Product of Bruhat Cells 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.
 A: Here is a (probably equivalent) way to see this. I'll give an argument for $(BsB)(BwB)$, where $s$ is a simple reflection. You should be able to expand this argument to $(BvB)(BwB)$ as requested in the question. The expansion will use the fact that $\ell(vw) = \ell(v) + \ell(w)$ implies $\ell(s_t \cdots s_1 w) = t + \ell(w)$, where $s_n \cdots s_1 = v$ is a reduced decomposition for $v$ and $n \geq t$. 
Throughout, let $T\subset B$ be a fixed maximal torus and $U \subset B$ the unipotent subgroup of $B$. Recall that $B = TU$. For any positive root $\alpha$, let $U_\alpha$ be the corresponding root subgroup of $U$. Recall that for any ordering of positive roots, we can write $U$ as a produce of $U_\alpha$ in the chosen order.  Fix an odering of positive roots so that $U = U_{w{-1}} U_{w_0w^{-1}}$, where $w_0$ is the longest element of the Weyl group. 
Then, the elements of $BwB$ are uniquely written as $U_{w^{-1}}wB$, where 
\begin{equation}
U_{w^{-1}} = \prod_{\alpha \in R(w^{-1})} U_\alpha.
\end{equation}
Here, $R(w^{-1}) =\{\alpha \in R^+\;|\; w^{-1} \alpha <0\;\}$. This decomposition follows from the Bruhat decomposition because if $w^{-1} \alpha >0$, then $w^{-1} u_{\alpha} w = u_{w^{-1} \alpha}$, and so $ u_{\alpha} w  = wu_{w^{-1} \alpha}$ and $u_{w^{-1} \alpha} \in B$. So, this element is absorbed to the right. 
Then, every element $ b s b_1b_2 w \tilde{b} \in (BsB)(BwB)$ can be written in the form $BsuwB$, where $u \in U_{w^{-1}}$ as above. Explicitly, $b_1b_2 = t(uu') \in B$, where $t \in T$ and $(uu')\in U_{w{-1}} U_{w_0w^{-1}}$ according to the decomposition given above. Then, $ t(uu')w \tilde{b} = u b'$ for some $u \in U_{w^{-1}}$ and $b' \in B$. For each $(b_1b_2)$, we get a unique  $u b'$. 
If we can show that $sR(w^{-1}) \subset R^+$, then for $u = u_{\alpha_1} \cdots u_{\alpha_k}$ (i.e, $\alpha_i \in R(w^{-1})$) we have 
\begin{equation}
su = s (u_{\alpha_1} \cdots u_{\alpha_k})=(u_{s\alpha_1} \cdots u_{s\alpha_k})s,
\end{equation}
 and $(u_{s\alpha_1} \cdots u_{s\alpha_k}) \in B$. Then, every $b s b_1b_2 w \tilde{b}= b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$. This tells us that every $b_3 sw b_4 =  b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$ (sorry for the excessive indexing), where $(b_1b_2)$ can be arbitrary because we need only adjust $b$ and $b'$ to make this equation work. However, once $(b_1b_2)$ is fixed, $b$ and $b'$ are determined by this value. 
To show $s R(w^{-1}) \subset R^+$, we use the assumption that 
\begin{equation}
1+ \ell(w^{-1}) = 1 + \ell(w) = \ell(sw) = \ell(w^{-1}s) .
\end{equation}
So, $|R(w^{-1}s)| = |R(w^{-1})| + 1$ since the cardinality of $R(w)$ is just the length of $w$. Using a basic formula for $R(w^{-1}s)$ (see, e.g., Springer's Linear Algebraic Groups, section 8.3.1) this implies $s R(w^{-1}) \subset R^+$ as desired. 
A: 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.
