Proof. Suppose $y \in \mathcal {J}_L(m)$ is not join-irreducible. Then $y=\bigvee X$ for some set $X\subseteq L$ with $y \not \in X$ . Consider some $x\in X$ . We have $y> x$ , so $m^*=y \vee m \geq x\vee m \geq m$ . This implies that $x\vee m$ is either m or $m^*$ . But we know that $x\vee m\neq m^*$ because y is a minimal element of $\{z\in L: m\vee z = m^*\}$ . Hence, $x\leq m$ . As x was arbitrary, it follows that $y=\bigvee X\leq m$ . However, this implies that $y\vee m=m$ , which contradicts the fact that $y\vee m=m^*$ . This proves that $\mathcal {J}_L(m)\subseteq \mathcal {J}_L$ . A dual argument proves that $\mathcal {M}_L(j)\subseteq \mathcal {M}_L$ .

Proof. Let $\kappa \colon \mathcal {J}_L\to \mathcal {M}_L$ be a bijection. Let $j\in \mathcal {J}_L$ , and write $m=\kappa (j)$ . Suppose $m\geq j_*$ , $m^*\geq j$ , and $m\not \geq j$ . Since $j \geq j \wedge m \geq j_*$ and $m \not \geq j$ , we must have $j \wedge m=j_*$ . The element m is maximal among elements z with $j \wedge z = j_*$ because we have $j\wedge x\geq j \wedge m^* = j$ for all $x>m$ . Therefore, $m \in \mathcal {M}_L(j)$ . An analogous argument shows that $j \in \mathcal {J}_L(m)$ . Hence, $\kappa $ is a pairing.

Conversely, suppose that $\kappa $ is a pairing. Fix $j\in \mathcal {J}_L$ and let $m=\kappa (j)$ . Since $m \in \mathcal {M}_L(j)$ , we have $j \wedge m = j_*$ . This implies that $m \geq j_*$ and $m \not \geq j$ . We also have $m^*\wedge j\geq m\wedge j=j_*$ . The maximality of m coming from the definition of $\mathcal {M}_L(j)$ guarantees that $m^*\wedge j\neq j_*$ , so $m^*\wedge j\geq j$ . Hence, $m^* \geq j$ .

Proof. Since $m_0^* \not \in [\hat {0},m_0]$ , we have $m_0^* \in [j_0,\hat {1}]$ , so $m_0^* \geq j_0$ . It follows that $m_0 \vee j_0 = m_0^*$ . Since $m_0\vee (j_0)_*=m_0$ , we find that $j_0 \in \mathcal {J}_L(m_0)$ . Since $L=[\hat {0},m_0]\sqcup [j_0,\hat {1}]$ , any z satisfying $m_0^*=m_0\vee z $ must satisfy $z \geq j_0$ . Therefore, $\mathcal {J}_L(m) = \{j_0\}$ . By the definition of a pairing, we must have $\kappa (j_0)=m_0$ .

Proof. Suppose that j is join-prime and that $j'\in \mathcal {J}_L$ satisfies $j\to j'$ . Write $m=\kappa _L(j)$ and $m'=\kappa _L(j')$ . Then $j\not \leq m'$ . Theorem 3.4 tells us that $(j,m)$ is a prime pair, so $L=[\hat 0,m]\sqcup [j,\hat 1]$ . Consequently, $m'< m$ . Because $m'$ is a maximal element of $\{z\in L: j^{\prime }_*=j'\wedge z\}$ , we cannot have $j'\wedge m=j^{\prime }_*$ . We know that $j' \wedge m \geq j' \wedge m' = j^{\prime }_*$ , so we must have $j' \wedge m> j^{\prime }_*$ . Hence, $j'\wedge m=j'$ . This means that $j'\leq m$ , so there is no edge $j'\to j$ in $G_L$ .

Proof. Let L be a semidistributive lattice. Then $\mathcal {M}_L(j)$ and $\mathcal {J}_L(m)$ are singleton sets for all $j\in \mathcal {J}_L$ and $m\in \mathcal {M}_L$ . Furthermore, one readily checks that m is the unique element of $\mathcal {M}_L(j)$ if and only if j is the unique element of $\mathcal {J}_L(m)$ . Therefore, we obtain a pairing $\kappa \colon \mathcal {J}_L\to M_L$ by declaring $\kappa (j)$ to be the unique element of $\mathcal {M}_L(j)$ . This is the only pairing on L.

Proof. Let L be an extremal lattice and fix a maximum-length chain $\hat {0}=x_0\lessdot x_1\lessdot x_2\lessdot \cdots \lessdot x_n=\hat {1}$ . For each $i\in [n]$ , there is a unique $j_i\in \mathcal {J}_L$ such that $j_i \vee x_{i-1}=x_i$ , and there is a unique $m_i\in \mathcal {M}_L$ such that $m_i \wedge x_i = x_{i-1}$ . This gives rise to a bijection $\kappa \colon \mathcal {J}_L\to \mathcal {M}_L$ defined by $\kappa (j_i)=m_i$ . Let us check that $\kappa $ is a pairing. Fix $i\in [n]$ . First, suppose j is a join-irreducible with $j\leq (j_i)_*$ . Then $j\vee x_{i-1}\leq j_i\vee x_{i-1}=x_i$ . By the uniqueness of $j_i$ , we know that $j\vee x_{i-1}\neq x_i$ . This shows that $j\vee x_{i-1}=x_{i-1}$ , so $j\leq x_{i-1}$ . As this is true for every join-irreducible j satisfying $j\leq (j_i)_*$ , we conclude that $(j_i)_*\leq x_{i-1}$ . Because $j_i\vee x_{i-1}=x_i$ , we have $j_i\not \leq x_{i-1}$ , so

$$\begin{align*}j_i\neq x_{i-1} \wedge j_i = (m_i \wedge x_i) \wedge j_i = m_i \wedge (x_i \wedge j_i) = m_i \wedge j_i.\end{align*}$$

Consequently, $m_i\wedge j_i<j_i$ . But since $(j_i)_* \leq x_{i-1}$ , we have $m_i \wedge j_i = x_{i-1}\wedge j_i= (j_i)_*$ . This shows that $m_i\geq (j_i)_*$ and that $m_i\not \geq j_i$ . A similar argument shows that $m_i^*\geq j_i$ . As this is true for all i, Lemma 3.3 tells us that $\kappa $ is a pairing.

We now show that $\kappa $ is the only pairing on L. Suppose instead that there is some pairing $\kappa '\colon \mathcal {J}_L\to \mathcal {M}_L$ with $\kappa '\neq \kappa $ . There must be some $i,k\in [n]$ with $i<k$ such that $\kappa '(j_i)=m_k$ . Then $x_k> x_{k-1} \geq x_i \geq j_i$ , so $m_k \wedge j_i = m_k \wedge (x_k \wedge j_i) = (m_k\wedge x_k)\wedge j_i=x_{k-1} \wedge j_i = j_i$ . This contradicts the fact that $\kappa '(j_i)\wedge j_i=(j_i)_*$ by the definition of a pairing.

Proof. It is immediate from the definition of the edges in the Galois graph that $\{j\in \mathcal {J}_L:j_0\leq \kappa _L(j)\}=\mathcal {J}_L\setminus (\{j_0\}\cup \mathrm {Out}(j_0))$ . Now suppose $j',j"\in \mathcal {J}_L$ are such that $j_0\leq \kappa _L(j'),\kappa _L(j")$ . Then we have $j'\to j"$ in $G_{L}$ if and only if $j'\not \leq \kappa _L(j")$ , and this occurs if and only if $j_0\vee j'\not \leq \kappa _L(j")$ . The first compatibility condition in Theorem 4.1 tells us that $\kappa _L(j")=\kappa _{L^0}(j_0\vee j")$ , so it follows that $j'\to j"$ in $G_{L}$ if and only if $\alpha (j')\to \alpha (j")$ in $G_{L^0}$ .

Now observe that $\mathcal {J}_{L_0}=\{j\in \mathcal {J}_L: j\leq m_0\}=\{j\in \mathcal {J}_L:j\leq \kappa _L(j_0)\}=\mathcal {J}_L\setminus (\{j_0\}\cup \mathrm {In}(j_0))$ . Suppose $j',j"\in \mathcal {J}_{L_0}$ . We have $j'\to j"$ in $G_L$ if and only if $j'\not \leq \kappa _L(j")$ , and this occurs if and only if $j'\not \leq m_0\wedge \kappa _L(j")$ . The second compatibility condition in Theorem 4.1 tells us that $\kappa _{L_0}(j")=m_0\wedge \kappa _{L}(j")$ , so it follows that $j'\to j"$ in $G_{L}$ if and only if $j'\to j"$ in $G_{L_0}$ .

Proof. Let L be a semidistributive lattice of cardinality at least $2$ . We know by Theorem 3.8 that L is uniquely paired; let $\kappa _L\colon \mathcal {J}_L\to \mathcal {M}_L$ be its unique pairing. Let $j_0$ be an atom of L. By [Reference Gaskill and Nation25, Lemma 1], $j_0$ is necessarily join-prime. Let $L_0=[\hat 0,m_0]$ and $L^0=[j_0,\hat 1]$ , where $m_0=\kappa (j_0)$ so that $(j_0,m_0)$ is a prime pair. Since intervals of semidistributive lattices are semidistributive, it follows by induction on the size of L that $L_0$ and $L^0$ are both compatibly dismantlable. Suppose $j\in \mathcal {J}_L$ is such that $j_0\leq \kappa _L(j)$ , and let $m=\kappa _L(j)$ . We want to show that $j_0\vee j$ is join-irreducible in $L^0$ . Because L is semidistributive, j is the unique element of the singleton set $\mathcal {J}_L(m)=\min \{z\in L: m^*=m\vee z\}$ . Since $L^0$ is semidistributive, the set $\{z\in L^0 : m^*=m \vee z \}$ has a unique minimal element. Certainly $j_0 \vee j$ is in this set since $m \vee (j_0 \vee j) = m \vee j = m^*$ . Suppose there is some $x\in L^0$ with $x< j\vee j_0$ and $m\vee x=m^*$ . We have $m\vee x=j\vee x=m^*$ , so it follows from the fact that L is join-semidistributive that $m\vee (x\wedge j)= m^*$ . Now $j\in \mathcal {J}_L(m)$ , so we must have $x \wedge j = j$ , meaning $x\geq j$ . But $x\geq j_0$ , so this contradicts the assumption that $x<j\vee j_0$ . This shows that no such x exists, so $j_0\vee j$ is the unique minimal element of $\{z\in L^0 : m^*=m \vee z \}$ . In other words, $j_0\vee j\in \mathcal {J}_{L^0}(m)$ . By Lemma 3.1, $j_0\vee j\in \mathcal {J}_{L^0}$ .

We now have a map $\alpha \colon \{j\in \mathcal {J}_L:j_0\leq \kappa _L(j)\}\to \mathcal {J}_{L^0}$ given by $\alpha (j)=j_0\vee j$ . Let us show that $\alpha $ is injective. Suppose $j,j'\in \mathcal {J}_L$ are such that $j_0\leq \kappa _L(j),\kappa _L(j')$ and $j_0\vee j=j_0\vee j'$ . Let $m=\kappa _L(j)$ and $m'=\kappa _L(j')$ . Then $(j_0\vee j)\vee m=j\vee (j_0\vee m)=j\vee m=m^*$ , so $j'\vee m=j'\vee (j_0\vee m)=(j_0\vee j')\vee m=(j_0\vee j)\vee m=m^*$ . Since j is the unique minimal element of $\{z\in L:m^*=m\vee z\}$ , we must have $j'\geq j$ . Reversing the roles of j and $j'$ shows that $j\geq j'$ . Hence, $j=j'$ .

We now show that $\alpha $ is surjective and that $\kappa _{L^0}(\alpha (j))=\kappa _L(j)$ for every $j\in \mathcal {J}_L$ with $j_0\leq \kappa _L(j)$ . Suppose $j'\in \mathcal {J}_{L^0}$ . Let $m'=\kappa _{L^0}(j')$ . Then $m'$ is meet-irreducible in $L^0$ , so it is meet-irreducible in L. Let $j=\kappa _L^{-1}(m')$ . We aim to show that $j_0\vee j=j'$ . Since $m'\vee j'=(m')^*$ and j is the unique minimal element of $\{z\in L:(m')^*=m'\vee z\}$ , we have $j'\geq j$ . Thus, $j'\geq j_0\vee j$ . However, we have $(j_0\vee j)\vee m'=(m')^*$ . Since $j'$ is the unique minimal element of the set $\{z\in L^0:(m')^*=m'\vee z\}$ , we must have $j'\leq j_0\vee j$ . This proves that $j'=j_0\vee j$ . Moreover, $\kappa _L(j)=m'=\kappa _{L^0}(j')=\kappa _{L^0}(\alpha (j))$ .

Proof. By Theorem 3.9, trim lattices are uniquely paired. [Reference Thomas and Williams50, Theorem 2.4] states that the Galois graph $G_L$ of a trim lattice is acyclic. Let $j_0$ be a sink of $G_L$ . Then [Reference Markowsky31, Theorem 15] tells us that $j_0$ is a join-prime atom of L; let $m_0$ be the corresponding meet-prime element so that $(j_0,m_0)$ is a prime pair. The proof of [Reference Thomas and Williams50, Proposition 3.13] (following the proof of [Reference Thomas48, Theorem 1]) shows that the compatibility conditions in Theorem 4.1 hold for this choice of prime pair.

Proof. Suppose $\lambda \colon \mathcal {J}_{L\times L'}\to \mathcal {M}_{L\times L'}$ is a pairing. If $j\in \mathcal {J}_L$ , then since $(j,\hat 0')\not \leq \lambda (j,\hat 0')$ , it follows from (7.1) that $\lambda (j,\hat 0')=(\mu (j),\hat 1')$ for some $\mu (j)\in \mathcal {M}_L$ . Similarly, if $j'\in \mathcal {J}_{L'}$ , then $\lambda (\hat 0,j')=(\hat 1,\mu '(j'))$ for some $\mu '(j')\in \mathcal {M}_{L'}$ . It is straightforward to check that the resulting maps $\mu \colon \mathcal {J}_L\to \mathcal {M}_L$ and $\mu '\colon \mathcal {J}_{L'}\to \mathcal {M}_{L'}$ are pairings on L and $L'$ , respectively. Since L and $L'$ are uniquely paired, we have $\mu =\kappa _L$ and $\mu '=\kappa _{L'}$ . This shows that if $L\times L'$ is paired, then it is uniquely paired. However, one can readily check that the map $\kappa _{L\times L'}\colon \mathcal {J}_{L\times L'}\to \mathcal {M}_{L\times L'}$ given by $\kappa _{L\times L'}(j,\hat 0')=(\kappa _L(j),\hat 1')$ and $\kappa _{L\times L'}(\hat 0,j')=(\hat 1,\kappa _{L'}(j'))$ is indeed a pairing on $L\times L'$ .

Proof. Theorem 7.1 tells us that $L\times L'$ is uniquely paired, and the proof shows how to compute $\kappa _{L\times L'}$ in terms of $\kappa _L$ and $\kappa _{L'}$ . Let $(j_0,m_0)$ and $(j_0',m_0')$ be dismantling pairs for L and $L'$ , respectively. One readily checks that $((j_0,\hat 0'),(m_0,\hat 1'))$ is a dismantling pair for $L\times L'$ . Using the easily-verified decomposition

$$\begin{align*}\{(a,b)\in \mathcal{J}_{L\times L'}:(j_0,\hat 0')\leq \kappa_{L\times L'}(a,b)\}=\{(j,\hat 0'):j_0\leq\kappa_L(j)\}\sqcup(\{\hat 0\}\times\mathcal{J}_{L'}),\end{align*}$$

one can check that the map

$$\begin{align*}\alpha\colon\{(a,b)\in \mathcal{J}_{L\times L'}:(j_0,\hat 0')\leq \kappa_{L\times L'}(a,b)\}\to\mathcal{J}_{[(j_0,\hat 0'),(\hat 1,\hat 1')]}\end{align*}$$

given by $\alpha (a,b)=(j_0,\hat 0')\vee (a,b)$ is a bijection such that

$$\begin{align*}\kappa_{[(j_0,\hat 0'),(\hat 1,\hat 1')]}(\alpha(a,b))=\kappa_{L\times L'}(a,b)\end{align*}$$

for all $(a,b)\in \mathcal {J}_{L\times L'}$ with $(j_0,\hat 0')\leq \kappa _{L\times L'}(a,b)$ . In other words, the first compatibility condition in Theorem 4.1 holds for $L\times L'$ . The second compatibility condition also holds similarly.

Proof. Theorem 7.2 tells us that $L\times L'$ is compatibly dismantlable. Using (7.1) and the description of $\kappa _{L\times L'}$ given in the proof of Theorem 7.1, it is straightforward to show that the Galois graph $G_{L\times L'}$ is naturally isomorphic to $G_L\sqcup G_{L'}$ , the disjoint union of $G_L$ and $G_{L'}$ (with no vertices in $G_L$ adjacent to any vertices in $G_{L'}$ ). More precisely, the isomorphism sends $(j,\hat 0')\in \mathcal {J}_L\times \{\hat 0'\}$ to j and sends $(\hat 0,j')\in \{\hat 0\}\times \mathcal {J}_{L'}$ to $j'$ .

Consider $(y,y')\in L\times L'$ . The elements covered by $(y,y')$ in $L\times L'$ are precisely the elements of the form $(x,y')$ with $x\lessdot y$ in L or of the form $(y,x')$ with $x'\lessdot y'$ in $L'$ . Moreover, the label of $(x,y')\lessdot (y,y')$ is $(j_{xy},\hat 0')$ , while the label of $(y,x')\lessdot (y,y')$ is $(\hat 0,j_{x'y'})$ . It follows that under the aforementioned isomorphism between $G_{L\times L'}$ and $G_L\sqcup G_{L'}$ , the set $\mathcal {D}_{L\times L'}(y,y')$ corresponds to $\mathcal {D}_L(y)\sqcup \mathcal {D}_{L'}(y')$ . Since $\mathcal {D}_L(y)\in \mathrm {Ind}(G_L)$ and $\mathcal {D}_{L'}(y')\in \mathrm {Ind}(G_{L'})$ , we have $\mathcal {D}_{L\times L'}(y,y')\in \mathrm {Ind}(G_{L\times L'})$ . A similar argument shows that $\mathcal {U}_{L\times L'}(y,y')\in \mathrm {Ind}(G_{L\times L'})$ .

Proof. Let $(j_0,m_0)$ be a dismantling pair for L and let $L_0=[\hat 0,m_0]$ and $L^0=[j_0,\hat 1]$ . First, suppose $v\in L_0$ . Lemma 6.3 tells us that $L_0$ is semidistrim, and the second compatibility condition in Theorem 4.1 states that $\kappa _{L_0}(j)=m_0\wedge \kappa _L(j)$ for all $j\in \mathcal {J}_{L}$ with $j\leq m_0$ . In particular, for $j\in \mathcal {J}_{L_v}$ , we have $\kappa ^{\wedge v}(j)=\kappa _L(j)\wedge v=\kappa _{L_0}(j)\wedge v$ . Since $L_v$ is a lower interval in $L_0$ , we can use induction on the size of the lattice to see that $\kappa ^{\wedge v}$ is a pairing on $L_v$ .

Now assume $v\in L^0$ . Let $j\in \mathcal {J}_{L_v}$ . By Lemma 5.7, we have $j \in \mathcal {U}_{L}(\kappa ^{\wedge v}(j))$ . In other words, there is some cover relation $\kappa ^{\wedge v}(j)\lessdot y$ with label j. Note that $y=\kappa ^{\wedge v}(j)\vee j\leq v$ . Suppose by way of contradiction that there is some element $z\in L_v$ with $z\neq y$ and $\kappa ^{\wedge v}(j)\lessdot z$ . Let $j'$ be the label of the cover $\kappa ^{\wedge v}(j)\lessdot z$ . Then $j'\not \leq \kappa ^{\wedge v}(j)=\kappa _L(j)\wedge v$ . We have $j'\leq v$ , so $j'\not \leq \kappa _L(j)$ . This means that $j'\to j$ is an edge in $G_L$ , which contradicts the fact that $\mathcal {U}_L(\kappa ^{\wedge v}(j))$ is an independent set in $G_L$ (because L is semidistrim). This shows that no such element z exists, which implies that y is the only element of $L_v$ that covers $\kappa ^{\wedge v}(j)$ . Hence, $\kappa ^{\wedge v}(j)\in \mathcal {M}_{L_v}$ . This shows that $\kappa ^{\wedge v}$ maps $\mathcal {J}_{L_v}$ to $\mathcal {M}_{L_v}$ . We have also seen that $\kappa ^{\wedge v}$ is injective because we can recover j as the label of the cover $\kappa ^{\wedge v}\lessdot y$ .

To see that $\kappa ^{\wedge v}$ is surjective, consider some $m\in \mathcal {M}_{L_v}$ . Let $m^*$ be the unique element of $L_v$ that covers m. Consider the label $j_{mm^*}$ of the cover $m\lessdot m^*$ . Then $\kappa ^{\wedge v}(j_{mm^*})=\kappa _L(j_{mm^*})\wedge v\geq m$ . If we had $\kappa ^{\wedge v}(j_{mm^*})\geq m^*$ , then we would have $j_{mm^*}\leq m^*\leq \kappa ^{\wedge v}(j_{mm^*})\leq \kappa _L(j_{mm^*})$ , which is impossible. Therefore, $\kappa ^{\wedge v}(j_{mm^*})=m$ .

We now know that $\kappa ^{\wedge v}$ is a bijection, so we need to check that it is a pairing. To do this, we will use Lemma 3.3. Fix $j'\in \mathcal {J}_{L_v}$ and let $m'=\kappa ^{\wedge v}(j')$ . We need to show that $m'\geq j^{\prime }_*$ , $(m')^*\geq j'$ , and $m'\not \geq j'$ . We saw above that $j'$ is the label of the cover $m'\lessdot (m')^*$ , so $(m')^*=m'\vee j'$ . This proves that $(m')^*\geq j'$ and $m'\not \geq j'$ . Since $\kappa _L$ is a pairing, we have $\kappa _L(j')\geq j^{\prime }_*$ . We also know that $v\geq j^{\prime }_*$ , so $m'=\kappa _L(j')\wedge v\geq j^{\prime }_*$ .

Proof. We know by Lemma 7.4 that $\kappa ^{\wedge v}$ is a pairing on $L_v$ , so we just need to show that it is the unique pairing. Let $\lambda \colon \mathcal J_{L_v}\to \mathcal M_{L_v}$ be a pairing on $L_v$ and define $\sigma \colon \mathcal J_{L_v}\to \mathcal J_{L_v}$ by $\sigma =(\kappa ^{\wedge v})^{-1}\circ \lambda $ . Our goal is to show that $\lambda =\kappa ^{\wedge v}$ or, equivalently, that $\sigma $ is the identity map on $\mathcal J_{L_v}$ . Define a map $\widetilde \kappa \colon \mathcal J_L\to \mathcal M_L$ by

$$\begin{align*}\widetilde\kappa(j) = \begin{cases} \kappa_L(j) & \text{ if } j \not\in L_v \\ \kappa_L(\sigma(j)) & \text{ if } j\in L_{v}. \end{cases}\end{align*}$$

We will prove that $\widetilde \kappa $ is a pairing on L. Since L is uniquely paired, this will imply that $\widetilde \kappa =\kappa _L$ . It will then follow that $\kappa _L(j)=\kappa _L(\sigma (j))$ for all $j\in \mathcal J_{L_v}$ . Because $\kappa _L$ is injective, this will imply that $\sigma $ is the identity, as desired.

It is straightforward to check that $\widetilde \kappa $ is a bijection. Now fix $j\in \mathcal J_L$ and let $m=\widetilde \kappa (j)$ . According to Lemma 3.3, we need to check that $m\geq j_*$ , $m^*\geq j$ and $m\not \geq j$ . If $j\not \in L_v$ , then $m=\kappa _L(j)$ , so these conditions follow from the fact that $\kappa _L$ is a pairing on L. Thus, we may assume $j\in L_v$ .

Note that

$$\begin{align*}m\wedge v=\widetilde\kappa(j)\wedge v=\kappa_L((\kappa^{\wedge v})^{-1}(\lambda(j)))\wedge v=\kappa^{\wedge v}((\kappa^{\wedge v})^{-1}(\lambda(j)))=\lambda(j).\end{align*}$$

We have

$$\begin{align*}m\wedge j=m\wedge(v\wedge j)=(m\wedge v)\wedge j=\lambda(j)\wedge j=j_*,\end{align*}$$

where the last equality is due to the fact that $\lambda $ is a pairing on $L_v$ . This proves that $m\geq j_*$ and $m\not \geq j$ .

We are left to show that $m^*\geq j$ . As above, note that $m\wedge v =\lambda (j)$ . Because $\lambda $ is a pairing on $\mathcal {M}_{L_v}$ , the element $\lambda (j)$ is meet-irreducible in $L_v$ . Let y be the unique element of $L_v$ that covers $\lambda (j)$ . Since $\lambda (j)=\kappa ^{\wedge v}(\sigma (j))$ and $\kappa ^{\wedge v}$ is a pairing on $L_v$ , we have $y=\lambda (j)\vee \sigma (j)$ . Since $m=\kappa _L(\sigma (j))$ and $\kappa _L$ is a pairing on L, we have $m^*\geq \sigma (j)$ . We also have $m^*\gtrdot m\geq m\wedge v=\lambda (j)$ , so $m^*\geq \lambda (j)\vee \sigma (j)=y$ . Since $\lambda $ is a pairing on $L_v$ and y is the unique element of $L_v$ covering $\lambda (j)$ , we must have $y\geq j$ . Thus, $m^*\geq y\geq j$ , as desired.

Proof. We know by Lemma 7.5 that $L_v$ is uniquely paired with its pairing $\kappa _{L_v}\colon \mathcal J_{L_v}\to \mathcal M_{L_v}$ given by $\kappa _{L_v}(j)=\kappa _L(j)\wedge v$ . Let $(j_0,m_0)$ be a dismantling pair for L and let $L_0=[\hat 0,m_0]$ and $L^0=[j_0,\hat 1]$ . According to Lemma 6.3, the intervals $L_0$ and $L^0$ are semidistrim. Therefore, if $v\in L_0$ , then the desired result follows by induction on the size of L.

We may now assume $v\in L^0$ . Note that $j_0$ is also join-prime in $L^0$ . Let $m_0^v=m_0\wedge v$ so that $(j_0,m_0^v)$ is a dismantling pair for $L_v$ . Since $L_0$ and $L^0$ are semidistrim, they have unique pairings $\kappa _{L_0}\colon \mathcal J_{L_0}\to \mathcal M_{L_0}$ and $\kappa _{L^0}\colon \mathcal J_{L^0}\to \mathcal M_{L^0}$ . By induction, the intervals $(L_v)_0=[\hat {0},m_0^v]\subseteq L_0$ and $(L_v)^0=[j_0,v]\subseteq L^0$ are compatibly dismantlable. Moreover, it follows from Lemma 7.5 that the unique pairings on these intervals are the bijections $\kappa _{(L_v)_0}\colon \mathcal J_{(L_v)_0}\to \mathcal M_{(L_v)_0}$ and $\kappa _{(L_v)^0}\colon \mathcal J_{(L_v)^0}\to \mathcal M_{(L_v)^0}$ defined by $\kappa _{(L_v)_0}(j)=\kappa _{L_0}(j)\wedge m_0^v$ and $\kappa _{(L_v)^0}(j)=\kappa _{L^0}(j)\wedge v$ .

Because L is compatibly dismantlable, there is a bijection $\alpha _L\colon \{j\in \mathcal {J}_L:j_0\leq \kappa _L(j)\}\to \mathcal {J}_{L^0}$ given by $\alpha _L(j)=j_0\vee j$ . We want to show that there is a bijection $\alpha _{L_v}\colon \{j \in \mathcal {J}_{L_v}:j_0\leq \kappa _{L_v}(j)\}\to \mathcal {J}_{(L_v)^0}$ given by $\alpha _{L_v}(j)=j_0\vee j$ . To this end, suppose j is a join-irreducible element of $L_v$ with $j_0\leq \kappa _{L_v}(j)$ . Then we have $j\in \mathcal {J}_L$ and $j_0\leq \kappa _L(j)$ , so $j_0 \vee j=\alpha _L(j)\in \mathcal {J}_{L^0}$ . Because $j_0 \vee j \leq v$ , the element $j_0\vee j$ is actually in $\mathcal {J}_{(L_v)^0}$ . This shows that $\alpha _{L_v}$ does actually map $\{j \in \mathcal {J}_{L_v}:j_0\leq \kappa _{L_v}(j)\}$ to $\mathcal {J}_{(L_v)^0}$ . Noting that

$$\begin{align*}\{ j \in \mathcal{J}_{L_v} : j_0\leq \kappa_{L_v}(j)\}= \{j \in \mathcal{J}_L : j_0\leq \kappa_L(j)\}\cap L_v,\\[-15pt]\end{align*}$$

we see that the injectivity of $\alpha _{L_v}$ follows immediately from the injectivity of $\alpha _L$ .

To see that $\alpha _{L_v}$ is surjective, let us choose a join-irreducible element $j'$ of $\mathcal {J}_{(L_v)^0}$ . Then $j'$ is join-irreducible in $L^0$ , so it is of the form $\alpha _L(j")$ for some $j" \in \mathcal {J}_L$ with $j_0\leq \kappa _L(j")$ . But then $j"\leq j'\leq v$ , so $j"\in \{j \in \mathcal {J}_L : j_0\leq \kappa _L(j)\}\cap L_v=\{j \in \mathcal {J}_{L_v} : j_0\leq \kappa _{L_v}(j)\}$ .

We also need to check that if $j\in \mathcal J_{L_v}$ satisfies $j_0\leq \kappa _{L_v}(j)$ , then

$$\begin{align*}\kappa_{(L_v)^0}(\alpha_{L_v}(j))=\kappa_{L_v}(j).\\[-15pt]\end{align*}$$

We know by Theorem 4.1 that $\kappa _{L^0}(\alpha _L(j))=\kappa _L(j)$ . Hence,

$$\begin{align*}\kappa_{(L_v)^0}(\alpha_{L_v}(j))=\kappa_{L^0}(\alpha_{L_v}(j))\wedge v=\kappa_{L^0}(\alpha_{L}(j))\wedge v=\kappa_L(j)\wedge v=\kappa_{L_v}(j).\\[-15pt]\end{align*}$$

To finish the proof, we need to show that there is a bijection

$$\begin{align*}\beta_{L_v}\colon\{ m \in \mathcal{M}_{L_v} : \kappa_{L_v}^{-1}(m) \leq m_0^v\}\to\mathcal M_{(L_v)_0}\\[-15pt]\end{align*}$$

given by $\beta _{L_v}(m)=m_0^v\wedge m$ such that $\kappa _{(L_v)_0}^{-1}(\beta _{L_v}(m))=\kappa ^{-1}_{L_v}(m)$ for all $m\in \mathcal M_{L_v}$ with $\kappa _{L_v}^{-1}(m)\leq m_0^v$ . Choose $m\in \mathcal M_{L_v}$ with $\kappa _{L_v}^{-1}(m)\leq m_0^v$ and let $j=\kappa _{L_v}^{-1}(m)$ . Then $m = \kappa _L(j) \wedge v$ , so

(7.2) $$ \begin{align} m_0^v\wedge m = (m_0 \wedge v)\wedge (\kappa_L(j) \wedge v) = m_0^v\wedge \kappa_L(j)=\kappa_{(L_v)_0}(j)=\kappa_{(L_v)_0}(\kappa_{L_v}^{-1}(m)),\\[-15pt]\nonumber \end{align} $$

where the equality $m_0^v\wedge \kappa _L(j)=\kappa _{(L_v)_0}(j)$ follows from the description of the unique pairing $\kappa _{(L_v)_0}$ on the interval $(L_v)_0=[\hat 0,m_0^v]$ given by Lemma 7.5. This proves that $m_0^v\wedge m\in \mathcal M_{(L_v)_0}$ and that $\kappa _{(L_v)_0}^{-1}(m_0^v\wedge m)=\kappa _{L_v}^{-1}(m)$ . This yields the map $\beta _{L_v}$ ; we still need to prove that this map is a bijection.

The map $\beta _{L_v}$ is injective because we can use (7.2) to see that

$$\begin{align*}m=\kappa_{L_v}(\kappa_{(L_v)_0}^{-1}(\beta_{L_v}(m)))\\[-15pt]\end{align*}$$

for every $m\in \mathcal {M}_{L_v}$ with $\kappa _{L_v}^{-1}(m)\leq m_0^v$ . To prove that $\beta _{L_v}$ is surjective, consider $m'\in \mathcal {M}_{(L_v)_0}$ and let $m=\kappa _{L_v}(\kappa _{(L_v)_0}^{-1}(m'))$ . Then $m\in \mathcal {M}_{L_v}$ and $\kappa _{L_v}^{-1}(m)\leq m_0^v$ . According to (7.2), we have $\beta _{L_v}(m)=m_0^v\wedge m=\kappa _{(L_v)_0}(\kappa _v^{-1}(m))=m'$ , as desired.

Proof. Consider $j,j' \in \mathcal {J}_{L_v}$ . We must show that $j\to j'$ in $G_L$ if and only if $j\to j'$ in $G_{L_v}$ . This is equivalent to showing that $j\leq \kappa _L(j')$ if and only if $j\leq \kappa _{L_v}(j')$ . We know by Lemma 7.5 that $\kappa _{L_v}(j')=\kappa _L(j')\wedge v$ . Since $j'\leq v$ , the desired result follows.

Proof. We first argue that if $v\in L$ , then the interval $L_v=[\hat {0},v]$ is semidistrim. We know by Lemma 7.6 that $L_v$ is compatibly dismantlable. Now suppose $y\in L_v$ . We know by Lemma 7.7 that the Galois graph $G_{L_v}$ of $L_v$ is an induced subgraph of $G_L$ . The sets $\mathcal {D}_{L_v}(y)$ and $\mathcal {U}_{L_v}(y)$ are the intersections of $\mathcal {D}_L(y)$ and $\mathcal {U}_L(y)$ , respectively, with the vertex set of $G_{L_v}$ . Since $\mathcal {D}_L(y)$ and $\mathcal {U}_L(y)$ are independent sets in $G_L$ , the sets $\mathcal {D}_{L_v}(y)$ and $\mathcal {U}_{L_v}(y)$ must be independent sets in $G_{L_v}$ . Hence, $L_v$ is semidistrim.

Now fix an interval $[u,v]$ in L. It is immediate from the definition of a semidistrim lattice that a lattice is semidistrim if and only if its dual is semidistrim, so it follows from the preceding paragraph that $[u,\hat {1}]$ is semidistrim. We can now apply the preceding paragraph again to see that the interval $[u,v]$ of the semidistrim lattice $[u,\hat {1}]$ is semidistrim.

Proof. If $u=\hat 0$ , then this follows from Lemma 7.5. There is a natural dual version of Lemma 7.5 that yields the desired result when $v=\hat 1$ . The general result then follows from combining these two cases (for example, by viewing $[u,v]$ as a lower interval in $[u,\hat 1]$ ).

Proof. Suppose $j,j'\in J_L(v)\cap M_L(u)$ . This means that

$$\begin{align*}j,j'\leq v\quad\text{and}\quad\kappa_L(j),\kappa_L(j')\geq u.\end{align*}$$

We must show that $j\to j'$ in $G_L$ if and only if $\alpha _{u,v}(j)\to \alpha _{u,v}(j')$ in $G_{[u,v]}$ . Equivalently, we must show that $j\leq \kappa _L(j')$ if and only if $\alpha _{u,v}(j)\leq \kappa _{[u,v]}(\alpha _{u,v}(j'))$ . We have $\alpha _{u,v}(j)=u\vee j$ by definition, and we have $\kappa _{[u,v]}(\alpha _{u,v}(j'))=\beta _{u,v}(\kappa _L(j'))=v\wedge \kappa _L(j')$ by Theorem 7.9. The assumptions $j\leq v$ and $\kappa _L(j')\geq u$ imply that $j\leq \kappa _L(j')$ if and only if $u \vee j\leq v\wedge \kappa _L(j')$ .

To prove the last statement, note that $j_{xy}\leq y$ and $\kappa _L(j_{xy})\geq x$ by Theorem 5.1. Since $y\geq u$ , this implies that $\alpha _{u,v}(j_{xy})=u\vee j_{xy}\leq y$ . Also, since $x\leq v$ , we have (by Theorem 7.9) $\kappa _{[u,v]}(\alpha _{u,v}(j_{xy}))=v\wedge \kappa _L(j_{xy})\geq x$ . Hence, $\alpha _{u,v}(j_{xy})$ is the label of $x\lessdot y$ in $[u,v]$ .

Proof. The statements about rowmotion follow from Theorem 5.6. For the statements about pop-stack sorting and dual pop-stack sorting, we compute

$$\begin{align*}\mathsf{Pop}^\downarrow_L(x)=x\wedge\bigwedge\{y\in L:y\lessdot x\}=x\wedge\bigwedge_{y\lessdot x}(x\wedge \kappa_L(j_{xy}))=x\wedge\bigwedge\kappa_L(\mathcal{D}_L(x))\end{align*}$$

and

$$\begin{align*}\mathsf{Pop}^\uparrow_L(x)=x\vee\bigvee\{y\in L:x\lessdot y\}=x\vee\bigvee_{x\lessdot y}(x\vee j_{xy})=x\vee\bigvee\mathcal{U}_L(x). \\[-46pt]\end{align*}$$

Proof. We know by Lemma 7.4 that $\kappa _{L_v}(j)=\kappa _L(j)\wedge v$ for all $j\in \mathcal {J}_{L_v}$ . Now consider $x\in L_v$ . We have $\mathcal {D}_{L_v}(x)=\mathcal {D}_L(x)$ by Theorem 7.10. Using Theorem 9.1, we find that

$$\begin{align*}\mathsf{Row}_{L_v}(x)=\bigwedge\kappa_{L_v}(\mathcal{D}_L(x))=\bigwedge_{j\in\mathcal{D}_L(x)}(\kappa_L(j)\wedge v)=v\wedge\bigwedge_{j\in\mathcal{D}_L(x)}\kappa_L(j)=v\wedge\mathsf{Row}_L(x).\\[-36pt] \end{align*}$$

Proof. We prove that $\mathsf {Row}_L(x)\in \max \{z\in L:\mathsf {Pop}^\downarrow _L(x)= x\wedge z\}$ ; the second statement follows from this one by taking duals. Let $(j_0,m_0)$ be a dismantling pair for L and let $L_0=[\hat {0},m_0]$ and $L^0=[j_0,\hat {1}]$ . We consider three cases.