Class of lattices that excludes $M_3$? 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 $N_5$).
 A: This is a (slightly edited) copy of the answer I posted on math.SE to the question What do we call a lattice that does not have a sublattice the shape of the diamond $M_3$?:

Let $\mathbf K$ be the class of all lattices not containing $M_3$ (the $5$-element nondistributive modular lattice) as a sublattice; in other words, lattices in which every modular sublattice is distributive. It is easy to see that $\mathbf K$ can be characterized as the class of all lattices satisfying the following sentence $\varphi$:$$\forall u\forall v\forall a\forall b\forall c[(ab=ac=bc=u)\wedge(a+b=a+c=b+c=v)\rightarrow(u=v)]$$or equivalently$$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(ab=a+b)]$$ or
$$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(a=b)].$$
Inasmuch as $\varphi$ is a universal Horn sentence, it follows that $\mathbf K$ is closed under taking sublattices, direct products and reduced products; it is a quasivariety.
On the other hand, $\mathbf K$ can not be characterized by identities, and is not closed under taking homomorphic images. Bjarni Jónsson [Sublattices of a free lattice, Canad. J. Math. 13 (1961), 256-264, Lemma 2.6(i)] observed that, as a corollary of P. M. Whitman's work, elements $u,a,b,c$ of a free lattice satisfy the condition: if $u=ab=ac$, then $u=a(b+c)$. It follows that $M_3$ is not a sublattice of a free lattice, i.e, the class $\mathbf K$ contains all free lattices. Thus every lattice is a homomorphic image of a member of $\mathbf K$, and every identity which holds in all $M_3$-free lattices is a consequence of the lattice axioms. In view of this, we are not likely to find a much simpler algebraic characterization of $M_3$-free lattices than the one in the previous paragraph.

A: 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.
