}$ thanks to eqs. (115) & (217). Using the symmetric and skew-symmetric parts of $\mathbf{a}_\nu$ with respect to the $\mathfrak{emt}$-reversal, as defined in section 3.1, a straightforward computation gives that

$$ \bar{S} - S = \int_0^T \left[ \mathbf{a}_\nu \cdot \mathbf{D}^{-1} \dot{\mathbf{x}}_t - \mathrm{div}_\nu (\mathbf{a}_\nu) - \mathbf{a}_\nu \cdot \mathbf{D}^{-1} \mathbf{S} \mathbf{a}_\nu \right] dt. \quad (220) $$

Let us now show that eq. (220) is actually gauge-invariant. From the way $\mathbf{a}_\nu$ and $\bar{\mathbf{a}}_\nu$ transform upon a gauge change $\nu \to \lambda$ (see section 4.2) and using the hypothesis $\mathbf{m}_*\mathfrak{e}(\mathbf{D}) = \mathbf{D}$, it can easily be shown that

$$ \mathbf{s}\mathbf{a}_\nu \to \mathbf{s}\mathbf{a}_\lambda = \mathbf{s}\mathbf{a}_\nu + \mathbf{D} \cdot d \ln \frac{\nu}{\lambda}, \quad (221) $$

$$ \mathbf{a}_\nu \to \mathbf{a}_\lambda = \mathbf{a}_\nu \quad (222) $$

In particular, this implies that

$$ \mathrm{div}_\nu(\mathbf{a}_\nu) = \mathrm{div}_\nu(\mathbf{a}_\lambda) = \nu^{-1} \partial_i \left( \nu \lambda^{-1} \lambda [\mathbf{a}_\lambda]^i \right) = \mathrm{div}_\lambda(\mathbf{a}_\lambda) + \mathbf{a}_\lambda \cdot d \ln \frac{\nu}{\lambda}, \quad (223) $$

where the last equality follows from the chain rule. In turn, we can re-write eq. (220) in an arbitrary gauge $\lambda$ as

$$ \bar{S} - S = \int_0^T \left[ \mathbf{a}_{\lambda} \cdot \mathbf{D}^{-1} \dot{\mathbf{x}}_t - \mathrm{div}_{\lambda}(\mathbf{a}_{\lambda}) - \mathbf{a}_{\lambda} \cdot d \ln \frac{\nu}{\lambda} - \mathbf{a}_{\lambda} \cdot \mathbf{D}^{-1} \left( \mathbf{s}\mathbf{a}_{\lambda} + \mathbf{D} \cdot d \ln \frac{\lambda}{\nu} \right) \right] dt, \quad (224) $$

i.e.

$$ \bar{S} - S = \int_0^T \left[ \mathbf{a}_{\lambda} \cdot \mathbf{D}^{-1} \dot{\mathbf{x}}_t - \mathrm{div}_{\lambda}(\mathbf{a}_{\lambda}) - \mathbf{a}_{\lambda} \cdot \mathbf{D}^{-1} \mathbf{S} \mathbf{a}_{\lambda} \right] dt, \quad (225) $$

which proves the gauge-invariance of the expression (220) of $\bar{S} - S$.

Because the explicit knowledge of $\bar{\mathbf{a}}_\lambda$ requires that of the stationary measure, so do that of $\mathbf{s}^A\mathbf{a}_\lambda$. We thus need to replace the latter by terms that can be obtained analytically from the specification of the Markov process without also knowing its stationary measure.

To do so, we first note that, from the definition of $\mathbf{s}^A\mathbf{a}_\lambda$ and expression (30) of $\bar{\mathbf{a}}_\lambda$ (which also reads $\bar{\mathbf{a}}_\lambda = 2\mathbf{D} \cdot \ln \mathbf{m}^*(p_{ss}^\epsilon)^\lambda - \mathbf{m}_*\mathfrak{e}(\mathbf{a}_\lambda)$), it directly follows that

$$ \mathbf{s}\mathbf{a}_\lambda = \mathbf{a}_\lambda^A + \frac{1}{2}\mathbf{D} \cdot d \ln \frac{\mathbf{m}_*\lambda}{\lambda} + \mathbf{D} \cdot d \ln \mathbf{m}^*(p_{ss}^\epsilon)^\lambda, \quad (226) $$

$$ \mathbf{a}_\lambda^S = \mathbf{a}_\lambda^A - \frac{1}{2} \mathbf{D} \cdot d \ln \frac{\mathbf{m}_* \lambda}{\lambda} - \mathbf{D} \cdot d \ln \mathbf{m}^*(p_{ss}^\epsilon)^\lambda. \quad (227) $$

In eqs. (226) & (227), $\mathbf{a}_\lambda^S$ and $\mathbf{a}_\lambda^A$ are respectively the symmetric and skew-symmetric parts of $\mathbf{a}_\lambda$ under the map $\mathbf{m}_*\mathfrak{e}$, as defined in section 3.1, and $(p_{ss}^\epsilon)^\lambda$ is the $\lambda$-density of the stationary measure $\mathbf{P}_{ss}^\epsilon$ of dynamics (109) under replacement of $\mathbf{a}_\lambda$ and $\mathbf{D}$ by $\mathfrak{e}(\mathbf{a}_\lambda)$ and $\mathfrak{e}(\mathbf{D})$, respectively.

Besides, in appendix A, we showed that the MT-reversal of a process whose stationary measure is $\mathbf{P}_{ss}$ has itself a stationary measure given by $\mathbf{m}_*\mathbf{P}_{ss}$. Hence, since the $\mathfrak{emt}$-reversal can be obtained by first applying the map $\mathfrak{e}$ and then the $\mathfrak{mt}$-reversal, we conclude that the stationary measure of the $\mathfrak{emt}$-dual, denoted by $\bar{\mathbf{P}}_{ss}$, reads $\bar{\mathbf{P}}_{ss} = \mathbf{m}_*\mathbf{P}_{ss}^\epsilon$. In turn, the $\lambda$-densities are related through

$$ \bar{p}_{ss}^{\lambda} = \frac{\mathbf{m}_{*} \lambda}{\lambda} \mathbf{m}^{*}(p_{ss}^{\epsilon})^{\lambda}. \quad (228) $$
This allows to reformulate eqs. (226) & (227) as

$$ {}^{S}\mathbf{a}_{\lambda} = \mathbf{a}_{\lambda}^{A} - \frac{1}{2}\mathbf{D} \cdot d \ln \frac{\mathbf{m}_{*}\lambda}{\lambda} + \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}, \qquad (229) $$

$$ {}^{A}\mathbf{a}_{\lambda} = \mathbf{a}_{\lambda}^{S} + \frac{1}{2}\mathbf{D} \cdot d \ln \frac{\mathbf{m}_{*}\lambda}{\lambda} - \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}, \qquad (230) $$

or alternatively, using the vector fields $\mathcal{A}_{\lambda}^{\pm}$ defined in section 4.1:

$$ {}^{S}\mathbf{a}_{\lambda} = \mathcal{A}^{-} + \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}, \qquad (231) $$

$$ {}^{A}\mathbf{a}_{\lambda} = \mathcal{A}_{\lambda}^{+} - \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}. \qquad (232) $$

Injecting eqs. (231) & (232) into eq. (225) and rearranging the terms gives

$$ \bar{S} - S = - \ln \bar{p}_{ss}^{\lambda}(\mathbf{x}_{T}) + \ln \bar{p}_{ss}^{\lambda}(\mathbf{x}_{0}) + \int_{0}^{T} dt \{ \mathcal{A}_{\lambda}^{+} \cdot \mathbf{D}^{-1} \dot{\mathbf{x}} - \operatorname{div}_{\lambda} \mathcal{A}^{-} - \mathcal{A}^{-} \cdot \mathbf{D}^{-1} \mathcal{A}_{\lambda}^{+} \\ + \operatorname{div}_{\lambda} [ \mathcal{A}^{-} - \mathcal{A}_{\lambda}^{+} + \mathbf{D} \cdot d \ln \bar{p}_{ss}^{\lambda} ] + [ \mathcal{A}^{-} - \mathcal{A}_{\lambda}^{+} + \mathbf{D} \cdot d \ln \bar{p}_{ss}^{\lambda} ] \cdot d \ln \bar{p}_{ss}^{\lambda} \}. \qquad (233) $$

The last ingredient that we need follows from the fact that, by definition, the stationary probability current $\bar{\mathbf{J}}_{ss}^{\lambda}$ of the emt-dual is divergence free: $\operatorname{div}_{\lambda} \bar{\mathbf{J}}_{ss}^{\lambda} = 0$. This also reads

$$ 0 = \operatorname{div}_{\lambda} [({}^{S}\mathbf{a} - {}^{A}\mathbf{a}_{\lambda}) \bar{\mathbf{p}}_{ss}^{\lambda} - \mathbf{D} \cdot d \bar{\mathbf{p}}_{ss}^{\lambda}] . \qquad (234) $$

Factorizing $\bar{\mathbf{p}}_{ss}^{\lambda}$ in the squared brackets, using relations (231) & (232), then using the chain rule and dividing by $\bar{\mathbf{p}}_{ss}^{\lambda}$ gives

$$ \operatorname{div}_{\lambda} [\mathcal{A}^{-} - \mathcal{A}_{\lambda}^{+} + \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}] + [\mathcal{A}^{-} - \mathcal{A}_{\lambda}^{+} + \mathbf{D} \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda}] \cdot d \ln \bar{\mathbf{p}}_{ss}^{\lambda} = 0. \qquad (235) $$

In turn, the second line in eq. (233) vanishes and the entropy production rate reads

$$ \sigma \equiv \lim_{T \to \infty} \frac{\bar{S} - S}{T} = \int_{M} \mathbf{J}_{ss}^{\lambda} \cdot \mathbf{D}^{-1} \mathcal{A}_{\lambda}^{+} d\lambda - \int_{M} p_{ss}^{\lambda} [\operatorname{div}_{\lambda} \mathcal{A}^{-} + \mathcal{A}^{-} \cdot \mathbf{D}^{-1} \mathcal{A}_{\lambda}^{+}] d\lambda \qquad (236) $$

where we used the ergodicity of the dynamics together with relation (237) below.

Note that, to compute the entropy production rate in this appendix, we started from the action functional (218). Starting from the so-called Cameron-Martin-Girsanov theorem [12] instead would have been more rigorous mathematically but would have led to the same entropy production rate (236).

## C Stratonovitch average dynamics

In this section, we show that, for any one form $\alpha$ on $M$, its steady-state average against the (Stratonovitch) velocity $\dot{\mathbf{x}}_t$ of dynamics (109) is equal to the spatial integral of this observable against the stationary probability current $\mathbf{J}_{ss}$:

$$ \langle \alpha(\mathbf{x}_t) \cdot \dot{\mathbf{x}}_t \rangle = \int_{\mathbb{R}^d} \alpha(\mathbf{x}) \cdot \mathbf{J}_{ss}^{\lambda}(\mathbf{x}) d\lambda, \qquad (237) $$

where the left-hand side is defined as

$$ \langle \alpha(\mathbf{x}_t) \cdot \dot{\mathbf{x}}_t \rangle \equiv \lim_{\Delta t \to 0} \left\langle \alpha_k \left( \mathbf{x}_t + \frac{\Delta \mathbf{x}_t}{2} \right) \frac{\Delta x_t^k}{\Delta t} \right\rangle, \qquad (238) $$

$\langle \cdot \rangle$ standing for the steady-state average. But Taylor expanding $\alpha_k(\mathbf{x}_t + \Delta \mathbf{x}_t/2)$ around $\mathbf{x}_t$ and using the Ito prescription in dynamics (109) during a time increment $\Delta t$, we get:

$$ \langle \alpha_k \left( \mathbf{x}_t + \frac{\Delta \mathbf{x}_t}{2} \right) \Delta x_t^k \rangle = \langle \left[ \alpha_k(\mathbf{x}_t) + \partial_t \alpha_k(\mathbf{x}_t) \frac{\Delta x_t^k}{2} \right] \left[ A_{(0)}^k(\mathbf{x}_t) \Delta t + B^{kj}(\mathbf{x}_t) \Delta \eta_t^j \right] \rangle + \mathcal{O}(\Delta t^{3/2}) \qquad (239) $$

where $A_{(0)}$ is the Ito-total drift of dynamics (109) (see the beginning of section 2.1). Since the random increment $\Delta \eta_t$ between time $t$ and $t + \Delta t$ and $\mathbf{x}_t$ are independent, and $\langle \Delta x_t^i \Delta \eta_t^j \rangle =$
$\langle b_{\gamma}^{i}(\mathbf{x}_{t})\langle\Delta\eta_{t}^{\gamma}\Delta\eta_{t}^{\beta}|\mathbf{x}_{t}\rangle\rangle = 2\Delta t\langle b_{\beta}^{i}\rangle$ (the notation $\langle\cdot|\mathbf{x}_{t}\rangle$ here stands for the conditional average), we have

$$ \left\langle \alpha_k \left( \mathbf{x}_t + \frac{\Delta \mathbf{x}_t}{2} \right) \Delta x_t^k \right\rangle = \langle \alpha_k(\mathbf{x}_t) A^k(\mathbf{x}_t) + b_{\beta}^i b_{\beta}^k \partial_i \alpha_k(\mathbf{x}_t) \rangle \Delta t + \mathcal{O}(\Delta t^{3/2}) . \quad (240) $$

Finally, we divide each side by $\Delta t$ and take the limit $\Delta t \to 0$. Writing the average explicitly and integrating by parts gives the required result (237):

$$ \langle \boldsymbol{\alpha}(\mathbf{x}_t) \cdot \dot{\mathbf{x}}_t \rangle = \int \boldsymbol{\alpha}(\mathbf{x}) \cdot \left[ p_{\mathrm{ss}}^{\lambda} \mathbf{A}_{(0)}(\mathbf{x}_t) - \mathrm{div}_{\boldsymbol{\lambda}}(\mathbf{D} p_{\mathrm{ss}}^{\lambda}) \right] \boldsymbol{\lambda}(\mathbf{x}) \mathrm{d}\mathbf{x} = \int \boldsymbol{\alpha}(\mathbf{x}) \cdot \mathbf{J}_{\mathrm{ss}}^{\lambda}(\mathbf{x}) \mathrm{d}\boldsymbol{\lambda} . \quad (241) $$

## D Exterior derivative and vorticity operator in curved space

Let $(\mathcal{M}, g)$ be a Riemannian manifold, with Levi-Civita connection denoted by $\nabla$. Let $\mathbf{u}$ be a vector field over $\mathcal{M}$, with dual 1-form $\mathbf{u}^b$. In this appendix we prove that

$$ g(\nabla_{\mathbf{v}}^{A} \mathbf{u}, \mathbf{w}) = \frac{1}{2} \mathrm{d} \mathbf{u}^{b}(\mathbf{v}, \mathbf{w}), \quad (242) $$

where $\nabla^A \mathbf{u}$ is the skew-symmetric part (with respect to $g$) of $\nabla \mathbf{u}$ seen as an operator from the space of vector fields to itself.

Let $\boldsymbol{\alpha}$ be a one-form and $\mathbf{v}, \mathbf{w}$ two vector fields over $\mathcal{M}$. Let us first prove that

$$ (\nabla_{\mathbf{v}} \boldsymbol{\alpha})(\mathbf{w}) - (\nabla_{\mathbf{w}} \boldsymbol{\alpha})(\mathbf{v}) = d\boldsymbol{\alpha}(\mathbf{v}, \mathbf{w}). \quad (243) $$

By compatibility of the connection with respect tensor product, we have

$$ \nabla_{\mathbf{w}}(\boldsymbol{\alpha} \otimes \mathbf{v}) = \nabla_{\mathbf{w}}\boldsymbol{\alpha} \otimes \mathbf{v} + \boldsymbol{\alpha} \otimes \nabla_{\mathbf{w}}\mathbf{v}. \quad (244) $$

Then the compatibility of $\nabla$ with tensor contraction gives

$$ \nabla_{\mathbf{w}}(\boldsymbol{\alpha}(\mathbf{v})) = (\nabla_{\mathbf{w}}\boldsymbol{\alpha})(\mathbf{v}) + \boldsymbol{\alpha}(\nabla_{\mathbf{w}}\mathbf{v}). \quad (245) $$

Since $\boldsymbol{\alpha}(\mathbf{v})$ is just a function on $\mathcal{M}$, its covariant derivative in the direction of $\mathbf{w}$ is simply its Lie derivative along $\mathbf{w}$: $\nabla_{\mathbf{w}}(\boldsymbol{\alpha}(\mathbf{v})) = \mathbf{w}(\boldsymbol{\alpha}(\mathbf{v}))$. The above equation thus reads

$$ (\nabla_{\mathbf{w}}\boldsymbol{\alpha})(\mathbf{v}) = \mathbf{w}(\boldsymbol{\alpha}(\mathbf{v})) - \boldsymbol{\alpha}(\nabla_{\mathbf{w}}\mathbf{v}). \quad (246) $$

This implies that

$$ (\nabla_{\mathbf{v}}\boldsymbol{\alpha})(\mathbf{w}) - (\nabla_{\mathbf{w}}\boldsymbol{\alpha})(\mathbf{v}) = \mathbf{v}(\boldsymbol{\alpha}(\mathbf{w})) - \mathbf{w}(\boldsymbol{\alpha}(\mathbf{v})) - (\boldsymbol{\alpha}(\nabla_{\mathbf{v}}\mathbf{w}) - \boldsymbol{\alpha}(\nabla_{\mathbf{w}}\mathbf{v})) \quad (247) $$

$$ = \mathbf{v}(\boldsymbol{\alpha}(\mathbf{w})) - \mathbf{w}(\boldsymbol{\alpha}(\mathbf{v})) - \boldsymbol{\alpha}(\nabla_{\mathbf{v}}\mathbf{w} - \nabla_{\mathbf{w}}\mathbf{v}) \quad (248) $$

where the second line follows from the linearity of $\boldsymbol{\alpha}$. But the Levi-Civita connection is torsion-free, i.e. $\nabla_{\mathbf{v}}\mathbf{w} - \nabla_{\mathbf{w}}\mathbf{v} = [\mathbf{v}, \mathbf{w}]$, where $[\cdot, \cdot]$ stands for the Lie bracket. Hence we have that

$$ (\nabla_{\mathbf{v}}\boldsymbol{\alpha})(\mathbf{w}) - (\nabla_{\mathbf{w}}\boldsymbol{\alpha})(\mathbf{v}) = \mathbf{v}(\boldsymbol{\alpha}(\mathbf{w})) - \mathbf{w}(\boldsymbol{\alpha}(\mathbf{v})) - \boldsymbol{\alpha}([\mathbf{v}, \mathbf{w}]). \quad (249) $$

We then recognize the covariant expression of the exterior derivative of $\boldsymbol{\alpha}$ applied to $\mathbf{v}, \mathbf{w}$ on the right-hand side of the latter equation, which thus proves (243).

Now, note that $\nabla$ commutes with the operator $\flat$. Indeed, by compatibility of $\nabla$ with the tensor product:

$$ \nabla_{\mathbf{v}}(g \otimes \mathbf{u}) = \nabla_{\mathbf{v}}g \otimes \mathbf{u} + g \otimes \nabla_{\mathbf{v}}\mathbf{u}, \quad (250) $$

But the Levi-Civita connection preserves the metric tensor, i.e. $\nabla_{\mathbf{v}}g = 0$, hence $\nabla_{\mathbf{v}}(g \otimes \mathbf{u}) = g \otimes \nabla_{\mathbf{v}}\mathbf{u}$. Hence, the definition of $\flat$ (i.e. the contraction of the vector field with the metric), together with the compatibility of $\nabla$ with tensor contraction, gives

$$ \nabla_{\mathbf{v}}(\mathbf{u}^{\flat}) = (\nabla_{\mathbf{v}}\mathbf{u})^{\flat}. \quad (251) $$
Upon denoting by $(\nabla \mathbf{u})^\top$ the adjoint of $\nabla \mathbf{u}$ with respect to $g$, and $\nabla_\mathbf{v}\mathbf{u}^\top$ its action on a vector $\mathbf{v}$, we have

$$g(\nabla_{\mathbf{v}}^{A} \mathbf{u}, \mathbf{w}) = \frac{1}{2} g(\nabla_{\mathbf{v}} \mathbf{u} - \nabla_{\mathbf{v}} \mathbf{u}^{\top}, \mathbf{w}) \quad (252)$$

$$= \frac{1}{2} [g(\nabla_{\mathbf{v}} \mathbf{u}, \mathbf{w}) - g(\mathbf{v}, \nabla_{\mathbf{w}} \mathbf{u})] \quad (253)$$

$$= \frac{1}{2} [ (\nabla_{\mathbf{v}} \mathbf{u})^{\flat}(\mathbf{w}) - (\nabla_{\mathbf{w}} \mathbf{u})^{\flat}(\mathbf{v}) ], \quad (254)$$

where we have just used the definition of $\nabla \mathbf{u}^\top$ and that of $\flat$. Then, using (254) together with (251), we have that

$$g(\nabla_{\mathbf{v}}^{A} \mathbf{u}, \mathbf{w}) = \frac{1}{2} [ (\nabla_{\mathbf{v}} \mathbf{u})^{\flat}(\mathbf{w}) - (\nabla_{\mathbf{w}} \mathbf{u})^{\flat}(\mathbf{v}) ] . \quad (255)$$

Finally, combining (255) and (243) gives the result we wanted to prove (242).

## E Pushforward and divergence of skew-symmetric contravariant tensors

Let $(\mathcal{M}, g)$ be a smooth oriented Riemannian manifold without boundary. We denote by $\flat$ the isomorphism that lowers all indices with $g$, i.e. that maps a contravariant tensors $\mathbf{a} = a^{i_1 \dots i_n} \partial_{i_1} \otimes \dots \otimes \partial_{i_n}$ to the covariant tensor $\mathbf{a}^{\flat} = a_{i_1 \dots i_n} \mathrm{d}x^1 \otimes \dots \otimes \mathrm{d}x^d$, with $a_{i_1 \dots i_n} = a^{j_1 \dots j_n} g_{j_1 i_1} \dots g_{j_n i_n}$. Finally, we denote by $\sharp$ the inverse map of $\flat$, i.e. the map that raises all indices.

The metric $g$ induces for all $\mathbf{x} \in \mathcal{M}$ an inner product on the space of covariant tensors of order $n$ at $\mathbf{x}$:

$$\langle \alpha, \beta \rangle_{T^n} = \alpha_{i_1 \dots i_n} g^{i_1 j_1}(\mathbf{x}) \dots g^{i_n j_n}(\mathbf{x}) \beta_{j_1 \dots j_n} . \quad (256)$$

(Note that all angular brackets in this appendix denote inner products and must not be confused with averages.) For skew-symmetric covariant tensors of order $n$, it is also useful to define the rescaled scalar product:

$$\langle \alpha, \beta \rangle_{\Omega^n} = \frac{1}{n!} \langle \alpha, \beta \rangle_{T^n} . \quad (257)$$

Denoting by $\lambda$ the Riemannian volume measure associating to $g$, we can define a (global) scalar product on skew-symmetric covariant tensor fields of order $n$ by integrating their scalar product a each point:

$$\langle\langle\alpha, \beta\rangle\rangle \equiv \int_{\mathcal{M}} \langle\alpha(\mathbf{x}), \beta(\mathbf{x})\rangle_{\Omega^n} \lambda(\mathbf{x}) \mathrm{d}\mathbf{x} . \quad (258)$$

The adjoint of the exterior derivative with respect to the scalar product (258) is denoted by $d^\dagger$. They both act on differential forms i.e. on skew-symmetric covariant tensor fields.

Let us now pick a skew-symmetric contravariant tensor field $\mathbf{a}$ of order $n$. We recall that

$$d^\dagger(\mathbf{a}^\flat) = -(\mathrm{div}_\lambda \mathbf{a})^\flat, \quad (259)$$

where $[\mathrm{div}_\lambda \mathbf{a}]^{i_1 \dots i_{n-1}} = \lambda^{-1} \partial_j (\lambda a^{j_1 \dots i_{n-1}})$. Let $\mathfrak{m}$ be a diffeomorphism on $\mathcal{M}$ that preserves the measure $\lambda$, i.e. such that $\mathfrak{m}_*\lambda = \lambda$. The objective of this appendix is to prove that

$$\mathrm{div}_\lambda(\mathfrak{m}_*\mathbf{a}) = \mathfrak{m}_*\mathrm{div}_\lambda\mathbf{a}, \quad (260)$$

i.e. that the pushforward by $\mathfrak{m}$ commutes with the divergence operator on skew-symmetric contravariant tensor fields. Let $\alpha$ be a skew-symmetric covariant tensor field of order $n-1$. For any contravariant tensor field $\mathbf{b}$ of order $n-1$ we have that $\alpha(\mathbf{b}) = \alpha_{i_1 \dots i_{n-1}} b^{i_1 \dots i_{n-1}} = \langle \alpha, \mathbf{b}^\flat \rangle_{T^{n-1}}$ at every point $\mathbf{x} \in \mathcal{M}$. Hence, using definition (257), we get

$$\int_{\mathcal{M}} \alpha \left( \left[ d^\dagger(\langle \mathfrak{m}_* \mathbf{a} \rangle^\flat) \right]^\sharp \right) d\lambda = \int_{\mathcal{M}} \langle \alpha, d^\dagger(\langle \mathfrak{m}_* \mathbf{a} \rangle^\flat) \rangle_{T^{n-1}} d\lambda = (n-1)! \int_{\mathcal{M}} \langle \alpha, d^\dagger(\langle \mathfrak{m}_* \mathbf{a} \rangle^\flat) \rangle_{\Omega^{n-1}} d\lambda . \quad (261)$$
But the last term corresponds to $(n-1)!(\langle\langle\alpha, d^{\dagger}[(m_*a)^b]\rangle\rangle)$, which is the scalar product for which $d^\dagger$ is the adjoint of $d$. We thus have

$$ 
\begin{aligned}
\int_M \alpha \left( \left[ d^{\dagger}((m_*a)^b) \right]^{\sharp} \right) d\lambda &= (n-1)! \int_M \left\langle d\alpha, (m_*a)^b \right\rangle_{\Omega^n} d\lambda \\
&= \frac{(n-1)!}{n!} \int_M \left\langle d\alpha, (m_*a)^b \right\rangle_{T^n} d\lambda \\
&= \frac{1}{n} \int_M d\alpha (m_*a) d\lambda,
\end{aligned}
$$

where once again we used definition (257) together with the relation between $\flat$ and $\langle, \rangle_{T^n}$. The definition of the pullback of differential forms gives

$$ d\alpha_x(m_*a) = [m^*d\alpha]_{m^{-1}(x)}(a) = [m^*d\alpha]_{m(x)}(a), \qquad (262) $$

where the fact that $m^{-1} = m$ is due to $m$ being involutive. Further, we assumed that $\lambda$ is $m$-invariant, which means that

$$ d\lambda(x) = d\mathbf{m}_*\lambda(x) = \lambda(\mathbf{m}(x))|\det[(\partial_j\mathbf{m}^i)(x)]|d\mathbf{x}. \qquad (263) $$

Relations (262) & (263) then give

$$ \int_M d\alpha_x(m_*a)\lambda(x)d\mathbf{x} = \int_M [m^*d\alpha]_{m(x)}(a)\lambda(m(x))|\det[(\partial_j m^i)(x)]|d\mathbf{x} = \int_M [m^*d\alpha]_y(a)\lambda(y)dy, (264) $$

where the last equality is obtained by a change of variable $y = m(x)$. In turn we have

$$ \int_M \alpha \left( \left[ d^{\dagger}((m_*a)^b) \right]^{\sharp} \right) d\lambda = \frac{1}{n} \int_M m^* d\alpha(a) d\lambda. \qquad (265) $$

Now, we need to use the well-known commutativity between the pullback and the exterior derivative, $m^*d\alpha = dm_*\alpha$, and then do the same transformations in the integral as the ones we have just performed, but in the opposite direction. The outcome reads

$$ 
\begin{aligned}
\int_M \alpha \left( \left[ d^{\dagger}((m_*a)^b) \right]^{\sharp} \right) d\lambda &= \frac{1}{n} \int_M dm^* \alpha(a) d\lambda \\
&= \frac{1}{n} \int_M \left\langle dm^* \alpha, a^b \right\rangle_{T^n} d\lambda \\
&= \frac{n!}{n} \int_M \left\langle dm^* \alpha, a^b \right\rangle_{\Omega^n} d\lambda \\
&= (n-1)! \int_M \left\langle m^* \alpha, d^{\dagger}(a^b) \right\rangle_{\Omega^{n-1}} d\lambda \\
&= \int_M \left\langle m^* \alpha, d^{\dagger}(a^b) \right\rangle_{T^{n-1}} d\lambda \\
&= \int_M m^* \alpha \left( \left[ d^{\dagger}(a^b) \right]^{\sharp} \right) d\lambda \\
&= \int_M \alpha \left( m_* \left[ d^{\dagger}(a^b) \right]^{\sharp} \right) dm_* \lambda \\
&= \int_M \alpha \left( m_* \left[ d^{\dagger}(a^b) \right]^{\sharp} \right) d\lambda.
\end{aligned}
$$

Since this is true for all $\alpha$, we conclude that $[d^{\dagger}((m_*a)^b)]^{\sharp} = m_*[d^{\dagger}(a^b)]^{\sharp}$, i.e. $\mathrm{div}_{\lambda}(m_*a) = m_*\mathrm{div}_{\lambda}a$.
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