$E$ 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.



Position of complete flags
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.


