the Masters' Thesis of S. Sundar

Proposition 4. Let τ be a nonzero complex number such that Pk(τ) �= 0
for k = 1,2, · · · ,n. Define fk in TL(τ) recursively as follows.
Then,
f0 = 1 = f1
fk+1 = fk − Pk−1(τ)
Pk(τ) fkekfk, 1 ≤ k ≤ n.
1. fk ∈ Tk(τ) for 1 ≤ k ≤ n + 1.
2. 1−fk is in the algebra generated by {e1,e2, · · · ,ek−1} for 2 ≤ k ≤ n+1.
3. (ekfk) 2 = Pk(τ)
Pk−1(τ) ekfk , (fkek) 2 = Pk(τ)
Pk−1(τ) fkek for 1 ≤ k ≤ n + 1.
4. fk is an idempotent for 1 ≤ k ≤ n + 1.
5. fkei = 0 , eifk = 0 if i ≤ k − 1 where 1 ≤ k ≤ n + 1
6. tr(fk) = Pk(τ) for 1 ≤ k ≤ n + 1.
When τ > 0, fk is selfadjoint.
Proof. This is due to Wenzl and we include a proof here for completeness.
The proof is by induction on k. 1,2··· ,6 are clearly true for k ≤ 2. Now
assume that 1,2··· ,6 are true for 1 ≤ k ≤ l where l ≥ 2. We will show the
result is true for k = l + 1.
Since fl is in the algebra generated by 1,e1,e2, · · · ,el−1 by definition it
follows that fl+1 is in the algebra generated by 1,e1,e2, · · · ,el. Hence
fl+1 ∈ Tl+1(τ). Since 1−fl is in the algebra genrated by e1,e2, · · · ,el−1 , by
definition, it follows that 1−fl+1 is in the algebra generated by e1,e2, · · · ,el.
Now note that fl+1fl = fl+1 and flfl+1 = fl+1 since fl is an idempotent.
Since fl ∈ Tl(τ), el+1 commutes with fl. Hence we have,
el+1fl+1el+1 = el+1fl − Pl−1(τ)
Pl(τ) flel+1elel+1fl
= Pl+1(τ)
Pl(τ) el+1fl
Hence (el+1fl+1) 2 = Pl+1(τ)
Pl(τ) el+1fl+1.
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