TITLE: Class of lattices that excludes $M_3$?
QUESTION [10 upvotes]: It is well known that a lattice is distributive iff it excludes as a sublattice $N_5$ (the pentagon) and $M_3$ (three unordered elements with a top and bottom). Further, a lattice that only excludes $N_5$ is modular, and consequently a lattice excluding $N_5$ but including $M_3$ is modular but not distributive.
What, if anything, is known about the class of lattices that excludes $M_3$? Does it relate to classes of structures of any particular interest? (Note this is not the same as asking about "nonmodular varieties", since those are varieties which include $N_5$).

REPLY [10 votes]: As bof says, the class $\mathcal K$ of lattices omitting $\mathbf M_3$ as a sublattice is a quasivariety that is not a variety. As bof also says, this implies that $\mathcal K$ is closed under the formation of sublattices, products, and reduced products, but the class is not closed under the formation of homomorphic images.
But $\mathcal K$ is closed under some types of homomorphic images. 
A. (Projection onto a direct factor)
If $L\times L'\in \mathcal K$, then $L, L' \in \mathcal K$. Every quasivariety of lattices has this property, but there exist quasivarieties of algebraic structures that do not have this property.
B. (Quotients of finite members) If $L\in \mathcal K$ is finite, then any homomorphic image of $L$ is also in $\mathcal K$. This property typically fails for quasivarieties, even quasivarieties of lattices. It holds for $\mathcal K$ because $\mathbf M_3$ is projective in the class of finite lattices.