2026-10-15 · chapter Yang–Mills and the mass gapPart 12 The second roaddriven insertion, Crooks and Jarzynski, the flow lemma · Kiichi, Shiori and Rin

Driven insertion and the flow lemma

1. The insertion of 12-05 was an operation that flips at once the signs of the couplings on a vortex sheet. Its cost $\Delta S$ is, for a thin vortex sheet, proportional to the number of plaquettes on the sheet (the area $N_T^2$ of the $T^2$), and the factor of the bound $\tanh(\Delta S/2)$ is almost 1.
2. So we split the coupling into $m$ steps and flip it little by little, updating only the inside of the container by heat bath at each step. The surrounding fluctuations relax to fit the new coupling, so the work paid $W$ goes down.
3. On the space of paths, the map "follow time backwards and flip the sector" is an involution, and the weight ratio is $e^{-W}$ (the Crooks relation) Cited. Applying Identity 3 of 12-05 to this involution gives $\mathbb E[H]=\mathbb E[H\tanh(W/2)]$ Proved here.
4. From the same relation, independently of the number of steps $m$ of the procedure, $\mathbb E[e^{-W}\,|\,\text{frame}]=Z_-/Z_+$ (the Jarzynski equality) Cited. With 0 steps, $W=\Delta S$ and we are back to 12-05.
5. For any stationary kernel that flips the sector, the probability $a$ that the sector flips in one step is at most $1-\theta$. In other words $\theta\le1-a$ (the flow lemma) Proved here. There is also a weaker form, $a\le1-\theta^2$ Proved here.
6. With the frame fixed the containers are independent, so running kernels in all containers simultaneously gives $\mathbb E\prod_t\theta_t\le\mathbb P(\text{no container flips})$ Proved here. The right-hand side is a quantity measurable by Monte Carlo, and it estimates an upper bound on the Wilson loop in the correct direction.
7. However, for a one-step kernel $a$ levels off at about $\tfrac12$, and what this guarantee can say goes only up to $\theta\lesssim\tfrac12$ Reading. It cannot be used to test the form of the band, $\theta\le e^{-cL^2}$. Its proper use is a lower bound on the slab tension with the $T^2$ fixed.

We use the sector-summed ensemble of 12-03 and 12-04. The weight on the $SU(2)$ link variables $U$ and the $\pm1$ plaquette variables $z$ (coclosed) is $w(U,z)=\exp\bigl(\tfrac\beta2\sum_pz_p\,\mathrm{tr}\,U_p\bigr)$. For a plaquette $p$ of the $(3,4)$ plane of the slab $T^2_{N_T}\times(\mathbb Z/M)^2$, flipping all the $z$ of the vortex sheet $S_p=\{p\}\times T^2$ flips the sign of the sector seen by a Wilson loop enclosing $p$ (12-04). We take a container $D_p\times T^2$ around $p$ ($D_p$ is a disc in the $(3,4)$ plane), and call the variables outside the container the frame $\Phi$.

Throughout this chapter we place the following three.

• (F) A finite configuration space $X$. The continuous case is as noted in 12-05 §02.
• (W) Positive weights $w=e^{-S}\gt0$.
• (K) Kernels that move only the inside of the container: the kernels $K$ below update only the variables inside the container and do not change the frame.

Interpolating the coupling. For $\lambda\in[0,1]$, let $S_\lambda$ be the action in which only the couplings of the plaquettes on $S_p$ are multiplied by $1-2\lambda$. $S_0=S$, and $S_1(x)=S(Rx)$ ($R$ is the flip of $z$ on $S_p$). $S_\lambda$ is gauge invariant.

Kernels. Let $K_\lambda$ be a heat-bath transition kernel that moves only the links inside the container and is reversible (detailed balance) with respect to $e^{-S_\lambda}$:

The forward procedure. $\lambda_k=k/m$ ($k=0,\dots,m$). Starting from a configuration $x_0$ in sector $+$, at each step $k=1,\dots,m$: "change the coupling $\lambda_{k-1}\to\lambda_k$, add the resulting change of action to the work, then update $x_{k-1}\to x_k$ by $K_{\lambda_k}$". The work is

$Rx_m$, obtained by finally applying $R$, is a configuration in sector $-$. From configurations in sector $-$, the same is done by the reverse procedure mapped by $R$ ($\lambda$ going $1\to0$). If $m=0$ there is no update, and $W=S_1(x_0)-S_0(x_0)=\Delta S$, the single insertion of 12-05.

The space of paths. Let $\Omega=\{(s,x_0,\dots,x_m)\}$ ($s=\pm$ is the sector of the starting point), and set the weight of a path starting in sector $+$ to be

The weight of a path starting in sector $-$ is defined by the product of the same form for the reverse procedure. The rows of the kernels sum to 1, so projecting the measure on $\Omega$ to the starting point gives back the original measure. Hence the expectation of a quantity determined by the starting point alone does not change when computed on $\Omega$ Proved here.

The involution. $\iota(+,x_0,\dots,x_m)=(-,Rx_m,\dots,Rx_0)$: follow the path backwards and flip the sector (and conversely for paths starting at $-$). $\iota\circ\iota=\mathrm{id}$. By (1), paths of positive weight are mapped to paths of positive weight.

Cited A relation of this form is due to Crooks (source table); here we write out the standard computation adapted to the lattice procedure. Proof. Let $\omega=(+,x_0,\dots,x_m)$. The weight of the reversed path, brought back to the coordinates of $+$ using that $S_\lambda$ maps to $S_{1-\lambda}$ under $R$, is $e^{-S_1(x_m)}\prod_{k=1}^mK_{\lambda_k}(x_k,x_{k-1})$. Replacing each factor by $K_{\lambda_k}(x_{k-1},x_k)\,e^{S_{\lambda_k}(x_k)-S_{\lambda_k}(x_{k-1})}$ using (1), the exponent of the ratio is

(shift the summation index by one, and let $S_0(x_0)$ cancel against the $k=1$ term and $-S_1(x_m)$ against the $k=m$ term). Computing (2) along the reversed path, the same terms appear with opposite signs, so $W\circ\iota=-W$. ∎

Lemma 1 says that "the involution with weight ratio $e^{-\Delta S}$" of (5) of 12-05 has been replaced, on the space of paths, by "an involution with weight ratio $e^{-W}$". So Identity 3 of 12-05 can be used as it is.

Proved here Proof. Regard $H$ as a function on $\Omega$ evaluated at the starting point. The kernels and $R$ do not change the frame, so $h$ is invariant under $\iota$ and $\sigma$ flips, giving $H\circ\iota=-H$. For the weight $\hat W$ and the involution $\iota$ on $\Omega$, the factor of (5) of 12-05 is $(\hat W-\hat W\circ\iota)/(\hat W+\hat W\circ\iota)=(1-e^{-W})/(1+e^{-W})=\tanh(W/2)$ by (4). From Identity 3 of 12-05, $\mathbb E_\Omega[H]=\mathbb E_\Omega[H\tanh(W/2)]$. The left-hand side is determined by the starting point alone, so it equals $\mathbb E[H]$ (§02). ∎

Lean: the core is BoundedInsertion (Bounded.single) (hypotheses: a finite set, an involution, $w\gt0$, $H$ odd. Conclusion: $\sum wHt=\sum wH$) applied to the finite space of paths $\Omega$ (restricted to paths of positive weight). The construction of the space of paths and Lemma 1 are outside Lean.

Cited Due to Jarzynski (source table). Proof. From (4), $\sum_\omega\hat W(\omega)e^{-W(\omega)}=\sum_\omega\hat W(\iota\omega)$. The right-hand side is the total weight of the paths starting in sector $-$, which is $Z_-(\Phi)$ since the rows of the kernels sum to 1. Dividing the left-hand side by the total weight $Z_+(\Phi)$ of paths in sector $+$ gives (6). ∎

(6) is a quantity that does not depend on the speed of the procedure (the number of steps $m$), and can be used as a numerical check. On the other hand, the average of $e^{-W}$ is dominated by rare small $W$, so with few samples the estimate fluctuates strongly (that is why some rows of the table in 12-08 have negative $-\ln\langle e^{-W}\rangle$).

Proved here One just takes the expectation of $1-\tanh(W/2)=2/(1+e^{W})$. It is Lemma 4 of 12-05 with $\Delta S$ replaced by $W$.

Many cells. For each of mutually separated cells $c$ (containers $D_c\times T^2$), choose a position $P_c$ uniformly and independently from a set of positions $P$ (a non-empty finite set), and perform driven insertion at the chosen position. The path of each cell reads only its own container and its surroundings in the starting configuration.

Proved here Proof. Build the enlarged space consisting of the starting point $x_0$, the choice of positions $(P_c)$ and the path of each cell. The involution $\iota_d$ of cell $d$ follows the path of cell $d$ backwards and replaces the inside of container $d$ of the starting point by the flip of the endpoint of the path. Since the containers are separated, the weight with the frame fixed splits into a product over containers, and $\iota_d$ does not change the path weights or works of other cells. So the weight ratio of $\iota_d$ is $e^{-W_d}$, and the factors $\tanh(W_c/2)$ of the other cells ($c\ne d$) do not move under $\iota_d$. The proof of Theorem 5 of 12-05 (the induction adding cells one at a time) goes through as it is, giving $\mathbb E[H]=\mathbb E[H\prod_c\tanh(W_{c,P_c}/2)]$. Taking absolute values, and using that with the starting point fixed the paths of the cells and the choices of positions are independent, gives (7). ∎

Unlike Theorem 5 of 12-05, (7) has the absolute value inside the average over positions. It is weaker by the amount that it does not use cancellation of signs between positions.

Zero steps. At $m=0$, $W=\Delta S$, and Theorem 2 is Identity 3 of 12-05 Proved here. For thin vortices $\Delta S$ is proportional to the area of the $T^2$, and $\tanh(W/2)$ is almost 1.

The quasi-static limit. As $m\to\infty$ (driving slowly), $W$ is expected to approach the conditional free energy of the container with the frame fixed, $\Delta F_D(\Phi)=\ln\bigl(Z_+(\Phi)/Z_-(\Phi)\bigr)$ Reading (physics; the standard estimate of thermodynamic integration, which requires the kernel to mix well enough at each $\lambda$. Not proved in this book). The factor of Theorem 2 then becomes $\tanh(\Delta F_D/2)=\theta_D(\Phi)$, and we return to the container identity of 12-03 (the Mack–Petkova identity). The reading is that the length $m$ of the procedure connects the crude bound of inserting at once with the exact conditional free energy.

The numerical picture. For $SU(2)$, $\beta=2.3$, the slab $2\times2\times8\times8$, twist $n_{12}=1$, disc radius 2, 6 configurations × 4 positions, the mean work at $m=0,4,16$ steps was 10.7, 6.5, 2.9, and the mean of $\tanh(W/2)$ was 0.990, 0.968, 0.688 Reading. Increasing the number of steps lowers the work and moves the factor away from 1. The records for other $\beta$, $N_T$ and container radii are collected in 12-08. If the container is smaller than the vortex, the work does not come down fully even as the number of steps increases (same table).

Inside the container $D\times T^2$, let $\mu_{\rm cut}$ be the measure (with the frame $\Phi$ fixed) in which only the plaquette terms on the flip sheet $S=\{p_0\}\times T^2$ are removed from the action, and set $X:=\sum_{p\in S}z_p\,\mathrm{tr}\,U_p$ ($z$ is the representative of sector $+$). The weight on the sheet is $e^{(\beta/2)z_p\mathrm{tr}U_p}$, and in sector $-$ the $z$ on $S$ are flipped, so

Proved here The first formula is from the definition; the second just puts the first into $\theta=|Z_+-Z_-|/(Z_++Z_-)$ and uses $e^{a}\mp e^{-a}=2\sinh a,\ 2\cosh a$.

Lean FlipLemma (FlipLemma.cut_formula) (hypotheses: real-valued weights $v$ on a finite set (non-negativity not used either), real-valued $X$, a real $b$. Conclusion: $\sum ve^{bX/2}-\sum ve^{-bX/2}=2\sum v\sinh(bX/2)$, and the equality in sum form for $\cosh$). What is guaranteed is only the algebra of sums.

$\mu_{\rm cut}$ has no sectors. $\theta(\Phi)$ can be read as "the strength with which the information of the frame reaches the sheet, crossing the cut measure over the size of the container". Driven insertion can also be set up in the symmetric form $+\to\text{cut measure}\leftarrow-$ (compatible with the involution of Lemma 1).

Let $\mu_\Phi$ be the conditional measure with the frame $\Phi$ fixed, $\sigma\in\{\pm1\}$ the sign of the sector of the container, $m_\Phi=\mathbb E[\sigma\,|\,\Phi]$, $\theta(\Phi)=|m_\Phi|$. Let $K$ be a Markov kernel that moves only the inside of the container, and $(X,X')\sim\mu_\Phi\otimes K$ ($X$ drawn from $\mu_\Phi$ and advanced one step by $K$ gives $X'$).

Proved here Proof. Since $\sigma=\pm1$, $\mathbb P(\sigma(X')\ne\sigma(X))=\sum_{x,y}\mu_\Phi(x)K(x,y)(\sigma(x)-\sigma(y))^2/4$. Bound it above by $(a-b)^2\le2(a-m)^2+2(b-m)^2$, and use that the rows sum to 1 and stationarity to turn both sums into $\sum_x\mu_\Phi(x)(\sigma(x)-m)^2=1-m^2$, giving $\tfrac12(1-m^2)+\tfrac12(1-m^2)$. ∎ Reversibility is not used.

Lean FlipLemma (FlipLemma.flip_le) (hypotheses: a finite set, $w\ge0$, $\sum w=1$, $K\ge0$, rows sum to 1, stationarity $\sum_xw_xK_{xy}=w_y$, $\sigma^2=1$. Conclusion: $\sum_{x,y}w_xK_{xy}(\sigma_x-\sigma_y)^2/4\le1-(\sum w\sigma)^2$).

Proved here Proof. $\mathbb P(\text{flip})=(1-\langle\sigma,K\sigma\rangle)/2$ (inner product in $\mu_\Phi$). Setting $\tilde\sigma=\sigma-m$, since the rows of $K$ sum to 1 and it is reversible, $\langle\sigma,K\sigma\rangle=\langle\tilde\sigma,K\tilde\sigma\rangle+m^2\ge m^2$. ∎ This corollary has no Lean check.

The next form is sharper, and is the one mainly used in this chapter.

Proved here Proof. Take the flow $A=\mathbb P(\sigma(X)=+,\sigma(X')=-)$ from $+\to-$ and the flow $B$ from $-\to+$. By stationarity the distribution of the sector of $X'$ is the same as that of $X$, so $A=B$. Also $A\le\mathbb P(\sigma(X)=+)=(1+m)/2$, $B\le(1-m)/2$. Hence $a=A+B=2A\le2\min\bigl((1+m)/2,(1-m)/2\bigr)=1-|m|$. ∎ Since $\theta\le1$, $1-\theta\le1-\theta^2$, and (10) is a sharper form than (9).

Lean FluxLemma (FluxLemma.flip_le_one_sub_abs) (hypotheses: a finite set, $w\ge0$, $\sum w=1$, $K\ge0$, rows sum to 1, stationarity, the values of $\sigma$ are $1$ or $-1$. Conclusion: $\sum_{x,y}w_xK_{xy}(\sigma_x-\sigma_y)^2/4\le1-|\sum w\sigma|$).

How to use it. If one can build a stationary kernel that flips the sector with flip probability at least $c$, then $\theta\le1-c$. A kernel that proposes a driven insertion and accepts the endpoint with probability $\min(1,e^{-W})$ is reversible (hence stationary) with respect to $\mu_\Phi$ by Lemma 1, and its flip probability is the acceptance rate. The acceptance rate is at least $\mathbb P(W\le w_0)\,e^{-w_0}$, so there is no need to control the average of $e^{-W}$; it is enough that "cheap work occurs with positive probability" Proved here.

Take a finite set $S$ of mutually separated containers, and let $\theta_t(\Phi)$ be the bias of the sector of each container $t\in S$ (the frame $\Phi$ is the variables outside all containers).

Proved here Proof. With the frame fixed, the insides of the containers are conditionally independent (the conditional independence of 12-03), and the kernels also move independently per container, so $\mathbb P(\text{none flips}\,|\,\Phi)=\prod_t(1-a_t(\Phi))$. By Theorem 9 each factor is at least $\theta_t(\Phi)$. Take expectations. ∎

The container identity of 12-03 had the form $|\langle W(C)\rangle|\le\mathbb E\prod_t\theta_t$ (for the containers inside $C$). Together with (11), an upper bound on the Wilson loop is bounded in the correct direction by a quantity measurable by Monte Carlo, $\mathbb P(\text{none flips})$ Proved here.

Limits (an important caution). What (10) gives is $\theta\le1-a$. The flip probability $a$ of a one-step kernel (a kernel that proposes one driven insertion and decides whether to accept it) was at most 0.46 in the range measured, and in the judgement of the record it levels off at about $\tfrac12$ Reading (Tables 1, 2, 8 and 9 of 12-08). So what can be said for containers of size $L$ is $\theta\le\theta_0$ with $\theta_0\sim\tfrac12$ at best. Even if this held uniformly in $L$, all that follows is

and this is exactly the "weak form" of 13-01 (a form that $U(1)$ also satisfies) Proved here. To show the correct form of the band $\theta(L,L)\le e^{-cL^2}$ by (10), one needs a kernel with $a\ge1-e^{-cL^2}$, that is, a kernel that flips the sector almost surely Proved here (a rewriting of (10)). There is no prospect of building that with a one-step kernel.

Therefore, the guarantee from simultaneous flips cannot be used as a numerical test of the crossover band. There is a measured record of this tool as a numerical picture of the band, but that positioning has been withdrawn (12-08 §06). Its proper uses are a lower bound on the slab tension with the size $L$ of the $T^2$ fixed, and an estimate from above of the chessboard density $\kappa_{cb}$ (13-05).

(a) Confinement. None of the identities and inequalities of this chapter implies confinement. One would need the factors of Theorems 2 and 5 to stay away from 1 uniformly in $L$ and the lattice spacing, or a kernel with $a$ close to 1, and this book shows neither.

(b) The quasi-static limit. That $W\to\Delta F_D$ as $m\to\infty$ is the estimate of §04; this book has no proof of it, including bounds on how the kernels mix.

(c) The continuous configuration space. All theorems are finite versions under (F). The version on the Haar measure of $SU(2)$ needs the reversibility of the heat-bath kernels and the measure on the space of paths to be rewritten for the continuous case, and is outside Lean.

(d) The levelling off of one-step kernels. "$a$ levels off at about $\tfrac12$" is a measurement and a judgement of the record, not a theorem. The conclusion that only up to $\theta_0\sim\tfrac12$ can be said rests on this reading. Conversely, if there were a kernel with $a$ close to 1, (10) could be used as it is, but no construction of one is known.

(e) Novelty. The Crooks relation and the Jarzynski equality are known, and the application to the sector-summed ensemble and Theorems 7, 9 and 10 are elementary. Techniques that raise the acceptance rate with non-equilibrium proposals that flip a sector are known in the Monte Carlo literature as tempered transitions and non-equilibrium candidate methods (not yet checked). No novelty is claimed for the results of this chapter.

The bundles in the machine-checking column are those confirmed, by re-checking the main text, to contain no sorry and to use axioms forming a subset of the three standard ones. What is guaranteed is only the algebra of sums over finite sets (10-13). flip_le and cut_formula are in the file FlipLemma, and flip_le_one_sub_abs is in the file FluxLemma.

Dependency graph (the numbers of premises are smaller than the number of the conclusion):

[01] PRV MARG    marginal of start on paths = original measure   rows sum to 1
[02] CIT CROOKS  weight ratio e^(−W), W∘ι = −W                   detailed balance (1)
[03] PRV DRIVEN  E[H] = E[H tanh(W/2)] ⟸ [01][02]               Lean Bounded.single on Ω
[04] CIT JARZ    E[e^(−W) | frame] = Z₋/Z₊ ⟸ [02]
[05] PRV ACC     1 − E[tanh(W/2)] = 2 × acceptance ⟸ [03]
[06] PRV MULTI   many cells (7) ⟸ [02]                           containers separated
[07] RDG QSTAT   m → ∞ gives W → ΔF_D (physics) ⟸ [04]
[08] RDG NUM     work goes down with the number of steps          β = 2.3, 2×2×8×8
[09] PRV CUT     formula of the cut measure                      Lean cut_formula
[10] PRV FLIP    flip probability ≤ 1 − θ²                       Lean flip_le
[11] PRV POS     (1 − θ²)/2 for positive kernels ⟸ [10]
[12] PRV FLUX    θ ≤ 1 − a                                       Lean flip_le_one_sub_abs
[13] PRV MH      min(1, e^(−W)) kernel stationary, lower bound ⟸ [02]
[14] PRV SIMUL   E∏θ ≤ P(no flip) ⟸ [12]                         containers separated
[15] RDG HALF    a of one-step kernels levels off at about ½     range measured
[16] PRV WEAK    only the weak form; band needs a ≥ 1 − e^(−cL²) ⟸ [12]

The way to check is to see by eye that every number to the right of "⟸" is smaller than the number on the left. There are no circles. This chapter has no assumption rows. The reading [15] is not made a premise of [16] ([16] is a rewriting that follows from $\theta\le1-a$ alone, and [15] is a judgement of how far it can be used). There is neither a node of confinement nor a node of the mass gap.

What this chapter can say

Sources and reproduction

Words used in this chapter — driven insertion, the number of steps $m$, the work $W$, the space of paths $\Omega$, the Crooks involution $\iota$, quasi-static, the cut measure, the flip probability, the flow lemma, simultaneous flips. Their meanings are in A-02 Glossary (the words of Part 12 are to be added).

Next chapter to read: 12-07 Why fixed maps fail — why driven insertion is cheap and a fixed map applied at once is expensive, explained at the one-loop level as the difference of "whether the shape of the fluctuations is changed".

← 12-05 Vortex-insertion identitiesContents12-07 Why fixed maps fail →