The projective dimension of the simple modules corresponding to the minimum of L in the incidence algebra of the lattice L.

The lattice of antichains of maximal size in a poset.
The set of antichains of maximal size can be ordered by setting $A \leq B \leftrightarrow \mathop{\downarrow} A \subseteq \mathop{\downarrow} B$, where $\mathop{\downarrow} A$ is the order ideal generated by $A$.
This is a sublattice of the lattice of all antichains with respect to the same order relation. In particular, it is distributive.

The Hessenberg poset of a Dyck path.
Let $D$ be a Dyck path of semilength $n$, regarded as a subdiagonal path from $(0,0)$ to $(n,n)$, and let $\boldsymbol{m}_i$ be the $x$-coordinate of the $i$-th up step.
Then the Hessenberg poset (or natural unit interval order) corresponding to $D$ has elements $\{1,\dots,n\}$ with $i < j$ if $j < \boldsymbol{m}_i$.

The zeta map on Dyck paths.
The zeta map $\zeta$ is a bijection on Dyck paths of semilength $n$.
It was defined in [1, Theorem 1], see also [2, Theorem 3.15] and sends the bistatistic (area, dinv) to the bistatistic (bounce, area). It is defined by sending a Dyck path $D$ with corresponding area sequence $a=(a_1,\ldots,a_n)$ to a Dyck path as follows:

• First, build an intermediate Dyck path consisting of $d_1$ north steps, followed by $d_1$ east steps, followed by $d_2$ north steps and $d_2$ east steps, and so on, where $d_i$ is the number of $i-1$'s within the sequence $a$.
  For example, given $a=(0,1,2,2,2,3,1,2)$, we build the path
  $$NE\ NNEE\ NNNNEEEE\ NE.$$
• Next, the rectangles between two consecutive peaks are filled. Observe that such the rectangle between the $k$th and the $(k+1)$st peak must be filled by $d_k$ east steps and $d_{k+1}$ north steps. In the above example, the rectangle between the second and the third peak must be filled by $2$ east and $4$ north steps, the $2$ being the number of $1$'s in $a$, and $4$ being the number of $2$'s. To fill such a rectangle, scan through the sequence a from left to right, and add east or north steps whenever you see a $k-1$ or $k$, respectively. So to fill the $2\times 4$ rectangle, we look for $1$'s and $2$'s in the sequence and see $122212$, so this rectangle gets filled with $ENNNEN$.
  The complete path we obtain in thus
  $$NENNENNNENEEENEE.$$