# 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.

• Probably not without further hypotheses, because the centralizers of your $y,z$ could be non-trivial. For example, in $G=SL_5$, with $y$ the $(1,2)$ transposition in the Weyl group $S_4$ and $z$ the $(2,3)$ transposition, all diagonal elements with diagonal entries of the form $(1,1,1,t,t^{-1})$ commute with both $y,z$ and $yz$, creating considerable ambiguity in the expression. May 31, 2016 at 18:27
• @paulgarrett But this ambiguity can be included in the right $B$-multiple of $y$ and the left $B$ multiple of $z$, right?
– Daps
May 31, 2016 at 18:48
• Maybe... It wouldn't surprise me if there's a reformulation of your question that avoids the obvious problems... but I think a reasonable literal reading of it as it stands makes it false ... for reasons that are possibly irrelevant to your ulterior purposes, etc. May 31, 2016 at 19:19
• @paulgarrett Hmm... I just rewrite a sentence... Can you please see if the question makes sense now?
– Daps
May 31, 2016 at 19:33
• Yes, this avoids the nuisance objections of earlier. Now maybe it suffices to use the fact that the length is the number positive roots sent to negative, and perhaps induction by adjoining one reflection at a time... May 31, 2016 at 20:12

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 $$U_{w^{-1}} = \prod_{\alpha \in R(w^{-1})} U_\alpha.$$ 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 $$su = s (u_{\alpha_1} \cdots u_{\alpha_k})=(u_{s\alpha_1} \cdots u_{s\alpha_k})s,$$ 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 $$1+ \ell(w^{-1}) = 1 + \ell(w) = \ell(sw) = \ell(w^{-1}s) .$$ 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.

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.