# Lecture 4. On q-analogs.

In class I introduced the $q$-analog of $n$ and $n!$:

$[n]_q = 1+q+q^2 + \cdots +q^{n-1}$,

$[n]!_q = [1] \cdot [2] \cdot \cdots \cdot [n]$.

Let $q=p^a$ be a power of a prime $p$ and let  $F_q$ be the field of $q$ elements. In several senses, the $q$-analog of the set $\{1, \ldots, n\}$ is the vector space $F_q^n$. Here is one sense:

1. Prove that the number of sequences of sets $\emptyset = S_0 \subsetneq S_1 \subsetneq \cdots \subsetneq S_n = \{1, \ldots, n\}$ is $n!$.

2. Prove that the number of sequences of subspaces $\emptyset = V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_n = F_q^n$ is $[n]_q!$.

1. 1. The number of sequences of sets $\emptyset=S_0\subsetneq \dots\subsetneq S_n=[n]$ is equal to the number of permutations of $[n]$, consider the following bijection: for a permutation $\sigma$ you can get the sequence given by $S_i=\{\sigma(1)\dots\sigma(i)\}$ ($S_0=\emptyset$). Therefore, the number of such sequences is $n!$
2. Note that given a subspace in the sequence $V_i, i, to get $V_{i+1}$ we only need to add any nonzero vector in $V_i^\perp$ and take the generated space of the union. $V_i^\perp$ has dimension $n-i$ and therefore $q^{n-i}-1$ nonzero vectors, and two of these give the same subspace $V_{i+1}$ if and only if they generate the same space, i.e. if and only if one is a nonzero-multiple of the other. So given $V_i, i, there are $q^{n-i}-1/q-1=(1+q+q^2+\dots +q^{n-i-1})=[n-i]_q$ possibilities for $V_{i+1}$.
Then, there are $[n]_q$ possibilities for $V_1$, $[n-1]_q$ for $V_2$, …, $[2]_q$ for $V_{n-1}$ and $1=[1]_q$ for $V_n=\mathbb{F}_q^n$. So the number of sequences is:
$[n]_q[n-1]_q\cdots[2]_q[1]_q=[n]_q!$.