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Q-analogues of the binomial theorem:

• Each cell or inversion has weight {$q$}.
• Subspaces of a vector space over a finite field.
• Simplexes - the {$k$} vertices have weights from {$1,q,q^2,\dots\,q^n$} and each edge has weight {$q^{-1}$}.

The sum of the weights of the vertices of an n-simplex is given by:

{$$\left [ n \right ]_q = 1 + q + q^2 + \dots + q^{n-1} = \frac{q^n-1}{q-1}$$}

Gaussian binomial coefficients are given by the induction:

{$$\begin{bmatrix}n\\k\end{bmatrix}_q = \begin{bmatrix}n-1\\k\end{bmatrix}_q + q^{n-k}\begin{bmatrix}n-1\\k-1\end{bmatrix}_q$$}

{$$\begin{bmatrix}n\\k\end{bmatrix}_q = \frac{\left [ n \right ]_q!}{\left [ k \right ]_q! \left [ n-k \right ]_q!}$$}

Given a prime {$q$}, the number of {$k$}-dimensional subspaces of {$\mathbb{F}^n_q$} is {$\begin{bmatrix}n\\k\end{bmatrix}_q$}.

{$\mathbb{F}^n_q$} consists of {$q^n$} vectors of which {$q^n-1$} are nonzero vectors. The number of k-tuples of linear independent vectors is given by the product of {$q^n-1$} possibilities for the first choice, {$q^n-q$} for the second choice, {$q^n-q^2$} for the third choice, and so on, up to {$q^n-q^2$} possiblities for the {$k$}th choice. This gives the product:

{$$(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{k-1})$$}

Then we have overcounted because some of these k-tuples generate the same vector subspace. By the same argument we find that any {$k$}-dimensional vector space is spanned by the following number of k-tuples of linear independent vectors:

{$$(q^k-1)(q^k-q)(q^k-q^2)\cdots(q^k-q^{k-1})$$}

Thus we divide, yielding the answer.

Challenge: I want to reconstruct an argument that I had. It went something as follows. Given a basis {$x_1, x_2, \dots, x_n$} there is only one choice for the first element, but then there are q choices for {$qx_1 + x_2$} and {$q^2$} choices for {$qx_1 + qx_2 + x_3$} and so on. But when q=1, then there is no real choice because we are choosing out of one. And that is the essence of {$F_1$}.

Peter Cameron: Every k-dimensional subspace has a unique basis consisting of k vectors in reduced echelon form. The number of matrices in reduced echelon form satisfies the recurrence, which adds the numbers of matrices where the leading 1 in the last row is:

• in the last position (so deleting the last row and column gives a reduced echelon matrix) or
• in an earlier position (so the last column is arbitrary, and deleting it gives a reduced echelon matrix).

Henry Cohn: Given a prime {$q$}, the number of {$k$}-dimensional subspaces of {$\mathbb{P}^n(\mathbb{F}^n_q)$} is {$\begin{bmatrix}n+1\\k+1\end{bmatrix}_q$}.

The weights of the Gaussian binomial coefficients count the number of cells to the left of the paths in Pascal's triangle.

Bruce Sagan via Peter Cameron: The binomial coefficient Bin(n,k) counts the k-element subsets of an n-element set. Now suppose we have a cyclic permutation σ of the set, and we wish to count its orbits on k-subsets. By the Orbit-counting Lemma, we have to count the subsets fixed by any power of σ. Now it turns out that the number of sets fixed by a power of σ of order d is the result of substituting a primitive dth root of unity for q in Gauss(n,k)q.

See Bruce Sagan: Words

inversions {$\sum_{w\in W_{n,k}}q^{\textrm{Inv} \; w}$} where {$\textrm{Inv} \; w = \left | \left \{ (i,j):i<j \; \mathrm{and} \; a_i>a_j \right \} \right |$}

major index {$\sum_{w\in W_{n,k}}q^{\textrm{Maj} \; w}$} where {$\textrm{Maj} \; w = \sum_{a_i>a_{i+1}}i$}

In what sense are the major index and the inversions duals of each other?

I'm studying the q-analogue of simplexes. I have a combinatorial interpretation (giving successive vertices weights 1, q, q2, q3... and giving all the edges weight 1/q). I found an algebraic version of this at this page which cites this book Quantum calculus about doing calculus without taking limits. May relate to Kirby's "delta-calculus" and "lambda-calculus" distinction.

Choice framework

• Three perspectives lets you define a coordinate system as a choice framework.
• Totally independent dimensions: Cartesian
• Totally dependent dimensions: simplex
• We "understand" the simplex in Cartesian dimensions but think it in simplex dimensions. We can imagine a simplex in higher dimensions by simply adding externally instead of internally. Consider (implicit + explicit)to infinity; and also (unlabelled + labelled) to infinity. Also consider (unlabelable + labelable). And (definitively labeled + definitively unlabeled).
• Gaussian binomial coefficients interpretation related to Young tableaux
• https://oeis.org/A013609 triangle for hypercubes. (1 + 2x)^n unsigned coefficients of chebyshev polynomials of the second kind
• nLab: Geometric shape for higher structures
• What do "globe", "globe category", "globular set" mean?
• What's the canonical embedding of the globe category into Top?
• They are the maps of the closed n-ball to the "northern" and "southern" hemispheres of the surface of the (n+1)-ball.
• This defines a functor G→Top that extends along the Yoneda embedding, yielding a geometric representation of any globular set Gˆ→Top.
• Ar savybių visuma yra simpleksas? Ar savybės skaidomos (koordinačių sistema).
• The empty set, or the center, has dimension -1.
• Terrence Tao: Twisted Convolution and the Sensitivity Conjecture
• Choices - polytopes, reflections - root systems. How are the Weyl groups related?
• Polytopes. edge = difference
• Esminis pasirinkimas yra: kurią pasirinkimo sampratą rinksimės?
• Kaip suvokiame {$x_i$}? Koks tai per pasirinkimas?
• Kada pasirinkimo samprata keičiasi, visgi už visų sampratų slypi bendresnis, pirmesnis suvokimų suvokimas, taip kad renkamės pačią sampratą. Folding is the basis for substitution.
• Keturios pasirinkimo sampratos (apimtys) visos reikalingos norint išskirti vieną paskirą pasirinkimą.

An simplexes allow gaps because they have choice between "is" and "not". But all the other frameworks lack an explicit gap and so we get the explicit second counting. But:

• for Bn hypercubes we divide the "not" into two halves, preserving the "is" intact.
• for Cn cross-polytopes we divide the "is" into two halves, preserving the "not" intact.
• for Dn we have simply "this" and "that" (not-this).
• Use "this" and "that" as unmarked opposites - conjugates.

Coordinate system

• Coordinate system is that which has an origin, a zero.
• How are coordinate rings in algebraic geometry related to coordinate systems?
• Coordinate systems are observers and they shouldn't affect what they observe. (Relate this to the kinds of polytopes.)

Readings

• Projective geometry over {$F_1$} and the Gaussian binomial coefficients, Henry Cohn, his research
• Peter Cameron: The field with one element
• Bruce Sagan: Two binomial coefficient analogues
• Bruce Sagan: The cyclic sieving phenomenon - an introduction
• Bruce Sagan: The cyclic sieving phenomenon: a survey
• J. Konvalina, Generalized binomial coefficients and the subset-subspace problem, Adv. in Appl. Math.21(1998) 228–240.
• J. Wang, Quotient sets and subset-subspace analogy,Adv. in Appl.Math.23 (1999) 333–339.
• J.R.Goldman and G.-C.Rota, On the foundations of combinatorial theory IV: Finite vector spaces and Eulerian generating functions, Stud. Appl. Math.49 (1970) 239–258; reprinted in Gian-Carlo Rota onCombinatorics, J. P. S. Kung, ed., Birkh ̈auser, Boston, 1995, pp. 226–245.