1
$\begingroup$

Fix two numbers $n,k\in \mathbb{N}$ and abbreviate $[n]:= \{0,1,\dots,n\}$ (as linearly ordered set). Consider diagrams $(I,\varphi)$ of the form $$[n]\supset I\xrightarrow{\varphi}[k]$$ where $\varphi$ is surjective and weakly monotone. Denote by $A(n,k)$ the set of such diagrams. An easy inspection shows that $$|A(n,k)|= \sum_{i=0}^n {n+1\choose i+1}{i\choose k}$$ where the $i$ corresponds to the dimension of $I$ (i.e. $I\cong [i]$).

Now introduce the following relation such diagrams: $(I,\varphi)\leq(I',\varphi')$ if $I\supset I'$ and $\varphi$ is the unique surjective weakly monotone map $I\to[k]$ that extends $\varphi'$. It is clear that $\leq$ is a partial order on $A(n,k)$.

Now my question: what is known about the structure of this partially ordered set and has it appeared in the literature before?

Some specific questions: how many connected components does it have? What are its maximal elements?

$\endgroup$
  • $\begingroup$ I don't see why $\varphi$ has to be unique. For example with $n=3, k=2$ the diagram $(\{1, 3\}, 1 \mapsto1, 3 \mapsto 2)$ can be extended in two ways that is $2 \mapsto 1$ or $2 \mapsto 2$. Or did I miss something ? $\endgroup$ – hivert Dec 17 '17 at 21:59
  • $\begingroup$ of course $\varphi$ is usually not unique. but if it is, then we define $(I,\varphi)\leq (I',\varphi')$. So $(I,\varphi)\leq(I',\varphi')$ means that $\varphi$ is a surjective weakly monotone extension of $\varphi'$ to $I$ AND that there is no other such extenison. $\endgroup$ – Tashi Walde Dec 17 '17 at 22:06
  • 1
    $\begingroup$ Might be an example of a tableau complex, as in Allen Knutson, Ezra Miller, Alexander Yong, Tableau complexes, arXiv:math/0510487. (Imagine your diagram $\left[n\right] \supseteq I \to \left[k\right]$ as a set-valued tableau with a single row of length $k$, where the $i$-th box contains the elements of $I$ that map to the element $i$ of $\left[k\right]$.) $\endgroup$ – darij grinberg Dec 17 '17 at 23:29

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.