How to construct the Kazhdan-Luztig basis

Upon re-reading Section 3.3.2 of Introduction to Soergel Bimodules, I feel that the chapter authors might have done a better job at explaining the algorithm behind constructing (and therefore proving the existence of) the Kazhdan-Lusztig basis of the Hecke algebra, as well as some of the symmetry properties and the \mu-coefficients. Right-multiplication by a generator partitions a Coxeter group into pairs, and right multiplication by \delta_s or b_s upon the standard basis just mixes coefficients inside of each pair. This affords a nice diagrammatic schematic of what is actually going on.

Inductively constructing basis elements

For the base case, we have b_{\id} = 1. We could also throw in b_s = \delta_s + v for each s \in S. Now, let x \in W be an element such that we know all b_w for w < x.

First, locate a right descent s \in S such that xs < x. The basic idea is to form the product b_{xs} b_s, which is self-dual and has a top term of \delta_x, but may violate the degree bound condition. We then show that we can subtract off various b_w for w < x in order to correct the degree bound condition; then the resulting element must be b_x.

Firstly, consider the terms of b_{xs} = \delta_{xs} + \sum_{y < xs} h_{y, xs} \delta_y. Right multiplication by s partitions the whole group W into pairs \set{y, ys}. Let us relabel so that y is the top element in each pair, ys < y. The coefficient h_{ys, xs} is nonzero only when ys \leq xs, and so the pairs where ys \leq xs fall into these three cases:

Case 1: the top pair xs < x
Case 2: a pair ys < y, with ys < xs but y \not < xs
Case 3: a pair ys < y, with y < xs
The element b_{xs}

Case 1 is the very top pair, which we treat separately for illustrative purposes. Case 2 is where ys < xs but y \not < xs (we have drawn this on its side to try and illustrate this). Although b_{xs} has no \delta_{y} term in case 2, b_{xs} b_s will have a \delta_y term. Case 3 is the case where both y and ys are already less than xs (the vast majority of elements will fall into this case). In cases 2 and 3, all coefficients lie in the ideal v \bbZ[v] by the degree bound condition.

We now right-multiply by b_s. The right multiplication formula is \delta_{w} b_s = \begin{cases} \delta_{ws} + v \delta_{w} & \text{when } w < ws, \\ \delta_{ws} + v^{-1} \delta & \text{when } w > ws, \end{cases} so a coefficient either moves up and leaves v times itself behind, or moves down and leaves v^{-1} times itself behind. Multiplying our three cases above by b_s on the right, we arrive at

Case 1
Case 2
Case 3
The element b_{xs} b_s

Now we indeed have a top coefficient \delta_x of 1, and cases 1 and 2 automatically satisfy the degree bound condition. The only place something could have gone wrong is in case 3: if h_{y, xs} has a nonzero v^1 term, then v^{-1} h_{y, xs} has a constant term. Define the \mu-coefficients \mu_{y, w} = \text{coefficient of } v^1 \text{ in } h_{y, w}. Since h_{y, y} = 1, we can maintain self-duality and the degree bound condition by subtracting multiples of b_y where appropriate, by subtracting \mu_{y, xs} b_y from each case 3 pair. Since b_y has a constant coefficient only at \delta_y, and each other term lies in the ideal v \bbZ[v], by doing this subtraction we do not affect the degree bound condition anywhere else. Hence we arrive at the element b_{x} = b_{xs} b_s - \sum_{y \colon ys < y < xs} \mu_{y, xs} b_y which is self-dual, has a \delta_{x} coefficient of 1, and satisfies the degree bound condition. Therefore it is indeed the canonical basis element.

A symmetry property

A key symmetry property of the KL polynomials is that for every pair x \in W, s \in S with xs < x, we have h_{ys, x} = v h_{y, x} whenever ys < y. This is almost self-evident from the diagrams above: already when looking at the element b_{xs} b_s we can see that every bottom triangle is v times the top triangle. Furthermore, we are only subtracting off some multiples of b_y for y in the top triangles; so the property holds by induction since it holds for the generators b_s in particular.

After acknowledging this property, it is straightforward to show the following complete multiplication formula: b_x b_s = \begin{cases} b_{xs} + \sum_{y \colon ys < y < x} \mu_{y, x} b_y & \text{if } x < xs, \\ (v + v^{-1})b_y & \text{if } x > xs. \end{cases} The formula for b_s b_x is analagous (note that while the \mu_{y, x} coefficients do not depend on whether multiplication is on the right or left, in one case we are summing over the y with ys < y, and in the other case we are summing over the y with sy < y). This shows that the \mu coefficients control the structure constants of the canonical basis, and indeed the \mu-coefficients are used to define the left, right, and two-sided cells of W.