$E$ is a vector space of dimension $n \geq 2$. $\mathbb{F}=(F_1,F_2,\dots,F_n)$ and $\mathbb{G}=(G_1,G_2,\dots,G_n)$ are two complete flags of $E$.

We say that $(\mathbb{F},\mathbb{G})$ is in position $\sigma \in \mathfrak{S}_n$ if and only if there exists a basis $(e_1,\dots,e_n)$ of $E$ such that for each $(i,j) \in [1,n]^2$, the sub-space $F_i$ is generated by $\{e_1,\dots,e_i\}$ and the sub-space $G_j$ by $\{e_{\sigma(1)},\dots,e_{\sigma(j)}\}$.

Suppose that $(\mathbb{F},\mathbb{G})$ is an ordered pair of complete flags in position $\sigma \in \mathfrak{S}_n$, that $k$ is an integer $1 \leq k \leq n-1$ and that $\sigma^{-1}(k) < \sigma^{-1}(k+1)$. $\mathbb{F^\prime}=(F^\prime_1,F^\prime_2,\dots,F^\prime_n)$ is another complete flag differing from $\mathbb{F}$ only for the sub-space of index $k$, i.e. $F_i=F^\prime_i$ for $i \in [1,n] \setminus \{k\}$ and $F_k \neq F^\prime_k$.

How can we prove that $(\mathbb{F^\prime},\mathbb{G})$ is in position $\tau_k \circ \sigma$ where $\tau_k$ is the transposition which swaps $k$ and $k+1$?

I processed the case $n=2$, $k=1$... but I'm not able to extrapolate to the general case.


Another way to characterize relative position is by looking at intersections: it is equivalent asking that $\dim(F_i\cap G_j)=\#\big([1,i]\cap\sigma([1,j])\big)$. How does this number change when we replace $\sigma$ by $\tau_k\sigma$? The RHS stays the same unless $i=k$, and in that case it drops if $k\in \sigma([1,j])$ and $k+1\notin \sigma([1,j])$. (It would increase if $k+1\in \sigma([1,j])$ and $k\notin \sigma([1,j])$ but this can't happen by assumption).

So, we just have to see that the behavior of the LHS is the same. If $i\neq k$, then $F_i=F_i'$, so that's good. On the other hand, $F_k\neq F_k'$, and in fact $F_k+F_k'=F_{k+1}$ and $F_k\cap F_k'=F_{k-1}$. Thus,

$\dim(F_k\cap G_j)+\dim(F_k'\cap G_j)=\dim(F_{k-1}\cap G_j)+\dim(F_{k+1}\cap G_j),\qquad (*)$

since $(F_k\cap G_j)\cap(F_k'\cap G_j)=F_{k-1}\cap G_j$ and $(F_k\cap G_j)+(F_k'\cap G_j)=F_{k+1}\cap G_j$. Let $m=\dim(F_{k-1}\cap G_j)$; in this case $m\leq \dim(F_{k+1}\cap G_j)\leq m+2$. This naturally divides us into 3 cases:

  • $k,k+1\notin \sigma([1,j])$: in this case, $m=\dim(F_{k+1}\cap G_j)$. Necessarily, $\dim(F_k\cap G_j)=\dim(F_k'\cap G_j)=m$.
  • $k,k+1\in \sigma([1,j])$: in this case, $m+2=\dim(F_{k+1}\cap G_j)$. This is only possible if $\dim(F_k\cap G_j)=\dim(F_k'\cap G_j)=m+1$.
  • $k\in \sigma([1,j])$ and $k+1\notin \sigma([1,j])$: in this case $\dim(F_{k+1}\cap G_j)=m+1$; by assumption, we must have $\dim(F_{k+1}\cap G_j)=\dim(F_{k}\cap G_j)$, so $\dim(F_{k}'\cap G_j)=\dim(F_{k-1}\cap G_j)$, and the dimension has dropped, as desired.

EDIT: I should mention that there's also an abstract way of thinking about this. If we let $B$ be the group of matrices preserving $G_j$ for all $j$. We may as well assume, $G_j$ is the span of the first $j$ coordinates, so these are upper triangular matrices. In this case $F_\bullet$ has relative position $\sigma$ if and only if $\sigma b(F_j)=G_j$, there $b\in B$ and $\sigma$ stands for the permutation matrix. Now, consider $G'_\bullet =\sigma b(F_\bullet')$. This is a flag that agrees with $G_\bullet$, except that $G_k\neq G_k'$. You can easily check that $\tau_kb'(G_j')=G_j$ for some $b'\in B$. Thus $\tau_kb'\sigma b(F_j')=G_j$. It's a standard calculation that $B\tau_kB\sigma B=B\tau_k\sigma B$, so for some $b''\in B$, we have $\tau_k\sigma b''(F_j')=G_j$, and we're done.

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  • $\begingroup$ Thanks Ben. Were you studying flag positions for a special intent? $\endgroup$ – mathcounterexamples.net May 31 '14 at 6:12
  • $\begingroup$ @jpmjpmjpm They play an important role in geometric representation theory. See, for example, Chriss and Ginzburg's book "Representation Theory and Complex Geometry." $\endgroup$ – Ben Webster Jun 1 '14 at 8:19
  • $\begingroup$ Ben, in fact there is a step in your proof with which I'm having difficulties. It is with your first equality with dimensions which is not always true. Am I missing something? $\endgroup$ – mathcounterexamples.net Jun 1 '14 at 9:35
  • $\begingroup$ @Jean-PierreMerx You mean the one now marked $(*)$? I'd like to see your counterexample. Of course, it's only true if $F_k \neq F_k'$. That's an important and necessary hypothesis. $\endgroup$ – Ben Webster Jun 3 '14 at 8:53
  • $\begingroup$ Hi Ben, yes I mean the one marked $(\ast)$. Take a space with dimension $n=3$ and $(e_1,e_2,e_3)$ a basis. $F_1=\mbox{span}\{e_1\}$, $F_2=\mbox{span}\{e_1,e_2\}$, $F^\prime_2=\mbox{span}\{e_1,e_3\}$, $F_3=\mbox{span}\{e_1,e_2,e_3\}$ and finally $G_j=\mbox{span}\{e_1,e_2+e_3\}$ with $k=2$. In fact, based on the aother elements of your elements of your proof, I can finalize without using (*). Enjoy your stay in France! $\endgroup$ – mathcounterexamples.net Jun 3 '14 at 11:46

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