January 01, 1970
The LYM inequality is a fundamental result concerning the sizes of subsets in a Sperner family. Subsequent studies on the LYM inequality have been generalized to families of \(r\)-decompositions, where all components are required to avoid chains of the same length. In this paper, we relax this constraint by allowing components of a family of \(r\)-decompositions to avoid chains of distinct lengths, and derive generalized LYM inequalities across all the relevant settings, including set-theoretic, \(q\)-analog, continuous analog, and arithmetic analog frameworks. Notably, the bound in our LYM inequalities does not depend on the maximal length of all forbidden chains. Moreover, we extend our approach beyond \(r\)-decompositions to \(r\)-multichains, and establish analogous LYM inequalities.
Mathematics Subject Classification: 05D05
Keywords: LYM inequality, Sperner’s theorem, Meshalkin’s theorem, antichain
Several New Generalizations of LYM Inequality
Zihao Huang\(^{1}\), Weikang Liang\(^{2}\), Yujiao Ma\(^{*3}\) and Suijie Wang\(^{4}\)
\(^{1,2,3,4}\)School of Mathematics
Hunan University
Changsha 410082, Hunan, P. R. China
Emails: \(^{1}\)zihaoh@hnu.edu.cn, \(^{2}\)kangkang@hnu.edu.cn, \(^{3}\)yujiaoma@hnu.edu.cn, \(^{4}\)wangsuijie@hnu.edu.cn
Let \(B(n)\) be the Boolean lattice, which consists of all subsets of \([n] = \{1, \ldots, n\}\) and is ordered by inclusion. An antichain in \(B(n)\) is a family of subsets of \([n]\) where no two contain each other. A fundamental question—what is the maximum size of an antichain in \(B(n)\)?—is answered by Sperner’s Theorem:
This result initiated a rich line of research on the Sperner property, which has since evolved into a well-developed theory within extremal set systems [2].
The LYM inequality, which naturally implies Sperner’s theorem as a special case, was independently obtained by Bollobás[3], Lubell[4], Yamamoto[5], and Meshalkin[6], and has since become a more general tool for analyzing antichains in \(B(n)\).
An \(r\)-decomposition of \([n]\) is an \(r\)-tuple \((D_1, \dots, D_r)\) of disjoint subsets of \([n]\) such that \(\bigcup_{i=1}^r D_i = [n]\). Given a family \(\mathcal{D}\) of \(r\)-decompositions of \([n]\), we define \[\mathcal{D}_k = \{ D_k \mid (D_1, \ldots,D_k,\ldots, D_r) \in \mathcal{D} \}\] for each \(k \in \{1,\ldots,r\}\). Building on Sperner’s Theorem and the LYM inequality, Meshalkin[6] generalized Sperner’s Theorem by extending the idea of antichain properties on single subsets to families of \(r\)-decompositions of \([n]\).
Hochberg and Hirsch[7] further extended the classic LYM inequality to a multinomial version, which is tailored to the setting of \(r\)-decompositions and directly implies Meshalkin’s theorem. Their result is stated as follows:
A chain of length \(t\) in \(B(n)\) is a collection of \(t+1\) subsets of \([n]\) where every pair of subsets is comparable. Recall that an antichain is precisely a family of subsets avoiding chains of length \(1\). A family \(\mathcal{A} \subseteq B(n)\) is said to be \(t\)-chain free if \(\mathcal{A}\) contains no chain of length \(t\). Erdős [8], Rota and HarperRota?, and?, Harper? further extended the classical Sperner’s Theorem and LYM inequality to the setting of \(t\)-chain free families, yielding the following results:
Erdős’ Theorem[8]. If \(\mathcal{A} \subseteq B(n)\) is \(t\)-chain free, then \[|\mathcal{A}| \leq \binom{n}{\left \lfloor \frac{n+1}{2} \right \rfloor } + \cdots + \binom{n}{\left \lfloor \frac{n+t}{2} \right \rfloor }.\]
Rota-Harper TheoremRota?, and?, Harper?. If \(\mathcal{A} \subseteq B(n)\) is \(t\)-chain free, then \[\label{b} \sum_{A\in \mathcal{A} }^{} \frac{1}{\binom{n}{\left | A \right |}}\leq t.\tag{3}\]
Beck and ZaslavskyBeck?, and?, Zaslavsky1? generalized these results further: they considered the families of \(r\)-decompositions with each \(\mathcal{D}_k\) is \(t\)-chain free, and thus extended these results to a more general setting.
Up to this point, we have focused on results in the Boolean lattice \(B(n)\). The corresponding \(q\)-analogue is given by the subspace lattice \(L_q(n)\) of \(\mathbb{F}_q^n\) over a finite field \(\mathbb{F}_q\). Paralleling the definition of \(t\)-chain free families in \(B(n)\), a family \(\mathcal{A}\subseteq L_q(n)\) is called \(t\)-chain free if \(\mathcal{A}\) contains no chain of length \(t\) in \(L_q(n)\). The number of \(k\)-dimensional subspaces of \(\mathbb{F}_q^n\) is denoted by \(\genfrac[]{0pt}{}{n}{k}_{q}\), called \(q\)-Gaussian coefficient. In the setting of the subspace lattice \(L_q(n)\), Rota and Harper Rota?, and?, Harper? established the \(q\)-analog of the LYM inequality as follows:
Similarly, the concept of \(r\)-decompositions of \([n]\) can be extended to the setting of \(L_q(n)\). An \(r\)-decomposition (also called an \(r\)-Meshalkin sequence) of \(\mathbb{F}_q^n\) is a tuple \((D_1, \ldots, D_r)\) where each \(D_i \in L_q(n)\) and \(D_1 \oplus \cdots \oplus D_r =\mathbb{F}_q^n\). Given a family \(\mathcal{D}\) of \(r\)-decompositions of \(\mathbb{F}_q^n\), define \[\mathcal{D}_k := \left\{ D_k \mid (D_1, \ldots,D_k,\ldots, D_r) \in \mathcal{D} \right\}\] for each \(k \in \{1, \ldots, r\}\). Beck and ZaslavskyBeck?, and?, Zaslavsky2? further proved the \(q\)-analog of Meshalkin’s theorem, stated as follows:
The continuous analogs of the LYM inequality and Meshalkin’s theorem were first introduced by Klain and Rota [9], Klain?, and?, Rota?. Denote by \(\mathrm{Mod}(n)\) the subspace lattice of \(\mathbb{R}^n\). A family \(\mathcal{A} \subseteq \mathrm{Mod}(n)\) is called \(t\)-chain free if \(\mathcal{A}\) contains no chain of length \(t\) in \(\mathrm{Mod}(n)\). Let \(\mathrm{Gr}(n, k)\) denote the Grassmannian of all \(k\)-dimensional linear subspaces of \(\mathbb{R}^n\). Notably, \(\mathrm{Gr}(n, k)\) forms a compact smooth manifold equipped with an orthogonal invariant measure \(\nu_k^n\). The total measure of \(\mathrm{Gr}(n,k)\) is denoted by \(\genfrac[]{0pt}{}{n}{k}_{\mathbb{R}}\). Klain and Rota established the following continuous analog of the LYM inequality:
An \(r\)-decomposition of \(\mathbb{R}^n\) is a tuple \((D_1, \ldots, D_r)\) of nonzero subspaces such that \(\mathbb{R}^n = \bigoplus_{i=1}^r D_i\), with \(D_i \perp D_j\) for all \(i \ne j\). Let \(\mathcal{D}(\mathbb{R}^n, r)\) denote the set of all \(r\)-decompositions of \(\mathbb{R}^n\). Given positive integers \(a_1, \ldots, a_r\) with \(a_1 + \cdots + a_r = n\) and \(\mathcal{D} \subseteq \mathcal{D}(\mathbb{R}^n, r)\), define \(\mathcal{D}_{a_1, \ldots, a_r}\) to be the subset of all \(r\)-decompositions \((D_1, \ldots, D_r) \in \mathcal{D}\) such that \(\dim(D_i) = a_i\) for all \(i = 1, \ldots, r\). Clearly, we have the disjoint union \[\mathcal{D}(\mathbb{R}^n, r) = \bigsqcup_{a_1 + \cdots + a_r = n} \mathcal{D}(\mathbb{R}^n, r)_{a_1, \ldots, a_r}.\] Each \(\mathcal{D}(\mathbb{R}^n, r)_{a_1, \ldots, a_r}\) is a compact smooth manifold equipped with a measure \(\nu_{a_1, \ldots, a_r}^n\), which is induced naturally from the measure \(\nu_k^n\) on \(\mathrm{Gr}(n,k)\). We denote by \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}\) the total measure of \(\mathcal{D}(\mathbb{R}^n, r)_{a_1, \ldots, a_r}\) with respect to \(\nu_{a_1, \ldots, a_r}^n\). The measure \(\nu_{n;r}\) on \(\mathcal{D}(\mathbb{R}^n, r)\) is defined for any \(\mathcal{D} \subseteq \mathcal{D}(\mathbb{R}^n, r)\) by \[\nu_{n;r}(\mathcal{D}) = \sum_{a_1 + \cdots + a_r = n} \nu_{a_1, \ldots, a_r}^n(\mathcal{D}_{a_1, \ldots, a_r}).\] Let \(\mathcal{D} \subseteq \mathcal{D}(\mathbb{R}^n, r)\), we define \[\mathcal{D}_k := \{D_k \mid (D_1, \ldots,D_k,\ldots, D_r) \in \mathcal{D} \}\] for each \(k \in \{1,\ldots,r\}\). Klain and Rota also established the following continuous analog of Meshalkin’s theorem:
Beyond the \(q\)-analogs and continuous analogs, one can also consider the arithmetic analogs of the LYM inequality. Let \(n\) be a positive integer, and let \(\mathrm{Div}(n)\) denote the lattice of all positive divisors of \(n\), ordered by divisibility. For a divisor \(x \in \mathrm{Div}(n)\), its rank is the total number of prime factors of \(x\), counting multiplicities. Let \(W_i(n)\) be the number of elements in \(\mathrm{Div}(n)\) of rank \(i\). Anderson[10] proved the following arithmetic analog of the LYM inequality:
In all known generalizations of the LYM inequality and its various analogs, the results have been established only for families \(\mathcal{D}\) of \(r\)-decompositions in which every component avoids chains of the same fixed length, i.e., all \(\mathcal{D}_k\) are \(t\)-chain free for a single integer \(t\).
One of the central contributions of this paper is to lift this restriction: we extend these results to the more general and flexible setting where different components of the family \(\mathcal{D}\) may avoid chains of distinct lengths. Specifically, we allow its \(k\)-th component family \(\mathcal{D}_k\) to be \(t_k\)-chain free, where \(t_1, \ldots, t_r\) are arbitrary given positive integers. We establish such extensions across four fundamental lattice structures: the Boolean lattice \(B(n)\), the subspace lattice \(L_q(n)\) of \(\mathbb{F}_q^n\), the subspace lattice \(\mathrm{Mod}(n)\) of \(\mathbb{R}^n\), and the divisor lattice \(\mathrm{Div}(n)\). We derive two types of results: the corresponding generalized LYM inequalities, and the corresponding upper bounds on the size or measure of the \(r\)-decomposition family \(\mathcal{D}\). These upper bounds rely on a key parameter \(\sigma = \frac{t_1 t_2 \cdots t_r}{\mathrm{max}\{t_1, \ldots, t_r\}}\), with the bound taking the form of the sum of the \(\sigma\) largest relevant summands. A summary of all these main results is provided in Table 1.
Remark 1. It is noted that the value of \(\sigma = \frac{t_1 t_2 \cdots t_r}{\mathrm{max}\{t_1, \ldots, t_r\}}\) is independent of the specific value of \(\rm max\{t_1, \ldots, t_r\}\). In other words, the component associated with the longest forbidden chain plays no essential role in the derived LYM inequalities and the corresponding bounds, which remain unchanged no matter how large \(\rm max\{t_1, \ldots, t_r\}\) is. For \(r=2\), this phenomenon admits a straightforward interpretation: by the definition of the \(2\)-decomposition, if \(\mathcal{D}_1\) is \(t_1\)-chain free, then \(\mathcal{D}_2\) is also \(t_1\)-chain free; consequently, \(\mathcal{D}_2\) is trivially \(t_2\)-chain free when \(t_1 \leq t_2\). When \(r\geq 3\), however, a clear interpretation of this phenomenon remains elusive.
11pt
| Lattice | LYM Inequality | Upper bound |
|---|---|---|
| \(B(n)\) | \(\displaystyle \sum_{a_{1}+\cdots+a_{r}=n}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1, \ldots, a_r}} } \leq \sigma\) | \(|\mathcal{D}| \leq \sum\binom{n}{a_1, \ldots, a_r}\) |
| \(L_q(n)\) | \(\displaystyle \;\sum_{a_1 + \cdots + a_r = n}\frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod\limits_{i<j}q^{a_ia_j}}\le \sigma\;\;\) | \(|\mathcal{D}| \leq \sum \genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod\limits_{i<j}q^{a_ia_j}\) |
| \(\mathrm{Mod}(n)\) | \(\displaystyle \sum_{a_{1}+\cdots+a_{r}=n}\dfrac{\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}}\le \sigma\) | \(\nu_{n;r}(\mathcal{D}) \leq \sum\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}\) |
| \(\mathrm{Div}(n)\) | \(\displaystyle \sum_{a_{1}+\cdots+a_{r}=\operatorname{rk}(n)}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{N_{a_1, \ldots,a_r}(n)} \leq \sigma\) | \(|\mathcal{D}| \leq \sum N_{a_1, \ldots,a_r}(n)\) |
In addition to extending results for \(r\)-decompositions, we further generalize our framework to a closely related structure: \(r\)-multichains. Formally, an \(r\)-multichain of a lattice is a tuple \((C_1, \ldots, C_r)\) of elements of the lattice such that \(C_1 \leq C_2 \leq \cdots \leq C_r\). For a family \(\mathcal{C}\) of \(r\)-multichains, we define \(\mathcal{C}_k = \{C_k \mid (C_1, \ldots, C_k, \ldots, C_r) \in \mathcal{C}\}\) for each \(k\in \{1, \ldots, r\}\), and \[\mathcal{C}_{a_1,\dots,a_r} = \{(C_1, \ldots, C_r) \in \mathcal{C} \mid \operatorname{rk}(C_i) = a_1 + \cdots + a_i \text{ for all } i = 1, \ldots, r\},\] where \(\operatorname{rk}(\cdot)\) denotes the rank function of this lattice, and \(a_1,\ldots,a_r\) are non-negative integers.
Following the same research paradigm as our \(r\)-decomposition results, another key central contribution of this paper is to establish two types of results for \(r\)-multichains \(\mathcal{C}\) such that each \(\mathcal{C}_k\) is \(t_k\)-chain free for arbitrary positive integers \(t_1,\ldots,t_r\): the generalized LYM inequalities, and the corresponding upper bounds on the size or measure of \(\mathcal{C}\). A summary of these results is provided in Table 2, where \(\tau\) denotes the product \(t_1 t_2 \cdots t_r\). In the last column, each summation is taken over the \(\tau\) largest terms among all summands.
12.3pt
| Lattices | LYM Inequality | Upper bound |
|---|---|---|
| \(B(n)\) | \(\displaystyle \sum_{a_1 + \cdots + a_{r+1} = n } \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_{r+1}}} }\leq \tau\) | \(|\mathcal{C}| \leq \sum\binom{n}{a_1, \ldots, a_{r+1}}\) |
| \(L_q(n)\) | \(\displaystyle \sum_{a_1 + \cdots + a_{r+1} = n}\frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}}\le \tau\) | \(|\mathcal{C}| \leq \sum \genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}\) |
| \(\mathrm{Mod}(n)\) | \(\;\;\displaystyle \sum_{a_{1}+\cdots+a_{r+1}=n}\dfrac{\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_{r}})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}}\le \tau\) | \(\mu_{n;r}(\mathcal{C}) \leq \sum\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}\) |
| \(\mathrm{Div}(n)\) | \(\displaystyle \sum_{a_{1}+\cdots+a_{r+1}=\operatorname{rk}(n)}^{} \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_{r+1}}(n)} \leq \tau\) | \(|\mathcal{C}| \leq \sum N_{a_1, \ldots,a_{r+1}}(n)\) |
Recall that an \(r\)-decomposition of \([n]\) is a tuple \(\left ( D_{1},\dots ,D_{r} \right )\) of subsets of \([n]\) where \(\bigcup_{i=1}^r D_i = [n]\) and \(D_i \cap D_j = \emptyset\) for all \(i \neq j\). Given a family \(\mathcal{D}\) of \(r\)-decompositions of \([n]\) and non-negative integers \(a_1,\ldots,a_r\) with \(a_1 + \cdots + a_r =n\), we write \[\mathcal{D}_k = \{D_k \mid (D_1, \ldots, D_k, \ldots, D_r) \in \mathcal{D} \}\] for each \(k \in \{1, \ldots, r\}\), and \[\mathcal{D}_{a_{1},\dots,a_{r}}=\{(D_{1},\dots ,D_{r}) \in \mathcal{D} \mid |D_i| = a_i,\; i=1,\ldots,r\}.\]
Theorem 2. Let \(t_1, \ldots, t_r\) be positive integers and \(\sigma = \frac{t_1t_2\cdots t_r}{\mathrm{max}\{t_1, \ldots, t_r\}}\). Suppose \(\mathcal{D}\) is a family of \(r\)-decompositions of \([n]\) such that \(\mathcal{D}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{1461} \sum_{a_{1}+\cdots+a_{r}=n}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1, \ldots, a_r}} } \leq \sigma.\qquad{(1)}\] Consequently, \[|\mathcal{D}| \leq m_1+\cdots+m_\sigma,\] where \(m_1,\ldots,m_\sigma\) are the \(\sigma\) largest \(\binom{n}{{a_1, \ldots, a_r}}\) for non-negative integers \(a_1, \ldots, a_r\) with \(a_1+\cdots+a_r = n\).
Before moving to the proof of Theorem 2, we state a crucial lemma that plays a key role in our proof.
Lemma 3. Suppose that \(c_{1} \ge \cdots \ge c_{n} > 0\). If \(c_{i}\ge x_{i}\ge 0\) for \(i \in \{1, \ldots, n\}\), and if \[\label{lem} x_{1} + \cdots + x_{n} > c_{1}+ \cdots +c_{t},\qquad{(2)}\] then \[\sum_{k=1}^{n}\frac{x_{k}}{c_{k}} > t.\]
Proof. Let \(y_i=c_i-x_i\) for each \(i\in \{1, \ldots, n\}\). Then the inequality ?? reduces to \[x_{t+1}+\cdots+x_n>y_1+\cdots+y_{t}.\] Therefore, we have \[\begin{align} \sum_{k=1}^{n}\frac{x_{k}}{c_{k}} &=t-\sum_{k=1}^{t}\frac{y_k}{c_{k}}+\sum_{k=t+1}^{n}\frac{x_{k}}{c_{k}}\\ &\geq t+\sum_{k=t+1}^{n}\frac{x_{k}}{c_{k}}-\frac{y_1+\cdots+y_{t}}{c_{t+1}}\\ &> t+\sum_{k=t+1}^{n}\frac{x_{k}}{c_{k}}-\frac{x_{t+1}+\cdots+x_n}{c_{t+1}}\\ &\geq t. \end{align}\] ◻
Proof of Theorem 2. Without loss of generality, assume that \(t_r\) is the maximal integer among \(\{t_1, \cdots, t_r\}\). We proceed by induction on \(r\) to prove the inequality ?? . For \(r =2\), note that any \(2\)-decomposition \((D_1, D_2)\) of \([n]\) satisfies \(D_2 = [n]\setminus D_1\). Then, if \(\mathcal{D}_1\) is \(t_1\)-chain free, it follows that \(\mathcal{D}_2\) is also \(t_1\)-chain free, and hence \(t_2\)-chain free. Therefore, the inequality ?? is equivalent to 3 in Rota-Harper Theorem by choosing the collection \(\mathcal{A}\) to be \(\mathcal{D}_1\). Now suppose that \(r>2\) and that the inequality ?? holds for \(r-1\). Then \[\label{c1} \begin{align} \sum_{a_{1}+\cdots+a_{r}=n}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_r}} }&=\sum_{(D_1,\ldots, D_r)\in \mathcal{D} }^{} \frac{1}{\binom{n}{{|D_1|,\ldots,|D_r|}} }\\ &=\sum_{(D_1,\ldots, D_r)\in \mathcal{D} }^{} \frac{1}{\binom{n}{|D_1|}}\frac{1}{\binom{n-|D_1|}{{|D_2|,\ldots,|D_r|}} } \\ &=\sum_{A\in \mathcal{D}_1}\frac{1}{\binom{n}{|A|}}\sum\limits_{\substack{{ (D_1,\ldots, D_r)\in \mathcal{D}}\\{D_{1} = A}}}\frac{1}{\binom{n-|A|}{{|D_2|,\ldots,|D_r|}}}. \end{align}\tag{10}\] It follows from the induction hypothesis that \[\sum\limits_{\substack{{ (D_1,\ldots, D_r)\in \mathcal{D}}\\{D_{1} = A}}}\frac{1}{\binom{n-|A|}{{|D_2|,\ldots,|D_r|}}} \leq t_2\cdots t_{r-1} .\] Therefore, the last expression in 10 is at most \(\sigma = t_1t_2\cdots t_{r-1}\) by 3 , which completes the proof of ?? .
The number of multinomial coefficients \(\binom{n}{a_{1},\ldots,a_{r}}\) that satisfy \(a_1+\cdots + a_r=n\) is \(\binom{n+r-1}{r-1}\). Note that \[\mathcal{D} = \bigsqcup_{a_{1}+\cdots+a_{r}=n}\mathcal{D}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{D}_{a_{1},\dots,a_{r}}|\le \binom{n}{{a_1,\ldots,a_r}}\) for each \(\mathcal{D}_{a_{1},\dots,a_{r}}\). When \(\sigma > \binom{n+r-1}{r-1}\), we assume that \(m_i = 0\) for all \(i > \binom{n+r-1}{r-1}\) and then \(|\mathcal{D}| \leq m_1+\cdots+m_\sigma\) trivially holds. For the case \(\sigma \leq \binom{n+r-1}{r-1}\), suppose for contradiction that \(|\mathcal{D}| > m_1+\cdots+m_\sigma\). Then by Lemma 3 we have \[\sum_{a_{1}+\cdots+a_{r}=n}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_r}} } > \sigma,\] which contradicts the inequality ?? . This completes the proof. ◻
Remark 4. The upper bound given in Theorem 2 is not tight in general. Consider the case where \(n=4\) and \(r = 3\). When \(t_1 = 1\) and \(t_2 = t_3 = 2\), Theorem 2 provides the upper bound \[\binom{4}{1,1,2} + \binom{4}{1,2,1} = 24\] on the size of \(\mathcal{D}\). This bound can be attained by the following collection: \[\left\{ \begin{array}{llll} (\{1\}, \{2\}, \{3,4\}), & (\{2\}, \{1\}, \{3,4\}), & (\{3\}, \{1\}, \{2,4\}), & (\{4\}, \{1\}, \{2,3\}), \\ (\{1\}, \{3\}, \{2,4\}), & (\{2\}, \{3\}, \{1,4\}), & (\{3\}, \{2\}, \{1,4\}), & (\{4\}, \{2\}, \{1,3\}), \\ (\{1\}, \{4\}, \{2,3\}), & (\{2\}, \{4\}, \{1,3\}), & (\{3\}, \{4\}, \{1,2\}), & (\{4\}, \{3\}, \{1,2\}), \\ (\{1\}, \{3,4\}, \{2\}), & (\{2\}, \{3,4\}, \{1\}), & (\{3\}, \{2,4\}, \{1\}), & (\{4\}, \{2,3\}, \{1\}), \\ (\{1\}, \{2,4\}, \{3\}), & (\{2\}, \{1,4\}, \{3\}), & (\{3\}, \{1,4\}, \{2\}), & (\{4\}, \{1,3\}, \{2\}), \\ (\{1\}, \{2,3\}, \{4\}), & (\{2\}, \{1,3\}, \{4\}), & (\{3\}, \{1,2\}, \{4\}), & (\{4\}, \{1,2\}, \{3\}) \end{array} \right\}.\] However, when \(t_1 = 1\) and \(t_2 = t_3 = 3\), a computational enumeration shows that the maximum size of \(\mathcal{D}\) is 28, which is strictly smaller than the upper bound \[\binom{4}{2,1,1} + \binom{4}{1,2,1} + \binom{4}{1,1,2} = 36\] given by Theorem 2.
Next, we turn to establishing the LYM inequality for \(r\)-multichains. An \(r\)-multichain in \(B(n)\) is an \(r\)-tuple \((C_1, \ldots, C_r)\) of elements of \(B(n)\) such that \(C_1 \subseteq C_2 \subseteq \cdots \subseteq C_r\). Let \(\mathcal{C}\) be a family of \(r\)-multichains, and let \(a_1, \ldots, a_r\) be non-negative integers with \(a_1 + \cdots + a_r \leq n\). Define \(\mathcal{C}_k = \{C_k \mid (C_1, \ldots, C_k, \ldots, C_r) \in \mathcal{C}\}\) for each \(k \in \{1, \ldots, r\}\), and \[\mathcal{C}_{a_1,\dots,a_r} = \{(C_1, \ldots, C_r) \in \mathcal{C} \mid |C_i| = a_1 + \cdots + a_i \text{ for all } i = 1, \ldots, r\}.\]
Theorem 5. Let \(t_1, \ldots, t_r\) be positive integers and \(\tau = t_1t_2\cdots t_r\). Suppose \(\mathcal{C}\) is a family of \(r\)-multichains in \(B(n)\) such that \(\mathcal{C}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{1462} \sum_{a_1 + \cdots + a_{r+1} = n } \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_{r+1}}} }\leq \tau.\qquad{(3)}\] Consequently, \[|\mathcal{C}| \leq m_1 + \cdots + m_\tau,\] where \(m_1,\ldots,m_\tau\) are the \(\tau\) largest \(\binom{n}{{a_1, \ldots, a_{r+1}}}\) for non-negative integers \(a_1, \ldots, a_r\) with \(a_1+\cdots+a_{r+1} = n\).
Remark 6. Denote by \(\mathcal{D}([n], r)\) and \(\mathcal{C}([n],r)\) the set of all \(r\)-decompositions and \(r\)-multichains of \([n]\), respectively. It can be easy to verify that the map \(\phi : \mathcal{D}([n], r) \to \mathcal{C}([n],r-1)\), defined by \[\phi(D_1, \ldots, D_r) = (D_1, D_1 \cup D_2, \ldots, D_{1} \cup \cdots \cup D_{r-1}),\] is a bijection. However, the image \(\phi(\mathcal{D})\) of the family \(\mathcal{D}\) appearing in Theorem 2 does not necessarily satisfy the condition that each \(\phi(\mathcal{D})_k\) is also \(t_k\)-chain-free. A concrete counterexample is given as follows: Let \(n = 4\) and \(r= 3\). Consider the collection \(\mathcal{D} = \{(\{1\},\{3\},\{2,4\}),(\{2,3\},\{1\},\{4\})\}\). Then \(\phi(\mathcal{D}) = \{(\{1\},\{1,3\}),(\{2,3\},\{1,2,3\})\}\). We observe that \(\mathcal{D}_2\) is \(1\)-chain free, while \(\phi(\mathcal{D})_2\) fails to preserve the \(1\)-chain free property. Therefore, Theorem 5 cannot be deduced from Theorem 2 via this bijection.
Proof of theorem 5. We proceed by induction on \(r\) to prove the inequality ?? . For \(r=1\), ?? is actually the generalized LYM inequality 3 . Suppose that \(r>1\) and that the inequality ?? holds for \(r-1\). Then \[\begin{align} \sum_{a_1 + \cdots + a_{r+1} = n } \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_{r+1}}} } &=\sum_{(C_1, \ldots, C_r) \in \mathcal{C} } \frac{1}{\binom{n}{{|C_1|,|C_2-C_1|,\ldots,|C_r-C_{r-1}|,|[n]-C_r|}}}\\ &=\sum_{(C_1, \ldots, C_r)\in \mathcal{C} } \frac{1}{\binom{n}{|C_1|}}\frac{1}{\binom{n-|C_1|}{{|C_2-C_1|,\ldots,|C_r-C_{r-1}|,|[n]-C_r|}} } \\ &=\sum_{{A\in \mathcal{C}_1}}^{}\frac{1}{\binom{n}{|A|}}\sum\limits_{\substack{{ (C_1, \ldots, C_r)\in \mathcal{C}}\\{C_{1} = A}}}\frac{1}{\binom{n-|A|}{{|C_2-C_1|,\ldots,|C_r-C_{r-1}|,|[n]-C_r|}}}. \end{align}\] It follows from the induction hypothesis and 3 that the last expression is at most \(\tau=t_1t_2\cdots t_r\).
The number of multinomial coefficients \(\binom{n}{a_{1},\ldots,a_{r+1}}\) that satisfy \(a_1+\cdots + a_{r+1}=n\) is \(\binom{n+r}{r}\). Note that \[\mathcal{C} = \bigsqcup_{a_{1}+\cdots+a_{r+1}=n}\mathcal{C}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{C}_{a_{1},\dots,a_{r}}|\le \binom{n}{{a_1,\ldots,a_{r+1}}}\) for each \(\mathcal{C}_{a_{1},\dots,a_{r}}\). The inequality \(|\mathcal{C}| \leq m_1 + \cdots + m_\tau\) trivially holds when \(\tau > \binom{n+r}{r}\), assuming that \(m_i = 0\) for all \(i > \binom{n+r}{r}\). For the case \(\tau \leq \binom{n+r}{r}\), suppose for contradiction that \(|\mathcal{C}| > m_1 + \cdots + m_\tau\). Then, by Lemma 3 we have \[\sum_{a_{1}+\cdots+a_{r+1}=n} \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\binom{n}{{a_1,\ldots,a_{r+1}}} } > \tau,\] which contradicts the inequality ?? . This completes the proof. ◻
Let \(\mathbb{F}_q\) be a finite field with \(q\) elements, and let \(L_q(n)=L(\mathbb{F}_q^n)\) denote the subspace lattice of \(\mathbb{F}_q^n\). Within this lattice, it is known that any \(k\)-dimensional subspace has exactly \(q^{k(n-k)}\) complements; see Beck?, and?, Zaslavsky2?. Notably, the number of \(k\)-dimensional subspaces in \(L_q(n)\) is captured by the \(q\)-analog of the binomial coefficients, which is called the \(q\)-Gaussian coefficients and defined as \[\begin{bmatrix}n\\k\end{bmatrix}_{q}=\frac{[n]_{q}!}{[k]_q![n-k]_q!},\;\;\text{where}\;\;[n]_{q}!=(q^n-1)(q^{n-1}-1)\cdots(q-1).\] Generalizing this notion, consider non-negative integers \(a_1, a_2, \ldots , a_r\) such that \(a_1 + \cdots + a_r = n\). The \(q\)-analog of the multinomial coefficients is \[\label{q-Gaussian} \genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}=\frac{[n]_q!}{[a_1]_{q}!\cdots [a_r]_{q}!}.\tag{11}\] Recall that an \(r\)-decomposition of \(\mathbb{F}_q^n\) is a tuple \((D_1,\ldots,D_r)\) of subspaces such that \(D_1 \oplus \cdots \oplus D_r = \mathbb{F}_q^n\). Given a family \(\mathcal{D}\) of \(r\)-decompositions of \(\mathbb{F}_q^n\), we write \[\mathcal{D}_k=\left\{D_{k}\mid (D_{1},\ldots,D_k, \ldots, D_{r})\in\mathcal{D}\right\}\] for each \(k \in \{1, \ldots, r\}\), and \[\mathcal{D}_{a_{1},\dots,a_{r}}= \{(D_1, \ldots, D_r)\in \mathcal{D}\mid \dim(D_i) = a_i\text{ for all } i=1,\ldots, r\}.\] It follows from Beck?, and?, Zaslavsky2? that when \(\mathcal{D}\) is the set of all \(r\)-decompositions, we have \[\label{11} |\mathcal{D}_{a_1, \ldots, a_r}| ={\textstyle \genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}}\prod\limits_{i<j}q^{a_ia_j} .\tag{12}\]
Theorem 7. Let \(t_1, \ldots, t_r\) be positive integers and \(\sigma = \frac{t_1t_2\cdots t_r}{\mathrm{max}\{t_1, \ldots, t_r\}}\). Suppose \(\mathcal{D}\) is a family of \(r\)-decompositions of \(\mathbb{F}_q^n\) such that \(\mathcal{D}_k\) is \(t_k\)-chain free* for each \(k\in \{1, \ldots, r\}\). Then \[\label{4461} \sum_{a_1 + \cdots + a_r = n}\frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod\limits_{i<j}q^{a_ia_j}}\le \sigma.\tag{13}\] Consequently, \[|\mathcal{D}|\le m_1+\cdots+m_\sigma,\] where \(m_1,\ldots,m_\sigma\) are the \(\sigma\) largest \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod\limits_{i<j}q^{a_ia_j}\) for non-negative integers \(a_1,\ldots,a_r\) with \(a_1+\cdots + a_r=n\).*
Proof of theorem 7. Without loss of generality, assume \(t_{r}=\mathrm{max}\{t_1, \ldots, t_r\}\). We proceed by induction on \(r\). For \(r = 2\), the inequality 13 reduces to \[\label{d1} \begin{align} \sum_{a_1 + a_2 = n}\frac{|\mathcal{D}_{a_{1},a_{2}}|}{\genfrac[]{0pt}{}{n}{a_1,a_2}_{q}q^{a_1a_2} } &=\sum_{(D_1, D_2)\in\mathcal{D}}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(D_1), n-\dim(D_1)}_{q}q^{\dim(D_1)(n-\dim(D_1)}}\\ &=\sum_{D\in \mathcal{D}_1}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(D)}_{q}q^{\dim(D)(n-\dim(D))}}\sum\limits_{\substack{{ (D, D_2)\in \mathcal{D}}}}1. \end{align}\tag{14}\] As any \(k\)-dimensional subspace in \(L_q(n)\) has \(q^{k(n-k)}\) complements, it follows that \[\sum\limits_{\substack{{ (D, D_2)\in \mathcal{D}}}}1 \leq q^{\dim(D)(n-\dim(D))}.\] Therefore, by the inequality 5 in the Rota-Harper theorem, the last expression in 14 is at most \(t_1\). Now suppose that \(r> 2\). Note that for any \(D \in \mathcal{D}_1\), by the induction hypothesis we have \[\begin{align} &\sum\limits_{\substack{{ (D, D_2, \ldots, D_r)\in \mathcal{D}} }}\frac{1}{\genfrac[]{0pt}{}{n-\dim(D)}{\dim(D_2), \ldots, \dim(D_r)}_{q}\prod\limits_{2\le i<j}q^{\dim(D_i)\dim(D_j)}}\\ &=\sum\limits_{F \oplus D = \mathbb{F}_q^n}\sum\limits_{\substack{ (D, D_2, \ldots, D_r)\in \mathcal{D} \\ D_2 \oplus \cdots \oplus D_r = F}}\frac{1}{\genfrac[]{0pt}{}{\dim(F)}{\dim(D_2), \ldots, \dim(D_r)}_{q}\prod\limits_{2\le i<j}q^{\dim(D_i)\dim(D_j)}}\\ &\leq q^{\dim(D)(n-\dim(D))} t_2\cdots t_{r-1}. \end{align}\] Therefore, \[\begin{align} &\sum_{a_1 + \cdots + a_r = n}\frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q} \prod\limits_{i<j}q^{a_ia_j}}\\ &=\sum_{(D_1, \ldots, D_r)\in\mathcal{D}}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(D_1), \ldots, \dim(D_r)}_{q}\prod\limits_{i<j}q^{\dim(D_i)\dim(D_j)}}\\ &=\sum_{D\in \mathcal{D}_1}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(D)}_{q}q^{\dim(D)(n-\dim(D))} }\sum\limits_{{ (D,D_2, \ldots, D_r)\in \mathcal{D}}}\frac{1}{\genfrac[]{0pt}{}{n-\dim(D)}{\dim(D_2), \ldots, \dim(D_r)}_{q}\prod\limits_{2\le i<j}q^{\dim(D_i)\dim(D_j)}}\\ &\leq t_2\cdots t_{r-1}\sum_{D\in \mathcal{D}_1}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(D)}_{q}}. \end{align}\] Again by 5 in the Rota-Harper Theorem, the last expression is at most \(\sigma = t_1\cdots t_{r-1}\), which completes the proof of the inequality 13 .
Note that \[\mathcal{D} = \bigsqcup_{a_{1}+\cdots+a_{r}=n}\mathcal{D}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{D}_{a_{1},\dots,a_{r}}|\le \genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod_{i<j}q^{a_ia_j}\) for each \(\mathcal{D}_{a_{1},\dots,a_{r}}\). When \(\sigma>\binom{n+r-1}{r-1}\), we assume that \(m_i = 0\) for all \(i > \binom{n+r-1}{r-1}\) and then \(|\mathcal{D}|\le m_1+\cdots+m_\sigma\) trivially holds. For the case \(\sigma \leq \binom{n+r-1}{r-1}\), suppose for contradiction that \(|\mathcal{D}| > m_1+\cdots+m_\sigma\). Then, by Lemma 3 we have \[\sum_{a_{1}+\cdots+a_{r}=n} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{q}\prod\limits_{i<j}q^{a_ia_j}} > \sigma,\] which contradicts the inequality 13 . This completes the proof. ◻
An \(r\)-multichain in the subspaces lattice \(L_q(n)\) is a tuple \((C_1, \ldots, C_r)\) of elements of \(L_q(n)\) such that \(C_1 \subseteq \cdots \subseteq C_r\). Let \(\mathcal{C}\) be a family of such \(r\)-multichains, and let \(a_1, \ldots, a_r\) be non-negative integers with \(a_1 + \cdots + a_r \leq n\). We define \[\mathcal{C}_k := \{C_k \mid (C_1, \ldots, C_k, \ldots, C_r) \in \mathcal{C}\}\] for each \(k \in \{1, \ldots, r\}\), and \[\mathcal{C}_{a_1,\dots,a_r} := \left\{ (C_1, \ldots, C_r) \in \mathcal{C} \mid \operatorname{rk}(C_i) = a_1 + \cdots + a_i \text{ for all } i = 1, \ldots, r \right\}.\]
Theorem 8. Let \(t_1, \ldots, t_r\) be positive integers and \(\tau= t_1t_2\cdots t_r\). Suppose \(\mathcal{C}\) is a family of \(r\)-multichains in \(L_q(n)\) such that \(\mathcal{C}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{chq1} \sum_{a_1 + \cdots + a_{r+1} = n}\frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}}\le \tau.\qquad{(4)}\] Consequently, \[|\mathcal{C}| \leq m_1+\cdots+m_\tau,\] where \(m_1,\ldots,m_\tau\) are the \(\tau\) largest \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}\) for non-negative integers \(a_1,\ldots,a_{r+1}\) with \(a_1+\cdots + a_{r+1}=n\).
Proof of Theorem 8. We proceed by induction on \(r\). For \(r=1\), the inequality ?? reduces to the inequality 5 stated in the Rota-Harper Theorem. Suppose that \(r>1\) and that the inequality ?? holds for \(r-1\). Then \[\begin{align} \sum_{a_1 + \cdots + a_{r+1} = n}\frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}} &=\sum_{(C_1, \ldots, C_r)\in\mathcal{C}}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(C_1),\dim(C_2)-\dim(C_1) ,\ldots,n-\dim(C_r)}_{q}}\\ &=\sum_{C\in \mathcal{C}_1}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(C)}_{q}}\sum\limits_{\substack{{ (C,C_2, \ldots, C_r)\in \mathcal{C}} }}\frac{1}{\genfrac[]{0pt}{}{n-\dim(C)}{\dim(C_2)-\dim(C_1), \ldots,n-\dim(C_r)}_{q}}\\ &\leq \sum_{C\in \mathcal{C}_1}\frac{1}{\genfrac[]{0pt}{}{n}{\dim(C)}_{q}} t_2\cdots t_{r}.\\ \end{align}\] Again, by 5 , the last expression is at most \(\tau = t_1\cdots t_{r}\).
Note that \[\mathcal{C} = \bigsqcup_{a_{1}+\cdots+a_{r+1}=n}\mathcal{C}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{C}_{a_{1},\dots,a_{r}}|\le \genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}\) for each \(\mathcal{C}_{a_{1},\dots,a_{r}}\). When \(\tau >\binom{n+r}{r}\), we assume that \(m_i = 0\) for all \(i > \binom{n+r}{r}\) and then \(|\mathcal{C}|\le m_1+\cdots+m_\tau\) trivially holds. For the case \(\tau \leq \binom{n+r}{r}\), suppose for contradiction that \(|\mathcal{C}| > m_1+\cdots+m_\tau\). Then, by Lemma 3 we have \[\sum_{a_{1}+\cdots+a_{r+1}=n} \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{q}} > \tau,\] which contradicts the inequality ?? . This completes the proof. ◻
Let \(\mathrm{Mod}(n)\) denote the subspace lattice of \(\mathbb{R}^n\). For each \(k \in \{0,\ldots, n\}\), the Grassmannian \(\mathrm{Gr}(n, k)\) is the set of all \(k\)-dimensional subspaces in \(\mathrm{Mod}(n)\), which is known as the Grassmann manifold. A flag in \(\mathrm{Mod}(n)\) is a tuple \(F = (F_0, \ldots, F_n)\) of elements of \(\mathrm{Mod}(n)\), such that \(F_k \in \mathrm{Gr}(n,k)\) for \(k \in \{0,\ldots, n\}\) and \(F_0 \subseteq F_1 \subseteq \cdots \subseteq F_n.\) Denote by \(\mathrm{Flag}(n)\) the set of all flags in \(\mathrm{Mod}(n)\), which is also called the flag manifold.
Let \(O(n)\) denote the \(n\)-dimensional orthogonal group, which naturally acts on \(\mathrm{Gr}(n,k)\) and \(\mathrm{Flag}(n)\) via matrix multiplication on \(\mathbb{R}^n\). Both \(\mathrm{Gr}(n,k)(k \geq 1)\) and \(\mathrm{Flag}(n)\) carry a unique \(O(n)\)-invariant measure, known as the Haar measures [11], [12] and denoted by \(\nu_k^n\) and \(\phi_n\), respectively. To make this concrete, let \(\omega_{n}\) denote the volume of the unit ball in \(\mathbb{R}^n\), and let \(S^{n-1}\) be the unit sphere with its standard \(O(n)\)-invariant measure denoted by \(\rho_{n-1}\). The measure \(\nu_1^n\) on \(\mathrm{Gr}(n,1)\) is defined by \[\nu_1^n(A) = \frac{\rho_{n-1}(\bigcup_{x \in A}x \cap S^{n-1})}{\omega_{n-1}},\] for any subset \(A \subseteq \mathrm{Gr}(n,1)\). By orthogonal duality, \(\nu_1^n\) induces the measure \(\nu_{n-1}^n\) on \(\mathrm{Gr}(n, n-1)\). Inductively, this duality extends to a measure \(\phi_n\) on \(\mathrm{Flag}(n)\): for any simple function \(f(F_0, F_1, \ldots, F_n)\) on \(\mathrm{Flag}(n)\), \[\int f d\phi_n = \int\int f(F_0, F_1, \ldots, F_n)d\phi_{n-1}(F_0, \ldots, F_{n-1})d\nu_{n-1}^n(F_{n-1}).\] Define \([n]_{\mathbb{R}} := \dfrac{n\omega_{n}}{2\omega_{n-1}}\). The total measure of \(\mathrm{Flag}(n)\) turns out to be \[\phi_n\big(\mathrm{Flag}(n)\big) = [n]_{\mathbb{R}}! := [n]_{\mathbb{R}}[n-1]_{\mathbb{R}}\cdots[1]_{\mathbb{R}} = \dfrac{n! \omega_n}{2^n}.\] Using \(\phi_n\), we induce the measure \(\nu_{k}^{n}\) on \(\mathrm{Gr}(n,k)\) by, for any subset \(A \subseteq \mathrm{Gr}(n, k)\), \[\nu_{k}^{n}(A) = \frac{1}{[k]_{\mathbb{R}}![n-k]_{\mathbb{R}}!}\phi_n\big(\mathrm{Flag}(A)\big),\] where \(\mathrm{Flag}(A) \subseteq \mathrm{Flag}(n)\) denotes the subset consisting of all flags \((F_0, \ldots, F_n)\) such that \(F_k \in A\). In particular, we have \[\label{con95nk} \nu_{k}^{n}\big(\mathrm{Gr}(n, k)\big) = \dfrac{[n]_{\mathbb{R}}!}{[k]_{\mathbb{R}}![n-k]_{\mathbb{R}}!} = \binom{n}{k} \frac{\omega_{n}}{\omega_{k}\omega_{n-k}},\tag{15}\] which is also denoted by \(\genfrac[]{0pt}{}{n}{k}_{\mathbb{R}}\).
We now turn to the continuous analog of the LYM inequality. An \(r\)-decomposition of \(\mathbb{R}^{n}\) is a tuple \((D_1, \ldots, D_r)\) of nonzero subspaces in \(\mathrm{Mod}(n)\) satisfying \(\bigoplus_{i=1}^r D_i = \mathbb{R}^n\) and \(D_i \perp D_j\) for all \(i \neq j\). We explicitly exclude the zero subspace from this definition, since \(\mathrm{Gr}(n,0)\) is not compatible with the recursive definition of the Haar measure on \(\mathrm{Gr}(n,k)\) for \(k \geq 1\). Denote by \(\mathcal{D}(\mathbb{R}^n,r)\) the set of all \(r\)-decompositions of \(\mathbb{R}^{n}\). Given positive integers \(a_1, \ldots, a_r\) such that \(a_1 + \cdots + a_r =n\), let \(\mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}\) be the set of all \(r\)-decompositions \((D_1, \ldots, D_r) \in \mathcal{D}(\mathbb{R}^n,r)\) such that \(\dim(D_i)= a_i\) for \(i= 1, \ldots, r\). Evidently we have the finite disjoint union \[\mathcal{D}(\mathbb{R}^n,r)=\bigsqcup_{a_{1}+\cdots+a_{r}=n}{\mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}}.\] Each \(\mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}\) is also a compact smooth manifold, which carries the unique Haar measure and is denoted by \(\nu_{a_1,a_2,\ldots,a_r}^{n}\). To construct this measure explicitly, we call a flag \((F_0, \ldots, F_n)\) compatible with \((D_1, \ldots, D_r) \in\mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}\) if \(F_{a_1+\cdots+a_i} = D_1 \oplus \cdots \oplus D_i\) for all \(i \in \{1, \ldots, r\}\). Then the measure \(\nu_{a_1,a_2,\ldots,a_r}^{n}\) is induced from \(\phi_n\) on \(\mathrm{Flag}(n)\) by, for any \(\mathcal{D} \subseteq \mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}\), \[\label{val} \nu_{a_1,a_2,\ldots,a_r}^{n}(\mathcal{D})=\frac{1}{[a_1]_{\mathbb{R}}![a_2]_{\mathbb{R}}!\cdots[a_r]_{\mathbb{R}}!}\phi_n\big(\mathrm{Flag}(\mathcal{D})\big),\tag{16}\] where \(\mathrm{Flag}(\mathcal{D})\) is the set of all flags compatible with some \(r\)-decomposition of \(\mathcal{D}\). In particular, the total measure is \[\label{evi} \nu_{a_1,a_2,\ldots,a_r}^{n}\big(\mathcal{D}(\mathbb{R}^n,r)_{a_1,\ldots,a_r}\big) = \frac{[n]_{\mathbb{R}}!}{[a_1]_{\mathbb{R}}!\cdots[a_r]_{\mathbb{R}}!}.\tag{17}\] These values are called multiflag coefficients and denoted by \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}\). Furthermore, the Haar measure \(\nu_{n;r}\) on \(\mathcal{D}(\mathbb{R}^n,r)\) is induced from \(\nu_{a_1,a_2,\ldots,a_r}^{n}\) by aggregating the measures of its disjoint components. That is, for any subset \(\mathcal{D} \subseteq \mathcal{D}(\mathbb{R}^n,r)\), \[\nu_{n;r}(\mathcal{D})=\sum_{a_{1}+\cdots+a_{r}=n}\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r}),\] where \(\mathcal{D}_{a_1,\ldots,a_r}=\mathcal{D} \cap \mathcal{D}(\mathbb{R}^n,r)_{a_1,\ldots,a_r}\). For more details, we refer the reader to Klain?, and?, Rota?.
Theorem 9. Let \(t_1, \ldots, t_r\) be positive integers and \(\sigma = \frac{t_1t_2\cdots t_r}{\mathrm{max} \{ t_{1} ,t_2,\ldots,t_r \}}\). Suppose \(\mathcal{D}\) is a family of \(r\)-decompositions of \(\mathbb{R}^n\) such that \(\mathcal{D}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{2461} \sum_{a_{1}+\cdots+a_{r}=n}\dfrac{\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}}\le \sigma.\qquad{(5)}\] Consequently, \[\nu_{n;r}(\mathcal{D}) \leq m_1 + \cdots + m_\sigma,\] where \(m_1, \ldots, m_\sigma\) are the \(\sigma\) largest multiflag coefficients \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}\) for positive integers \(a_1, \ldots, a_r\) with \(a_1+\cdots + a_r=n\).
Proof. It is known that the flag manifold \(\mathrm{Flag}(n)\) is a bundle over \(\mathrm{Gr}(n,k)\) with projection map \(\pi_k\), where \(\pi_k(F_0,F_1,\dots,F_n)=F_k\) for all \((F_0,F_1,\dots,F_n)\in\mathrm{Flag}(n)\). See [13] and [14]. The fibre of the bundle is \(\mathrm{Flag}(k)\times\mathrm{Flag}(n-k)\). For each \(D\in\mathrm{Gr}(n,k)\), there exists an open neighborhood \(U\subseteq\mathrm{Gr}(n,k)\) and a local trivialization homeomorphism \(\phi: \pi_{k}^{-1}(U) \longrightarrow U \times \mathrm{Flag}(k)\times\mathrm{Flag}(n-k)\) such that \(\pi_k = \text{proj}_U \circ \phi\), where \(\text{proj}_U\) denotes the projection onto the first coordinate. The measure on \(\pi_{k}^{-1}(U)\), which is homeomorphic to \(U \times \mathrm{Flag}(k)\times\mathrm{Flag}(n-k)\), is induced by the measure \(\nu_{k}^{n}\times\phi_k\times\phi_{n-k}\) on \(\mathrm{Gr}(n,k) \times \mathrm{Flag}(k)\times\mathrm{Flag}(n-k)\).
Given positive integers \(a_1,\cdots,a_r\) satisfying \(a_1+\cdots+a_r=n\), recall that \[(\mathcal{D}_{a_{1},\ldots,a_{r}})_1 = \{ D \in \mathcal{D}_1 \mid \mathrm{dim} (X)=a_1 \}.\] Taking \(D \in (\mathcal{D}_{a_{1},\ldots,a_{r}})_1\), we define \[\mathcal{D}_{D, a_2, \ldots, a_r} = \{(D_2, \ldots, D_r) \mid (D, D_2, \ldots, D_r) \in \mathcal{D}_{a_{1},\ldots,a_{r}}\}.\] Noting that \(\big(\mathcal{D}(\mathbb{R}^n, r)\big)_{D, a_2, \ldots, a_r}\) is homeomorphic to \(\big(\mathcal{D}(\mathbb{R}^{n-a_1}, r-1)\big)_{a_2, \ldots, a_r}\), we regard \(\mathcal{D}_{D, a_2, \ldots,a_r}\) as a subset of \(\big(\mathcal{D}(\mathbb{R}^{n-a_1}, r-1)\big)_{a_2, \ldots, a_r}\). Therefore, \[\begin{align} &\phi_n\big(\mathrm{Flag}(\mathcal{D}_{a_1,\ldots,a_r})\big)\\ &=\int_{\mathrm{Gr}(n,a_1)}\chi _{(\mathcal{D}_{a_{1},\ldots,a_{r}})_1}(D)\Big(\int_{\mathrm{Flag}(a_1)} 1 d\phi_{a_1}\Big)\phi_{n-a_1}\big(\mathrm{Flag}(\mathcal{D}_{D, a_2, \ldots, a_r})\big) d\nu_{a_1}^{n}\\[3pt] &=[a_1]!\int_{\mathrm{Gr}(n,a_1)}\chi _{(\mathcal{D}_{a_{1},\ldots,a_{r}})_1}(D)\phi_{n-a_1}\big(\mathrm{Flag}(\mathcal{D}_{D, a_2, \ldots, a_r})\big)d\nu_{a_1}^{n}, \end{align}\] where \(\chi_A\) denotes the indicator function of the set \(A\). It follows from 16 that \[\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r})=\int_{\mathrm{Gr}(n,a_1)}\chi_{(\mathcal{D}_{a_{1},\ldots,a_{r}})_1}(D)\nu_{a_2,\ldots,a_r}^{n-a_1}\big(\mathcal{D}_{D, a_2, \ldots, a_r}\big)d\nu_{a_1}^{n}.\] Without loss of generality, assume that \(t_{r}=\mathrm{max} \left \{ t_{1} ,t_2,\ldots,t_r \right \}\). For \(r = 2\), the inequality ?? reduces to \[\label{d2} \begin{align} \sum_{a_1 + a_2 = n}\dfrac{\nu_{a_1,a_2}^n(\mathcal{D}_{a_1,a_2})}{\genfrac[]{0pt}{}{n}{a_1,a_2}_{\mathbb{R}}} &=\sum_{a_1=1}^{n-1}\dfrac{\int_{\mathrm{Gr}(n,a_1)}\chi_{(\mathcal{D}_{a_{1},n-a_{1}})_1}(D)\nu_{n-a_1}^{n-a_1}(\mathcal{D}_{D,n-a_1})d\nu_{a_1}^{n}}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}}. \end{align}\tag{18}\] Since \(\nu_{k}^{k}(\mathbb{R}^k)=1\) for any \(k \geq 1\). It follows that for each \(D \in (\mathcal{D}_{a_{1},n-a_{1}})_1\), we have \(\nu_{n-a_1}^{n-a_1}(\mathcal{D}_{D,n-a_1})=1\). Hence 18 equals \[\sum_{a_1=1}^{n-1}\dfrac{\nu_{a_1}^{n}\big((\mathcal{D}_{a_1, n-a_1})_1\big)}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}},\] which is at most \(t_1\) by 7 in the Klain–Rota Theorem.
Now suppose that \(r>2\). In this case, \[\begin{align} &\sum_{a_{1}+\cdots+a_{r}=n}\dfrac{\nu_{a_1,\ldots,a_r}(\mathcal{D}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}}\\ &=\sum_{a_1=1}^{n-r+1}\sum_{a_2+\cdots+a_r=n-a_1}\dfrac{\int_{\mathrm{Gr}(n,a_1)}\chi_{(\mathcal{D}_{a_{1},\ldots,a_{r}})_1}(D)\nu_{a_2,\ldots,a_r}^{n-a_1}(\mathcal{D}_{D, a_2, \ldots, a_r})d\nu_{a_1}^{n}}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}\genfrac[]{0pt}{}{n-a_1}{a_2,\ldots,a_r}_{\mathbb{R}}}. \end{align}\] It follows from the induction hypothesis that for every \(D \in (\mathcal{D}_{a_1,\ldots,a_r})_1\), \[\sum_{a_{2}+\cdots+a_{r}=n-a_{1}}\dfrac{\nu_{a_2,\ldots,a_r}^{n-a_1}(\mathcal{D}_{D, a_2, \ldots, a_r})}{\genfrac[]{0pt}{}{n-a_1}{a_2,\ldots,a_r}_{\mathbb{R}}}\le t_2\cdots t_{r-1}.\] Thus, we have \[\sum_{a_{1}+\cdots+a_{r}=n}\dfrac{\nu_{a_1,\ldots,a_r}(\mathcal{D}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}} \le\sum_{a_1=1}^{n-r+1}\dfrac{\nu_{a_1}^{n}\big((\mathcal{D}_{a_1, \ldots, a_r})_1\big)}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}}t_2\cdots t_{r-1} \leq t_1\cdots t_{r-1} = \sigma.\]
Note that \[\mathcal{D}(\mathbb{R}^n,r)=\bigsqcup_{a_{1}+\cdots+a_{r}=n}\mathcal{D}(\mathbb{R}^n, r)_{a_1,a_2,\ldots,a_r}\] and \(0\le\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r})\le \genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}\) for each \(\mathcal{D}_{a_{1},\dots,a_{r}}\) by 17 . When \(\sigma > \binom{n-1}{r-1}\), we assume that \(m_i = 0\) for all \(i > \binom{n-1}{r-1}\) and then \(\nu_{n;r}(\mathcal{D}) \le m_1+\cdots+m_\sigma\) trivially holds. For the case \(\sigma \leq \binom{n-1}{r-1}\), suppose for contradiction that \(\nu_{n;r}(\mathcal{D}) > m_1 + \cdots + m_\sigma\). Then, by Lemma 3, \[\sum_{a_{1}+\cdots+a_{r}=n}\dfrac{\nu_{a_1,\ldots,a_r}^n(\mathcal{D}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_r}_{\mathbb{R}}} > \sigma,\] which contradicts the inequality ?? . This completes the proof. ◻
An \(r\)-chain of \(\mathrm{Mod}(n)\) is a tuple \((C_1, \ldots, C_r)\) of nonzero subspaces in \(\mathbb{R}^n\) such that \(C_1 \subsetneq \cdots \subsetneq C_r\). We denote by \(\mathcal{C}(\mathbb{R} ^{n},r)\) the set of all \(r\)-chains of \(\mathrm{Mod}(n)\). Given positive integers \(a_1, \ldots, a_r\) with \(a_1 + \cdots + a_r < n\), denote by \(\mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\) the subset consisting of all \(r\)-chains \((C_1, \ldots, C_r)\) such that \(\dim(C_i)=a_1+a_2+\cdots+a_i\) for all \(i= 1, \ldots, r\). \(\mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\) is also a compact smooth manifold equipped with a unique \(O(n)\)-invariant measure \(\mu_{a_1,\ldots,a_r}^n\). More precisely, we call a flag \((F_0, \ldots, F_n)\) of \(\mathrm{Mod}(n)\) compatible with \((C_1,\ldots,C_r) \in \mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\) if \(F_{a_1+\cdots+a_i} = C_i\) for all \(i \in \{1, \ldots, r\}\). Then the measure \(\mu_{a_1,\ldots,a_r}^n\) is induced from the measure \(\phi_n\) on \(\mathrm{Flag}(n)\) by \[\label{chain} \mu_{a_1,a_2,\ldots,a_r}^{n}(\mathcal{C})=\frac{1}{[a_1]_{\mathbb{R}}!\cdots[a_r]_{\mathbb{R}}![a_{r+1}]_{\mathbb{R}}!}\phi_n\big(\mathrm{Flag}(\mathcal{C})\big)\tag{19}\] for any \(\mathcal{C} \subseteq \mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\), where \(a_{r+1} = n-\sum_{i=1}^{r}a_i\), and \(\mathrm{Flag}(\mathcal{C})\) represents the subset of \(\mathrm{Flag}(n)\) consisting of all flags compatible with some \(r\)-chain of \(\mathcal{C}\). In particular, The total measure of \(\mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\) is given by \[\mu_{a_1,a_2,\ldots,a_r}^{n}\big(\mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\big) = \genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r}, a_{r+1}}_{\mathbb{R}}.\] For any \(\mathcal{C} \subseteq \mathcal{C}(\mathbb{R}^n, r)\), let \(\mathcal{C}_{a_1,\ldots,a_r} = \mathcal{C} \cap \mathcal{C}(\mathbb{R} ^{n},r)_{a_1,\ldots,a_r}\) and \[\mathcal{C}_k = \{C_k \mid (C_1,\ldots,C_k,\ldots,C_r) \in \mathcal{C}\}.\] These measures \(\mu_{a_1,a_2,\ldots,a_r}^{n}\) induce the measure \(\mu_{n;r}\) on \(\mathcal{C}(\mathbb{R}^n,r)\) by \[\mu_{n;r}(\mathcal{C})=\sum_{a_{1}+\cdots+a_{r+1}=n}\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_r})\] for any subset \(\mathcal{C} \subseteq \mathcal{C}(\mathbb{R}^n,r)\).
Theorem 10. Let \(t_1, \ldots, t_r\) be positive integers and \(\tau = t_1t_2 \cdots t_r\). Suppose \(\mathcal{C}\) is a family of \(r\)-chains in \(\mathrm{Mod}(n)\) such that \(\mathcal{C}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{2462} \sum_{a_{1}+\cdots+a_{r+1}=n}\dfrac{\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}}\le \tau.\qquad{(6)}\] Consequently, \[\mu_{n;r}(\mathcal{C})\leq m_1 +\cdots + m_\tau,\] where \(m_1, \ldots, m_\tau\) are the \(\tau\) largest multiflag coefficents \(\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}\) for positive integers \(a_1, \ldots, a_{r+1}\) with \(a_1+\cdots + a_{r+1}=n\).
Proof. Given positive integers \(a_1,\cdots,a_r\) satisfying \(a_1+\cdots+a_r < n\), recall that \[(\mathcal{C}_{a_{1},\ldots,a_{r}})_1 = \left\{ C \in \mathcal{C}_1 \mid \mathrm{dim} (C)=a_1 \right\}.\] For any \(C \in (\mathcal{C}_{a_{1},\ldots,a_{r}})_1\), we define \[\mathcal{C}_{C, a_2, \ldots, a_r} = \{(C_2, \ldots, C_r) \mid (C, C_2, \ldots, C_r) \in \mathcal{C}_{a_{1},\ldots,a_{r}}\}.\] Then \(\mathcal{C}(\mathbb{R}^n, r)_{C, a_2, \ldots, a_r}\) is homeomorphic to \(\mathcal{C}(\mathbb{R}^{n-a_1}, r-1)_{a_2, \ldots, a_r}\). Therefore, by regarding \(\mathcal{C}_{C, a_2, \ldots,a_r}\) as a subset of \(\mathcal{C}(\mathbb{R}^{n-a_1}, r-1)_{a_2, \ldots, a_r}\), we deduce from 19 that \[\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_r})=\int_{\mathrm{Gr}(n,a_1)}\chi_{(\mathcal{C}_{a_{1},\ldots,a_{r}})_1}(C)\mu_{a_2,\ldots,a_r}^{n-a_1}(\mathcal{C}_{C, a_2, \ldots, a_r})d\nu_{a_1}^{n}.\] Similar to the proof of Theorem 9. We proceed by induction on \(r\). For \(r=1\), the inequality ?? reduces to the inequality 7 in the Klain-Rota Theorem. Suppose that \(r>1\) and that the inequality ?? holds for \(r-1\). Then, \[\begin{align} &\sum_{a_{1}+\cdots+a_{r+1}=n}\dfrac{\mu_{a_1,\ldots,a_r}(\mathcal{C}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}}\\ &=\sum_{a_1=1}^{n-r}\sum_{a_2+\cdots+a_{r+1}=n-a_1} \dfrac{\int_{\mathrm{Gr}(n,a_1)}\chi_{(\mathcal{C}_{a_{1},\ldots,a_{r}})_1}(C)\mu_{a_2,\ldots,a_r}^{n-a_1}(\mathcal{C}_{C, a_2, \ldots, a_r})d\nu_{a_1}^{n}}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}\genfrac[]{0pt}{}{n-a_1}{a_2,\ldots,a_{r+1}}_{\mathbb{R}}}. \end{align}\] It follows from the induction hypothesis that \[\sum_{a_{2}+\cdots+a_{r+1}=n-a_{1}}\dfrac{\mu_{a_2,\ldots,a_r}^{n-a_1}(\mathcal{C}_{C, a_2, \ldots, a_r})}{\genfrac[]{0pt}{}{n-a_1}{a_2,\ldots,a_{r+1}}_{\mathbb{R}}}\le t_2\cdots t_{r}\] for any \(C \in (\mathcal{C}_{a_{1},\ldots,a_{r}})_1\). Thus we have \[\sum_{a_{1}+\cdots+a_{r+1}=n}\dfrac{\mu_{a_1,\ldots,a_r}(\mathcal{C}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}} \le\sum_{a_1=1}^{n-r}\dfrac{\nu_{a_1}^{n}\big((\mathcal{C}_{a_1, \ldots, a_r})_1\big)}{\genfrac[]{0pt}{}{n}{a_{1}}_{\mathbb{R}}}t_2\cdots t_{r} \leq t_1\cdots t_{r} = \tau.\]
Note that \[\mathcal{C}(\mathbb{R}^n,r)=\bigsqcup_{a_{1}+\cdots+a_{r+1}=n}{\mathcal{C}(\mathbb{R}^n,r)_{a_1,\ldots,a_r}}\] and \(0\le\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_r})\le \genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}\) for each \(\mathcal{C}_{a_{1},\dots,a_{r}}\). The inequality \(\mu_{n;r}(\mathcal{C}) \leq m_1+\cdots+m_\tau\) clearly holds when \(\tau > \binom{n-1}{r}\), assuming that \(m_i = 0\) for all \(i > \binom{n-1}{r}\). For the case \(\tau \leq \binom{n-1}{r}\), suppose for contradiction that \(\mu_{n;r}(\mathcal{C}) > m_1 + \cdots + m_\tau\). Thus, by Lemma 3, \[\sum_{a_{1}+\cdots+a_{r+1}=n}\dfrac{\mu_{a_1,\ldots,a_r}^n(\mathcal{C}_{a_1,\ldots,a_r})}{\genfrac[]{0pt}{}{n}{a_1,\ldots,a_{r+1}}_{\mathbb{R}}} >\tau,\] which contradicts the inequality ?? . This completes the proof. ◻
Let \(n\) be a positive integer. By the fundamental theorem of arithmetic, there exists a unique expression \(n = p_1^{e_1} \cdots p_m^{e_m}\), where \(p_1, \ldots, p_m\) are distinct primes and \(e_1, \ldots, e_m\) are positive integers. Let \(\mathrm{Div}(n)\) denote the set of all positive divisors of \(n\), partially ordered by divisibility. Then \(\mathrm{Div}(n)\) forms a lattice. For any \(x = p_1^{\ell_1} \cdots p_m^{\ell_m} \in \mathrm{Div}(n)\), the rank of \(x\) is given by \(\operatorname{rk}(x) = \sum_{i=1}^{m} \ell_i\). Building on the inequality 9 in Anderson’s theorem, the following lemma serves as a crucial step in proving our main result.
Lemma 11. For any \(t\)-chain free family \(\mathcal{A}\) of elements of \(\mathrm{Div}(n)\), we have \[\sum_{i=0}^{\operatorname{rk}(n)} \frac{|\mathcal{A}_i|}{W_i(n)} \le t,\] where \(\mathcal{A}_i\) denotes the elements of \(\mathcal{A}\) of rank \(i\).
Proof. According to Mirsky’s Theorem [15], which states that any \(t\)-chain free family of a poset can be expressed as the disjoint union of \(t\) antichains, we write \(\mathcal{A} = \mathcal{A}^{(1)} \cup \cdots \cup \mathcal{A}^{(t)}\) such that each \(\mathcal{A}^{(i)}\) is an antichain. Therefore, by 9 in Anderson’s Theorem, \[\sum_{i=0}^{\operatorname{rk}(n)} \frac{|\mathcal{A}_i|}{W_i(n)} = \sum_{x\in \mathcal{A}}\frac{1}{W_{\operatorname{rk}(x)}(n)} =\sum_{i=1}^{t}\sum_{x\in \mathcal{A}^{(i)}}\frac{1}{W_{\operatorname{rk}(x)}(n)} \le \sum_{i=1}^{t}1 =t.\] ◻
Recall that an \(r\)-decomposition of \(n\) is an \(r\)-tuple \(\left(x_1, \ldots, x_r\right)\) of \(\mathrm{Div}(n)\) such that \(x_1\cdots x_r=n\). Denote by \(\mathcal{D}(n,r)\) the set of all \(r\)-decompositions of \(n\). Let \(\mathcal{D} \subseteq \mathcal{D}(n,r)\) be a subset, and \(a_1, \ldots, a_r\) be non-negative integers such that \(a_1 + \cdots + a_r = \operatorname{rk}(n)\). We write \[\mathcal{D}_k:=\left\{x_k\mid \left(x_1, \ldots, x_r\right)\in\mathcal{D}\right\}\] for each \(k \in \{1, \ldots, r\}\), and \[\mathcal{D}_{a_1, \ldots, a_r}=\left\{\left(x_1, \ldots, x_r\right)\in\mathcal{D}\mid \operatorname{rk}(x_i)=a_i,\;i=1, \ldots, r\right\}.\] Furthermore, we denote by \(N_{a_1,\ldots,a_r}(n) = |\mathcal{D}(n, r)_{a_1, \ldots, a_r}|\).
Theorem 12. Let \(t_1, \ldots, t_r\) be positive integers and \(\sigma = \frac{t_1t_2\cdots t_r}{\mathrm{max}\{t_1, \ldots, t_r\}}\). Suppose \(\mathcal{D}\) is a family of \(r\)-decompositions of \(n\) such that \(\mathcal{D}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{5461} \sum_{a_{1}+\cdots+a_{r}=\operatorname{rk}(n)}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{N_{a_1, \ldots,a_r}(n)} \leq \sigma.\qquad{(7)}\] Consequently, \[|\mathcal{D}| \leq m_1 + \cdots + m_\sigma,\] where \(m_1,\ldots,m_\sigma\) are the \(\sigma\) largest \(N_{a_1,\ldots,a_r}(n)\) for non-negative integers \(a_1,\ldots,a_r\) with \(a_1+\cdots + a_r=\operatorname{rk}(n)\).
Proof of Theorem 12. Without loss of generality, assume that \(t_r=\mathrm{max}\{t_1, \cdots, t_r\}\). We proceed by induction on \(r\) to prove the inequality ?? . For \(r =2\), since any \(2\)-decomposition \((x_1, x_2)\) of \(\mathrm{Div}(n)\) satisfies \(x_2 = n/x_1\), \(\mathcal{D}_1\) being \(t_1\)-chain free implies \(\mathcal{D}_2\) is \(t_1\)-chain free, and hence \(t_2\)-chain free. Therefore, the inequality ?? is equivalent to the LYM inequality in Lemma 11 by choosing the collection \(\mathcal{A}\) to be \(\mathcal{D}_1\). Now we suppose that \(r>2\) and that the inequality ?? holds for \(r-1\). Then \[\label{ap1} \begin{align} \sum_{a_{1}+\cdots+a_{r}=\operatorname{rk}(n)}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_r}(n)} &=\sum_{(x_1,\ldots, x_r)\in \mathcal{D} }^{} \frac{1}{N_{\operatorname{rk}(x_1), \ldots, \operatorname{rk}(x_r)}(n)}\\ &=\sum_{x\in\mathcal{D}_1}\frac{1}{N_{\operatorname{rk}(x),\operatorname{rk}(n)-\operatorname{rk}(x)}(n)}\sum\limits_{\substack{{ (x_1,\ldots, x_r)\in \mathcal{D}}\\{x_{1} = x}}}\frac{1}{N_{\operatorname{rk}(x_2),\ldots,\operatorname{rk}(x_r)}(n/x)}. \end{align}\tag{20}\] It follows from the induction hypothesis that \[\sum\limits_{\substack{{ (x_1,\ldots, x_r)\in \mathcal{D}}\\{x_{1} = x}}}\frac{1}{N_{\operatorname{rk}(x_2),\ldots,\operatorname{rk}(x_r)}(n/x)} \leq t_2\cdots t_{r-1} .\] Therefore, by Lemma 11, the last expression in 20 is at most \(\sigma= t_1t_2\cdots t_{r-1}\), which completes the proof of the inequality ?? .
Note that \[\mathcal{D} = \bigsqcup_{a_{1}+\cdots+a_{r}=\operatorname{rk}(n)}\mathcal{D}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{D}_{a_{1},\dots,a_{r}}|\le N_{a_1,\ldots,a_r}(n)\). Note that \(|\mathcal{D}| \leq m_1 + \cdots + m_\sigma\) clearly holds when \(\sigma > \binom{n+r-1}{r-1}\), assuming that \(m_i = 0\) for all \(i > \binom{n+r-1}{r-1}\). For the case \(\sigma \leq \binom{n+r-1}{r-1}\), suppose for contradiction that \(|\mathcal{D}| > m_1 + \cdots + m_\sigma\). Then, by Lemma 3 we have \[\sum_{a_{1}+\cdots+a_{r}=\operatorname{rk}(n)}^{} \frac{|\mathcal{D}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_r}(n)} > \sigma,\] which contradicts the inequality ?? . This completes the proof. ◻
An \(r\)-multichain in \(\mathrm{Div}(n)\) is a sequenece \((x_1, \ldots, x_r)\) of \(\mathrm{Div}(n)\) with \(x_1 \leq x_2 \leq \cdots \leq x_r\). We denote by \(\mathcal{C}(n, r)\) the collection of all \(r\)-multichains in \(\mathrm{Div}(n)\). Let \(\mathcal{C} \subseteq \mathcal{C}(n,r)\) and \(a_1, \ldots, a_r\) be non-negative integers such that \(a_1 + \cdots + a_r \leq \operatorname{rk}(n)\). Denote by \[\mathcal{C}_k = \{x_k \mid (x_1, \ldots, x_k, \ldots, x_r) \in \mathcal{C}\}\] for each \(k \in \{1, \ldots, r\}\), and denote by \[\mathcal{C}_{a_{1},\dots,a_r} = \{(x_1, \ldots, x_r) \in \mathcal{C}\mid \operatorname{rk}(x_i) = a_1 + \cdots + a_i, \; i=1, \ldots, r\}.\] Through a bijection similar to that in Remark 6, we can quickly observe that \[|\mathcal{C}(n, r)_{a_1, \ldots, a_r}| = |\mathcal{D}(n, r)_{a_1, \ldots, a_r, a_{r+1}}| = N_{a_1, \ldots, a_r, a_{r+1}}(n),\] where \(a_{r+1} = \operatorname{rk}(n) - \sum_{i=1}^{r}a_i\).
Theorem 13. Let \(t_1, \ldots, t_r\) be positive integers and \(\tau = t_1t_2\cdots t_r\). Suppose \(\mathcal{C}\) is a family of \(r\)-multichains of \(\mathrm{Div}(n)\) such that \(\mathcal{C}_k\) is \(t_k\)-chain free for each \(k \in \{1, \ldots, r\}\). Then \[\label{5466} \sum_{a_{1}+\cdots+a_{r+1}=\operatorname{rk}(n)}^{} \frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_{r+1}}(n)} \leq \tau.\qquad{(8)}\] Consequently, \[|\mathcal{C}| \leq m_1 + \cdots + m_\tau,\] where \(m_1, \ldots, m_\tau\) are the \(\tau\) largest \(N_{a_1,\ldots,a_{r+1}}(n)\) for non-negative integers \(a_1, \ldots, a_{r+1}\) with \(a_1+\cdots + a_{r+1}=\operatorname{rk}(n)\).
Proof of Theorem 13. We proceed by induction on \(r\). For \(r=1\), the inequality ?? reduces to the inequality stated in Lemma 11. Suppose that \(r>1\) and that the inequality ?? holds for \(r-1\). Then \[\begin{align} &\sum_{a_1 + \cdots + a_{r+1} = \operatorname{rk}(n)}\frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_{r+1}}(n)}\\ &=\sum_{(x_1, \ldots, x_r)\in\mathcal{C}}\frac{1}{N_{\operatorname{rk}(x_1),\operatorname{rk}(x_2)-\operatorname{rk}(x_1),\ldots,\operatorname{rk}(x_{r})-\operatorname{rk}(x_{r-1}),\operatorname{rk}(n)-\operatorname{rk}(x_{r})}(n)}\\ &=\sum_{x\in\mathcal{C}_1}\frac{1}{N_{\operatorname{rk}(x),\operatorname{rk}(n)-\operatorname{rk}(x)}(n)}\sum\limits_{\substack{{ (x_1,\ldots, x_r)\in \mathcal{C}}\\{x_{1} = x}}}\frac{1}{N_{\operatorname{rk}(x_2)-\operatorname{rk}(x_1),\ldots,\operatorname{rk}(x_{r})-\operatorname{rk}(x_{r-1}),\operatorname{rk}(n)-\operatorname{rk}(x_{r})}(n/x)}\\ &\leq \sum_{x\in\mathcal{C}_1}\frac{1}{N_{\operatorname{rk}(x),\operatorname{rk}(n)-\operatorname{rk}(x)}(n)} t_2\cdots t_{r}. \end{align}\] It follows from the induction hypothesis and Lemma 11 that the last expression is at most \(\tau = t_1t_2\cdots t_r\).
Note that \[\mathcal{C} = \bigsqcup_{a_{1}+\cdots+a_{r+1}=\operatorname{rk}(n)}\mathcal{C}_{a_{1},\dots,a_{r}}\] and \(0\le|\mathcal{C}_{a_{1},\dots,a_{r}}|\le N_{a_1, \ldots, a_{r+1}}(n)\) for each \(\mathcal{C}_{a_{1},\dots,a_{r}}\). Then \(|\mathcal{C}| \leq m_1 + \cdots + m_\tau\) clearly holds when \(\tau > \binom{n+r}{r}\), assuming that \(m_i = 0\) for all \(i > \binom{n+r}{r}\). For the case \(\tau\leq \binom{n+r}{r}\), suppose for contradiction that \(|\mathcal{C}| > m_1 + \cdots + m_\tau\). Then, by Lemma 3 we have \[\sum_{a_1 + \cdots + a_{r+1} = \operatorname{rk}(n)}\frac{|\mathcal{C}_{a_{1},\dots,a_{r}}|}{N_{a_1,\ldots,a_{r+1}}(n)} > \tau\] which contradicts the inequality ?? . This completes the proof. ◻
This paper is supported by the National Natural Science Foundation of China (Grant No. 12571350) and the Guangdong Basic and Applied Basic Research Foundation (Grant No. 2025A1515010457).