Proof of the quantum Steinberg identity

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 26 July 2013

Proof of the quantum Steinberg identity

Use the relations qq-1=q-1q=1 and the notation

[n]= qn-q-n q-q-1 =qn-1+ qn-3++ q-(n-3) +q-(n-1), forn, [n]!=[n] [n-1][2] [1],forn 0, [mn]= [m]q [m-1] [m-n+1] [n]! ,formand n0, (1.5)

Use the relations qE=Eq, qF=Fq, qK=Kq, KK-1=K-1K=1,

KEK-1=q2E, KFK-1= q-2F,EF =FE+ K-K-1 q-q-1 . (1.6)


E(r)= Er[r]! and F(r)= Fr[r]!, forr>0,

and for c let

[qcK]= qcK- q-cK-1 q-q-1 , [qcK0]= 1,and [qcKk]= [qcK] [qc-1K] [qc-(k-1)K] [1][2][k] ,

for k>0.

Let r,s0. Then

E(r)F(s)= k=0min(r,s) F(s-k) [ q-(s-k+r-k)K k ] E(r-k).


The case r=1 and s=1 is the first identity in (1.6). If r=1 and 1<s then

EF(s) = EF(s-1) F[s]= ( F(s-1)E+ F(s-2) [q-(s-2)K1] ) F[s] = F(s-1)[s] (FE+[K1])+ F(s-2)F[s] [q-sK1] = F(s)E+ F(s-1)[s] ( [K1]+ [s-1] [q-sK1] ) = F(s)E+ F(s-1)[s] [s][q-(s-1)K1] =F(s)E+ F(s-1) [q-(s-1)K1],

where the second equality is the induction hypothesis and the third equality uses the relations in (1.6).

For 1<rs,

E(r)F(s) = E[r] E(r-1)F(s) = E[r] ( F(s) E(r-1)+ F(s-1) [q-(s+r-1)+2K1] E(r-2)+F(s-2) [q-(s+r-1)+4K2] E(r-3) ++ F(s-r+1) [q-(s+r-1)+2(r-1)Kr-1] ) = 1[r] ( F(s)E+ F(s-1) [q-(s-1)K1] ) E(r-1) + 1[r] ( F(s-1)E+ F(s-2) [q-(s-2)K1] ) [q-(s+r-1)+2K2] E(r-2) ++1[r] ( F(s-r+1)E+ F(s-r) [q-(s-r)K1] ) [q-(s+r-1)+2(r-1)Kr-1],

where the first equality is the induction hypothesis and the second equality is the r=1 case. Then the coefficient of F(s-i) in [r]E(r)F(s) is

[q-(s-i)K1] [q-(s+r-1-2(i-1))Ki-1] E(r-i)+E [q-(s+r-1-2i)Ki] E(r-i-1) = ( [q-(s-i)K1] [q-(s+r-2i+1)Ki-1] + [q-(s+r-1-2i)q-2Ki] [r-i] ) E(r-i) =[r] [q-(s+r-2i)Ki] E(r-i),

where the first equality follows from EK=q-2KE and the second equality is the computation

[q-(s-i)K1] [q-(s+r-2i+1)Ki-1] + [q-(s+r+1-2i)Ki] [r-i] = [q-(s+r-2i+1)Ki-1] ( [q-(s-i)K1] + [q-(s+r+1-2i)-(i-1)] [i] [r-i] ) = [q-(s+r-2i+1)Ki-1] · ( q-s+iK- qs-iK-1 q-q-1 + q-s-r+iK- qs+r-iK-1 q-q-1 · qr-i- q-r+i qi-q-i ) = [q-(s+r-2i+1)Ki-1] · ( q-s+iK- qs-iK-1 ) (qi-q-i) + ( q-s-r+iK- qs+r-i K-1 ) (qr-i-q-r+i) (q-q-1) (qi-q-i) = [q-(s+r-2i+1)Ki-1] · ( q-s-r+2iK- qs+r-2iK-1 ) (qr-q-r) (q-q-1) (qi-q-i) = [q-(s+r-2i+1)Ki-1] ·[q-s-r+2iK] ·[r][i]=[r] [q-(s+r-2i)Ki].

This completes the proof for the cases 1rs. The cases 1sr are done analogously, first by induction on r with s=1 and then by induction on s for a fixed r.

Notes and References

The identity of Proposition 1.1 is a generalization of the identity in [Ste1967, Lemma 5].

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