Fix a $(B,N)$ pair (Tits system) of a semisimple Lie group $G$. Let $u$ and $v$ be two Weyl group elements such that $l(uv)=l(u)+l(v)$. It is known that $BuvB=(BuB)(BvB)$ (see for example Humphreys's Linear Algebraic Groups Section 29.3 Lemma A). Let $x$ be an element of the Bruhat cell $BuvB$ and I would like to factor $x$ as a product $yz$ with $y\in BuB$ and $z\in BvB$. Is it true that $y$ is unique up to a right multiple of an element in $B$ and $z$ is unique up to a left multiple of an element in $B$? If so, how do I see it? Thanks a lot.



Product of Bruhat Cells
Many thanks to Paul Garrett's comment above. Inspired by his comment I come up with a proof of the following equivalent statement.


Proposition. *If $x\in BwB$ where $w=s\_{i(1)}s\_{i(2)}\dots s\_{i(l)}$ is a reduced word for $w$, then there exist $x\_k\in Bs\_{i(k)}B$ such that $x=x\_1x\_2\dots x\_l$; further if $x=x'\_1x'\_2\dots x'\_l$ is another such factorization then $x\_k^{-1}x'\_k\in B$ for all $1\leq k\leq l-1$ and $x'\_kx\_k^{-1}\in B$ for all $2\leq k\leq l$.*


Proof. The existence part is a direct corollary of the fact cited in the original post. So this proof is only about the uniqueness part. We will do an induction on $l$. There is nothing to show for the base case $l=1$. Suppose $l>1$. Let $x=yx\_l=y'x'\_l$ where both $y$ and $y'$ are in $Bs\_{i(1)}\dots s\_{i(l-1)}B$ and both $x\_l$ and $x'\_l$ are in $Bs\_{i(l)}B$. Then from the fact that $(Bs\_{i(l)}B)^2\subset B\cup Bs\_{i(l)}B$ we know that $x'\_lx\_l^{-1}$ is in either $B$ or $Bs\_{i(l)}B$. To rule out the latter possibility, note that if $x'\_lx\_l^{-1}\in Bs\_{i(l)}B$ then $y'x'\_lx\_l^{-1}=xx\_l^{-1}=y$ is in both Bruhat cells $Bs\_{i(1)}\dots s\_{i(l-1)}B$ and $BwB$, which is a contradiction. Thus $x'\_lx\_l^{-1}\in B$.


The fact that $x'\_lx\_l^{-1}\in B$ implies that $y^{-1}y'=x\_lx^{-1}x{x'\_l}^{-1}=x\_l{x'\_l}^{-1}\in B$. Thus $y$ and $y'$ can only differ by a right multiple of $B$. This difference can be absorbed into the right ambiguity of $x\_{l-1}$, and hence without loss of generality one can assume that $y=y'$, and the proof is finished by induction.


