Contact calibrators under explicit join coherence

Work in [F8]'s common affine mathematical energy orientation, without redefining any physical energy or calibration. Apply [F1] and [F2] to each simple type, including every finite join, using the positive normalization by its fixed weight in [A1]. Throughout, entropy at a state means its represented-state value: $w_{\mathrm{\Gamma }}$ times its unit-point-mass representation. This normalization preserves order and finite scalar amounts. The convex-combination access hypothesis in [A1] makes each resulting entropy concave, since the comparison of a scaled pair with its convex-combination state translates under the representation to the concavity inequality. By [F4], each entropy is strictly increasing with energy at fixed work coordinates.

Use T5 to choose $\bar{x}\in {\mathrm{\Gamma }}_1$ and $\bar{y}\in {\mathrm{\Gamma }}_2$ with $\bar{x}{\sim }_T\bar{y}$. Let $\varphi =J_{12}(\bar{x},\bar{y})$ and consider its adiabat $\mathrm{\partial }{A}_{\varphi }$. T2 gives every $Z$ on that adiabat at least one split $Z=J_{12}(x_Z,y_Z)$ with $x_Z{\sim }_T{y}_Z$.

Suppose a join state $Z$ on $\mathrm{\partial }{A}_{\varphi }$ has two distinct equilibrium splits $(x^{(1)},y^{(1)})$ and $(x^{(2)},y^{(2)})$. By the definition of ${\sim }_T$, both pairs are reversible with $Z$, hence mutually accessible compounds. Their work coordinates are fixed by $Z$ and their total energies equal its energy. If the first-component energies agreed, both component states in the splits would agree, so distinctness forces unequal first-component energies. If $u_{x^{(1)}}<u_{x^{(2)}}$, fixed-work comparison [F4] gives $x^{(1)}\prec \prec x^{(2)}$; conservation of total energy gives $u_{y^{(2)}}<u_{y^{(1)}}$, hence $y^{(2)}\prec \prec y^{(1)}$. Taking $(x_0,x_1,y_0,y_1)=(x^{(1)},x^{(2)},y^{(2)},y^{(1)})$ gives $(x_0,y_1)\sim (x_1,y_0)$ because both split compounds are reversible with $Z$. If the first energies occur in the opposite order, swap the two splits and obtain the same calibrator form. Therefore, if no calibrator exists, every $Z\in \mathrm{\partial }{A}_{\varphi }$ has a unique equilibrium split.

Assume no calibrator exists. For any equilibrium split $(x_Z,y_Z)$ of $Z\in \mathrm{\partial }{A}_{\varphi }$, reversibility with $Z$ and the common adiabat give $(x_Z,y_Z)\sim (\bar{x},\bar{y})$ as compounds. By [F1], $x_Z$ and $\bar{x}$ are comparable; [F4] says their sectors coincide exactly when they are equivalent. If $x_Z\prec \prec \bar{x}$, pair reversibility gives $(\bar{x},\bar{y})\prec (x_Z,y_Z)$. Consistency adds $\bar{y}$ to $x_Z\prec \bar{x}$, and cancellation [F5] gives $\bar{y}\prec y_Z$. If $y_Z\prec \bar{y}$, consistency and the pair comparison instead cancel to give $\bar{x}\prec x_Z$, a contradiction. Thus $\bar{y}\prec \prec y_Z$, making a calibrator. The case $\bar{x}\prec \prec x_Z$ is symmetric: it gives $y_Z\prec \prec \bar{y}$, again a calibrator. Since neither is allowed under the assumption, $x_Z\sim \bar{x}$. Adding spectators to the two directions and cancelling by [F5] then gives both $\bar{y}\prec y_Z$ and $y_Z\prec \bar{y}$, so $y_Z\sim \bar{y}$. Thus every split on this adiabat has components in the corresponding component adiabats.

Under the no-calibrator assumption, fix a first-component work coordinate $v_1$ in the projection of $\mathrm{\partial }{A}_{\bar{x}}$. The boundary-graph result [F4] gives exactly one state $x_{\ast }$ on that component adiabat at $v_1$. Every split of a join state on $\mathrm{\partial }{A}_{\varphi }$ with first work coordinate $v_1$ therefore has first component $x_{\ast }$. Each split is thermal, so $x_{\ast }{\sim }_T{y}_Z$ for its second component. T3 makes all such $y_Z$ thermally equivalent, and C1 makes each joined state $J(x_{\ast },y_Z)=Z$ thermally equivalent to $x_{\ast }$. Hence all join states on this adiabat over fixed $v_1$ lie in one thermal class. Fixing a second-component work coordinate instead gives the corresponding conclusion for first components.

Let $\mathrm{\Omega }$ be the open connected work-coordinate projection of $\mathrm{\partial }{A}_{\varphi }$ from [F4]. From any point of $\mathrm{\Omega }$, the set reachable by finitely many coordinate-axis segments contained in open boxes in $\mathrm{\Omega }$ is relatively open, and its complement is also relatively open; connectedness makes the reachable set all of $\mathrm{\Omega }$. Along each such segment, step 4.1 gives equivalence of the corresponding join states; T3 composes these finitely many equivalences. Thus the whole adiabat $\mathrm{\partial }{A}_{\varphi }$ is one thermal class. Since the argument works for every starting join state, every join adiabat is one thermal class.

Still assuming no calibrator, apply [F3] to each join system. Its alternative with two thermally inequivalent states on one adiabat is excluded by step 5.1, so all states of each join system are thermally equivalent. For any $x,x^{\mathrm{\prime }}\in {\mathrm{\Gamma }}_1$, use T5 with one allowed second work coordinate to choose $y,y^{\mathrm{\prime }}$ with $x{\sim }_{T}y$ and $x^{\mathrm{\prime }}{\sim }_T{y}^{\mathrm{\prime }}$. C1 relates each component to its joined state, and all join states are in one thermal class; T3 therefore gives $x{\sim }_T{x}^{\mathrm{\prime }}$. The same argument gives one thermal class in ${\mathrm{\Gamma }}_2$. Now take $Z=\varphi \in \mathrm{\partial }{A}_{\varphi }$ and a split $(x,y)$ supplied by T2. Choose nonzero sufficiently small $\delta$ so $(u_x+\delta ,v_1)\in {\mathrm{\Gamma }}_1$ and $(u_y-\delta ,v_2)\in {\mathrm{\Gamma }}_2$, possible because the domains are open. These states form a distinct split of the same $Z$. Since the original split has $x{\sim }_{T}y$ and each component type is one thermal class, T3 gives thermal equivalence of the changed pair as well. This contradicts uniqueness of the split from step 2.1. Therefore a calibrator exists.

Fix a calibrator $x_0\prec \prec x_1$, $y_0\prec \prec y_1$, whose existence is step 6.1. Choose same-type entropy representatives $S_1,S_2$ from [F2], with positive gaps $h_1=S_1(x_1)-S_1(x_0)$ and $h_2=S_2(y_1)-S_2(y_0)$. Normalize ${\tilde{S}}_i=S_i\mathrm{/}{h}_i$, so both calibrator increments are one, and put $s(x)={\tilde{S}}_1(x)-{\tilde{S}}_1(x_0)$ and $t(y)={\tilde{S}}_2(y)-{\tilde{S}}_2(y_0)$. Write $u=\mathsf{J}(x_0)+\mathsf{J}(y_1)$ and $v=\mathsf{J}(x_1)+\mathsf{J}(y_0)$. The calibrator says $u\sim v$, so $v-u=(\mathsf{J}(x_1)-\mathsf{J}(x_0))-(\mathsf{J}(y_1)-\mathsf{J}(y_0))$ is equivalent to zero; translation and cancellation [F5] give $(\mathsf{J}(x_1)-\mathsf{J}(x_0)){\sim }_s(\mathsf{J}(y_1)-\mathsf{J}(y_0))$. For any state $x$, the signed map $d_x=\mathsf{J}(x)-(1-s(x))\mathsf{J}(x_0)-s(x)\mathsf{J}(x_1)$ has total mass and ${\tilde{S}}_1$-value zero. Its positive and negative parts have equal mass and entropy, so [F2] makes them equivalent. Hence $\mathsf{J}(x){\sim }_s(1-s(x))\mathsf{J}(x_0)+s(x)\mathsf{J}(x_1)$; the same argument applies to states of the second type. These are formal signed maps when a coefficient lies outside $[0,1]$.

For a state pair $(x,y)$, put $\lambda =(s(x)+t(y))\mathrm{/}2$ and $\delta =(s(x)-t(y))\mathrm{/}2$. Adding the signed reductions from step 7.1 makes the pair equivalent to $(1-\lambda )(\mathsf{J}(x_0)+\mathsf{J}(y_0))+\lambda (\mathsf{J}(x_1)+\mathsf{J}(y_1))$ plus $\delta [(\mathsf{J}(x_1)-\mathsf{J}(x_0))-(\mathsf{J}(y_1)-\mathsf{J}(y_0))]$. The calibrator increment equivalence from step 7.1 makes the latter term equivalent to zero for either sign of $\delta$. Thus each pair reduces to its reference map at parameter $\lambda$. For any finite compound $A$ with amounts $M_1,M_2$ of the two types, add and scale these single-state reductions. Writing $q_1(A),q_2(A)$ for the amount-weighted sums of $s(x),t(y)$, respectively, gives $A{\sim }_s(M_1-q_1(A))\mathsf{J}(x_0)+q_1(A)\mathsf{J}(x_1)+(M_2-q_2(A))\mathsf{J}(y_0)+q_2(A)\mathsf{J}(y_1)$. Using $(\mathsf{J}(x_1)-\mathsf{J}(x_0)){\sim }_s(\mathsf{J}(y_1)-\mathsf{J}(y_0))$, this is equivalent to $M_1\mathsf{J}(x_0)+M_2\mathsf{J}(y_0)+Q(A)(\mathsf{J}(x_1)-\mathsf{J}(x_0))$, where $Q(A)=q_1(A)+q_2(A)$. If the total mass is zero, both compounds are empty and reflexivity compares them. Otherwise, for same-composition $A,B$, their base terms agree. If $Q(A)<Q(B)$, the positive and negative parts of the reference-map difference are $(Q(B)-Q(A))\mathsf{J}(x_1)$ and $(Q(B)-Q(A))\mathsf{J}(x_0)$, so $x_0\prec x_1$ and positive scaling compares the maps forward. If the Q values agree the maps coincide; if $Q(A)>Q(B)$ the reverse comparison holds. Translation, addition, and ordinary-order compatibility [F5] transfer the result to $A,B$. This proves CH for every fixed finite composition of the two types and their scaled copies.