Exercise 1 in Chapter 4, Section 1 of Bourbaki, *Lie Groups and Lie Algebras*

Question

This is the first exercise in Bourbaki, Lie Groups and Lie Algebras, Chapter IV.

Answer 1

(a): There must exist a $2 \leq i \leq r$ such that $\ell(s_1 \cdots s_i) = \ell(s_1 \cdots s_{i-1}) - 1$. Otherwise,

$$\ell(s_1s_2) = \ell(s_1) + 1 = 2$$

$$\ell(s_1s_2s_3) = \ell(s_1s_2) + 1 = 3$$

$$ \vdots$$

$$\ell(s_1 \cdots s_r) = r$$

contradicting the hypothesis. So let $q$ be a number satisfying this property. So $\ell(s_1 \cdots s_{q-1}s_{q}) = \ell(s_1 \cdots s_{q-1}) - 1$. The exchange condition implies that there exists a $1 \leq p \leq q-1$ such that

$$s_1 \cdots s_q = s_1 \cdots s_{p-1}s_{p+1} \cdots s_{q-1}$$

Then $$w = s_1 \cdots s_r = (s_1 \cdots s_q)(s_{q+1} \cdots s_r) = s_1 \cdots s_{p-1}s_{p+1} \cdots s_{q-1}s_{q+1} \cdots s_r$$

If this is a reduced decomposition, we are done. Otherwise, we can iterate the same procedure to find a decomposition of smaller length, etc.

(b): For the first assertion, it is clear that $(W_X \cap W_Y).(W_X \cap W_Z)$ is contained in $W_X.W_X = W_X$ as well as in $W_Y.W_Z$. So the right hand side is contained in the left.

Conversely, take a $w \in W_X \cap (W_Y.W_Z)$. Write $w = w_1w_2$ for $w_1 \in W_Y, w_2 \in W_Z$. Let $(s_1, ... , s_q)$ be a reduced decomposition for $w_1$, and $(s_1', ... , s_r')$ one for $w_2$. We know that each $s_i$ is in $X$ and each $s_i'$ is in $Y$. Then by part (a), we can remove various pairs of generators from $(s_1, ... , s_q, s_1', ... , s_r')$ until the resulting thing is a reduced decomposition for $w_1w_2$. Since $w_1w_2 \in W_X$, this tells us that the remaining generators are all in $X$. So the remaining $s_i$ are in $X \cap Y$, and the remaining $s_i'$ are in $X \cap Z$. Thus the product of the remaining $s_i$ is in $W_{X \cap Y} = W_X \cap W_Y$, and the product of the remaining $s_i'$ is in $W_{X \cap Z} = W_X \cap W_Z$.

Second assertion: (to do)