Let $G=\operatorname{GL}(n,\mathbb C)$. What follows can be put into a more general context, but I would like to first understand it for this case, the generalization is a second step.

For Zariski-almost all $x\in G$, there is a unique decomposition $x=lu$ with $l$ lower triangular and $u$ upper unipotent triangular. Now, I can precompose $x$ with some permutation (matrix) $\pi$ (an element of the Weyl group of $G$), i.e. permute the rows of $x$. For generic $x$, the matrix $\pi x$ will again have some LU-Decomposition $\pi x = l_\pi u_\pi$ with $l_\pi$ lower triangular and $u_\pi$ upper unipotent triangular.

We have an open immersion $\iota:L\times U\hookrightarrow G$ mapping $(l,u)\mapsto lu$, where $L$ is the Borel of lower triangular matrices and $U$ the unipotent radical of its opposite Borel, the variety of upper unipotent triangular matrices. Restricting to some open subset $V\subseteq L\times U$, we can consider the map $(l,u)\mapsto \pi.(l,u) := (l_\pi,u_\pi)$ for all $\pi\in\mathfrak S_n$ such that $\iota(\pi.(l,u))=l_\pi u_\pi= \pi lu$.

Since $\iota(\pi.\sigma.(l,u))= (l_\pi)_\sigma (u_\pi)_\sigma = \sigma l_\pi u_\pi = \sigma\pi lu = l_{\sigma\pi} u_{\sigma\pi} = \iota(\pi\sigma.(l,u))$ and because $\iota$ is an immersion, this defines a Group action on $V$, turning $\iota|_V$ into a $\mathfrak S_n$-equivariant map. Hence, the action of $\mathfrak S_n$ on $V$ is algebraic.

My question is: Is there any way to describe the open sets $L\cap V$ (resp. $U\cap V$) and the action $l\mapsto l_\pi$ (resp. $u\mapsto u_\pi$) of the permutation group $\mathfrak S_n$ on them? I have no intuition what-so-ever what they look like, and calculating examples was not very insightful.

Thanks a lot in advance for any pointers.

Edit: The $n=2$ case might still explain better what is going on. Observe that $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ ca^{-1} & 1 \end{pmatrix} \cdot \begin{pmatrix} a & b \\ 0 & d-ba^{-1}c\end{pmatrix} $$ Now, switching the rows maps \begin{align*} \begin{pmatrix} 1 & 0 \\ ca^{-1} & 1 \end{pmatrix} &\mapsto \begin{pmatrix} 1 & 0 \\ ac^{-1} & 1 \end{pmatrix} \\ \begin{pmatrix} a & b \\ 0 & d-ba^{-1}c\end{pmatrix} &\mapsto \begin{pmatrix} c & d \\ 0 & b-dc^{-1}a\end{pmatrix} \end{align*} so on the lower unipotent triangular matrices, in the nonvanishing locus of the lower left entry, we map \begin{align*} \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} &\mapsto \begin{pmatrix} 1 & 0 \\ a^{-1} & 1 \end{pmatrix} \end{align*} I don't see immediately what the action on the upper triangular matrices is.

  • $\begingroup$ Have you tried to compute this for $n=2$? The open locus $L \times U \subset {\rm{GL}}_2$ is given by non-vanishing of the upper-left entry if I'm not mistaken, but if you consider points of either $L \times \{1\}$ or $\{1\} \times U$ you see that nothing in either of these retains that non-vanishing property after swapping the two rows. Thus, $L \cap V$ and $U \cap V$ are empty when $n=2$. $\endgroup$
    – user76758
    Dec 13, 2013 at 9:22
  • $\begingroup$ @user76758: I did not claim that the action of $\mathfrak S_2$ on $L$ or $U$ is by swapping rows. $\endgroup$ Dec 15, 2013 at 21:25
  • $\begingroup$ Since you wrote "i.e., permute the rows of $x$" I thought you meant that $\pi$ is the operation of permuting rows, so $V$ is contained in the open locus of matrices with non-vanishing upper-left entry (that much gives $L \times U$) which moreover retain that property after swapping of the rows. That $V$ has empty intersection with $L$ and $U$. Please describe an explicit $V$ that meets $L$ and $U$. $\endgroup$
    – user76758
    Dec 15, 2013 at 23:28
  • $\begingroup$ @user76758: No, see, it permutes the rows when acting on $G$. But certainly $l_\pi$ is not $\pi l$, nor is $u_\pi$ the same as $\pi u$, because why would $\pi l u = \pi l \pi u$? I do not know how $\mathfrak S_n$ acts on $L$ and $U$, this is what I am asking. $\endgroup$ Dec 16, 2013 at 15:42
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    $\begingroup$ Are you conceivably wanting something like the cell multiplication rules for a Bruhat decomposition? $\endgroup$ Dec 17, 2013 at 17:06

1 Answer 1


In lieu of any other answers till now: the question reminded me somewhat vaguely of "cell multiplication" in a Bruhat decomposition, namely, in a Bruhat decomposition $G=\bigsqcup_{w\in W} BwB$ of $G=GL_n(k)$ with $B$ a minimal parabolie (Borel), and $W$ (representatives for) the (spherical) Weyl group, the "cells" $BwB$ have a kind of multiplication rule. For $S\subset W$ the collection of reflections attached to simple roots (determined by choice of $B$), and for $\ell(w)$ the length-of-word function associated to generators $S$ for $W$, for $s\in S$, $w\in W$, $BsB\cdot BwB=BswB$ when $\ell(sw)>\ell(w)$, and $BsB\cdot BwB=BswB\cup BwB$ when $\ell(sw)<\ell(w)$.

One exposition of such stuff, minimizing reliance upon general Coxeter-group stuff, is at http://www.math.umn.edu/~garrett/m/v/bldgs.pdf

The standard treatment, going back many decades, and abstracted by Bruhat-Tits and others, does rely on the combinatorial group theory of Coxeter groups (and for not-split algebraic groups, on the more serious theory of algebraic groups).

  • $\begingroup$ Hm. I am still not sure how to connect the dots. I am looking at $BsBs$, where $s$ is the order-reversing permutation, so $sBs$ is the transpose of $B$. Now I am asking what $wBsBs\cap BsBs$ looks like. I hope this translation is correct. Does this relate somehow? $\endgroup$ Dec 18, 2013 at 22:33
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    $\begingroup$ Yes, your "s" is the "longest Weyl element", more often denoted by $w_o$ in the relevant literature, and in that literature "s" is a reflection. In your notation, the "s" can be cancelled from the right, of course, and asking what is $BwB\cdot Bw_oB\cap Bw_oB$ differs from your question just by the leading $B$. This variant question has an answer, by writing $w$ as a minimal-length word in reflections, and apply the cell multiplication rule inductively. Even the "extra" left multiplication by $B$ can be accounted for, pretty much. Is this plausibly going in a direction helpful to you? $\endgroup$ Dec 18, 2013 at 23:40
  • $\begingroup$ Yes! It very much is. Thanks for the patience, I will dig into it. And merry Christmas! $\endgroup$ Dec 19, 2013 at 11:24
  • $\begingroup$ Good. You're welcome. And happy holidays, for sure! $\endgroup$ Dec 19, 2013 at 13:42

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