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Author |
hbghlyj
Posted at 2023-5-8 23:10:04
A well-known generalization of the binomial coefficients $\left(\begin{array}{c}n \\ k\end{array}\right)$ are the q-binomial coefficients $\left[\begin{array}{c}n \\ k\end{array}\right]$, defined for $0 \leq k \leq n$ by
$$
\left[\begin{array}{l}
n \\
k
\end{array}\right]=\frac{[n] !}{[k] ![n-k] !},
$$
where
\begin{aligned}
{[j] ! } &=[1][2] \cdots[j] \\
{[i ] } &=1-q^{i} .
\end{aligned}Here $q$ may be regarded as an indeterminate (in which case $\left[\begin{array}{l}n \\ k\end{array}\right]$ is a polynomial in $q$ of degree $k(n-k))$ or as a number. If $q=1$, then $\left[\begin{array}{l}n \\ k\end{array}\right]$ reduces to the binomial coefficient $\left(\begin{array}{l}n \\ k\end{array}\right)$. If $q$ is a prime power, then $\left[\begin{array}{c}n \\ k\end{array}\right]$ is equal to the number of $k$-dimensional subspaces of an $n$-dimensional vector space $V_{n}$ over the finite field $\mathfrak{F}_{q}$ with $q$ elements.
$\left[\begin{array}{l}n \\ 1\end{array}\right],\left[\begin{array}{l}n \\ 0\end{array}\right], \ldots,\left[\begin{array}{l}n \\ n\end{array}\right]$ 是对数凹序列, $q \geq 0$.
证明:$\left[\begin{array}{l}n \\ k\end{array}\right]^2 /\left[\begin{array}{c}n \\ k-1\end{array}\right]\left[\begin{array}{c}n \\ k+1\end{array}\right]=\dfrac{[k+1][n-k+1]}{[k][n-k]}>1$
组合证明 见Stanley (1989)第2页
A combinatorial proof that $\left[\begin{array}{c}n \\ 0\end{array}\right],\left[\begin{array}{l}n \\ 0\end{array}\right], \ldots,\left[\begin{array}{l}n \\ n\end{array}\right]$ is log-concave was given by L. Butler (private communication). The unimodality of the sequence $\left[\begin{array}{l}n \\ 0\end{array}\right],\left[\begin{array}{l}n \\ 1\end{array}\right], \ldots,\left[\begin{array}{l}n \\ n\end{array}\right]$ for $q \geq 0$ should not be confused with the problem of showing that for fixed $k$ and $n$, the coefficients $a_{0}, a_{1}, \ldots, a_{k(n-k)}$ of the polynomial
$$
\left[\begin{array}{l}
n \\
k
\end{array}\right]=\sum_{i=0}^{k(n-k)} a_{i} q^{i}
$$
are unimodal. This much more difficult problem is discussed in later sections. (The example $\left[\begin{array}{l}4 \\ 2\end{array}\right]=1+q+2 q^{2}+q^{3}+q^{4}$ shows that the coefficients of $\left[\begin{array}{l}n \\ k\end{array}\right]$ need not be log-concave.)
An extraordinary example of a combinatorial proof of unimodality was given recently by L. Butler [19, chap. 2], [20], and is a generalization of the unimodality of $\left[\begin{array}{l}n \\ 0\end{array}\right],\left[\begin{array}{l}n \\ 1\end{array}\right], \ldots,\left[\begin{array}{l}n \\ n\end{array}\right]$. To state Butler's result, it is convenient to use terminology from the theory of posets (partially ordered sets). A finite poset $P$ is graded of rank $n$ if every maximal chain of $P$ has length $n$ (or cardinality $n+1$ ). In this case, we define the rank $\rho(x)$ of $x \in P$ to be the length $i$ of the longest chain $x_{0}<x_{1}<\cdots<x_{i}=x$ of $P$ with top element $x$. Let $a_{i}$ be the number of elements of $P$ of rank $i$, and call $P$ rank-unimodal if the sequence $a_{0}, a_{1}, \ldots, a_{n}$ is unimodal. Now consider the case where $P$ is the lattice of subgroups of a finite Abelian group $G$. If the prime factorization of $|G|$ is given by $p_{1}^{b_{1}} p_{2}^{b_{2}} \cdots \cdot$, then $P$ is graded of rank $b_{1}+b_{2}+\cdots$ If $H \in P$ and $|H|=p_{1}^{c_{1}} p_{2}^{c_{2}} \cdots \cdot$ then $\rho(H)=c_{1}+c_{2}+\cdots$ in $P$ |
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kuing
Posted at 2023-5-8 23:04:21
