Taken from: [1, Exercise 12.26].
Exercise 12.26
Let $\Gamma$ and $\Delta$ be simplicial complexes on disjoint vertex sets $V$ and $W$, respectively. Define the join $\Gamma * \Delta$ to be the simplicial complex on the vertex set $V \cup W$ with faces $F \cup G$, where $F \in \Gamma$ and $G \in \Delta$. (If $\Gamma$ consists of a single point, then $\Gamma * \Delta$ is the cone over $\Delta$. If $\Gamma$ consists of two disjoint points, then $\Gamma * \Delta$ is the suspension of $\Delta$.)
- Compute the $h$-vector $h(\Gamma* \Delta)$ in terms of $h(\Gamma)$ and $h(\Delta)$.
- Show that if $\Gamma$ and $\Delta$ are Cohen–Macaulay, then so is $\Gamma* \Delta$.
- Generalizing Exercise 7, show that if $\Gamma$ and $\Delta$ are shellable, then so is $\Gamma* \Delta$.
Solution
Let $ d = \dim\Gamma + 1 $ and $ e = \dim\Delta + 1 $.
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The union of an $ i $-dimensional face in $\Gamma$ with a $ j $-dimensional face in $\Delta$ results in an $ (i + j + 1 )$-dimensional face in $\Gamma * \Delta$. Therefore,
\[f_k(\Gamma * \Delta) = \sum_{i + j = k - 1} f_i(\Gamma) f_j(\Delta),\]for every $k=-1,\ldots,d+e-1$; where $ f_i(\Gamma) = 0 $ if $ i < -1 $ or $ i \geq d $, and similarly for $ f_j(\Delta) $. Thus,
\[\begin{align*} \left(\sum_{i=0}^d f_{i-1}(\Gamma)x^{d-i}\right)\left(\sum_{j=0}^e f_{j-1}(\Delta)x^{e-j}\right) &=\sum_{k=0}^{d+e}\sum_{i+j=k-2}f_i(\Gamma)f_j(\Delta)x^{d+e-k}\\ &=\sum_{k=0}^{d+e}f_{k-1}(\Gamma*\Delta)x^{d+e-k}. \end{align*}\]By substituting into [1, Eq. 12.5] with $ x \to x + 1 $, we obtain
\[\begin{align*} \left(\sum_{i=0}^d h_i(\Gamma)(x+1)^{d-i}\right)\left(\sum_{j=0}^e h_j(\Delta)(x+1)^{e-j}\right) &=\sum_{k=0}^{d+e}h_k(\Gamma*\Delta)(x+1)^{d+e-k}. \end{align*}\]The left-hand side of this equation becomes
\[\sum_{k=0}^{d+e}\sum_{i+j=k}h_i(\Gamma)h_j(\Delta)(x+1)^{d+e-k}.\]Comparing coefficients, we finally obtain
\[h_k(\Gamma*\Delta)=\sum_{i+j=k}h_i(\Gamma)h_j(\Delta)\]for every $k=0,\ldots,d+e$.
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Since the vertex sets of $\Gamma$ and $\Delta$ are disjoint, the nonfaces of $\Gamma * \Delta$ are precisely those of $\Gamma$ and $\Delta$, making the Stanley–Reisner ideal decomposable as $I_\Gamma + I_\Delta$. This naturally leads to $K[\Gamma * \Delta] \cong K[\Gamma] \otimes_K K[\Delta],$ whence $ \dim K[\Gamma * \Delta] = \dim K[\Gamma] + \dim K[\Delta] = d + e.$ Now let $\theta_1,\ldots, \theta_d\in K[\Gamma]_1$ and $\eta_1,\ldots, \eta_e \in K[\Delta]_1$ be maximal regular sequences. We also have
\[\begin{align*} K[\Gamma * \Delta] / (\theta_1, \ldots, \theta_d, \eta_1, \ldots, \eta_e) \cong \big(K[\Gamma] / (\theta_1, \ldots, \theta_d)\big) \otimes_K \big(K[\Delta] / (\eta_1, \ldots, \eta_e)\big). \end{align*}\]The Hilbert series of the quotient is therefore
\[\begin{align*} &L\left(K[\Gamma * \Delta] / (\theta_1, \ldots, \theta_d, \eta_1, \ldots, \eta_e), \lambda\right)\\ &\quad\quad= L\left(K[\Gamma] / (\theta_1, \ldots, \theta_d), \lambda\right) \cdot L\left(K[\Delta] / (\eta_1, \ldots, \eta_e), \lambda\right). \end{align*}\]By [1, Th. 12.10],
\[\begin{align*} \frac{L(K[\Gamma * \Delta] / (\theta_1, \ldots, \theta_d, \eta_1, \ldots, \eta_e), \lambda)}{(1-\lambda)^{d+e}} =L(K[\Gamma], \lambda) \cdot L(K[\Delta], \lambda) = L(K[\Gamma * \Delta], \lambda), \end{align*}\]confirming that $\theta_1, \ldots, \theta_d, \eta_1, \ldots, \eta_e$ is a regular sequence in $K[\Gamma * \Delta]_1$, i.e., $\dim K[\Gamma * \Delta] = \operatorname{depth} K[\Gamma * \Delta] = d + e$. Thus, $\Gamma*\Delta$ is Cohen–Macaulay as well. $\blacksquare$
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Let $m = f_{d-1}(\Gamma)$ and $n = f_{e-1}(\Delta)$; and let $F_1, \ldots, F_m$ be a shelling order for $\Gamma$, and $G_1, \ldots, G_n$ a shelling order for $\Delta$. As we proved in item (a), $f_{d+e-1}(\Gamma*\Delta)=mn$.
We define $H_i = F_{\lceil i/n \rceil} \cup G_{i \bmod n}$ for $i = 1, \ldots, mn$, that is, the unions of the faces of the shellings ordered lexicographically. For practical purposes, let $\alpha_i = \lceil i/n \rceil$ and $\beta_i = i \bmod n$.
Let $r(F_i)$ and $r(G_j)$ be the restrictions of $F_i$ and $G_j$ in their respective shelling orders.
Now, for each $i = 1, \ldots, mn$, if $H_i = F_{\alpha_i} \cup G_{\beta_i}$, then the set $\langle H_i \rangle \setminus \langle H_1, \ldots, H_{i-1} \rangle$ must contain the face $r(F_{\alpha_i}) \cup r(G_{\beta_i})$, and this is indeed the restriction of $H_i$. Otherwise, either $r(F_{\alpha_i})$ or $r(G_{\beta_i})$ would not be unique.
Thus, $\Gamma * \Delta$ has the shelling order $H_1, \ldots, H_{mn}$. $\blacksquare$
References
- Stanley. Richard P. (2018). Algebraic Combinatorics: Walks, Trees, Tableaux, and More (2nd ed.). Springer International Publishing. https://doi.org/10.1007/978-3-319-77173-1