Counting number of posets of fixed dimension

Question

What should be a reasonable upper bound on the number of posets on $\{1, \ldots, n\}$ with dimension $d$?

Note that $d=2$ is a very special case and the count is then $n! / 2$ asymptotically.

Answer 1

This is a long comment.

The order dimension of a poset $\mathbb P = \langle P; \leq \rangle$ is the least $d$ such that $\mathbb P$ is embeddable in a product of $d$ chains.

Comment. It is not true that the number of posets with domain $\{1,\ldots,n\}$ which have order dimension $2$ is $n!/2$ asymptotically. For example, suppose that $n = k^2$ for some $k>1$. Then the number of posets of size $n$ that are embeddable in a product of two $k$-element chains is exactly $n!/2$.

Reason: let $C_k$ be a $k$-element chain. Each bijection from $\{1,\ldots,n\}$ to the poset $C_k\times C_k$ defines a $2$-dimensional order on $\{1,\ldots,n\}$, and two bijections define the same order iff one is obtained from the other by swapping the factors of $C_k\times C_k$. This yields exactly $n!/2$ different $2$-dimensional orderings on $\{1,\ldots,n\}$.

But there are a lot of other posets of order dimension $2$ on $\{1,\ldots,n\}$ that are not so symmetrically diamond shaped. In fact, you have roughly $n!$ different $2$-dimensional orderings of $\{1,\ldots,n\}$ that come from 1-1 near-bijections of $\{1,\ldots,n\}$ with $C_{\ell}\times C_{\lceil n/\ell\rceil}$ for each $1<\ell\leq\sqrt{n}$. This construction, as $\ell$ ranges from $2$ to $\sqrt{n}$, produces roughly $(\sqrt{n}\cdot n!)$-many different $2$-dimensional orderings of $\{1,\ldots,n\}$. I don't know how close this estimate is to an optimal one.