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I am interested in the field with one element. I am thus interested in combinatorial interpretations of the Gaussian binomial coefficients. Richard Stanley's "Enumerative combinatorics" mentions that they count the number of $k$-dimensional vector subspaces of an $n$-dimensional vector space over a finite field $F_q$ of characteristic $q$. He mentions that they also count the number of Young tableaux that fit in a $k \times (n-k)$ rectangle where each cell has weight $q$. However, he doesn't mention the following interpretation in terms of simplexes and I would like to know if there is any related literature or if it seems original.

Give the vertices of an $n$-simplex weights $1,$ $q,$ $q^2,$ $\ldots,$ $q^{n-1}$ and give weight $1/q$ to each edge. Define the weight of a $k$-simplex to be the product of the weights of its vertices and edges. Then the Gaussian binomial coefficients enumerate the $k$-simplexes within the $n$-simplex. This can be seen by induction from the recurrence relation where given a new vertex of weight $q^{n-1}$, and calculating the polynomial for the $k$-simplexes, we sum the "existing solutions" $[n-1 \:\: k]$ and the "newly possible solutions" $[n-1 \:\: k-1]$ times the weight $q^{n-k}$. The $n$-th vertex, when used in a newly possible solution, contributes weight $q^{n-1}$ and its $k-1$ edges to the other vertices of the $k$-simplex contribute weight $q^{-(k-1)}$ for a total contribution of $q^{n-1}/q^{k-1} = q^{n-k}$.

The $n$-simplex models a total order, that is, an ordered set. The $k$-simplex models an ordered subset. When $q=1$, then we get unordered subsets of an unordered set. This sheds light on a vector space as intrinsically having, in practice, an ordered canonical basis which is constructed or deconstructed. Typically, in constructing the basis, we create elements $e_1$, then $e_2 + f_1\cdot e_1$, $e_3 + f_2\cdot e_2 + f_1\cdot e_1$ where each field element $f_i$ contributes weight $q$. The exponents of these weights indicate an order on the basis elements. When $q=1$, then there is only one scalar, there is no real choice and so there is no way to order the basis elements based on that choice. I wish to know what is known about all of this and where I might take this further? Thank you!

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    $\begingroup$ Welcome to MathOverflow! You can use LaTeX in your question and title: simply use \$ ... \$ for displaying inline maths, and \$\$ ... \$\$ for displaying block equations (centred and on a separate line). $\endgroup$ – R. van Dobben de Bruyn Jun 22 '16 at 2:00
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    $\begingroup$ (I think you mean weights $1,q,q^2,...$ for the vertices.) This can easily be seen to be equivalent to the Young diagram interpretation: Young diagrams inside a $k \times (n-k)$ rectangle are in bijection with $k$-subsets of $\{1,2,\ldots,n\}$, and the bijection is such that the size of the diagram corresponds to the sum of the subset (minus $1+2+\cdots+k$, the minimum possible sum). $\endgroup$ – Sam Hopkins Jun 22 '16 at 13:52
  • $\begingroup$ R, thank you! and thank you especially to Mee Seong Im for converting my post to LaTex. That makes me feel I can learn it. :) $\endgroup$ – Andrius Kulikauskas Jun 22 '16 at 19:15
  • $\begingroup$ Sam, thank you. I like your bijection. That's helpful. $\endgroup$ – Andrius Kulikauskas Jun 22 '16 at 19:18
  • $\begingroup$ You can also make sure individuals are notified of your responses by prefixing their names with '@'. I can't demonstrate here, because one can't '@'-notify the original poster (but you will see that you as OP get the notification anyway). $\endgroup$ – LSpice Aug 15 '16 at 2:17

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