Prove One, Get Infinitely Many Free!

This is a better deal than any deal you can find on the market.

Bailey’s lemma is an amazing tool. At its core it is a really simple observation but in practice it is a beast. I can (and will) state the actual lemma and you will think to yourself “is this it?”. I am serious. It is as easy as teaching a kid that they can roll a snowball on snow to get it larger to make a snowman. Yet, in practice the same action of snowball rolling down a hill can lead to huge avalanches.

Here is the lemma: Wilfrid N. Bailey noticed that if you start two identities of the form

$\displaystyle \beta_L = \sum_{r=0}^L \alpha_r u_{L-r} v_{L+r}$ and $\displaystyle \gamma_L = \sum_{r=L}^\infty \delta_r u_{L-r} v_{L+r}$

for some sequences $\alpha_j,\ \beta_j,\ \gamma_j, \delta_j,\ u_j,$ and $v_j$ , then we have

$\displaystyle \sum_{L=0}^\infty \alpha_L \gamma_L = \sum_{L=0}^\infty \beta_L \delta_L.$

OK, so I can hear you say that “Ali, you lied, this doesn’t look all that easy.” Well, it is not so easy to roll a snowball for a kid at their first try. Yes, Bailey’s observation is sophisticated but let me give you the proof. It will make you agree that this implication is actually easy. Follow from left to right:

$\displaystyle \sum_{L=0}^\infty \alpha_L \gamma_L = \sum_{L=0}^\infty \alpha_L \sum_{r=L}^\infty \delta_r u_{L-r} v_{L+r} = \sum_{r=0}^\infty \delta_r \sum_{L=0}^r \alpha_L u_{r-L} v_{r+L} = \sum_{r=0}^\infty \beta_r \delta_r.$

Of course, the most rigorous among us must be telling me that we need many convergence conditions to justify the magic trick I did above. You would be right, but this is supposed to be a general writing that I am failing to keep light. Please excuse me. I will not justify them here.

Now back to the main point. There are standard choices for $u_j,\ v_j,\ \delta_j,$ and $\gamma_j$ which makes Bailey’s lemma work. With these standard choices, we are left with free choices for $\alpha_j$ and $\beta_j$ . A pair $(\alpha_j,\beta_j)$ that satisfies Bailey’s lemma is conveniently called a Bailey pair.

Bailey and later his student Lucy Slater did amazing things with this result. I am talking about hundreds of novel sum-product identities by merely applying Bailey’s lemma. Skipping ahead, two big visionaries, Peter Paule and George E. Andrews independently observed something Bailey and Slater missed completely. They observed that if you start with a Bailey Pair $(\alpha_j,\beta_j)$ then you can find a new pair $(\alpha'_j,\beta'_j)$ that is also a Bailey pair. So you can apply the lemma again and generate a new-new Bailey pair $(\alpha''_j,\beta''_j)$ … That is the avalanche. You start with a snowball -the initial seed identity that defines your Bailey pair- then you apply Bailey’s lemma as many times as you like. This creates an infinite hierarchy on top of your seed identity.

Andrews-Gordon identities is one of the most famous q-series identities that builds on the Rogers-Ramanujan identities with this very lemma. These identities are the main content of Andrews’ American Mathematical Society Presidential Signature:

I am sure you also noticed Srinivasa Ramanujan‘s silhouette watching over this infinite hierarchy. He sow the case $k=2$ seed that turned into this infinite family of sum-product identities by Bailey’s lemma in the hands of Andrews.

One remark I need to make is that at each step of the application of the lemma you are summing both sides of your seed identity in a special way. It is the reason that you see an implicit $\underline{n} = (n_1,n_2,\dots,n_{k-1})$ vector on the left-hand side of the above identity. That is actually a $k-1$ -fold sum.

To sum it up, Bailey’s lemma is an incredible tool. Every day there are more and more being discovered using this technique. In my recent paper with Alexander Berkovich, we reaped the seeds we sowed. For example, with a special form of Bailey’s lemma, we saw that

$\displaystyle \sum_{m,n\geq 0} \frac{q^{2m^2+6m n+6n^2}(q^3;q^3)_M}{(q;q)_m (q^3;q^3)_n(q^3;q^3)_{M-2n-m}} = \sum_{j=-M}^M q^{3j^2+j} {2M\brack M-j}_{q^3}$

(which we proved in an earlier joint paper) implies

$\begin{aligned} \sum_{m,n,n_1,n_2,\dots,n_f\geq 0 } &\frac{q^{2m^2+6mn+6n^2+3(N_1^2 + N_2^2+\dots+N_f^2)} (q^3;q^3)_{n_f}}{(q;q)_m (q^3;q^3)_n (q^3;q^3)_{n_f-2n-m }(q^3;q^3)_{n_1}(q^3;q^3)_{n_2}\dots(q^3;q^3)_{n_{f-1}}(q^3;q^3)_{2n_f}}\\&\hspace{9cm} = \frac{(q^{6f+6}, -q^{3f+2}, -q^{3f+4}; q^{6f+6})_\infty}{(q^3;q^3)_\infty},\end{aligned}$