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This question is inspired by explicit bijection between ${[2n+1]\choose n+1}$ and ${[2n+1]\choose n}$ mapping $A$ to a subset of $A$, or equivalently, Is there an explicit construction of this bijection?. Fix a finite field $\mathbf{F}_q$, positive integers $n > 2k$ and consider the flag manifold $$F(k, n-k)= \{(A, B) \mid A \subset B\} \subset G(k) \times G(n-k)\,,$$ where $G(r) = G(r, \mathbf{F}_q^n)$ is the Grassmannian of $r$-dimensional subspaces in $\mathbf{F}_q^n$. There are two obvious projections $F(k, n-k) \to G(k)$ and $F(k, n-k) \to G(n-k)$.

What is an explicit section $G(k) \to F(k, n-k)$ so that the composition $G(k) \to F(k, n-k) \to G(n-k)$ is a bijection? Equivalently, what is an explicit bijection $f: G(k) \to G(n-k)$ such that $A \subset f(A)$ for each $A$?

This is also equivalent to finding an explicit subset of $F(k, n-k)$ restricted to which each projection is a bijection. Note that $G(k) \cong G(n-k)$, by either counting directly, or fixing an isomorphism $\mathbf{F}_q^n \cong (\mathbf{F}_q^n)^*$ and sending $A$ to $A^{\perp}$, but it is not clear why these bijections would satisfy the above condition. As Jyrki Lahtonen notes in the comments, for $k = 1$ there is a bijection of the form $v \mapsto v^\perp$ iff $n$ is even, since this is equivalent to a non-degenerate anti-symmetric bilinear form.

There always exists a bijection satisfying $A \subset f(A)$ by Hall's marriage theorem (since the bipartite graph $(G(k) \sqcup G(n-k), F(k, n-k))$ is regular), similar to the case in the linked questions.

I tried to linearly extend the combinatorial construction(s) from the linked questions but it is not at all clear to me that this is well defined with respect to the choice of bases.

ronno
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  • If I knew what $\mathbf{F}_1$ is, I would probably say that the linked questions are the $\mathbf{F}_1$ case. One can also ask a similar question about $\mathbf{R}$ or $\mathbf{C}$ and homeo/diffeomorphisms; some of the cases are discussed in this MO post. – ronno Apr 24 '23 at 15:34
  • Elements of $G(k)$ correspond to $k\times(n-k)$ RREFs, which in turn yield $k$-subsets of ${1,\cdots,n}$ (pivot columns). In light of this, we can ask if the linked question's $\Bbb F_q$-version (finding a bijection $f$ of $G(k)$ so $U,f(U)$ always intersect trivially) admits a solution which "restricts" to an $\Bbb F_1$ solution, i.e. plays nice with RREFs, and I think the answer to that is no (reasoning with the corresponding Grassmanian cell sizes). Which means there may be no obvious combinatorial algorithm to build off of, or $\Bbb F_1$ thing to "linearize." – coiso Apr 25 '23 at 00:37
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    That said, Hall's marriage theorem for bipartite graphs has an effective proof using an augmentation algorithm, and we can enumerate $G(k)$ and $G(n-k)$ by lexicographically ordering RREFs so as to use this algorithm. An algorithm is not an "explicit formula," but the solution in Artimis's answer (I haven't read Gregory's yet) is an algorithm so who knows. – coiso Apr 25 '23 at 00:39
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    An easy special case is $k=1$, $n$ even. With $n=4$ we can then do the following. If $A$ is spanned by the vector $(a_0,a_1,a_2,a_3)\in\mathbf{F}_q^4$, we can use $$B={(x_0,x_1,x_2,x_3)\in\mathbf{F}_q^4\mid a_1x_0-a_0x_1+a_3x_2-a_2x_3=0}.$$ Clearly then $A\subset B$, and equally clearly all the 3-dimensional subspaces appear in the role of $B$. This extends to all even $n$ in a hopefully obvious way. – Jyrki Lahtonen Apr 25 '23 at 05:01
  • But I don't immediately see how to tweak this even to the simplest case $k=1$, $n=3$. If $3\mid q-1$, then we have third roots of unity in $\mathbf{F}_q$, and it may work... – Jyrki Lahtonen Apr 25 '23 at 05:03
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    Nope. The $(k=1,n=4)$-trick above requires an antisymmetric full rank $n\times n$ matrix (with zeros along the diagonal even in characteristic two), and those can only exist when $n$ is even. – Jyrki Lahtonen Apr 25 '23 at 05:33
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    After clicking at your link to MO I realized that my construction is just the textbook counterexample to Hairy Ball theorem in even dimensions :-) – Jyrki Lahtonen Apr 25 '23 at 06:26

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