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!.

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One comment

  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<n, 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<n, 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!.

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