Solvability and the Fredholm alternative

Each order of the expansion is an equation $Q w = f$. It has a solution only when $f$ averages to zero, and that condition gives the averaged model.

Every expansion in the examples reduces, at each order, to one kind of equation: $L w = f$, where $L$ is the generator of the fast process.

The generator of the fast process

For a Markov chain with states $1, \dots, n$ the generator is a matrix $Q$. Its off-diagonal entry $Q_{ij}$ is the rate of jumping from $i$ to $j$, and each diagonal entry makes its row sum to zero. For the two-state switch that leaves each state at rate $\lambda$,

Applied to a vector $f$, the generator gives the expected rate of change of $f(y_t)$:

For a diffusion the generator is a differential operator. The fast Ornstein–Uhlenbeck factor in the examples has $\mathcal L = -y\,\partial_y + \partial_{yy}$. Everything below applies to both.

Two null vectors

Every generator has two special vectors.

• Constants are killed on the right: $Q\mathbf 1 = 0$. A constant function does not change as the chain moves, so its expected rate of change is zero. This is why the rows sum to zero.
• The stationary distribution $\pi$ is killed on the left: $\pi Q = 0$. If the chain starts in its stationary distribution it stays there, so the average of $Qf$ under $\pi$ is always zero.

For the two-state switch, $\pi = (\tfrac12, \tfrac12)$. For the Ornstein–Uhlenbeck factor, $\pi$ is the standard normal law.

The Fredholm alternative

Consider the equation

Multiply on the left by $\pi$. Since $\pi Q = 0$, the left side vanishes, so a solution can exist only if

The converse also holds when the chain is irreducible: then the null space of $Q$ is one-dimensional, the range of $Q$ has dimension $n - 1$, and it is exactly the set of vectors with $\pi\cdot f = 0$. So

This is the Fredholm alternative: either the right side is orthogonal to the left null vector and the equation can be solved, or it is not and no solution exists. When a solution exists it is unique up to adding a constant, because constants are in the null space. The group inverse selects the particular solution whose stationary mean is zero. In an asymptotic expansion this centers only the range part of a coefficient; its full coefficient can still have a null-space amplitude that the next solvability condition determines.

The intuition

Read $f$ as a source that pays $f_i$ per unit time while the chain sits in state $i$. The solution of $Q w = -f$ is the expected total payment still to come, measured relative to the average:

This integral converges only if the payments average to zero in the long run. If they do not, the running total grows forever and no finite $w$ exists.

So the solvability condition removes the average first. Whatever the fast process averages to is handled separately, on the slow time scale, by the averaged equation.

The averaged equation

Suppose a slow quantity $u$ depends on the state of the fast process and satisfies

where $G$ describes the slow dynamics. Try $u = u_0 + \varepsilon u_1 + \varepsilon^2 u_2 + \cdots$ and match powers of $\varepsilon$.

• At order $\varepsilon^{-1}$: $Q u_0 = 0$. So $u_0$ is in the null space, which means it does not depend on the fast state. The leading answer has already forgotten the fast variable.
• At order $1$: $Q u_1 = \partial_t u_0 - G u_0$. By the Fredholm alternative this has a solution only if the right side averages to zero:

That is the averaged equation. The fast variable has been replaced by its stationary average, and the step that forced it was the solvability condition.

With the averaged equation in hand, the order-1 equation determines the state-dependent part of $u_1$. It does not yet determine the part of $u_1$ shared by all fast states. The next order fixes that remaining slow amplitude, and so on.

The full hierarchy: average plus shape

Put $P=\mathbf1\pi$ and decompose every coefficient as

Here $\chi_m=(I-P)u_m$ is the centered shape, while $a_m=\pi u_m$ is the null-space amplitude. If $u_0=a_0\mathbf1$, take $\chi_0=0$. At order $m$ the equation for the next shape is

Its solvability condition and centered solution are respectively

Thus $Q^{\#}$ determines only $\chi_{m+1}$. The scalar or slow-variable function $a_{m+1}$ remains free until the next order. Its initial or terminal condition comes from matching the original data, including an initial or terminal layer when one is present. Centering the whole $u_{m+1}$ would set $a_{m+1}=0$ prematurely and can make the next Poisson equation insoluble.

A moving generator adds geometric transport

If the fast generator itself varies on the slow time scale, its stationary projection also moves. Let $Q=Q(t)$ be a $C^1$ family of irreducible finite-state generators, let $\pi(t)Q(t)=0$, and assume its nonzero spectrum stays uniformly away from zero on the fixed time interval. Put $P(t)=\mathbf1\pi(t)$ and $R(t)=Q(t)^{\#}$. The decomposition is now

The Poisson equation is unchanged pointwise, but its solvability condition is not. Differentiating the moving gauge gives $\pi\chi_m'=-\pi'\chi_m$, hence

The extra term is geometric transport of the centered shape by the moving stationary subspace. It vanishes for fixed $Q$, but not merely because the forcing is scalar or because every instantaneous corrector is centered.

For $G(t)=\operatorname{diag}g(t)$, write $\bar g=\pi g$, $f=g-\bar g\mathbf1$. Then

and the first null-space amplitude satisfies

Only $K_{\rm dyn}$ is the instantaneous Green–Kubo contraction. The second coefficient records motion of $\pi(t)$ and is generally signed.

Exact two-state protocol integral

The word geometric is literal in the two-state case. Write

and let $\delta=g_0-g_1$. The centered forcing is $f=((1-p)\delta,-p\delta)^\top$, so $Qf=-sf$ and $Rf=-f/s$. Therefore

On a closed protocol $\mathcal C:t\mapsto(p(t),h(t))$, with $h=\delta/s$, the accumulated geometric contribution is the oriented line integral

Consequently $\Gamma_{\rm geo}$ is unchanged by every orientation-preserving $C^1$ reparameterization of the same protocol, and changes sign when the loop is traversed backwards. It vanishes whenever $h$ is a single-valued function of $p$ along the loop, because then $h(p)\,dp$ is exact. Thus a nonzero periodic correction requires a loop in the two protocol coordinates $(p,h)$ that is not confined to a graph $h=H(p)$. In contrast, $\int K_{\rm dyn}\,dt$ depends on traversal time and remains nonnegative.

This is the same path-geometric mechanism emphasized for stochastic pumps by Sinitsyn and Nemenman (2007), but the observable here is different: it is a null-space amplitude in a backward-equation corrector, not a transported-current cumulant. No pump-current identity is being asserted.

Finite-state connection, curvature, and Stokes formula

The line-integral structure is not special to two states. Let $Q=Q(\theta)$ and $g=g(\theta)$ be $C^2$ on a parameter patch, with $Q(\theta)$ an irreducible finite-state generator whose nonzero spectrum remains uniformly separated from zero. Write $\pi=\pi(\theta)$, $R=Q(\theta)^{\#}$, and $f=g-(\pi g)\mathbf1$. Differentiating $\pi Q=0$ and $\pi\mathbf1=1$ gives $\partial_a\pi=-\pi(\partial_aQ)R$. Hence the geometric coefficient along a piecewise $C^1$ protocol $\theta(t)$ is the pullback of the one-form $\mathcal A=\sum_a\mathcal A_a\,d\theta_a$, where

Thus the closed-protocol correction is $\Gamma_{\rm geo}(\mathcal C)=\oint_{\mathcal C}\mathcal A$. It depends on the oriented path, not its traversal speed. For two parameters, Stokes' theorem gives

Writing $\pi_a=\partial_a\pi$, $R_a=\partial_aR$, and $f_a=\partial_af$, cancellation of the mixed derivatives of $\pi$ yields the explicit curvature

The one-form itself depends on the choice of particular Poisson corrector, but its closed-loop content does not. Indeed, replace the centered first shape $-a_0Rf$ by $-a_0Rf+a_0c(\theta)\mathbf1$. To leave the full first corrector unchanged, its normalized null amplitude $b=a_1/a_0$ must be replaced by $\widetilde b=b-c(\theta)$. Therefore

For an open protocol the line integral changes by $-\{c(\theta_{\mathrm{final}})-c(\theta_{\mathrm{initial}})\}$, exactly compensated by the endpoint change in the null amplitude. Thus the connection is gauge covariant, while its curvature and closed-protocol holonomy are gauge invariant. The centered convention $\pi\chi_1=0$ simply fixes $c=0$. This is the same exact-differential gauge mechanism as in Berry's original geometric-phase construction (1984), although the transported object and governing equation here are different.

Nonzero curvature is therefore a local obstruction to writing $\mathcal A$ as an exact differential. On a simply connected patch, vanishing curvature makes every contractible-loop correction zero; nonzero curvature produces a nontrivial correction for sufficiently small loops. This is a statement about the geometric part alone: the instantaneous Green–Kubo term is still traversal-time dependent.

The certificate tests this theorem on an affine, nonreversible three-state family over a rectangle where every jump rate is at least $0.425$. It verifies $\mathcal A_a=-\pi_aRf=\pi(\partial_aQ)R^2f$ exactly to displayed precision, finds center curvature $2.90856\times10^{-3}$, and obtains the same nonzero correction $4.801835316245\times10^{-3}$ from the boundary integral and a $40\times40$ Gauss–Legendre curvature integral, with residual $8.7\times10^{-19}$. A nonconstant polynomial corrector gauge independently leaves the boundary integral and curvature unchanged to numerical precision, while its open-edge shift agrees with the predicted endpoint difference.

For the smooth periodic example used here, the certificate finds $\int_0^{2\pi}K_{\rm geo}(t)\,dt=-0.00639952$. At the one-period endpoint the complete first-order outer approximation converges with observed order $1.990$, whereas deleting $K_{\rm geo}$ leaves observed order $0.957$. It also verifies the geometric formula to $5.3\times10^{-18}$, invariance under a nonuniform traversal to $8.7\times10^{-19}$, sign reversal to $5.3\times10^{-18}$, and zero contribution for a graph-valued loop to $3.5\times10^{-18}$. This is a fixed-positive-time test: an initial layer is still needed for a bound uniform down to $t=0$, and no long-maturity uniformity is claimed.

This moving-projection recursion is the finite-state backward-equation counterpart of the asymptotic-series machinery in Khasminskii and Yin (1996) and the singularly perturbed chain framework of Yin and Zhang (1998).

The missing amplitude is the Green–Kubo correction

For a time-independent scalar forcing $G=\operatorname{diag}g$, put $\bar g=\pi g$ and $\widetilde g=g-\bar g\mathbf1$. Then $a_0=e^{\bar g t}$ and the first centered shape is

The next solvability condition becomes

With regime-free initial data, first-order matching gives $a_1(0)=0$, so $a_1=tK a_0$. Therefore the null-space amplitude that centering would discard is exactly the first Green–Kubo correction to the slow eigenvalue.

The next coefficient is a three-time cumulant

The same recursion gives one more coefficient without assuming reversibility. Write $R=Q^{\#}$, $f=\widetilde g$, and let $D_f$ denote multiplication by $f$. Normalizing the slow right eigenvector by $\pi h_\varepsilon=1$ gives

and the slow eigenvalue of $Q/\varepsilon+\operatorname{diag}g$ is

Multiplying the eigen-equation order by order proves the formula: the first centered equation is $Qh_1=-f$, the second is $Qh_2=D_fRf+K\mathbf1$, and its next solvability condition is $L=\pi D_fh_2$. Analyticity follows locally because zero is a simple eigenvalue of an irreducible finite-state generator, the finite-dimensional setting of Kato's perturbation theory.

Since $R=-\int_0^\infty(e^{Qt}-\mathbf1\pi)\,dt$, the second coefficient also has the ordered-correlation form

Thus $K$ is the variance-rate coefficient and is nonnegative, whereas $L$ is signed third-order information. It is not a Gram matrix and carries no positive-semidefinite or rank claim. These coefficients are the constant-forcing specialization of the normalized average/shape recursion on the engine page.

An all-orders recursion and the fourth cumulant

The same coefficient matching gives a finite-dimensional recursion to every order. Write

with $h_0=\mathbf1$ and $h_1=-Rf$. Then, for $m\ge1$,

Indeed, the coefficient of $\varepsilon^{m+1}$ in the eigen-equation has the expression in parentheses as its Poisson right side. Its stationary mean vanishes precisely when $\kappa_m=\pi D_fh_m$; applying $R$ then selects the normalized centered solution. This proves the recursion by induction for every order inside the local analytic neighborhood.

The first new term after $K=\kappa_1$ and $L=\kappa_2$ is

so

Equivalently, $M$ is one twenty-fourth of the long-run fourth-cumulant rate of $\int_0^T f(Y_s)\,ds$. The second term in the displayed formula is essential: fourth order uses a connected cumulant, not the raw ordered four-time moment. Like $L$, $M$ is signed and has no positive-semidefinite or rank interpretation.

A computable convergence disk and truncation bound

The analytic neighborhood can be certified directly from the unperturbed generator, including for a nonnormal chain. Choose $r>0$ so that the circle $|w|=r$ separates the simple eigenvalue zero of $Q$ from every other eigenvalue, and define

If $|z|<\rho_r$, then on this circle

The Neumann resolvent identity therefore keeps the circle in the resolvent set of $Q+zD_f$. Its Riesz projection has the same rank one throughout the disk, so the slow eigenvalue $\mu(z)$ of $Q+zD_f$ is analytic there and remains inside $|w|<r$. This supplies a quantitative version of the local perturbation argument.

For any $0<\rho<\rho_r$, Cauchy's estimate applied to $\mu$ turns the recursion into the explicit remainder bound

This is conservative but fully computable. It bounds the eigenvalue series, not a whole time-dependent semigroup or a Fourier inversion. In particular it does not replace the separate initial-layer and transform-tail estimates needed elsewhere on the site.

The slow amplitude has a certified series too

The same contour also controls the spectral projection, which is the missing object when initial data are not already aligned with the slow mode. Let

Here adding the scalar $\bar g I$ changes the slow eigenvalue but not this projection. In particular,

where $R=Q^\#$. For every $0<\rho<\rho_r$, the singular-value perturbation inequality gives, uniformly on $|z|=\rho$,

Applying the vector-valued Cauchy formula coefficient by coefficient therefore yields the explicit operator remainder

Consequently the same right-hand side, multiplied by $\|v\|_2$, certifies the error in the slow amplitude $P(\varepsilon)v$ selected by arbitrary initial data $v$. This isolates amplitude matching from eigenvalue truncation.

For the nonreversible three-state certificate below, the same rigorously sampled contour gives $\rho_r\ge2.103817190523$. The exact projector is computed independently from paired left and right eigenvectors. Its first-order remainder converges with measured order $1.998987$, and the largest observed error is only $0.020318$ of the displayed Cauchy bound.

The complementary fast semigroup has a separate bound

The slow contour alone cannot control the initial layer, but a second contour can. Let $\Gamma$ be a simple positively oriented contour that encloses every nonzero eigenvalue of $Q$, excludes zero, and lies in $\operatorname{Re}w\le-\gamma<0$. Put

If $0<\varepsilon\|D_f\|_2<s_\Gamma$, the singular-value perturbation inequality keeps $\Gamma$ in the resolvent set of $B_\varepsilon=Q+\varepsilon D_f$. The contour still encloses exactly the fast eigenvalues, so Dunford's formula gives

Taking the operator norm proves the explicit all-time estimate

Restoring the scalar mean forcing multiplies the right side by $e^{\bar g t}$. The prefactor retains the possible nonnormal transient amplification; it is not silently set to one. This controls the exact fast component for arbitrary initial data, but it does not make a truncated slow eigenvalue expansion uniformly accurate at arbitrarily long maturities: any slow-eigenvalue truncation error still accumulates in its exponential.

For the nonreversible three-state certificate, the circle centered at $-3.15$ with radius $2.4$ has $\gamma=0.75$ and a rigorously sampled validity radius $s_\Gamma/\|D_f\|_2\ge0.898895533832$. Over all tested $\varepsilon\le0.2$ and $0\le t/\varepsilon\le5$, the largest exact-fast-semigroup to proved-bound ratio is $0.305868$.

The three bounds combine into an all-time semigroup certificate

The eigenvalue, projector, and fast-contour estimates can be assembled without another asymptotic step. For integers $p,q\ge0$, write

and denote the two Cauchy bounds above by $E_{\lambda,p}(\varepsilon)$ and $E_{P,q}(\varepsilon)$. Also put

For real $0<\varepsilon<\min(\rho,s_\Gamma/\|D_f\|_2)$, the exact spectral split is

Subtracting $e^{t\lambda_p}P_q$, then using $|e^z-1|\le e^{|z|}-1$, gives the explicit bound

This is an all-time inequality, not a claim that a fixed truncation is uniformly accurate for all maturity. On the slow exponential scale its two algebraic terms are $O(\varepsilon^{q+1}+t\varepsilon^{p+1})$ while $tE_{\lambda,p}=o(1)$; the third term is the certified initial layer. Thus, on a window $t=O(\varepsilon^{-\alpha})$, the normalized algebraic error has order $\min(q+1,p+1-\alpha)$ when $\alpha<p+1$. The formula makes the maturity loss from eigenvalue truncation explicit instead of absorbing it into an unspecified constant.

In the three-state certificate, $p=3$, $q=1$, and $t=\varepsilon^{-2}/4$. The normalized full-semigroup error converges with measured order $1.998990$; its proved bound has measured order $2.030627$, and the largest observed error is $0.020181$ of that bound. The evaluation uses the exact paired spectral projectors mode by mode, avoiding overflow from the common slow exponential at these growing maturities.

A uniform first-order composite

The initial layer matters if the approximation is required uniformly all the way down to $t=0$. For the two-state example below with $u(0)=\mathbf1$, write $u=m\mathbf1+z(1,-1)^\top$ and put

The exact solution splits into its slow and fast modes:

The first-order outer approximation is

At $t=0$ its centered error is exactly $\varepsilon/4$, so it cannot have a uniform $O(\varepsilon^2)$ remainder on an interval containing zero. Restoring the leading fast layer gives

For every $T\ge0$ and $0<\varepsilon\le1$, direct estimates of the exact modal formula give the fully uniform bound

Indeed, $0\le\lambda_s-1/2\le\varepsilon/8$, $|\lambda_s-1/2-\varepsilon/8|\le\varepsilon^3/128$, and the exact fast amplitude differs from $-\varepsilon/4$ only at the order needed by the bound. Taylor's inequality for the slow exponential and $t e^{-at}\le a^{-1}$ for the fast exponential yield the displayed $C_T$. This is a finite-time theorem uniform over $[0,T]$; because $C_T$ grows with $T$, it is not a long-maturity-uniform claim.

A two-state obstruction

For a minimal example, take

Writing $u_0=a(t)\mathbf1$, leading solvability gives $a'=a/2$. The centered first shape is

If one incorrectly sets the full coefficient $u_1=\chi_1$, then

so no $u_2$ can solve the next equation. The missing term is $a_1(t)\mathbf1$:

For $u(0)=\mathbf1$, matching gives $a_1(0)=0$, hence $a_1=ta/8$. At any fixed positive time the centered-only approximation has an order-$\varepsilon$ error, whereas including $a_1\mathbf1$ reduces it to order $\varepsilon^2$ at fixed positive time. The certificate checks the Poisson equations, the obstruction, the exact modal formula, the finite-time uniform bound, and the matrix-exponential convergence orders. The general recursion remains a formal outer hierarchy; beyond the exact two-state theorem above, rigorous remainder bounds require the usual regularity and layer hypotheses. In particular, the primary two-time-scale analysis of Khasminskii and Yin (1996) constructs the outer and initial-layer pieces together and proves remainder bounds under its stated assumptions.

The group inverse

When $\mathbb{E}_\pi[f] = 0$, the solution of $Qw = f$ with $\mathbb{E}_\pi[w] = 0$ is written $w = Q^{\#}f$. The matrix $Q^{\#}$ is the group inverse of $Q$.

It inverts $Q$ on vectors that average to zero and returns vectors that average to zero. It can be computed as

For the two-state switch, a vector with $\mathbb{E}_\pi[f] = 0$ has the form $f = (c, -c)$, and

So solving costs a division by the rate $2\lambda$ at which the switch forgets its state. That is why each order of the expansion is smaller by a factor of order $1/\lambda$.

Hermite coordinates

For the Ornstein–Uhlenbeck factor write functions as sums of Hermite polynomials, $f = \sum_n f_n\,\mathrm{He}_n(y)$, with $\mathrm{He}_0 = 1$, $\mathrm{He}_1 = y$, $\mathrm{He}_2 = y^2 - 1$ and so on. The generator acts on each polynomial separately:

The stationary average of $f$ is its coefficient $f_0$, because every other Hermite polynomial averages to zero under the standard normal law. So the solvability condition is $f_0 = 0$, and the solution divides each coefficient by its rate:

The fast-factor example uses this.

Layers and the engine

The expansion that results from repeating these two steps describes the answer once the fast process has mixed. Near $t = 0$ it needs one more ingredient, described on the initial layers page. The engine page assembles the whole recursion.