Recall that , and any -module has a ‘standard’ basis (corresponding to the choice of generators ), as described here. Now if is an irreducible root system, and is the corresponding simple Lie algebra, then various subspaces of are -modules in many ways, but the message here is that there exists a basis that is standard (up to multiplication by ) for *all* these module structures simultaneously. More precisely,

**Theorem:** There exist vectors such that

- , where are defined by .
- For any and any the set is a standard basis, up to , of the module over .
- For any let be the corresponding reflection. Then .

To clarify the ideas underlying the proof, I should state a few lemmas.

**Lemma 1:** Let be the standard basis for an -module. Then the standard reflection acts as follows:

I don’t see a better way then to do this ‘by hands’, using the known formulas for the action of and . The calculation boils down to a certain binomial sum, which can be shown equivalent to the Vandermonde identity. Anyway, it can be found in Graham-Knuth-Patashnik…

Let be a basis for . Let’s define a *path* between and a root to be any sequence , such that , and . Let’s define a *homotopy* between two paths and to be a sequence of paths , such that for any the paths and differ only in one place , in the only way possible: .

**Lemma 2:** Any two paths are homotopic.

**Proof:** Proceed by induction on the length of the paths. Suppose that and are two paths joining and . If , then we are done by induction hypothesis. If not, observe that is also a root. This can be shown directly by examining the -dimensional root systems. Now by induction hypothesis, is homotopic to , and is homotopic to . Now the parts are homotopic, and finally, is a homotopy. **QED.**

**Proof of Theorem:** Let be a basis, let be arbitrary vectors, and define , , where are chosen to satisfy the ‘standard basis’ requirement for both - and -module structures. To check that this is possible, it is sufficient to consider two-dimensional root systems. Now we should also check that this definition of does not depend on the way to write as a sum of simple roots. This can be done via homotopy: to verify each step of the homotopy, it is sufficient to consider -dimensional root systems. Finally, choose via in an appropriate way. **QED.**