## Proof of the quantum Steinberg identity

Last update: 26 July 2013

## Proof of the quantum Steinberg identity

Use the relations $q{q}^{-1}={q}^{-1}q=1$ and the notation

$[n]= qn-q-n q-q-1 =qn-1+ qn-3+…+ q-(n-3) +q-(n-1), for n∈ℤ, [n]!=[n] [n-1]…[2] [1],for n∈ ℤ≥0, [mn]= [m]q [m-1]… [m-n+1] [n]! ,for m∈ℤ and n∈ℤ≥0, (1.5)$

Use the relations $qE=Eq,$ $qF=Fq,$ $qK=Kq,$ $K{K}^{-1}={K}^{-1}K=1,$

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

Let

$E(r)= Er[r]! and F(r)= Fr[r]!, for r∈ℤ>0,$

and for $c\in ℤ$ let

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

for $k\in {ℤ}_{>0}\text{.}$

Let $r,s\in {ℤ}_{\ge 0}\text{.}$ Then

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

 Proof. The case $r=1$ and $s=1$ is the first identity in (1.6). If $r=1$ and $1 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 $E(r)F(s) = E[r] E(r-1)F(s) = E[r] ( F(s) E(r-1)+ F(s-1) [q-(s+r-1)+2K1] E(r-2)+F(s-2) [q-(s+r-1)+4K2] E(r-3) +…+ F(s-r+1) [q-(s+r-1)+2(r-1)Kr-1] ) = 1[r] ( F(s)E+ F(s-1) [q-(s-1)K1] ) E(r-1) + 1[r] ( F(s-1)E+ F(s-2) [q-(s-2)K1] ) [q-(s+r-1)+2K2] E(r-2) +…+1[r] ( F(s-r+1)E+ F(s-r) [q-(s-r)K1] ) [q-(s+r-1)+2(r-1)Kr-1],$ where the first equality is the induction hypothesis and the second equality is the $r=1$ case. Then the coefficient of ${F}^{\left(s-i\right)}$ in $\left[r\right]{E}^{\left(r\right)}{F}^{\left(s\right)}$ 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 $EK={q}^{-2}KE$ 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 $1\le r\le s\text{.}$ The cases $1\le s\le r$ are done analogously, first by induction on $r$ with $s=1$ and then by induction on $s$ for a fixed $r\text{.}$ $\square$

## Notes and References

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