Epistemology
Introduction E9F5FC Questions FFFFC0 Software 
Qanalogues of the binomial theorem:
The sum of the weights of the vertices of an nsimplex is given by: {$$\left [ n \right ]_q = 1 + q + q^2 + \dots + q^{n1} = \frac{q^n1}{q1}$$} Gaussian binomial coefficients are given by the induction: {$$\begin{bmatrix}n\\k\end{bmatrix}_q = \begin{bmatrix}n1\\k\end{bmatrix}_q + q^{nk}\begin{bmatrix}n1\\k1\end{bmatrix}_q$$} {$$\begin{bmatrix}n\\k\end{bmatrix}_q = \frac{\left [ n \right ]_q!}{\left [ k \right ]_q! \left [ nk \right ]_q!}$$} 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 {$q^n1$} are nonzero vectors. The number of ktuples of linear independent vectors is given by the product of {$q^n1$} possibilities for the first choice, {$q^nq$} for the second choice, {$q^nq^2$} for the third choice, and so on, up to {$q^nq^2$} possiblities for the {$k$}th choice. This gives the product: {$$(q^n1)(q^nq)(q^nq^2)\cdots(q^nq^{k1})$$} Then we have overcounted because some of these ktuples generate the same vector subspace. By the same argument we find that any {$k$}dimensional vector space is spanned by the following number of ktuples of linear independent vectors: {$$(q^k1)(q^kq)(q^kq^2)\cdots(q^kq^{k1})$$} 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 kdimensional 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:
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 kelement subsets of an nelement set. Now suppose we have a cyclic permutation σ of the set, and we wish to count its orbits on ksubsets. By the Orbitcounting 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 qanalogue 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 "deltacalculus" and "lambdacalculus" distinction. Choice framework
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:
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