The aztec diamond is an area of a 2-dimensional square lattice. The quarter aztec diamond it's a part of this area, it can be seen in the following picture:

Triangular arrangement of a 2-dimensional square lattice

It would be something like this (just in case the picture it's not showed):



$* - *$

|$ \ \ \ \ \ \ |$

$* - * - *$

|$ \ \ \ \ \ \ $|$ \ \ \ \ \ $|

$* - * - * - *$

|$ \ \ \ \ \ \ $|$ \ \ \ \ \ $|$ \ \ \ \ \ \ $|

$* - * - * - * - *$

|$ \ \ \ \ \ \ $|$ \ \ \ \ \ $|$ \ \ \ \ \ \ $|$ \ \ \ \ \ \ $|

$* - * - * - * - * - *$

May be, rotating the diagram 45 degrees, one can have a better view of the lattice.

The question is related with the latticial properties of this grid, so it has to be viewed as a lattice, not as an area. The picture represents the Hasse diagram; where the top element has a unique edge (in-degree = 1), i.e., it is drawn at the top part of the diagram. And the bottom element also has a unique edge (out-degree = 1), but it is drawn at the bottom part of the diagram.

I'm interested on latticial properties of this triangular grid, i.e., if it is modular, distributive, orthomodular, boolean algebra, etc.

As far as I know, it must be modular, because it has the following properties (characterization of a modular lattice):

a) If $x$, $y$ cover $x \wedge y$, then $x \vee y$ covers $x$ and $y$.

b) If $x \vee y$ covers $x$ and $y$, then $x$ and $y$ cover $x \wedge y$.

I'm not sure if it is distributive. As regard to Boolean algebras, I'm almost sure that it is not one of them, because a totally ordered set (i.e. a linear order) it is not a Boolean algebra.

Any pointers would be greatly appreciated.

  • $\begingroup$ could the relevant tag be integer-lattices? I don't see how is that a Hasse diagram. Is it to be understood with a 45º rotation? If so, it's clearly distributive but not Boolean. You also claim that a totally ordered set is not Boolean (which is true if it has more than two elements; by the way, a totally ordered set is also always distributive), but how is that diagram totally ordered? $\endgroup$
    – amrsa
    Dec 19, 2021 at 11:02
  • $\begingroup$ Thanks for the reply amrsa. I've added the tag and a comment about the rotation. You say that it is distributive, could you say what characterization properties are need to demonstrate this claim? Regarding the linear ordering, I wanted to say a "chain", whose diagram is "---", so, you have answered my question very well. $\endgroup$
    – bport
    Dec 19, 2021 at 11:55

1 Answer 1


The lattice is clearly distributive, which can be seen in a number of ways.
Let me just mention a two of them. To start with, I'll refer to this lattice as $\mathbf L$.

  • The lattice $\mathbf L$ is a sub-lattice of the direct product of two chains (both with six elements), i.e., $\mathbf L \leq \mathbf6\times\mathbf6$, and since the chains are distributive (it's easy to check) and direct products of distributive lattices are distributive, and also sub-lattices of distributive lattices are distributive, it follows that $\mathbf L$ is distributive too.

  • The lattice $\mathbf L$ doesn't have a sub-lattice isomorphic to the diamond, $\mathbf M_3$, or to the pentagon, $\mathbf N_5$. As such, it's distributive, by the very well known characterization.

Being distributive, it's also modular, as you suspected, but notice that your "characterization" of modular lattices only applies the ones of finite length.

The lattice $\mathbf L$ is also not an ortho-lattice because the top element $\top$ is join-irreducible (and the bottom is meet-irreducible), i.e., if $x \vee y = \top$ then $x=\top$ or $y=\top$.
Thus it can't be ortho-modular (even though it is modular), and it can't be Boolean either, since a Boolean lattice is always an ortho-lattice.

Here, I'm using the definition that an ortho-lattice is a lattice $\mathbf K = \langle K, \wedge, \vee, ', \top, \bot \rangle$ where the equations $$x \vee x'=\top,\quad x\wedge x'=\bot, \quad x''=x$$ are satisfied and also $x\leq y$ implies $y' \leq x'$. An ortho-modular lattice is an ortho-lattice which is also modular.

  • $\begingroup$ Thank you very much amrsa, great answer! $\endgroup$
    – bport
    Dec 19, 2021 at 13:31

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