Recall: $\displaystyle e(P-y) = \sum_{C \in \mathcal{C}\colon y\in C} |\{f \in \mathcal{E}(P) : \Phi(f) = C\}|$.
Then,
$$\displaystyle
\begin{align*}
e(P-y) &= \sum_{C \in \mathcal{C}\colon y\in C} |\{f \in \mathcal{E}(P): \Phi(f) = C\}|\\\
&=\sum_{C \in \mathcal{C}\colon y\in C} \sum_{f \in \mathcal{E}(P)}\begin{cases}
1 \quad\text{ if } \Phi(f)=C\\\
0 \quad\text{ otherwise}
\end{cases}
\\\
&= \sum_{f \in \mathcal{E}(P)} \begin{cases}
1 \quad\text{ if } y \in \Phi(f)\\\
0 \quad\text{ otherwise}
\end{cases}
\end{align*}$$
Hence,
$$\begin{align*}
e(P-y)e(Q-y) &= \left(\sum_{f \in \mathcal{E}(P)}
\begin{cases}
1 \quad\text{ if } y \in \Phi(f)\\\
0 \quad\text{ otherwise}
\end{cases}
\right) \left(\sum_{g \in \mathcal{E}(Q)} \begin{cases}
1 \quad\text{ if } y \in \Phi(g)\\\
0 \quad\text{ otherwise}
\end{cases}
\right)\\\
&= \sum_{f \in \mathcal{E}(P)}\sum_{g \in \mathcal{E}(Q)} \begin{cases}
1 \quad\text{ if } y \in \Phi(f) \cap \Phi(g)\\\
0 \quad\text{ otherwise}
\end{cases}
\end{align*}$$
Now,
$$\begin{align*}
\sum_{y \in X}e(P-y)e(Q-y)
&= \sum_{y \in X}\sum_{f \in \mathcal{E}(P)}\sum_{g \in \mathcal{E}(Q)} \begin{cases}
1 \quad\text{ if } y \in \Phi(f) \cap \Phi(g)\\\
0 \quad\text{ otherwise}
\end{cases}\\\
&= \sum_{f \in \mathcal{E}(P)}\sum_{g \in \mathcal{E}(Q)}\sum_{y \in X} \begin{cases}
1 \quad\text{ if } y \in \Phi(f) \cap \Phi(g)\\\
0 \quad\text{ otherwise}
\end{cases}\\\
&= \sum_{f \in \mathcal{E}(P)}\sum_{g \in \mathcal{E}(Q)}|\Phi(f) \cap \Phi(g)|\\\
&\leq \sum_{f \in \mathcal{E}(P)}\sum_{g \in \mathcal{E}(Q)}k &\text{(Chains intersect at most $k$ elements.)}\\\
&= k \cdot e(P) e(Q)
\end{align*}$$
Induction:
base case $n=k$: $\displaystyle e(P)e(Q) \geq \frac{n!}{k!k^{n-k}} = \frac{n!}{n!n^0}=1$
inductive step $n>k$:
$$\begin{align*}
e(P)e(Q) &\geq \frac{1}{k} \sum_{y \in X}e(P-y)e(Q-y)\\\
&\geq \frac{1}{k} \sum_{y \in X}\frac{(n-1)!}{k! k^{(n-1)-k}}\\\
&= \frac{n}{k}\cdot\frac{(n-1)!}{k! k^{(n-1)-k}}\\\
&= \frac{n!}{k!k^{n-k}}
\end{align*}$$
- Chains in a poset form cliques in its comparability graph.
- If $G(P_1)$ is isomorphic to $G(P_2)$, then their cliques are also isomorphic.
- The chain polytope of a poset only depends on the set of its chains.
-
$\text{vol}(\mathcal{C}_{P_1}) = \text{vol}(\mathcal{C}_{P_2}) = \frac{e(P_1)}{n!} = \frac{e(P_2)}{n!}$.
All vertices have integral coordinates.
- Vertices of $\mathcal{O}_P$: ${f: X \to {0,1}\text{ order preserving}}$
- Vertices of $\mathcal{C}_P$: for each chain, at most $1$ axis with value $1$.
Not sure what to write.
Details: https://math.mit.edu/~rstan/pubs/pubfiles/66.pdf
- We sort the points by $x$-values to obtain rank $\mu_x: X \to [n]$
and similarly by $y$-values obtain rank $\mu_y: X \to [n]$.
- Ties are broken arbitrarily.
-
$\sigma$ is obtained by $\sigma(\mu_x(p))=\mu_y(p)$ for every $p \in X$.
Now, we show the isomorphism.
TBD
