Question
Let $(L,\leq)$ be a lattice of finite length and let its height function $h$ be supermodular, meaning that $$h(x \wedge y) + h(x \vee y) \geq h(x) + h(y) \quad \forall x,y\in L.$$ Does it always follow that $(L,\leq)$ is lower semimodular? If yes, how?
Background
This statement is covered by Theorem 4.14 4) (p. 107) in the book [Steven Roman: Lattices and Ordered Sets, 2008]. However, I'm not convinced of the stated proof, which indicates that the statement follows via dualization from the corresponding property of upper semimodular lattices. I'm concerned that in general the dual of the height function might not be sufficiently related to the actual height function, see below for details.
Definitions
Let $(L,\leq)$ be a lattice of finite length. For $a \leq b \in L$, denote the maximal length of a chain from $a$ to $b$ by $d(a,b)$, and denote the height by $h(a) = d(\bot,a)$, where $\bot$ is the bottom element of the lattice $(L,\leq)$ (which exists since it is of finite length).
Define the following properties of $(L,\leq)$:
- (USM) if $(L,\leq)$ is upper semimodular,
- (LSM) if $(L,\leq)$ is lower semimodular,
- (dSubM) if $d$ is submodular, i.e. $d(y,x \vee y) \leq d(x \wedge y, x)$ for all $x,y\in L$,
- (dSupM) if $d$ is supermodular, i.e. $d(y,x \vee y) \geq d(x \wedge y, x)$ for all $x,y\in L$,
- (hSubM) if $h$ is submodular, i.e. $h(x \wedge y) + h(x \vee y) \leq h(x) + h(y)$ for all $x,y\in L$,
- (hSupM) if $h$ is supermodular, i.e. $h(x \wedge y) + h(x \vee y) \geq h(x) + h(y)$ for all $x,y\in L$.
Question (again)
Is the following always true? $$\text{(hSupM)} \implies \text{(LSM)}$$ If yes, how can we prove it?
Remarks
- (dIneq) For $x \leq y \leq z$ we have $d(x,z) \geq d(x,y) + d(y,z)$ by the definition of $d$. In terms of $h$, this implies $h(y) \geq h(x) + d(x,y)$.
- (dEq) If $(L,\leq)$ is Jordan-Dedekind, then (dIneq) is in fact always an equality, i.e. $d(x,z) = d(x,y) + d(y,z)$ and $h(y) = h(x) + d(x,y)$.
- Any semimodular lattice is Jordan-Dedekind. Hence (USM) and (LSM) both independently imply (dEq).
- (USM) is the dual statement of (LSM), and (dSubM) is the dual statement of (dSupM).
- In general, (hSubM) is not the dual statement of (hSupM). The reason is that the dual notion of $h(x)$ is $h^{\perp}(x) := d(x,\top)$, and I don't see that $h$ determines $h^{\perp}$ and vice versa. All we have is $h(x) + h^{\perp}(x) \leq d(\bot,\top)$ by (dIneq), but this is too weak for this purpose. Important: I have the feeling that this might be the critical point of the whole discussion. If it turns out that (hSubM) actually is the dual statement of (hSupM) in a suitable sense, then this question is resolved.
- If $(L,\leq)$ is Jordan-Dedekind, then (hSubM) is indeed the dual of (hSupM). The reason is that now we can use (dEq) such that $h^{\perp}(x) = d(\bot,\top) - h(x)$.
- (CharacterizeUSM) One can show that the three properties (USM), (dSubM) and (hSubM) are equivalent.
- For our purpose, the critical implication in (CharacterizeUSM) is "(hSubM) $\implies$ (USM)". I found a convincing proof for it in the book [George Grätzer, General lattice theory, 2nd ed. 2003], where it is the proof of Theorem 2, "(iv) implies (ii)", on page 227, done by induction (my elaboration of this proof is given further below). However, I also came across two sources with a "proof" I don't understand and where I'm afraid that they might be incorrect. Surprisingly, one comes from the very same author, namely [George Grätzer, Lattice Theory: Foundation, 2011], proof of Theorem 375, "(iv) implies (ii)" on page 332. The other one is the above mentioned book by Steven Roman, proof of Theorem 4.14, Part 3). In both cases, my impression is that the authors use (dEq), where under the assumption of (hSubM) a priori we only have (dIneq).
- Dualizing (CharacterizeUSM) we get the equivalence "(LSM) $\iff$ (dSupM)". Since under the assumption of (LSM) we have that (hSupM) is the dual of (hSubM), moreover we get "(LSM) $\implies$ (hSupM)".
- As discussed above, I don't see that in general (hSupM) is the dual statement of (hSubM). Thus the dualization of (CharacterizeUSM) does not prove "(hSupM) $\implies$ (LSM)" (as indicated in the book of Steven Roman).
- I tried to show "(hSupM) $\implies$ (LSM)" similarly to George Grätzers induction proof of "(hSubM) $\implies$ (USM)", but I didn't succeed. Maybe I could have tried harder, but by now I'm pretty frustrated about the whole situation.
Addition
Here is my elaboration of Grätzer's proof of "(hSubM) $\implies$ (USM)".
Let $z\in L$. We show by induction over $h(z)$ that the upper covering condition holds in the interval $[\bot,z]$. This is enough since for all $x,y\in X$ the elements $x$, $y$, $x\wedge y$ and $x\vee y$ are all contained in the interval $[\bot,x\vee y]$.
For $h(z) = 0$ the claim is clearly true. So assume $h(z) \geq 1$ and let $x,y\in [\bot,z]$ with $x \wedge y \lessdot x$. Then $x\not\leq y$ and hence $x < x \vee y \leq z$. Therefore $d(x,z) \geq 1$ and using (dIneq) we get $h(x) \leq h(z) - d(x,z) \leq h(z) - 1$. By the induction hypothesis, the sublattice $[\bot,x]$ is (USM) and hence in particular Jordan-Dedekind, such that we may apply (dEq) in $[\bot,x]$ to get $h(x) = h(x\wedge y) + 1$. Now $$d(y,x\vee y) \leq h(x\vee y) - h(y) \leq h(x) - h(x\wedge y) = 1,$$ where we used (dIneq) in the first step and (hSubM) in the second. This implies $y = x\vee y$ or $y \lessdot x \vee y$, where the first case is not possible as otherwise $x = x\wedge y$ in contradiction to the assumption $x\wedge y \lessdot x$.
Can anyone do something similar to show "(hSupM) $\implies$ (LSM)"?
