Proof of the quantum Steinberg identity

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 26 July 2013

Proof of the quantum Steinberg identity

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

\begin{matrix} \begin{matrix} [n] = \frac{q^{n} - q^{- n}}{q - q^{- 1}} = q^{n - 1} + q^{n - 3} + \dots + q^{- (n - 3)} + q^{- (n - 1)}, for n \in ℤ, \\ [n]! = [n] [n - 1] \dots [2] [1], for n \in ℤ_{\geq 0}, \\ [\binom{m}{n}] = \frac{{[m]}_{q} [m - 1] \dots [m - n + 1]}{[n]!}, for m \in ℤ and n \in ℤ_{\geq 0}, \end{matrix} & (1.5) \end{matrix}

Use the relations $q E = E q,$ $q F = F q,$ $q K = K q,$ $K K^{- 1} = K^{- 1} K = 1,$

\begin{matrix} K E K^{- 1} = q^{2} E, K F K^{- 1} = q^{- 2} F, E F = F E + \frac{K - K^{- 1}}{q - q^{- 1}} . & (1.6) \end{matrix}

Let

E^{(r)} = \frac{E^{r}}{[r]!} and F^{(r)} = \frac{F^{r}}{[r]!}, for r \in ℤ_{> 0},

and for $c \in ℤ$ let

[q^{c} K] = \frac{q^{c} K - q^{- c} K^{- 1}}{q - q^{- 1}}, [\binom{q^{c} K}{0}] = 1, and [\binom{q^{c} K}{k}] = \frac{[q^{c} K] [q^{c - 1} K] \dots [q^{c - (k - 1)} K]}{[1] [2] \dots [k]},

for $k \in ℤ_{> 0} .$

Let $r, s \in ℤ_{\geq 0} .$ Then

E^{(r)} F^{(s)} = \sum_{k = 0}^{min (r, s)} F^{(s - k)} [\binom{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 $1 < s$ then

\begin{matrix} E F^{(s)} & = & E F^{(s - 1)} \frac{F}{[s]} = (F^{(s - 1)} E + F^{(s - 2)} [\binom{q - (s - 2) K}{1}]) \frac{F}{[s]} \\ = & \frac{F^{(s - 1)}}{[s]} (F E + [\binom{K}{1}]) + F^{(s - 2)} \frac{F}{[s]} [\binom{q^{- s} K}{1}] \\ = & F^{(s)} E + \frac{F^{(s - 1)}}{[s]} ([\binom{K}{1}] + [s - 1] [\binom{q^{- s} K}{1}]) \\ = & F^{(s)} E + \frac{F^{(s - 1)}}{[s]} [s] [\binom{q^{- (s - 1)} K}{1}] = F^{(s)} E + F^{(s - 1)} [\binom{q^{- (s - 1)} K}{1}], \end{matrix}

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

For $1 < r \leq s,$

\begin{matrix} E^{(r)} F^{(s)} & = & \frac{E}{[r]} E^{(r - 1)} F^{(s)} \\ = & \frac{E}{[r]} (\binom{F^{(s)} E^{(r - 1)} + F^{(s - 1)} [\binom{q^{- (s + r - 1) + 2} K}{1}] E^{(r - 2)} + F^{(s - 2)} [\binom{q^{- (s + r - 1) + 4} K}{2}] E^{(r - 3)}}{+ \dots + F^{(s - r + 1)} [\binom{q^{- (s + r - 1) + 2 (r - 1)} K}{r - 1}]}) \\ = & \frac{1}{[r]} (F^{(s)} E + F^{(s - 1)} [\binom{q^{- (s - 1)} K}{1}]) E^{(r - 1)} \\ + \frac{1}{[r]} (F^{(s - 1)} E + F^{(s - 2)} [\binom{q^{- (s - 2)} K}{1}]) [\binom{q^{- (s + r - 1) + 2} K}{2}] E^{(r - 2)} \\ + \dots + \frac{1}{[r]} (F^{(s - r + 1)} E + F^{(s - r)} [\binom{q^{- (s - r)} K}{1}]) [\binom{q^{- (s + r - 1) + 2 (r - 1)} K}{r - 1}], \end{matrix}

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

\begin{matrix} [\binom{q^{- (s - i)} K}{1}] [\binom{q^{- (s + r - 1 - 2 (i - 1))} K}{i - 1}] E^{(r - i)} + E [\binom{q^{- (s + r - 1 - 2 i)} K}{i}] E^{(r - i - 1)} \\ = ([\binom{q^{- (s - i)} K}{1}] [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] + [\binom{q^{- (s + r - 1 - 2 i)} q^{- 2} K}{i}] [r - i]) E^{(r - i)} \\ = [r] [\binom{q^{- (s + r - 2 i)} K}{i}] E^{(r - i)}, \end{matrix}

where the first equality follows from $E K = q^{- 2} K E$ and the second equality is the computation

\begin{matrix} [\binom{q^{- (s - i)} K}{1}] [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] + [\binom{q^{- (s + r + 1 - 2 i)} K}{i}] [r - i] \\ = [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] ([\binom{q^{- (s - i)} K}{1}] + \frac{[q^{- (s + r + 1 - 2 i) - (i - 1)}]}{[i]} [r - i]) \\ = [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] \cdot (\frac{q^{- s + i} K - q^{s - i} K^{- 1}}{q - q^{- 1}} + \frac{q^{- s - r + i} K - q^{s + r - i} K^{- 1}}{q - q^{- 1}} \cdot \frac{q^{r - i} - q^{- r + i}}{q^{i} - q^{- i}}) \\ = [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] \cdot \frac{(q^{- s + i} K - q^{s - i} K^{- 1}) (q^{i} - q^{- i}) + (q^{- s - r + i} K - q^{s + r - i} K^{- 1}) (q^{r - i} - q^{- r + i})}{(q - q^{- 1}) (q^{i} - q^{- i})} \\ = [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] \cdot \frac{(q^{- s - r + 2 i} K - q^{s + r - 2 i} K^{- 1}) (q^{r} - q^{- r})}{(q - q^{- 1}) (q^{i} - q^{- i})} \\ = [\binom{q^{- (s + r - 2 i + 1)} K}{i - 1}] \cdot [q^{- s - r + 2 i} K] \cdot \frac{[r]}{[i]} = [r] [\binom{q^{- (s + r - 2 i)} K}{i}] . \end{matrix}

This completes the proof for the cases $1 \leq r \leq s .$ The cases $1 \leq s \leq r$ 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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