Equivariant Divergence Formula for Hyperbolic Chaotic Flows

Abstract

We prove the equivariant divergence formula for axiom A flow attractors. It is a pointwisely-defined and recursive formula for perturbation of SRB measures along center-unstable manifolds. It depends on only the zeroth and first order derivatives of the map, the observable, and the perturbation. Hence, the linear response acquires an ‘ergodic theorem’, which means that it can be sampled by recursively computing a few vectors on one orbit.

1 Introduction

1.1 Literature Review

The long-time statistic of a hyperbolic chaotic system is given by the SRB measure, or physical measure, $\rho$, which is a model for fractal limiting invariant measures of chaos [10, 46, 51] (under the setting of this paper, the SRB and the physical measure are the same). The linear response is the derivative of the SRB measure with respect to the parameters of the system.

There are several different formulas for the linear response. The most well-known are the path perturbation formula, the divergence formula, and the kernel differentiation formula [8, 13, 20, 22, 26, 27, 43, 47, 48]. The path perturbation formula averages conventional adjoint formula over a lot of orbits, which is also known as the ensemble method or stochastic gradient method; however, it is cursed by the gradient explosion [14, 24, 29, 31]. The divergence formula, also known as the transfer operator formula, computes the perturbation operator, which is typically not pointwisely defined for hyperbolic systems, and we have to partition the full phase space to obtain some mollified approximation: this is very expensive in high dimensional spaces [3, 5, 11, 12, 15,16,17,18, 25, 28, 30, 42, 49, 50, 52]. The kernel differentiation formula is also known as the likelihood ratio method in probability context [21, 26, 37, 44, 45]; it works only for stochastic systems, and the cost is large when the noise is small. In this paper, we consider only the deterministic flow with hyperbolicity.

The system dimension M of most real-life dynamical systems is large; for example, a discretized 3-dimensional Navier–Stokes equation can easily have $M\ge 10^6$. The only way to overcome the curse of dimensionality is essentially the ergodic theorem: fix any $C^\infty$ observable function $\Phi :\mathcal {M}\rightarrow \mathbb {R}$, then for almost all x in the attractor basin U,

$$\begin{aligned} \lim _{T\rightarrow \infty } \frac{1}{T} \int _{t=0}^T \Phi (f^t x) = \rho (\Phi ): = \int \Phi (x) \rho (dx). \end{aligned}$$

This is both theoretically interesting and important for high dimensional numeric since, roughly speaking, $\Phi$ reduces to a 1-dimensional random variable, and with some decay of correlations, we can ignore the details of the random variables when performing averages. The trade-off here is that we have to confine our interest to the averaged observable, and we can not ask the details of $\rho$ at arbitrary locations. By ‘ergodic theorem’ type formulas, we refer to formulas that are averages of pointwise functions of a few vectors, which can be expressed by recursive relations on a orbit. This paper provides such a formula for the linear response of hyperbolic flows with perhaps the least number of vectors being tracked. Note that the first obstacle, is that although here $\Phi$ is pointwisely defined, in the case of linear response formulas, pointwisely definition is not easy.

A promising approach to obtain ‘ergodic theorem’ type formula is blending the path perturbation and divergence formulas, but the obstruction was that the split lacks smoothness. It is natural to apply the path perturbation formula to the stable or shadowing contribution of the linear response. The earliest recursive formula should be the stable part in the blended response formula, which evolves M many vectors [1]. The formula tracking the least number of vectors is the nonintrusive shadowing formula, which involves only u many vectors [35, 38, 40]. When the ratio of unstable directions is low, with some additional statistical assumptions, the unstable contribution could sometimes be small, so we sometimes might still get a useful linear response even if we do not compute the unstable contribution [34].

However, for hyperbolic systems, the unstable contribution much is harder to sample on an orbit. The first obstacle is lacking a pointwisely defined formula, since the most well-known formula is a divergence of oblique projections of the perturbation in the unstable submanifold, but the directional derivatives are distributions rather than functions. This seems to be an essential difficulty, for example, Proposition 8.1 in [23] seems to give a different formula via Livisic theorem. But that formula involves a divergence of oblique projections in the stable submanifold; so it still involves directional derivatives which are not pointwisely defined, and it has the same difficulty as us working in unstable manifolds. A hint to solve this difficulty was given in [47], but it did not give a formula.

Besides being pointwisely defined, we also wish that the new formula: (1) is recursive; (2) involves tracking minimum number of vectors or covectors, which can be computed by recursive relations on an orbit. The equivariant divergence formula solves the regularity difficulty by evolving only 2u many vectors per step, which is currently the most efficient [33, 36, 39]; but it does not yet have continuous-time versions.

The continuous-time cases are perhaps more important than discrete-time, because most physical and engineering systems are continuous-time. It is also more difficult due to the presence of the one-dimensional center direction parallel to the flow, where perturbations do not decay exponentially either forward or backward in time. But we know the expression of the centre direction, and the stable and unstable are differentiable in the flow direction; these extra information allow us to assemble the shadowing, unstable, and the center contributions into formulas similar to discrete-time.

This paper does not fully solve the practical problem of computing an approximated and useful linear response for most high-dimensional continuous-time systems, but it should be a building block of a good solution. In discrete time, it seems that the fast response algorithm, an algorithm based on the equivariant divergence formula, works whenever the existence of the linear response could be proved, including moderately nonuniform hyperbolic and moderately discontinuous case. Note that the linear response may or may not exist for the general case; for example, there is an explicit linear response formula in one dimensional non-uniformly expanding maps in [6]. However, there are many counter examples where the linear response can be disproved [7], and our algorithm explodes. In terms of continuous time case, most well-known systems, such as Lorenz systems, are not hyperbolic. There is significant progress on linear response of maps with singularities [4], indicating that the Poincaré map of the Lorenz 63 system is likely to have linear response. But whether linear response exists for continuous-time Lorenz 63 system, and moreover the Lorenz 96 system, is still unknown. To remedy this, we have to introduce some approximation, and we proposed a trinity program in [35]: for regions with bad hyperbolicity, we try to add local noise and occasionally switch to the kernel differentiation formula.

1.2 Main Results

This paper gives the equivariant divergence formula for the axiom A flow attractors, which is a recursive formula for perturbation of SRB measures along center-unstable manifolds. This further gives a recursive formula for the linear response, which has potential numerical applications.

For continuous-time systems, let f be the flow of $F+\gamma X|_{\gamma =0}$, where both F and X are $C^3$ vector fields, F is the vector field of the base flow, X is the vector field of the perturbation, and the parameter $\gamma$ has base value zero. Let $X^{cu}$ and $X^s$ denote the center-unstable and stable part of X with respect to the hyperbolic splitting. We assume that $X^s\ne 0$ and $X^{cu} \ne 0$ without loss of generality. We denote $\delta (\cdot ):=\partial (\cdot )/\partial \gamma |_{\gamma =0}$.

It it known that a transitive axiom A attractor has a unique physical SRB measure which has absolutely continuous conditional measure on center-unstable manifolds. Let $\sigma$ be the conditional density function of the unique physical SRB measure $\rho$ on a local center-unstable manifolds. We now let $\tilde{f}$ denote the perturbation map, which is the time $\gamma$ map of the flow generated by the perturbation vector field X, and denote $\xi$ as the local stable holonomy map. Then for $\gamma$ small enough, $\xi \tilde{f}$ is a well-defined operator from a center unstable manifold to itself. We mention that $\delta (\xi \tilde{f})=X^{cu}$, roughly because $\tilde{f}$ contribute an X and $\xi$ project it to the center-unstable direction. Let ${\tilde{L}}^{cu}$ be the local transfer operator of $\xi \tilde{f}$, so $\delta {\tilde{L}}^{cu}\sigma$ is the center-unstable perturbation of the conditional density function $\sigma$. (see Sect. 3.2 for detailed explanations).

Since the elements above are defined locally, we can take the ratio $-\frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }$, which does not depend on the choice of the neighborhood of the local foliation. $-\frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }$ may as well be denoted by the unstable submanifold divergence, $\text {div}^{cu}_\sigma X^{cu}$. Note that this is a priori only a distribution, since $X^{cu}$ is not differentiable: indeed, $X^{cu}$ involves both the stable and the center-unstable direction, and the stable subspace is not differentiable along center-unstable directions. This paper shows that $\text {div}^{cu}_\sigma X^{cu}$ has the equivariant divergence formula, from which we can see that $\text {div}^{cu}_\sigma X^{cu}$ is pointwisely defined. Hence, we could formally say that summing all the directional derivatives cancels the roughness in each directional derivative; but we need to run a different proof via transfer operators, which does not involve directional derivatives of $X^{cu}$ at all.

Theorem 1

(Equivariant divergence in continuous-time) Let $\eta : = \varepsilon ^c(X)$, then

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }= \text {div}^{cu}_\sigma X^{cu} = \mathcal {S}(\text {div}^v \nabla F, \text {div}^v F) X + \text {div}^v X + F(\eta ). $$

holds $\rho$ almost everywhere.

Here $\varepsilon ^c$ is the the covector in the center subspace such that $\varepsilon ^c (F)=1$, and $F(\cdot )$ means to differentiate in the direction of F. $\mathcal {S}$ is the adjoint shadowing operator. Roughly speaking it maps a (covector field, function) pair to a covector field with following expression:

$$\begin{aligned} \mathcal {S}(\omega ,\psi ) = \int _{t\ge 0} f^{*t} \omega ^s_t dt - \int _{t\le 0} f^{*t} \omega ^u_t dt - \psi \varepsilon ^c, \end{aligned}$$

It is the adjoint operator of the tangent shadowing operator S under a special inner product which combine a (covector field, function) pair with a (vector field, function) pair. The tangent shadowing operator S map a vector field to a (vector field, function) pair as below:

$$ S: X \mapsto [v, \eta ] \quad \text {where} \quad \eta F=X^c, v=\int _0^\infty f_*^tX^s_{-t}dt-\int _{-\infty }^0f^t_*X^u_{-t}dt. $$

the detail is reviewed in Sect. 2.2.

The equivariant unstable divergence $\text {div}^v$ is the contraction by the unstable vector and covectors. Roughly speaking, we define the u-vector and the u-covector by the wedge operation, $e:=e_1\wedge \cdots \wedge e_u, \varepsilon :=\varepsilon _1\wedge \cdots \wedge \varepsilon _u$,where $e_i$ is a basis of the unstable tangent space and $\varepsilon _i$ is the dual basis. Then we can extend some operations in the differential geometry naturally to the u-vector fields and the u-covector fields. Define the equivariant unstable divergence as $\text {div}^v X:= \tilde{\varepsilon }\nabla _{\tilde{e}} X$, where $\tilde{\varepsilon }$ is a local unit u-covector field and $\tilde{e}$ is a local unit u-vector field. In fact, this divergence operator calculates the unstable divergence in a dynamical way. The functions $\text {div}^v F$ and $\text {div}^v X$ are the equivariant divergences of the vector fields F and X.

The $\text {div}^v \nabla F$ is a covector field defined by the equivariant unstable divergence and the Hessian of F. It has the expression $\text {div}^v \nabla F(X):= \tilde{\varepsilon }\nabla ^2_{\tilde{e}, X} F$ where the Hessian is defined by $\nabla ^2_{X,Y} Z = \nabla _X \nabla _Y Z- \nabla _{ \nabla _X Y} Z$. All these notations will be explicitly defined later in Sect. 3.1. It is worth mentioning that all the ‘equivariant divergences’ are geometric quantities determined by the geometry of the manifold, the base flow, and the perturbation vector field in a quite direct and natural way.

Theorem 1 is a pointwisely defined formula, since no intermediate quantities are distributions. Moreover, it is an ‘ergodic theorem’ type formula, which means that it can be calculated by the information on one typical orbit pointwise recursively. This is of physical importance because in physical world we can only observe the phase space by one typical orbit. It is also of application importance because the algorithmic complexity of this type of formula avoids the curse of dimensionality.

With Theorem 1, we can give a new formula for the unstable contribution of the linear response of mixing axiom A attractors. Let $\rho$ be the SRB measure which coincide with the long-time statistic. Fix a smooth objective function $\Phi$. Denote $\delta :=\partial /\partial \gamma |_{\gamma =0}$, we decompose the linear response formula from [13, 48] into the shadowing and the unstable contribution,

$$ \delta \rho (\Phi ) = \int _{0}^{\infty }\rho (X(\Phi _t))dt = SC - UC, $$

where $\Phi _t:=\Phi \circ f^t$, and

$$ SC :=\int _{0}^{\infty }\rho (X^{s}(\Phi _t))dt-\int _{-\infty }^{0}\rho (X^{cu}(\Phi _t))dt, \quad UC := \int _{-\infty }^\infty \rho \left( X^{cu}(\Phi _t) \right) dt. $$

To have an ‘ergodic theorem’ formula for linear response, we just need to give a new interpretation of the unstable contribution by transfer operator $\delta {\tilde{L}}^{cu}$:

Proposition 2

(Fast response formula for linear response)

$$ SC = \rho (X\mathcal {S}(d\Phi ,\Phi -\rho (\Phi ))), \quad UC = \lim _{W\rightarrow \infty } \rho \Bigg (\varphi \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }\Bigg ), \quad \text {where}\quad \varphi :=\int _{-W}^W \Phi \circ f^t dt . $$

Since the SRB measure can be sampled by a typical orbit, using Theorem 1 and Proposition 2, the linear response can be expressed by recursively computing only O(u) many vectors on one typical orbit. In fact, in the real computing process, we can sample the SRB measure $\rho$ by an orbit. Then we can compute a basis of the unstable subspace by pushing forward u many randomly initialized vectors while performing occasional renormalizations, on the same orbit we used to sample $\rho$. Similarly, we can compute a basis of the adjoint unstable subspace. With these two basis we can compute the equivariant divergence $\text {div}^v$. Also these two basis are the main data required by the nonintrusive shadowing algorithm for computing $\mathcal {S}$ (see Sect. 2.2).

Another benefit of this new formula is that it exhibits explicit dependence on X and its derivative. This eases estimation on the norm of the linear response operator $R:X\mapsto \delta \rho (\Phi )$. This should be useful in the optimal response problem [2, 19].

This paper is organized as follows. First, Sect. 2 reviews the tangent linear response theory and the adjoint shadowing lemma. Then Sect. 3 proves the equivariant divergence formula. Finally, Sect. 4 uses the equivariant divergence formula to express the linear response.

2 Preliminaries

2.1 Hyperbolicity and Linear Response

Let f be the flow of $F+\gamma X$, where both F and X are $C^3$ vector fields, F is the vector field of the base flow, X is the vector field of the perturbation, and the parameter $\gamma$ has base value zero. Assume that

$$ F\ne 0 \quad \text {on } K. $$

Assume that the attractor K is compact, |F| is uniformly bounded away from zero and infinity. We further assume that K is a mixing axiom A attractor. Now $T_KM$ has a continuous $f_*$-invariant splitting into stable, unstable, and center subspaces, $T_KM = V^s \bigoplus V^u \bigoplus V^c$, where $V^c$ is the one-dimensional subspace spanned by F, and

$$ \begin{aligned} \max _{x\in K}|f_* ^{-t}|V^u(x)| , |f_* ^{t}|V^s(x)| \le C\lambda ^{n} \quad \text {for } t\ge 0, \\ \max _{x\in K}|f_* ^{t}|V^c(x)| \le C \quad \text {for } t\in \mathbb {R}. \end{aligned} $$

Here $f_*$ is the pushforward operator on vectors. Define the oblique projection operators $P^u$ and $P^s$, such that

$$ v = P^u v + P^s v, \quad \text {where} \quad P^u v\in V^u, P^s v\in V^s. $$

For an orbit $x_t=f^t(x_0)$, the so-called homogeneous tangent equation of $e_t\in T_{x_t}\mathcal {M}$ can be written by three equivalent notations:

$$ e_t =f_*^t e_0 \quad \Leftrightarrow \quad \mathcal {L}_F e = 0 \quad \Leftrightarrow \quad \nabla _F e = \nabla _e F. $$

Here $\mathcal {L}$ is the Lie-derivative and $\nabla$ is the Riemannian derivative. The last expression is an ODE, since $\nabla _F$ is $\partial /\partial t$ in $\mathbb {R}^M$.

A transitive axiom A attractor admits a physical SRB measure $\rho$. Furthermore, when the attractor is mixing, existence of linear responses and the particular formula we shall use was proved [13, 48]. Denote

$$ \delta (\cdot ):=\partial (\cdot )/\partial \gamma |_{\gamma =0} \quad \text {is the total derivative with all other parameters fixed.} $$

Here ‘total derivative’ means when $g=g(\gamma ,x(\gamma ,y))$, then $\delta g = \partial g/\partial \gamma + \partial g/\partial x \cdot \partial x /\partial \gamma |_{\gamma =0}$. We sometimes write $\delta (\cdot ):=\partial (\cdot )/\partial \gamma |_{\gamma =0}$ to emphasize that all other parameters are fixed, while sometimes write $\delta (\cdot ):=d (\cdot )/d \gamma |_{\gamma =0}$ to emphasize that we expand all levels of dependence on $\gamma$. The linear response can be given by a path perturbation formula,

$$\begin{aligned} \delta \rho (\Phi ) = \lim _{T\rightarrow \infty } \rho \left( \int _{0}^T X(\Phi _t) dt \right) , \end{aligned}$$

(1)

where $X(\cdot )$ is to differentiate in the direction of X.

In this paper we shall use the above formula, the so-called ‘ensemble formula’, as a starting point, and derive a linear response formula that can be sampled by 2u recursive relations on an orbit. First, we decompose the linear response into

$$ \delta \rho (\Phi ) = SC + UC, \quad SC=\int _{0}^{\infty }\rho (X^{s}(\Phi _t))dt-\int _{-\infty }^{0}\rho (X^{cu}(\Phi _t))dt $$

which we call the shadowing and the unstable contribution. When $u\ll M$, it might happen that $\delta \rho \approx SC$ [34].

2.2 Adjoint Shadowing Lemma

The adjoint shadowing lemma gives three equivalent characterizations for the adjoint shadowing operator $\mathcal {S}$ in continuous-time [35]. Let $f^*$ be the pullback operator on covectors. The adjoint flow $\{f^{*t}\}_{t\in \mathbb {R}}$ is also hyperbolic. More specifically, we can show that the dual space of $V^s, V^c$, and $V^u$, denoted by $V^{*s}, V^{*c}$, and $V^{*u}$, are also the stable, unstable, and center subspace. Define the image space of $\mathcal {P}^u$ and $\mathcal {P}^s$ as $V^{u*}$ and $V^{s*}$; we can show that they are in fact the stable, center, and unstable subspaces for the adjoint system [35, Lemma 3].

Denote $(\cdot )_t:=(\cdot )(f^tx)$. Let $\varepsilon ^c$ be the covector in the center subspace such that $\varepsilon ^c (F)=1$. Let $\mathcal {L}_{(\cdot )}(\cdot )$ be the Lie derivative, and $\nabla _{(\cdot )}(\cdot )$ be the Riemannian covariant derivative. Define the derivative of vector field F along a covector $\nu$ as

$$ \nabla _\nu F:= \nabla _F\nu -\mathcal {L}_F\nu , $$

so $\nabla _\nu F = -(\nabla F)^T \nu$ when $\mathcal {M}=\mathbb {R}^M$ and $\nu$ is denoted as a column vector [35, Section 6.1].

We first define the tangent shadowing operator. Denote the space of Holder vector and covector fields by $\mathfrak {X}^\alpha$ and $\mathfrak {X}^{*\alpha }$; define the quotient space

$$ A(K):=\{ (v,\eta ): v\in \mathfrak {X}^\alpha (K), \nabla _F v\in \mathfrak {X}^\alpha (K), \eta \in C^\alpha (K) \} \, / \sim . $$

where the equivalent relation $\sim$ is defined as

$$ (v_1,\eta _1)\sim (v_2,\eta _2) \quad \text {iff} \quad \mathcal {L}_F v_1 +\eta _1 F = \mathcal {L}_F v_2 +\eta _2 F. $$

We define the tangent shadowing operator $S:\mathfrak {X}^\alpha (K)\rightarrow A(K)$ as

$$ S: X \mapsto [v, \eta ], \quad \text {where} \quad \mathcal {L}_F v + \eta F = X . $$

Here $[v,\eta ]$ is the equivalent class of $(v,\eta )$ according to $\sim$. Roughly speaking, v is the location difference and $\eta$ is the time rescaling between two shadowing orbits. A particular v and $\eta$ can be given by the expression:

$$ \eta F=X^c, v=\int _0^\infty f_*^tX^s_{-t}dt-\int _{-\infty }^0f^t_*X^u_{-t}dt. $$

Then we define $\mathcal {A}$, the dual space of A. Let $F(\psi )$ be the derivative of $\psi$ along F; define the space of pairs of a covector field and a scalar function,

$$ \mathcal {A}(K):=\{(\omega , \psi ) \,|\, \omega \in \mathfrak {X}^{*\alpha }, \psi \in C^\alpha , F(\psi ) \in C^\alpha (K), F(\psi ) = \omega (F)\}. $$

Then we can show that, for $(\omega , \psi )\in \mathcal {A}(K)$, is well-defined, where

Hence we can well-define the shadowing contribution SC as

All these definitions can all be made pathwise.

Theorem 3

(Adjoint shadowing lemma for continuous-time [35]) On a compact mixing axiom A attractor with SRB measure $\rho$ and exponential mixing, the adjoint shadowing operator $\mathcal {S}: \mathcal {A}(K) \rightarrow \mathfrak {X}^{*}(K)$ is equivalently defined by the following characterizations:

• (1) $\mathcal {S}$ is the linear operator such that

  

  Hence, if $X=\delta f$, $\nu =\mathcal {S}(d\Phi , \Phi -\rho (\Phi ))$, then the shadowing contribution is

  

• (2) $\mathcal {S}$ has the ‘split-propagate’ expansion formula

  $$\begin{aligned} \mathcal {S}(\omega ,\psi ) = \int _{t\ge 0} f^{*t} \omega ^s_t dt - \int _{t\le 0} f^{*t} \omega ^u_t dt - \psi \varepsilon ^c, \end{aligned}$$

• (3) The shadowing covector $\nu = \mathcal {S}(\omega ,\psi )$ is the unique solution of the inhomogeneous adjoint ODE,

  $$ \nabla _F \nu - \nabla _\nu F = \mathcal {L}_F\nu = - \omega \; \text { on } K, \quad \quad \nu F (x) = - \psi (x) \; \text { at all or any } x\in K. $$

Moreover, $\mathcal {S}$ preserves Holder continuity.

Characterization (b) is the most powerful characterization for proving all properties of $\mathcal {S}$. In particular, the ‘split-propagate’ scheme is a very common trick in hyperbolic systems to keep boundedness when pushing vectors or pulling covectors. As we shall see, (b) appears in the unstable divergence, so we can apply the adjoint shadowing lemma. On the other hands, characterization (a) shows that SC can be expressed by $\mathcal {S}$, and (c) is useful for numerically computing $\mathcal {S}$ by the nonintrusive shadowing algorithm [32, 38, 40, 41].

3 Equivariant Divergence Formula in Continuous-Time

We derive a formula for the unstable contribution such that it can be sampled by 2u recursive relations on one orbit.

3.1 Definitions and Geometric Notations

The unstable contribution is defined as the other part of the linear response, and in [35] we showed that

$$\begin{aligned} UC := \delta \rho (\Phi ) - SC = \int _{-\infty }^\infty \rho \left( X^u(\Phi _t) \right) dt = \int _{-\infty }^\infty \rho \left( X^{cu}(\Phi _t) \right) dt. \end{aligned}$$

(2)

where $X^{cu}=X^c+X^u$. Without loss of generality, we can assume that $X^s \ne 0$ and $X^{cu} \ne 0$. We use $\mathcal {V}^{cu}(x)$ to denote the center-unstable manifold containing x, which is a $C^3$ immersed submanifold under our assumption, and $\mathcal {V}^{cu}_r(x)$ to denote the local center-unstable manifold of size r, which is a $C^3$ embedded submanifold. Similarly, by the classical invariant manifolds theory, the stable manifolds $\mathcal {V}^s(x)$, $\mathcal {V}^s_r(x)$ and the unstable manifolds $\mathcal {V}^u(x)$, $\mathcal {V}^u_r(x)$ are defined in the same way.

Define the Hessian of a map, $\nabla f_*$, by the Leibniz rule: for vector fields X, Y, Z,

$$ (\nabla _{X} f_*) Y := \nabla _{f_* X } (f_*Y) - f_*\nabla _{ X} Y. $$

Denote the Hessian of vector fields by $\nabla ^2$, so

$$ \nabla _X \nabla _Y Z = \nabla _{ \nabla _X Y} Z + \nabla ^2_{X,Y} Z. $$

We can verify that the Hessians are symmetric,

$$ (\nabla _{X} f_*) Y = (\nabla _{Y} f_*) X, \quad \quad \nabla ^2_{X,Y} Z = \nabla ^2_{Y,X} Z. $$

Denote a u-vector, which is basically a u-dimensional cube, by $e:=e_1\wedge \cdots \wedge e_u$. Recall that

$$ \mathcal {L}_Xe := \sum _{i=1}^u e_1\wedge \cdots \wedge \mathcal {L}_{X} e_i \wedge \cdots \wedge e_u. $$

Then we define some notations for derivatives, so that one slot in the derivative can take a cube, yet the rules still look like vectors.

$$\begin{aligned} \nabla _Xe := \sum _{i=1}^u e_1\wedge \cdots \wedge \nabla _{X} e_i \wedge \cdots \wedge e_u , \quad \quad \nabla _eX := \sum _{i=1}^u e_1\wedge \cdots \wedge \nabla _{e_i}X \wedge \cdots \wedge e_u . \end{aligned}$$

(3)

So we still have $\mathcal {L}_Xe = \nabla _Xe-\nabla _eX$. Also, the Hessian of a map is

$$\begin{aligned} (\nabla _{X} f_*) e&:= \nabla _{f_* X } (f_*e) - f_*\nabla _{ X} e = \sum _{i=1}^u f_*e_1\wedge \cdots \wedge (\nabla _{X} f_*) e_i \wedge \cdots \wedge f_*e_u \nonumber \\&= \sum _{i=1}^u f_*e_1\wedge \cdots \wedge (\nabla _{e_i} f_*) X \wedge \cdots \wedge f_* e_u =: (\nabla _{e} f_*) X . \end{aligned}$$

(4)

Then we define Hessian of a vector field, where one slot takes a cube. First recall that for 1-vector fields X, Y, and Z, we have

$$ \nabla ^2_{X,Y} Z:= \nabla _X\nabla _Y Z - \nabla _{\nabla _XY}Z. $$

Recursively apply Eq. (3), also note that the wedge product is anti-symmetric, so interchanging two entries in a wedge product changes the sign, we have

$$\begin{aligned} \nabla ^2_{X,Y} e&:= \nabla _X\nabla _Y e - \nabla _{\nabla _XY}e = \sum _{i\ne j}\cdots \nabla _Xe_i\cdots \nabla _Ye_j\cdots + \sum _i \cdots \nabla _X\nabla _Ye_i\cdots \nonumber \\ &\quad - \sum _i \cdots \nabla _{\nabla _XY}e_i\cdots = \sum _i \cdots (\nabla _X\nabla _Ye_i -\nabla _{\nabla _XY}e_i) \cdots \\ &= \sum _i e_1\wedge \cdots \wedge \nabla ^2_{X,Y}e_i \wedge \cdots \wedge e_u = \nabla ^2_{Y,X} e . \end{aligned}$$

$$\begin{aligned} \nabla ^2_{X,e} Y&:= \nabla _X\nabla _e Y - \nabla _{\nabla _Xe}Y = \sum _{i\ne j}\cdots \nabla _X e_j \cdots \nabla _{e_i} Y\cdots + \sum _i \cdots \nabla _X\nabla _{e_i}Y \cdots \nonumber \\&\quad - \sum _{i\ne j}\cdots \nabla _Xe_i\cdots \nabla _{e_j}Y \cdots - \sum _i \cdots \nabla _{\nabla _X e_i} Y \cdots = \sum _i e_1 \wedge \cdots \wedge \nabla ^2_{X,e_i} Y \wedge \cdots \wedge e_u . \end{aligned}$$

$$\begin{aligned} \nabla ^2_{e,X} Y&:= \nabla _e\nabla _X Y - \nabla _{\nabla _e X}Y = \sum _i \cdots \nabla {e_j} \nabla _{X} Y\cdots - \sum _i \nabla _{\cdots \nabla _{e_i} X \cdots } Y \nonumber \\&= \sum _i \cdots \nabla {e_j} \nabla _{X} Y\cdots - \sum _{i\ne j} \cdots \nabla _{e_i} X \cdots \nabla _{e_j} Y \cdots - \sum _i \cdots \nabla _{\nabla _{e_i} X } Y \cdots \nonumber \\&= \sum _i \cdots \nabla {e_j} \nabla _{X} Y - \nabla _{\nabla _{e_i} X } Y\cdots = \sum _i \cdots \nabla ^2_{e_i,X} Y \cdots = \nabla ^2_{X, e} Y. \end{aligned}$$

We would expect that the Hessian of a map of a flow is related to the Hessian of the vector field that generates the flow: this is confirmed by the next lemma.

Lemma 4

(Expression of $\nabla f_*$ by $\nabla ^2 F$) Let $f^t$ be the flow generated by the vector field F, for any x, any $X\in T_x\mathcal {M}$, any vector field e defined at x and differentiable along X,

$$ (\nabla _X f_*^t) e := \nabla _{f_*^t X} f_*^t e - f_*^t \nabla _Xe = \int _0^t f_*^{t-\tau } \nabla ^2_{f_*^\tau X,f_*^\tau e} F_\tau d\tau . $$

Proof

Just check that both expressions solve the same ODE,

$$ (\mathcal {L}_F v)_t = \nabla ^2_{f_*^t X,f_*^t e} F_t, \quad \quad v_0 = 0. $$

Let $v_t = \nabla _{f_*^t X} f_*^t e - f_*^t \nabla _Xe$ be the left side of the expression, then the Lie derivative of the second term along F is zero, so

$$\begin{aligned} \mathcal {L}_{F_t} v_t&= \mathcal {L}_{F_t} \nabla _{f_*^t X} f_*^t e = \nabla _{F_t} \nabla _{f_*^t X} f_*^t e - \nabla _{ \nabla _{f_*^t X} f_*^t e} F_t \\&= \nabla ^2_{F_t, f_*^t X} f_*^t e + \nabla _{\nabla _{F_t} f_*^t X} f_*^t e - \nabla _{ \nabla _{f_*^t X} f_*^t e} F_t . \end{aligned}$$

Since f is the flow of F, by definition of Lie derivative, $\mathcal {L}_{ F_t}f_*^t X = 0$, hence $\nabla _{ f_*^t X} F_t = \nabla _{ F_t}f_*^t X$, and

$$ \mathcal {L}_{F_t} v_t = \nabla ^2_{F_t, f_*^t X} f_*^t e + \nabla _{\nabla _{ f_*^t X} F_t} f_*^t e - \nabla _{ \nabla _{f_*^t X} f_*^t e} F_t = \nabla _{f_*^t X} \nabla _{F_t} f_*^t e - \nabla _{ \nabla _{f_*^t X} f_*^t e} F_t $$

Similar as above, $\nabla _{F_t} f_*^t e =\nabla _{f_*^t e}F_t$, so

$$ \mathcal {L}_{F_t} v_t = \nabla _{f_*^t X} \nabla _{f_*^t e} F_t - \nabla _{ \nabla _{f_*^t X} f_*^t e} F_t = \nabla ^2_{f_*^t X, f_*^t e} F_t $$

Then let $v_t$ be the right side, $\int _0^t f_*^{t-s} \nabla ^2_{f_*^s X,f_*^s e} F_s ds$, to verify the ODE. $\square$

The rules above are independent of the choice of e. Later in this paper, we mostly use e to denote the unstable cube, unless otherwise noted. We also define co-cube $\varepsilon$, the unit cubes $\tilde{e}$ and $\tilde{\varepsilon }$, and cubes in the center-unstable subspace,

$$ \begin{aligned} e:=e_1\wedge \cdots \wedge e_u,\quad \tilde{e}:= e/|e|, \quad e^{cu} := e\wedge F, \quad \tilde{e}^{cu} := e^{cu}/|e^{cu}|, \quad \text { where } e^i\in V^{u}; \\ \varepsilon := \varepsilon ^1 \wedge \cdots \wedge \varepsilon ^u, \quad \tilde{\varepsilon }:= \varepsilon /\varepsilon (\tilde{e}), \quad \varepsilon ^{cu} := \varepsilon \wedge \varepsilon ^c, \quad \tilde{\varepsilon }^{cu} := \varepsilon ^{cu}/|\varepsilon ^{cu}|, \quad \text { where } \varepsilon ^i\in V^{u*}. \end{aligned} $$

Here $\varepsilon ^c\in V^{*c}$ and $\varepsilon ^c F = 1$, $|\cdot |$ is the volume induced by $\left\langle \cdot , \cdot \right\rangle$, which is the inner product between u-vectors.

Define the equivariant unstable divergence as

$$ \text {div}^v X:= \tilde{\varepsilon }\nabla _{\tilde{e}} X, $$

so $\text {div}^v X$ is a real number. Define $\text {div}^v \nabla F$, which is a covector, as

$$ \text {div}^v \nabla F(X) := \tilde{\varepsilon }\nabla ^2_{\tilde{e}, X} F. $$

Similarly, we can define the center-unstable equivariant divergence as

$$ \begin{aligned} \text {div}^{cv} F := \tilde{\varepsilon }^{cu}\nabla _{\tilde{e}^{cu}} F\\ \text {div}^{cv} \nabla F(Y) := \tilde{\varepsilon }^{cu}\nabla ^2_{Y,\tilde{e}^{cu}}F \end{aligned} $$

The derivative in this divergence hits only smooth quantities such as X and $\nabla F$, so it is pointwisely defined. The equivariant divergence is Holder continuous since e and $\varepsilon$ are Holder.

3.2 Center-Unstable Transfer Operator ${\tilde{L}}^{cu}$

We present the proof of Theorem 1 by first deriving an equivariant divergence formula using $e^{cu}$. This derivation is longer than that in the appendix, but here we only need to use textbook theorems. Compared with the discrete-time case [39], the main difficulty here is to handle the extra center direction.

First note that if $x\in K$ then $\mathcal {V}^{cu}(x)\subseteq K$. This can be deduced from considering the time-one map $f^1$ of flow F. $T_KM$ has a $f^1_*$-invariant splitting $V^s\bigoplus V^c\bigoplus V^u$ where $V^s$ is uniformly contracting, $V^u$ is uniformly expanding and $V^c$ is the flow direction. Particularly, K is a partially hyperbolic attractor of $f^1$. It is a textbook theorem that under such condition we have $\mathcal {V}^u(x) \subseteq K$ if $x\in K$, where $\mathcal {V}^u(x)$ is the unstable manifold of both $f^1$ and F. Further note that K is invariant under the flow, we can get $\mathcal {V}^{cu}(x)\subseteq K$ if $x\in K$.

The perturbation $X^{cu}$ can be written as $X-X^s$. If we change the measure on a center-unstable manifold $\mathcal {V}^{cu}$ by the flow of $X^{cu}$ for a distance of $\gamma$, up to first order of $\gamma$, this perturbation equals to changing the measure by $\xi \circ \tilde{f}$, where $\tilde{f}$ is the flow of X for a distance of $\gamma$, and $\xi$ is the holonomy map projecting along the stable manifolds onto $\mathcal {V}^{cu}$. Note that the holonomy map $\xi$ is absolutely continuous with respect to the Lebesgue measure of the transverses. More precisely, $\delta (\xi \tilde{f})=X^{cu}$, so the linear response caused by $X^{cu}$ is the same as $\delta (\xi \tilde{f})$. See Fig. 2 for pictorial explanations. The following lemma, from [39], shows that the map $\xi \tilde{f}$ is locally well-defined.

Lemma 5

($\xi \tilde{f}$ is locally well defined [39]) Given $r>0$ small enough, $\exists \gamma _0>0$ such that for any $|\gamma |<\gamma _0$ and any $x\in K$, the point $y=\xi \tilde{f}x$ uniquely exists, and $|x-y|<0.1r$. From now on, we always assume $|\gamma |<\gamma _0$.

Now we define the conditional measure density $\sigma$ and the transfer operator ${\tilde{L}}^{cu}$. First we choose $r>0$ small enough such that Lemma 5 holds and for each $x\in K$ its neighbourhood $B(x,r)\bigcap K$ has an local foliation by local unstable-center manifolds. Since the physical SRB measure $\rho$ is absolutely continuous alone the unstable-center foliation, we can define $\sigma$ as the density of the conditional measure on a local unstable-center foliation; $\sigma$ is at least $C^1$ on a unstable leaf.

Define ${\tilde{L}}^{cu}$ as the local transfer operator of $\xi \tilde{f}:\mathcal {V}^{cu} \rightarrow \mathcal {V}^{cu}$; note that $\tilde{\cdot }$ indicates dependence on $\gamma$. Let r be small enough as above; for each $x\in K$, let $P:=(\xi \tilde{f})^{-1}\mathcal {V}^{cu}_{0.1r}(x)$, we define ${\tilde{L}}^{cu}$ as the transfer operator from $L^1(P)$ density function space to $L^1(\mathcal {V}_{0.1r}^{cu}(x))$. Since $\delta (\xi \tilde{f}) = X^{cu}$, $\delta {\tilde{L}}^{cu}$ is the perturbation by $X^{cu}$. Let $y_{-\Delta T} = (\xi \tilde{f})^{-1} x$ (the subscript $\Delta T$ will be defined later), then the pointwise definition of ${\tilde{L}}^{cu}$ on any density function $\sigma$ is

$$\begin{aligned} {\tilde{L}}^{cu}\sigma (x) := \frac{\sigma }{ |\xi _* \tilde{f}_{*}| } (y_{-\Delta T}) = \frac{\sigma (y_{-\Delta T})}{ |\tilde{f}_{*}(y_{-\Delta T})| \, |\xi _*(\tilde{f}y_{-\Delta T})| }, \end{aligned}$$

(5)

and we can interpret the last equality as dissecting $X^{cu}$ into X and $-X^s$. Equivalently, fix the factor measure and the foliation of $\mathcal {V}^{cu}_{0.1r}(x)$, then ${\tilde{L}}^{cu}\sigma$ is the conditional density of the new measure obtained by $\xi _* \tilde{f}_{*}$-pushforward the old measure on B(x, r).

Then we can define $\frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x)$ and $\frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x)$. Notice that the domain of $\sigma$ includes $\mathcal {V}_{r}^{cu}(x)$, and $P:=(\xi \tilde{f})^{-1}\mathcal {V}^{cu}_{0.1r}(x)\subset \mathcal {V}^{cu}_r(x) \subset K\cap B(x,r)$, so we can define $\frac{{\tilde{L}}^{cu}\sigma }{\sigma }(x)$ on the smaller leaf $\mathcal {V}^{cu}_{0.1r}(x)$ for the particular r. Moreover, notice that both $\sigma |_{\mathcal {V}_{0.1r}^{cu}(x)}$ and $\sigma |_P$, the source of ${\tilde{L}}^{cu} \sigma |_{\mathcal {V}_{0.1r}^{cu}(x)}$, are restrictions of the same $\sigma$ from the same larger leaf $\mathcal {V}_{r}^{cu}(x)$ of the same foliation. Hence, we can expect that, in $\frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x)$, the factor due to the selection of B(x, r), would cancel. More specifically, notice that $\sigma$, ${\tilde{L}}^{cu}\sigma$ and $\delta {\tilde{L}}^{cu}\sigma$ may differ by a constant factor, depending on the choice of the neighborhood of the local foliation, so they can only be defined locally; but the ratio $\frac{{\tilde{L}}^{cu}\sigma }{\sigma }$ and $\frac{\delta {\tilde{L}}^{cu} \sigma }{\sigma }$ does not depend on that choice, since the constant factors are cancelled, so they are globally well-defined.

Fix the point of interest x and denote $x_t:=f^tx$. We define $y_t$ as the pseudo-orbit shadowing $x_t$, with a discontinuity $\tilde{f}$ at $t=0$ when $\gamma \ne 0$. More specifically,

$$ \begin{aligned} y_0:=\xi ^{-1}x, \quad y_t:=f^t y_0 \quad \text {for}\quad t\ge 0, \\ y_{-\Delta T}:=\tilde{f}^{-1}y_0, \quad y_{t}=f^{t+\Delta T}(y_{-\Delta T}) \quad \text {for}\quad t<0, \end{aligned} $$

where $\Delta T$ is such that $f^{\Delta T}(y_{-\Delta T})\in \mathcal {V}^{u}_{r}(x)$. By the invariance of stable and unstable manifolds, we have $y_t\in \mathcal {V}^s(x_t)$ for $t\ge 0$ and $y_t\in \mathcal {V}^u(x_t)$ for $t\le 0$. Note that $y_t$ depends on $(\gamma ,t)$: when $t<0$, and $(\gamma ,t)$ change slightly, $y_t$ form a small neighbourhood in a $C^3$ two-dimensional submanifold of $\mathcal {V}^{cu}(x)$; when $t>0$, $y_t$ form a small submanifold of $\mathcal {V}^{cs}(x)$. Above definitions are illustrated in Fig. 1.

Fig. 1

Pseudo-orbit $y_t$ for different $\gamma$ (Color figure online)

We briefly show that, for fixed x, $y_0$ and $y_{-\Delta T}$ are $C^3$ functions of $\gamma$ when $\gamma$ is small enough, that is, the red dashed curves in Fig. 1 are $C^3$. Choose an local coordinate chart U around x which is small enough. Recall that $\mathcal {V}^{s}_r(x)$ and $\mathcal {V}^{cu}_r(x)$ intersect transversely, and by translate $\mathcal {V}^{s}_r(x)$ parallelly in U we can get a $C^3$ smooth local pseudo-stable foliation which is transverse to $\mathcal {V}^{cu}_r(x)$ in U. We denote the local pseudo-stable holonomy map onto $U\bigcap \mathcal {V}^{cu}_r(x)$ as $\xi '$. Then by definition, the dependency of $y_{\Delta T}$ on $\gamma$ can be characterized by the equation:

$$ \xi '\tilde{f}(y_{\Delta T},\gamma )=x, \ \text {where}\ \tilde{f}(\cdot ,\gamma ) \ \text {denote the time}\ \gamma \ \text {map of flow}\ \tilde{f}. $$

Because $\tilde{f}(y_{\Delta T},\gamma )$ is always on $\mathcal {V}^s$, where $\xi$ and $\xi '$ coincide. Since $X^{cu}\ne 0$, the derivative of $\xi '\tilde{f}$ with respect to the second variable $\gamma$ is non-zero when U and $\gamma$ are small enough. Notice that both $\xi '$ and $\tilde{f}$ are $C^3$ maps (this is why we use $\xi '$ instead of $\xi$). Finally, apply the implicit function theorem to show the desired smoothness.

We define the vector field $e(y_{t})$ as

$$\begin{aligned} e(y_t):= {\left\{ \begin{array}{ll} f_*^{t+T} \tilde{e}(y_{-T}), \quad \text {if } t<0; \\ f_*^{t}\tilde{f}_*f_*^{T-\Delta T} \tilde{e}(y_{-T}), \quad \text {if } t\ge 0. \end{array}\right. } \end{aligned}$$

(6)

When we use this notation in the following sections, T is a fixed big number. Similarly we can define $e^{cu}(y_t)$ as

$$\begin{aligned} e^{cu}(y_t):= {\left\{ \begin{array}{ll} f_*^{t+T} \tilde{e}^{cu}(y_{-T}), \quad \text {if } t<0; \\ f_*^{t}\tilde{f}_*f_*^{T-\Delta T} \tilde{e}^{cu}(y_{-T}), \quad \text {if } t\ge 0. \end{array}\right. } \end{aligned}$$

(7)

When $\gamma =0$, we have $\tilde{f}=I_d,\tilde{f}_*=I_d$ and $\Delta T=0$. Denote $e^{cu}_t:= f_*^{t+T} \tilde{e}^{cu}(x_{-T})$ and $\varepsilon ^{cu}_t:=f^{-(T+t)*} \tilde{\varepsilon }^{cu}(x_{-T})$ for convenience, so $e^{cu}:=e^{cu}_0=e^{cu}(y_{-\Delta T})|_{\gamma =0}$ and $\varepsilon ^{cu}=\varepsilon ^{cu}_0$. Because the two-dimensional submanifold is $C^3$ and the center-unstable manifolds are $C^3$, the $V^{cu}(x)$ is $C^2$ when restricted to the submanifold. As a result, the cu-vector field $e^{cu}$ and the cu-covector field $\varepsilon ^{cu}$ are both differentiable alone the tangent direction of the submanifold.

Fig. 2

Measure transfer on $\mathcal {V}^{cu}$. Here $y+X\gamma$ means to start from y and flow along the direction of X for a length of $\gamma$. We use e as short of $e^{cu}$, and $|e_1| = |e_2|$. Roughly speaking, $\delta {\tilde{L}}^u \sigma /\sigma (x) = l_2/l_1=|e_7-e_6| / |e_7|$

Note that SRB is both invariant and also the pushforward of Lebesgue; it is nontrivial for the two definitions to coincide [10]. If we accept both properties of SRB, then the conditional SRB measure on $\mathcal {V}^{cu}(x)$, close to x, is obtained by pushing forward a $u+1$ dimensional Lebesgue measure on $\mathcal {V}^u(f^{-T}x)$ for large T. This gives the following volume ratio formula for $\frac{{\tilde{L}}^{cu}\sigma }{\sigma }$, which solidifies the intuition in the caption of Fig. 2.

Lemma 6

(One volume ratio) Fix x and the conditional density $\sigma$, Let ${\tilde{L}}^{cu}$ be the transfer operator of $\xi \tilde{f}$, then we have the expression

$$ \frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{|f_*^{T'+2T}\tilde{e}^{cu}(x_{-T})|}{|f_*^{T'}f_*^{T}\tilde{f}_*f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})|}. $$

Proof

First, recall that for SRB measures, for $y_{-\Delta T}\in \mathcal {V}^{u}_r(x_{-\Delta T})$, the density $\sigma$ satisfies

$$\begin{aligned} \frac{\sigma (x_{-\Delta T})}{\sigma (y_{-\Delta T})} = \lim _{t \rightarrow \infty } \frac{\left| f_{*}^{t} \tilde{e}^{cu}(f^{-t} y_{-\Delta T}) \right| }{\left| f_{*}^{t} \tilde{e}^{cu}(f^{-t} x_{-\Delta T}) \right| }. \end{aligned}$$

(8)

This expression was stated and proved for example in [47, Proposition 1] using the notation of unstable Jacobians. We give this an intuitive explanation by considering how the Lebesgue measure on $\mathcal {V}^{cu}(x_{-t})$ is evolved. The mass contained in the cube $\tilde{e}^{cu}_{-t}$ is preserved via pushforwards, but the volume increased to $f_*^t \tilde{e}^{cu}_{-t}$. Hence,

$$ \frac{\sigma (x_{-\Delta T})}{\sigma (y_{-\Delta T})} =\frac{\left| f_{*}^{t} \tilde{e}^{cu}(f^{-t} y_{-\Delta T}) \right| }{\left| f_{*}^{t} \tilde{e}^{cu}(f^{-t} x_{-\Delta T}) \right| } \cdot \frac{\sigma (f^{-t}x_{-\Delta T})}{\sigma (f^{-t}y_{-\Delta T})} $$

Note that the conditional measure is determined up to a constant coefficient, which is canceled in the ratio. Since $\lim _{t \rightarrow \infty } d(f^{-t}x_{-\Delta T},f^{-t}y_{-\Delta T})=0$, $\sigma$ is positive and continuous and K is compact, so $\lim _{t \rightarrow \infty } \frac{\sigma (f^{-t}x_{-\Delta T})}{\sigma (f^{-t}y_{-\Delta T})}=1$ as $t\rightarrow \infty$, yielding Eq. (8).

Then we consider the ratio $\frac{\sigma (x)}{\sigma (x_{-\Delta T})}$, where $x_{-\Delta T}:=f^{-\Delta T}x$. Note that here the difference between x and $x_{-t}$ is along the center direction. In this case, due to the invariance of the SRB measure, we have

$$ \frac{\sigma (x)}{\sigma (x_{-\Delta T})} =\frac{|f^{\Delta T}_*e^{cu}(x_{-\Delta T})|}{|e^{cu}(x_{-\Delta T})|} =:|f^{\Delta T}_*(x_{-\Delta T})|, $$

for any $e^{cu}(x_{-\Delta T})$. Combining with Eq. (8), we get

$$ \frac{\sigma (y_{-\Delta T})}{\sigma (x)} = \frac{\sigma (x_{-\Delta T})}{\sigma (x)} \cdot \frac{\sigma (y_{-\Delta T})}{\sigma (x_{-\Delta T})} = \lim _{T\rightarrow \infty } \frac{|f_*^T\tilde{e}^{cu}(x_{-T})|}{|f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})| }. $$

Substituting into Eq. (5), we get

$$ \frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \frac{|f_*^T \tilde{e}^{cu}(x_{-T})|}{|f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})| } \frac{1}{|\tilde{f}_{*}(y_{-\Delta T})| \, |\xi _*(\tilde{f}(y_{-\Delta T}))|} . $$

Here $|\tilde{f}_{*}(y_{-\Delta T})|:= \frac{|\tilde{f}_* e^{cu}(y_{-\Delta T})|}{|e^{cu}(y_{-\Delta T})|}$, $|\xi _{*}(\tilde{f}(y_{-\Delta T}))|:= \frac{|\xi _*\tilde{f}_* e^{cu}(y_{-\Delta T})|}{|\tilde{f}_*e^{cu}(y_{-\Delta T})|}$, where $\tilde{f}_* e^{cu}(y_{-\Delta T})$ is a vector at $y_0=\tilde{f}y$, $\xi _*\tilde{f}_* e^{cu}(y_{-\Delta T})$ is a vector at x.

Then we give an expression of $|\xi _*|$. By a corollary of the absolute continuity of the holonomy map [9, Theorem 4.4.1],

$$\begin{aligned} \lim _{T'\rightarrow \infty } \frac{|f^{T'}_* \xi _*\tilde{f}_* e^{cu}(y_{-\Delta T}) |}{ |f_*^{T'} \tilde{f}_* e^{cu}(y_{-\Delta T})|} = 1. \end{aligned}$$

(9)

Hence,

$$\begin{aligned} \frac{1}{|\xi _*(\tilde{f}y_{-\Delta T})|} := \frac{|\tilde{f}_* e^{cu}(y_{-\Delta T})|}{|\xi _*\tilde{f}_* e^{cu}(y_{-\Delta T})|} = \lim _{T'\rightarrow \infty } \frac{|\tilde{f}_* e^{cu}(y_{-\Delta T})|}{ |f_*^{T'} \tilde{f}_* e^{cu}(y_{-\Delta T})|} \frac{|f^{T'}_* \xi _*\tilde{f}_* e^{cu}(y_{-\Delta T}) |}{|\xi _* \tilde{f}_* e^{cu}(y_{-\Delta T})|} . \end{aligned}$$

(10)

By substitution and cancellation,

$$ \frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty }\lim _{T'\rightarrow \infty } \frac{|f_*^Te^{cu}(x_{-T})|}{|f_*^{T-\Delta T}e^{cu}(y_{-T})| } \frac{|e^{cu}(y_{-\Delta T})|}{|f_*^{T'} \tilde{f}_*e^{cu}(y_{-\Delta T})|} \frac{|f_*^{T'}\xi _*\tilde{f}_* e^{cu}(y_{-\Delta T})|}{|\xi _*\tilde{f}_* e^{cu}(y_{-\Delta T})|} $$

To simplify the above expression, notice that both $f_*^T e^{cu}(x_{-T})$ and $\xi _*\tilde{f}_*e^{cu}(y_{-\Delta T})$ are in the one-dimensional subspace $\wedge ^{cu} V^{cu}(x)$, so the growth rate of their volumes are the same when pushing forward by $f_*$, hence

$$ \frac{|f^{T'}_* \xi _*\tilde{f}_* e^{cu}(y_{-\Delta T}) |}{|\xi _* \tilde{f}_* e^{cu}(y_{-\Delta T})|} = \frac{|f^{T'+T}_* e^{cu}(x_{-T}) |}{| f_*^T e^{cu}(x_{-T})|}. $$

Similarly,

$$ \frac{|e^{cu}(y_{-\Delta T})|}{|f_*^{T'} \tilde{f}_*e^{cu}(y_{-\Delta T})|} = \frac{|f_*^{T-\Delta T} e^{cu}(y_{-T})|}{|f_*^{T'} \tilde{f}_* f_*^{T-\Delta T} e^{cu}(y_{-T})|}. $$

Finally, by substitution and cancellation, we have

$$ \frac{{\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{|f_*^{T'+T}\tilde{e}^{cu}(x_{-T})|}{|f_*^{T'}\tilde{f}_*f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})|}. $$

For later convenience, we pass $T'$ to $T+T'$, yielding the expression in the statement. $\square$

3.3 Equivariant Divergence by $e^{cu}$

Then we take derivative of the volume-ratio formula. There is a formula by integration by parts of (2), which is

$$ \delta L^{cu}\sigma = - \text {div}^{cu} (\sigma X^{cu}), $$

where $\text {div}^{cu}$ is the submanifold divergence on $\mathcal {V}^{cu}$. The main issue is that $X^{cu}$ is not differentiable, not even in the unstable direction. This is because that $X^{cu}$ is an oblique projection which depends on both $V^u$ and $V^s$, and $V^s$ is not differentiable along unstable directions. We need more work to find a pointwisely defined formula, then we can obtain an ‘ergodic theorem’ for linear response, which can be sampled by progressively computing a few recursive relations on an orbit.

Roughly speaking, in Fig. 2, ${\delta {\tilde{L}}^{cu}\sigma }/{\sigma } = |e_7-e_6| / |e_7|$, where $e_7-e_6=\nabla _{f_*^TX^s} e_T$. Since $f_*^{T'}\tilde{f}_*f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})$ is almost in the unstable subspace after pushing forward for a long time, the difference $e_7-e_6$ is also in the unstable, so we can get the norm by applying the unstable co-cube. More specifically,

Lemma 7

(One volume ratio for $\delta {\tilde{L}}$) Using the vector field defined in Eq. (7),

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \varepsilon ^{cu}_T \nabla _{f_*^TX^s} e^{cu}(y_T). $$

Proof

Differentiate Lemma 6 (this formal differentiation will be justified by its uniform convergence), treat $e^{cu}(y_T) = f_*^{T}\tilde{f}_*f_*^{T-\Delta T}\tilde{e}^{cu}(y_{-T})$ as a whole when applying the Leibniz rule, and notice that ${e^{cu}(y_{T+T'})|_{\gamma =0}}= e^{cu}_{T+T'}$, we get

$$\begin{aligned} - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x)&= \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }-\delta \frac{|e^{cu}_{T+T'}|}{|e^{cu}(y_{T'+T})|}\\&=\lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{|e^{cu}_{T+T'}|\cdot (\delta |e^{cu}(y_{T+T'})|)}{|e^{cu}_{T'+T}|^2}\\&=\lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{|e^{cu}_{T+T'}|\cdot \delta \left\langle e^{cu}(y_{T+T'}),e^{cu}(y_{T+T'}) \right\rangle ^{\frac{1}{2}}}{|e^{cu}_{T'+T}|^2} \\&=\lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{|e^{cu}_{T+T'}|\cdot \frac{1}{2\left\langle e^{cu}_{T+T'},e^{cu}_{T+T'} \right\rangle } 2 \left\langle \delta e^{cu}(y_{T+T'}),e^{cu}_{T+T'} \right\rangle }{|e^{cu}_{T'+T}|^2} \\&= \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\frac{\left\langle e^{cu}_{T+T'},\delta e^{cu}(y_{T'+T}) \right\rangle }{|e^{cu}_{T'+T}|^2}\\ \end{aligned}$$

For the forth equality to hold, here $\delta e^{cu}(y_{T+T'})=\nabla _{f_*^{T+T'}X^s}e^{cu}_{y_{T+T'}}$ should be the Riemannian derivative. Hence,

$$\begin{aligned} - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x)&= \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\frac{\left\langle e^{cu}_{T+T'},\delta e^{cu}(y_{T'+T}) \right\rangle }{|e^{cu}_{T'+T}|^2}\\&= \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\frac{\left\langle e^{cu}_{T+T'},\nabla _{f_*^{T+T'}X^s}e^{cu}_{T'+T} \right\rangle }{|e^{cu}_{T'+T}|^2}\\&= \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \frac{\left\langle e^{cu}_{T'+T}, f_*^{T'}\nabla _{f_*^TX^s} e^{cu}_T + (\nabla _{f_*^TX^s} f_*^{T'}) e^{cu}_T \right\rangle }{|e^{cu}_{T'+T}|^2} \end{aligned}$$

For the first term in the product, the so-called ‘relative decay’ lemma in the appendix of [39] shows the uniform convergence of

$$ \lim _{T'\rightarrow \infty } \frac{\left\langle e^{cu}_{T'+T}, f_*^{T'}\nabla _{f_*^TX^s} e^{cu}_T \right\rangle }{|e^{cu}_{T'+T}|^2} = \frac{1}{|e^{cu}_{T}|} \tilde{\varepsilon }^{cu}_T\nabla _{f_*^TX^s} e^{cu}_T = \varepsilon ^{cu}_T \nabla _{f_*^TX^s} e^{cu}_T. $$

Intuitively, this lemma says that only the center-unstable part of $\nabla _{f_*^TX^s} e^{cu}_T$ grows at the same speed as $e^{cu}_{T'+T}$, and all other parts are relatively negligible.

For the other term, first use Lemma 4, then notice that the growth rate of any cu-vector is bounded by the unstable cu-vector by definition

$$\begin{aligned}&\lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\frac{\left\langle e^{cu}_{T'+T}, \big (\nabla _{f_*^TX^s} f_*^{T'}\big ) e^{cu}_T \right\rangle }{\big |e^{cu}_{T'+T}\big |^2} \\&\quad =\lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty } \left\langle \tilde{e}^{cu}_{T'+T}, \int _0^{T'} \frac{1}{ \big |f_*^{T'-t}\tilde{e}^{cu}_{T+t}\big | }f_*^{T'-t} \nabla ^2_{f_*^{T+t} X^s, \tilde{e}^{cu}_{T+t}} F_{T+t} dt \right\rangle \\&\quad \le \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\int _0^{T'} C \Big |\nabla ^2_{f_*^{T+t} X^s, \tilde{e}^{cu}_{T+t}} F_{T+t}\Big | dt \le \lim _{T\rightarrow \infty } \lim _{T'\rightarrow \infty }\int _0^{T'} C \lambda ^{t+T} dt \\&\quad \le \lim _{T\rightarrow \infty }\int _0^{\infty } C \lambda ^{t+T} dt \le \lim _{T\rightarrow \infty } C \lambda ^{T} = 0 \end{aligned}$$

where the values of the C’s above may change from expression to expression. $\square$

Lemma 8

(Expression of $\delta {\tilde{L}}$ by $\nabla f_*$)

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \varepsilon ^{cu} \nabla _{e^{cu}} X -\eta \varepsilon ^{cu}\nabla _{ e^{cu}} F - \varepsilon ^{cu} \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) e^{cu}_{-T} + \varepsilon ^{cu}_T \big (\nabla _{X^s} f_*^{T}\big ) e^{cu} . $$

Proof

Apply the Leibniz rule on $\nabla _{f_*^TX^s}e^{cu}(y_T) = \nabla _{f_*^TX^s}(f_*^{T}\tilde{f}_* e^{cu}(y_{-\Delta T}))$, and take value $\gamma =0$ we get

$$\begin{aligned} - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x)&= \lim _{T\rightarrow \infty } \varepsilon ^{cu}_T \left( \big (\delta f_*^{T}\big ) e^{cu} + f_*^{T}\big (\delta \tilde{f}_*\big ) e^{cu} + f_*^{T} \nabla _{-X^{cu}} e^{cu}(y_{-\Delta T}) \right) \nonumber \\&= \lim _{T\rightarrow \infty } \varepsilon ^{cu}_T \big (\delta f_*^{T}\big ) e^{cu} + \varepsilon ^{cu} \big (\delta \tilde{f}_*\big ) e^{cu} + \varepsilon ^{cu} \nabla _{-X^{cu}} e^{cu}(y_{-\Delta T}) \end{aligned}$$

(11)

This equation can be intuitively explained in Fig. 2 as:

$$\begin{aligned} \gamma \nabla _{f_*^TX^s}\big (f_*^{T}\tilde{f}_* e^{cu}\big (y_{-\Delta T}\big )\big )&\approx e_6-e_7 = \big [(e_6-e_7)-f_*^T(x)(e_4-e_5)\big ] \\ &\quad + \big [f_*^T(x)(e_4-e_3)\big ] + [f_*^T(x)(e_3-e_5)], \end{aligned}$$

if we multiply $\varepsilon ^{cu}$ on both sides, then the three terms on the right side correspond to the three terms in Eq. (11).

For the first term in Eq. (11), just use

$$ \big (\delta f_*^{T}\big ) e^{cu} := \big (\nabla _{X^s} f_*^{T}\big ) e^{cu} $$

For $(\delta \tilde{f}_*) e^{cu}$ in the second term in (11),

$$ \big (\delta \tilde{f}_*\big ) e^{cu} :=\left( \left. \frac{d }{d \gamma }\big (\tilde{f}_*(\gamma ,y_{-\Delta T}\big )e^{cu}(y_{-\Delta T})) -\tilde{f}_*(\gamma ,y_{-\Delta T})\frac{d }{d \gamma }e^{cu}(y_{-\Delta T}) \right) \right| _{\gamma =0} $$

The first term equals the sum of partial derivatives with respect to $\gamma$ and $y_{-\Delta T}$, where $y_{-\Delta T}$ further depends on $\gamma$. Notice that $\tilde{f}_*(\gamma =0,y_{-\Delta T})=Id$ and $(y_{-\Delta T})|_{\gamma =0}=x$, so

$$\begin{aligned} \begin{aligned} \big (\delta \tilde{f}_*\big ) e^{cu}&=\left( \frac{d }{d \gamma }\big (\tilde{f}_*(\gamma ,x)e^{cu}(x)\big ) +\frac{d }{d \gamma }\big (\tilde{f}_*(0,y_{-\Delta T})e^{cu}(y_{-\Delta T})\big )\right. \\&\quad \left. \left. -\tilde{f}_*(\gamma ,y_{-\Delta T})\frac{d }{d \gamma }e^{cu}(y_{-\Delta T}) \right) \right| _{\gamma =0}, \end{aligned} \end{aligned}$$

and the last two terms cancel. Since $\tilde{f}$ is the flow of X, we can use the Lie bracket statement $\mathcal {L}_X (\tilde{f}_* e^{cu})=0$, note that $e^{cu}$ is a fixed vector at x, to get

$$ \big (\delta \tilde{f}_*\big ) e^{cu} =\left( \left. \frac{d }{d \gamma }\big (\tilde{f}_*(\gamma ,x)e^{cu}(x)\big ) \right) \right| _{\gamma =0} =\nabla _{X}(\tilde{f}_*e^{cu}) =\nabla _{e^{cu}} X . $$

For the last term in Eq. (11), $\nabla _{X^{cu}} e^{cu}(y_{-\Delta T}) = \nabla _{X^{c}} e^{cu} + \nabla _{X^{u}} e^{cu}$. Then, $\nabla _{X^{c}} e^{cu} = \eta \nabla _F e^{cu} = \eta \nabla _{ e^{cu}} F$, where $\eta :=\tilde{\varepsilon }^c X$. This is because by definition, $e^{cu}_{t}$ is generated by the push-forward of the flow F when $t\le 0$, so we have $\mathcal {L}_Fe^{cu}=\nabla _Fe^{cu}-\nabla _{e^{cu}}F=0$. Also, $\nabla _{X^{u}} e^{cu} = f_*^T \nabla _{f_*^{-T}X^{u}} e^{cu}_{-T} + ( \nabla _{f_*^{-T}X^{u}} f_*^T) e^{cu}_{-T}$ by the definition of $\nabla _{f_*^{-T}X^{u}} f_*^T$. Since $f_*^{-T}X^{u} \rightarrow 0$,

$$ \lim _{T\rightarrow \infty } \varepsilon ^{cu} f_*^{T}\nabla _{f_*^{-T}X^{u}} e^{cu}_{-T} = \lim _{T\rightarrow \infty } \varepsilon ^{cu}_{-T} \nabla _{f_*^{-T}X^{u}} e^{cu}_{-T} =0, $$

Hence,

$$ \lim _{T\rightarrow \infty } \varepsilon ^{cu} \nabla _{X^{cu}} e^{cu} = \lim _{T\rightarrow \infty } \varepsilon ^{cu} \left( \eta \nabla _{ e^{cu}} F + \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) e^{cu}_{-T}. \right) $$

To summarize, we transformed (11) to the expression in the lemma. $\square$

Theorem 9

(Center-unstable equivariant divergence formula)

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x) = \text {div}^{cv} X + \mathcal {S}(\text {div}^{cv} \nabla F, \text {div}^{cv} F) X . $$

Proof

Applying Lemma 4, $\nabla f_*^T$ in the last two terms of Lemma 8 become

$$\begin{aligned} - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x)&= \lim _{T\rightarrow \infty } \Bigg (\varepsilon ^{cu} \nabla _{e^{cu}} X -\eta \varepsilon ^{cu}\nabla _{ e^{cu}} F \\&\quad - \varepsilon ^{cu} \int _0^T f_*^{T-t} \nabla ^2_{f_*^{t-T} X^{u}, e^{cu}_{t-T}} F_{t-T} dt \\&\quad + \varepsilon ^{cu}_T \int _0^T f_*^{T-t} \nabla ^2_{f_*^t X^s, e^{cu}_t } F_t dt \Bigg ) \\&= \tilde{\varepsilon }^{cu} \nabla _{\tilde{e}^{cu}} X -\eta \tilde{\varepsilon }^{cu}\nabla _{\tilde{e}^{cu}} F - \int _{-\infty }^0 \tilde{\varepsilon }^{cu}_{t} \nabla ^2_{f_*^{t} X^{u}, \tilde{e}^{cu}_{t}} F_{t} dt \\&\quad + \int _0^\infty \tilde{\varepsilon }^{cu}_{t} \nabla ^2_{f_*^t X^s, \tilde{e}^{cu}_t } F_t dt . \end{aligned}$$

Denote covector field $\omega$ and function $\psi$ on K such that

$$ \omega (Y):=\text {div}^{cv} \nabla F(Y) := \tilde{\varepsilon }^{cu}\nabla ^2_{Y,\tilde{e}^{cu}}F, \quad \quad \psi := \text {div}^{cv} F := \tilde{\varepsilon }^{cu}\nabla _{\tilde{e}^{cu}} F. $$

Then we can get the ‘split-propagate’ scheme in characterization (b) of the adjoint shadowing operator in Theorem 3,

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x) = \text {div}^{cv} X + \left( - \psi \varepsilon ^c - \int _{-\infty }^0 f_*^{t}\omega ^u_{t} dt + \int _0^\infty f_*^t \omega ^s_{t} dt \right) X . $$

To check $(\omega ,\psi ) \in \mathcal {A}$, once again notice that $\nabla _Fe^{cu} = \nabla _{e^{cu}} F$, and $\varepsilon ^{cu}_t$ is invariant under the push-forward of the flow F, we have

$$\begin{aligned} F(\psi )&= \mathcal {L}_F(\tilde{\varepsilon }^{cu}\nabla _{\tilde{e}^{cu}} F) = \varepsilon ^{cu} \mathcal {L}_F(\nabla _{e^{cu}} F) = \varepsilon ^{cu} (\nabla _F(\nabla _{e^{cu}} F) - \nabla _{\nabla _{e^{cu}} F}F) \nonumber \\&= \varepsilon ^{cu} \big (\nabla ^2_{F,e^{cu}}F + \nabla _{\nabla _Fe^{cu}}F - \nabla _{\nabla _{e^{cu}} F}F\big ) = \varepsilon ^{cu} \nabla ^2_{F,e^{cu}}F = \omega (F). \end{aligned}$$

(12)

Hence, the adjoint shadowing lemma applies and the theorem is proved. $\square$

3.4 Removing Center Direction from $e^{cu}$

This subsection seeks to further simplify the equivariant divergence formula by using the unstable cube e instead of the center-unstable cube $e^{cu}$. This formula might be beneficial for potential numerical applications.

Theorem 1

(Equivariant divergence in continuous-time) Let $\eta : = \varepsilon ^c(X)$, then

$$ - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }= \text {div}^{cu}_\sigma X^{cu} = \mathcal {S}(\text {div}^v \nabla F, \text {div}^v F) X + \text {div}^v X + F(\eta ). $$

holds $\rho$ almost everywhere.

Remark

(1) This formula is given by u many recursive relations on an orbit, which should be optimal. (2) $\rho (F(\eta ))=0$. To see this, choose an $x\in K$ such that $\lim \limits _{T\rightarrow \infty } \frac{1}{T}\int _0^T \Phi (f^t(x)) dt= \rho (\Phi )$ for every continuous function $\Phi$. Take $\Phi =F(\eta )$ and we have

$$ \rho (F(\eta )) = \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^TF(\eta )(f^t(x)) dt = \lim _{T\rightarrow \infty }\frac{1}{T}\big (\eta \big (f^Tx\big )-\eta (x)\big ) = 0. $$

Proof

We start from the formula in Lemma 8. In this proof we let $\tilde{e}^{cu}:=\tilde{e}\wedge F$, $\tilde{\varepsilon }^{cu}:=\tilde{\varepsilon }\wedge \varepsilon ^c$, where $\varepsilon ^c \in V^{*c}$ and $\varepsilon ^c(F)=1$. Then we define $e^{cu}_t$ and $\varepsilon ^{cu}_t$ same as Eq. (7); $e_t$ and $\varepsilon _t$ are the same as Eq. (6). Note that F and $\varepsilon ^c$ are invariant, or equivariant, under pushforwards. Now $|\tilde{e}^{cu}|\ne 1$ but still $\varepsilon ^{cu}_t(e^{cu})_t=\tilde{\varepsilon }^{cu}_t(\tilde{e}^{cu}_t)=1$. Notice that $e^{cu}$ and $\varepsilon ^{cu}$ always appear in pairs and the constant factors cancel, so the formula in Lemma 8 still holds with the new $\varepsilon ^{cu}$ and $e^{cu}$.

Since $F\perp V^{*u}$, $\varepsilon ^c\perp V^u$,

$$\begin{aligned}&\varepsilon ^{cu} \nabla _{e^{cu}} X \ = \varepsilon \wedge \varepsilon ^c(\nabla _{e\wedge F}X) \\&= \varepsilon \wedge \varepsilon ^c(e\wedge \nabla _FX ) + \sum _{i=1}^u \varepsilon \wedge \varepsilon ^c (e_1\wedge \cdots \wedge \nabla _{e_i} X \wedge \cdots \wedge e_u\wedge F) \\&= \varepsilon ^c\nabla _FX + \varepsilon \nabla _{e} X = \varepsilon ^c\nabla _FX + \text {div}^v X . \end{aligned}$$

Similarly,

$$ \varepsilon ^{cu} \nabla _{e^{cu}} F = \varepsilon ^c\nabla _FF + \text {div}^v F. $$

Also, by definition of the hessian of a map in (4), and due to invariance of $V^{c}$, $V^{u}$, and $V^s$, we have

$$ \varepsilon ^{cu} \big (\nabla _{{(\cdot )}} f_*^T\big ) e^{cu}_{-T} = \varepsilon \big (\nabla _{{(\cdot )}} f_*^T\big ) e_{-T} + \varepsilon ^c \big (\nabla _{{(\cdot )}} f_*^T\big ) F_{-T} . $$

The last term in Lemma 8 is handled similarly.

Substituting into Lemma 8, we get

$$\begin{aligned} \begin{aligned} - \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma } (x) = \lim _{T\rightarrow \infty } \big (\text {div}^v X -\eta \text {div}^v F - \varepsilon \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) e_{-T} + \varepsilon _T \big (\nabla _{X^s} f_*^{T}\big ) e \\ + \varepsilon ^c\nabla _FX - \eta \varepsilon ^c\nabla _F F - \varepsilon ^c \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) F_{-T} + \varepsilon ^c_T \big (\nabla _{X^s} f_*^{T}\big ) F). \end{aligned} \end{aligned}$$

(13)

By the same proof of Proposition 9, the first line in (13) can be expressed by an equivariant divergence in the unstable manifold, that is

$$ -\eta \text {div}^v F - \varepsilon \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) e_{-T} + \varepsilon _T \big (\nabla _{X^s} f_*^{T}\big ) e = \mathcal {S}\big (\text {div}^v \nabla F, \text {div}^v F\big ) X $$

For the terms in the second line of (13), when $T\rightarrow \infty$

$$ \begin{aligned} \varepsilon ^c \big ( \nabla _{f_*^{-T}X^{u}} f_*^T\big ) F_{-T} = \varepsilon ^c \nabla _{X^{u}} F - \varepsilon ^c_{-T} \nabla _{f_*^{-T}X^{u}} F_{-T} \rightarrow \varepsilon ^c \nabla _{X^{u}} F ; \\ \varepsilon ^c_T \big (\nabla _{X^s} f_*^{T}\big ) F = \varepsilon ^c_T \nabla _{f_*^{T}X^s} F_T - \varepsilon ^c\nabla _{X^s} F \rightarrow - \varepsilon ^c\nabla _{X^s} F. \end{aligned} $$

Also,

$$ \eta \varepsilon ^c\nabla _F F = \varepsilon ^c\nabla _{\eta F} F = \varepsilon ^c\nabla _{X^c} F. $$

Hence, the second line equals

$$ \varepsilon ^c\nabla _FX - \eta \varepsilon ^c\nabla _F F - \varepsilon ^c \nabla _{X^{u}} F - \varepsilon ^c\nabla _{X^s} F = \varepsilon ^c\big (\nabla _FX- \nabla _{X} F\big ) = \varepsilon ^c\mathcal {L}_FX = F(\eta ), $$

where the last equality is because $F(\eta )=\mathcal {L}_F(\varepsilon ^c X)$ and $\mathcal {L}_F \varepsilon ^c =0$. This proves the expression in the lemma. $\square$

4 Unstable Contribution of Linear Response

This section proves Proposition 2, which shows that we can use the equivariant divergence to express the unstable contribution of linear response.

Proposition 2

(Fast response formula for linear response)

$$ SC = \rho (X\mathcal {S}(d\Phi ,\Phi -\rho (\Phi ))), \quad UC = \lim _{W\rightarrow \infty } \rho \Bigg (\varphi \frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }\Bigg ), \quad \text {where}\quad \varphi :=\int _{-W}^W \Phi \circ f^t dt . $$

Proof

The first equation is due to Theorem 3. To prove the second equation, we only need to prove that for any t,

$$ \rho (X^{cu}(\Phi _t))=\rho \Bigg (\Phi \circ f^t\frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }\Bigg ), $$

and take integral $\int _{-\infty }^{\infty } dt$ of both sides of the equation. We shall prove this by integration by parts on local center-unstable foliation in sets from a partition R of K, whose construction is given below.

First define some notations for the local product structure. We use B(x, r) and $\bar{B}(x,r)$ to denote the open ball and the closed ball center at x of radium r. For each $x\in K$, we choose a $r'_x>0$ so small that the neighbourhood $B(x,r'_x)\bigcap K$ has a local center-unstable foliation. There is a local product structure in this foliation. For each $a\in \mathcal {V}^{cu}_{r}(x)$ and $b\in \mathcal {V}^{s}_r(x)\bigcap K$ we denote [a, b] as the unique transverse intersection of $\mathcal {V}^{s}_{loc}(a)$ and $\mathcal {V}^{cu}_{loc}(b)$ in $B(x,r'_x)\bigcap K$: this is well-defined when r is small enough. Conversely, for each $x'$ near x, by the continuity of the stable and center-unstable manifolds, there exist unique $a=\mathcal {V}^{cu}_r(x) \bigcap \mathcal {V}^{s}_{loc}(x')=: \pi ^{cu}_x(x')$ and $b=\mathcal {V}^{s}_r(x) \bigcap \mathcal {V}^{cu}_{loc}(x') =: \pi ^{s}_x(x')$ such that $[a,b]=x'$, and a and b both depend on $x'$ continuously: they are sometimes called the local coordinate of $x'$.

Then, for each point x, we define rectangles containing x. For two sets $A\subseteq \mathcal {V}^{cu}_{r}(x)$ and $B\subseteq \mathcal {V}^{s}_r(x)\bigcap K$, we call the set $[A,B]:=\{[a,b]: a\in A,b\in B\}$ a rectangle. Choose an $r_x<r'_x$ small enough. We take $A=B(x,r_x)\bigcap \mathcal {V}^{cu}_{loc}(x)$ and $B=B(x,r_x)\bigcap \mathcal {V}^{s}_{loc}(x)\bigcap K$ which are open in $\mathcal {V}^{cu}_{loc}(x)$ and $\mathcal {V}^{s}_{loc}(x)\bigcap K$, $\bar{A}=\bar{B}(x,r_x)\bigcap \mathcal {V}^{cu}_{loc}(x)$ and $\bar{B}=\bar{B}(x,r_x)\bigcap \mathcal {V}^{s}_{loc}(x)\bigcap K$ which are closed in $\mathcal {V}^{cu}_{loc}(x)$ and $\mathcal {V}^{s}_{loc}(x)\bigcap K$. Then $r_x$ can be so small that [A, B] and $[\bar{A},\bar{B}]$ are well-defined rectangles in $B(x,r'_x)$.

We claim that [A, B] is open in K and $[\bar{A},\bar{B}]$ is closed in K. To see that, first notice that if $y\in [A,B]$ then $\pi ^{cu}_x(y) \in A$. Then for any $y'\in K$ close enough to y, $\pi ^{cu}_x(y')\in A$, since A is open and $\pi ^{cu}_x$ is continuous. Also, for any $y'\in K$ close to y, $\pi ^{s}_x(y') \in B$. Hence $y'\in [A,B]$, which imply [A, B] is open in K. Similarly, $[\bar{A},\bar{B}]$ is closed in K.

Then we construct the partition R from an open cover $R'$, which consists of rectangles with negligible boundaries. Denote $R'(x)=[A,B]$ and $cl(R'(x))=[\bar{A},\bar{B}]$. Define $\partial R'(x)=cl(R'(x))-R'(x)$. Then $r_x$ can be chosen such that $\partial R'(x)$ has zero physical SRB measure. This is because by definition, $\partial R'(x)$ will not intersect for different $r_x$; and $r_x$ has uncountable options while the SRB measure is finite. Then $\{{R'(x):x\in K}\}$ forms a open cover of K. K is compact so we can choose a finite subcover ${R'_1,R'_2,\dots ,R'_m}$. We define $R_i= (R'_i-\bigcup _{k=1}^{i-1} cl(R'_k)),1\le i\le m$, then $R_i$ are pairwise disjoint and $\rho (K-\bigcup _{i=1}^{m}R_i)=0$. Define the partition $R:=\{R_i \}_{1\le i\le m}$, and this is the partition we want.

Then we can do integration by parts. Let $\sigma '$ denote the factor measure and $\sigma$ denote the conditional density of $\rho$ with respect to this foliation. Notice that $\partial R:= K-\bigcup _{i=1}^m R_m$ has zero SRB measure, we take the following integration on $\bigcup _{i=1}^mR_i$.

$$ \rho (X^{cu}(\Phi _t))=\iint \sigma X^{cu}\cdot {{\,\textrm{grad}\,}}\big (\Phi \circ f^t\big )dxd\sigma '(x). $$

Integrate-by-parts on unstable manifolds, note that the flux terms on the boundary of the partition cancel, so

$$ \rho (X^{cu}(\Phi _t)) =\iint -\text {div}_\sigma ^{cu}X^{cu}\big (\Phi \circ f^t\big )\sigma dxd\sigma '(x) =\rho \Bigg (\Phi \circ f^t\frac{\delta {\tilde{L}}^{cu}\sigma }{\sigma }\Bigg ). $$

Take it back to the formula and we can finish the proof. $\square$

Appendix A: Derivation Only in the Unstable

Our proof in the main body of the paper feels like taking a detour, since we started on the center-unstable manifold and then removed the center direction. In this appendix we do not take that detour, and prove a similar formula by working only in the unstable manifold. This proof is shorter and intuitive, but is only formal. Nevertheless, it is consolidating to see that two proofs give equivalent formulas.

We first mention the main difficulties that stop us from making the proof in the appendix rigorous. The first difficulty is the lack of smoothness. For example, notice that $\nabla _e X^c$ is not well-defined in general, since the oblique projection to $V^c$ also depends on $V^{su}$, which is not differentiable in general cases. However, here we only do the formal calculation, and we assume that every object we use is smooth enough and all the notions are well-defined. Another main difficulty is the ‘absolutely continuous property of the holonomy map’ we want to use here is non-standard: we shall explain this in the formal proof.

We add some new notations similar to those in the main body of the paper, but in unstable version instead of center-unstable version. We fix the point of interest x, choose a local unstable foliation, denote the density of the conditional SRB measure on $\mathcal {V}^u_r(x)$ at $\gamma =0$ by $\sigma ^u$. Similarly, we can define the local unstable transfer operator ${\tilde{L}}^u$, which is the transfer operator of $\tilde{f}\xi ': \mathcal {V}^u(x)\rightarrow \mathcal {V}^u(x)$, $\tilde{f}$ is still the time one map of the flow $\gamma X$ and $\xi '$ is the holonomy map of center-stable foliation. Then we can define the ratio $\frac{{\tilde{L}}^u\sigma ^u}{\sigma ^u}$ and $\frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u}$. We will calculate the latter one.

Theorem 10

$$ \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u} = -\text {div}^v X - \mathcal {S}(\text {div}^v\nabla F,\text {div}^vF)X. $$

An indirect proof for $\rho$ almost everywhere. We give a proof of the theorem, but only for $\rho$ almost everywhere, based on Theorem 1; it also shows that the two theorems do not contradict.

First notice that, in a local center-unstable foliation, each center-unstable manifold $\mathcal {V}^{cu}_{loc}$ is foliated by the unstable foliation $\mathcal {V}^u_{loc}$, and the map $\tilde{f}\xi ':\mathcal {V}^{cu}\rightarrow \mathcal {V}^{cu}$ preserve the unstable foliation $\mathcal {V}^u$. So ${\tilde{L}}^u$ can be understood as a local transfer operator of either the map $\tilde{f}\xi ':\mathcal {V}^u(x)\rightarrow \mathcal {V}^u(x)$, or the map $\tilde{f}\xi ':\mathcal {V}^{cu}(x)\rightarrow \mathcal {V}^{cu}(x)$; an interesting fact is that for $\rho$ almost everywhere,

$$\begin{aligned} \frac{{\tilde{L}}^u\sigma ^u}{\sigma ^u} = \frac{{\tilde{L}}^u\sigma ^{cu}}{\sigma ^{cu}}, \quad \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u} = \frac{\delta {\tilde{L}}^u\sigma ^{cu}}{\sigma ^{cu}}, \end{aligned}$$

(14)

where $\sigma ^{cu}$ is the conditional density in center-unstable manifold (the main body of the paper uses $\sigma$ to denote the same quantity). Intuitively, this is because in our framework, the measure change is entirely accounted for by the change of the conditional density, so they all must equal. By Eq. (14), we have

$$ \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u} = \frac{\delta {\tilde{L}}^u\sigma ^{cu}}{\sigma ^{cu}} = \frac{\delta {\tilde{L}}^{cu}\sigma ^{cu}}{\sigma ^{cu}} -\frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}}. $$

By substitution of Theorem 1 into the above equation, we see that now we only need to prove

$$\begin{aligned} \frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}} (x) =- F(\eta ) \quad \rho \text { almost everywhere.} \end{aligned}$$

(15)

To see this, pick any smooth observable $\Phi$, choose a regular point $x_0$, such that for any continuous function g, $\lim _{T\rightarrow \infty }\frac{1}{T} \int _0^T g(f^tx_0) dt=\rho (g)$, and we have

$$\begin{aligned} \rho (\Phi F(\eta )) + \rho (X^c (\Phi ))&= \rho (\Phi F(\eta ) + \eta F (\Phi )) = \rho ( F(\Phi \eta )) \\&= \lim _{T\rightarrow \infty }\frac{1}{T} \int _0^T F(\Phi \eta )(f^tx_0) dt\\&= \lim _{T\rightarrow \infty }\frac{1}{T} \big ((\Phi \eta )\big (f^{T}x_0\big ) - (\Phi \eta )(x_0)\big ) = 0. \end{aligned}$$

Hence

$$ - \rho (\Phi F(\eta )) = \rho (X^c (\Phi )) = \rho \left( \Phi \left( \frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}} \right) \right) $$

Since $\Phi$ is arbitrary, this proves Eq. (15) for $\rho$ almost everywhere. $\square$

Another way to prove Eq. (15) is to rerun the proof of Lemmas 6 and 7. Let $e^{cu}$ be the vector field $e^{cu}(y_t)$ defined on $\mathcal {V}^{cu}(x)$, then

$$\begin{aligned} \frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}}&= \varepsilon ^{cu}\mathcal {L}_{X^c}e^{cu} = \varepsilon ^{cu} \mathcal {L}_{\eta F} e^{cu}\\&= \varepsilon ^{cu} \sum _i e_1\wedge \cdots \wedge \mathcal {L}_{\eta F} e_{i} \wedge \cdots \wedge e_u \wedge F. \end{aligned}$$

Since $\mathcal {L}_{\eta F} e_{i} = -e_{i}(\eta )F + \eta \mathcal {L}_F e_i= -e_{i}(\eta )F$, we have

$$ \frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}} = - \varepsilon ^{cu} \sum _i e_1\wedge \cdots \wedge e_{i}(\eta ) F \wedge \cdots \wedge e_u \wedge F. $$

Since $F \wedge F=0$, all terms in the sum are zero except for the last, so

$$ \frac{\delta {\tilde{L}}^{c}\sigma ^{cu}}{\sigma ^{cu}} = - \varepsilon ^{cu}F(\eta ) e^{cu} = - F(\eta ). $$

Above is how to rigorously prove Theorem 10 via taking a detour through Theorem 1; then we show a non-rigorous but more straight-forward argument.

Formal but direct proof of Theorem 10. Fix an arbitrary large number T, define the u-covector $\varepsilon$ and the u-vector e field in a small neighbourhood of x on $\mathcal {V}^u(x)$ as

$$ \varepsilon (y) = f^{*-T} \tilde{\varepsilon }(y_{-T}), \quad \quad e(y) = f_*^{T} \tilde{e}(y_{-T}), \quad \quad y\in \mathcal {V}^u(x), $$

so $\varepsilon e=1$. In this proof let $\sigma ^u$ be the conditional density of the measure obtained by pushing forward the Lebesgue measure for time T, so

$$\frac{\sigma ^u(x)}{\sigma ^u(y)}=\frac{|e(y)|}{|e(x)|}=\frac{|\varepsilon (x)|}{|\varepsilon (y)|}$$

The re-distribution of the conditional measure caused by $X^{u}$ is

$$\begin{aligned} \begin{aligned} \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u} = \frac{(\mathcal {L}_{-X^{u}} \varepsilon ) e}{\varepsilon e} = \varepsilon (\mathcal {L}_{X^{u}} e) = \varepsilon \nabla _{X^{u}}e - \varepsilon \nabla _e X^u \\ = \varepsilon \left( \nabla _{X^{u}}e - \nabla _e X + \nabla _e X^c +\nabla _e X^s \right) \end{aligned} \end{aligned}$$

(16)

For the second term in (16), by definition of $\text {div}^v$,

$$ \varepsilon \nabla _e X =: \text {div}^v X. $$

For the third term in (16), we can formally write

$$ \varepsilon \nabla _e X^c = \varepsilon \nabla _e(\eta F) = \varepsilon \sum _{i=1}^u e_1\wedge \cdots \wedge \left( e_i(\eta ) F + \eta \nabla _{e_i}(F)\right) \wedge \cdots \wedge e_u = \eta \varepsilon \nabla _{e}F =: \eta \text {div}^v F, $$

where $\eta := \tilde{\varepsilon }^c(X).$ The third equality uses that $\varepsilon _i(F)=0$ for $\varepsilon _i\in V^{*u}$. The last expression is well defined since F is smooth.

For the first term in (16), use Lemma 4,

$$\begin{aligned} \varepsilon \nabla _{X^{u}} e&= \varepsilon f_*^T \nabla _{f_*^{-T}X^{u}} e_{-T} + \varepsilon \int _0^T f_*^{T-t} \nabla ^2_{f_*^{t-T}X^{u}, e_{t-T}}F_{t-T} dt \\&= \varepsilon _{-T} \nabla _{f_*^{-T}X^{u}} e_{-T} + \int _0^T \varepsilon _{t-T} \nabla ^2_{f_*^{t-T}X^{u}, e_{t-T}}F_{t-T} dt. \end{aligned}$$

The first terms goes to zero as $T\rightarrow 0$. Denote $\omega :=\varepsilon \nabla ^2_{{(\cdot )}, e}F$, then

$$ \varepsilon \nabla _{X^{u}} e = w \int _{-\infty }^0 \omega _{t} f_*^{t}X^{u} dt = X^u \int _{-\infty }^0 f^{*t} \omega _{t} dt. $$

Fig. 3

Intuitions for $\nabla _eX^s$. $x'-x=e_5=e(x)$, $y-x=X^s_1:=X^s(x)$, $y'-x'=X^s_2:=X^s(x')$, and $e_4:=y'-y$. Note that here $e_4$ is pushforward of e along $X^s$: this is different from Fig. 2

For the last term in (16), the intuitive explanation for $\nabla _eX^s$ is illustrated in Fig. 3. Intuitively, assuming $u=1$ and $\mathcal {M}=\mathbb {R}^M$,

$$ \nabla _eX^s = X^s_2 - X^s_1 = (y'-x') - (y-x) = (y'-y) - (x'-x) = e_4-e_5 $$

More specifically, if we denote the flow of $X^s$ by $\xi$, and let $\xi _*e$ be a vector field on $\mathcal {V}^s(x)$, which is the pushfoward vector field of e by $\xi ^\tau$ for a small interval of $\tau$. Notice that here $\xi _*e$ is different from $e_{t\ge 0}$ in the main body of the paper by a constant. Then $\mathcal {L}_{X^s}\xi _*e = 0$, so that $\nabla _{X^s}{e}=\nabla _{X^s}{\xi _*e|_{\tau =0}}$ is well defined and

$$ \nabla _{e} X^s = \nabla _{\xi _*e|_{\tau =0}} X^s = \nabla _{X^s}{\xi _*e}. $$

Apply Lemma 4, at $\tau =0$,

$$ \varepsilon \nabla _{X^s} \xi _*e = \varepsilon _T f_*^T \nabla _{X^s} \xi _*e = \varepsilon _T \nabla _{f_*^T X^s} f_*^T \xi _*e - \varepsilon _T \int _0^T f_*^{T-t} \nabla ^2_{f_*^t X^s, e_t}F_t dt. $$

Notice that, vaguely speaking, $\xi$ can be seen as the holonomy map of local stable foliation $\mathcal {V}^s$. Since the holonomy map formally tends to identity, we can formally have

$$\begin{aligned} \varepsilon _T \nabla _{f_*^T X^s} f_*^T \xi _* e \rightarrow 0. \end{aligned}$$

(17)

In Fig. 3, this means that $\varepsilon _T(e_6-e_7)\rightarrow 0$. Although very plausible, this equation was not really proved. The main reason is that e is only a u-vector, so $\xi _*e$ is not exactly the holonomy map in the textbook. Holonomy map requires transverse intersections between e and the stable, which is not true here. The detour we took in the main body of the paper is mainly because $e^{cu}$ and $\xi$ transversely intersect, so we can use the absolute continuity lemma offered in textbooks.

Nevertheless, if we accept the plausible formula in (17), then

$$ \varepsilon \nabla _{e} X^s = \varepsilon \nabla _{X^s} e = - \varepsilon _T \int _0^T f_*^{T-t} \nabla ^2_{f_*^t X^s, e_t}F_t dt = - \int _0^T \varepsilon _{t} \nabla ^2_{f_*^t X^s, e_t}F_t dt = - X^s \int _0^T f^{*t}\omega _t dt. $$

Summarizing, denote $\psi :=\text {div}^v F$, then $\eta \text {div}^v F=\psi \varepsilon ^c X$, and

$$\begin{aligned} \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u}&= \varepsilon \left( \nabla _{X^{u}}e - \nabla _e X + \nabla _e X^c +\nabla _e X^s \right) \\&= -\text {div}^v X + \psi \varepsilon ^c X - X^s \int _0^T f^{*t}\omega _t dt + X^u \int _{-\infty }^0 f^{*t} \omega _{t} dt. \end{aligned}$$

By the same argument as Eq. (12), we have $\omega F=F(\psi )$, so we can let $T\rightarrow \infty$ and apply the adjoint shadowing lemma to get,

$$ \frac{\delta {\tilde{L}}^u\sigma ^u}{\sigma ^u} = -\text {div}^v X - \mathcal {S}(\text {div}^v\nabla F,\text {div}^vF)X. $$

$\square$