Cauchy’s Series product identity, an exercise from Macdonald’s book on symmetric functions

After seeing how diminished my daily visitor count has gone, since the last series of rapid fire posting of technical posts on CS related issues, I decide to come back to mathematics. The result is motivated by an example/exercise from Macdonald’s book “Symmetric functions and Hall polynomials”, which I stumbled upon while reading his Chapter 6 on Macdonald polynomials ( of course he called it something like two-parameter generalized Hall polynomials).

After he defined the inner product {\langle , \rangle_{q,t}} on the space of symmetric functions {\Lambda}, he defines a new basis of symmetric functions {g_\lambda(x; q,t )} which are dual to the monomial symmetric functions (which are of course not orthonormal), in the sense that

\displaystyle  \begin{array}{rcl}  \langle g_\lambda, m_\mu \rangle_{q,t} = \delta_{\lambda,\mu}. \end{array}

It turns out that {g_n :=g_{(n)}} is the {y^n} coefficient of

\displaystyle   \prod_{i \ge 1} \frac{(tx_i y; q)_\infty}{(x_i y; q)_\infty} \ \ \ \ \ (1)

and {g_\lambda = \prod_{j=1}^{\ell(\lambda)} g_{\lambda_j}}. Here the symbol (for the lack of a standard name) {(a;q)_\infty:= \prod_{r=0}^\infty (1 -aq^r)}. (1) is a basic example of a hypergeometric series, which means the ratio of the {n}th and {(n+1)}st terms in the series is given by some rational function in {n}.

To evaluate {g_n}, we need to express (1) as a power series in {y}. We first do that to the individual factor {\frac{(t x_i y; q)_\infty}{(x_i y; q)_\infty}}. This is achieved by Cauchy’s formula (see Macdonald page 27, Chapter I section 2 example 5):

\displaystyle   \prod_{n=0}^\infty \frac{1 - t y q^n}{1-y q^n} = 1 + \sum_{n=1}^\infty \frac{(1-t) \ldots (1-y q^{n-1})}{(1-q) \ldots (1- q^n)} y^n \ \ \ \ \ (2)

In other words, {\frac{(t y; q)_\infty}{(y; q)_\infty} = \sum_{m \ge 0} \frac{(t;q)_m}{(q;q)_m} y^m}, where {(;)_m} has the obvious meaning analogous to {(;)_\infty}. Let’s use this to evaluate {g_n} first. We have

\displaystyle   g_n = [[y^n]] \prod_{i \ge 1} \sum_{m \ge 0} \frac{(t;q)_m}{(q;q)_m} (x_i y)^m \\ = \sum_{\lambda \vdash n} \prod_{j=1}^{\ell(\lambda)} \frac{(t;q)_{\lambda_j}}{(q;q)_{\lambda_j}} m_\lambda(x_1,x_2,\ldots) y^n \\ =: \sum_{\lambda \vdash n} \frac{(t;q)_\lambda}{(q;q)_\lambda} m_\lambda(x_1, x_2, \ldots) y^n. \ \ \ \ \ (3)

Here we used the fact that when we take the product over a sum like the left hand side, we can do a discrete change of variables and sum over partitions of {n} first and then sum over {n \in {\mathbb N}}. The set of partitions of {n} can be thought of as spheres of radius {n} in this infinite dimensional product space, much like in polar coordinates.

Also recall the definition of monomial symmetric polynomials, {m_\lambda(x_1, \ldots, x_n) = \sum_{w \in S_n} \prod_{i=1}^n x_i^{w(\lambda)_i}}, where {w(\lambda)_i} is the {w(i)}th component of {\lambda} in non-increasing order. (3) is pretty explicit. We just need to prove the Cauchy’s formula (2). The idea is to take advantage of the infinite product of the left hand side, and derive a recurrence on the coefficients of its power series (see G. E. Andrews’s Theory of Partitions Section 2.2): Write {\varphi(y):=\prod_{n=0}^\infty \frac{1- ty q^n}{1-y q^n} = \sum_{n=0}^\infty A_n y^n}, we have

\displaystyle   (1-y) \varphi(y) = (1-ty) \prod_{n=1}^\infty \frac{ 1 -ty q^n}{1-y q^n} \\ =(1-ty) \prod_{n=0}^\infty \frac{ 1 - t y q^{n+1}}{1 - y q^{n+1}} \\ = (1-ty) \varphi(yq). \ \ \ \ \ (4)

So we can write {\sum_n (1-y) A_n y^n = \sum_n (1-ty) A_n (qy)^n}. Extracting the {y^n} coefficient, we have

\displaystyle  \begin{array}{rcl}  A_n - A_{n-1} = A_n q^n - t A_{n-1} q^{n-1}, \end{array}

from which we easily obtain

\displaystyle  \begin{array}{rcl}  A_n = \frac{1 - tq^{n-1}}{1 - q^n} A_{n-1}. \end{array}

We still need to figure out {A_0}, but that’s simply equal to {\varphi(0) = 1}. Thus we have

\displaystyle  \begin{array}{rcl}  A_n = \prod_{j=1}^n \frac{1 - t q^{j-1}}{1 -q^j} = \frac{(t;q)_n}{(q;q)_n}. \end{array}

p.s. I spent hours trying to figure out the optimal way to convert ordinary latex to wordpress accepted form. Apparently something I learned a year ago has slipped out of my memory. I would still recommend using the python code latex2wp freely distributed online. But the instruction there does not tell me how to number equations in a multi-line setting, because only eqnarray* not eqnarray are supported. By playing around a bit, I discovered that in fact the environment slash begin{equation} slash end{equation} also supported multi-line equation typesetting. Futhermore, one can insert slash label{eq:1} as usual. The only drawback of this approach is that when you compile the tex code on your local machine, the equation environment would simply ignore that the line break symbol slash slash, so you can’t see the multi-line typesetting in action there.

p.p.s: It’s worthwhile quoting a user of latex2wp trying to address the issue of verbatim environment:

December 15, 2009 at 7:27 pm


Hi Luca,

Thanks for this converter. Currently, the lines enclosed in slash ifblog… slash fi are also processed and have their double returns converted to and their multiple spaces/single returns to a single space. If this structure is used to put source code in WordPress, this changes the formatting, which is undesirable. Is there a way to suppress this processing so that a verbatim sort of environment is provided by slash ifblog… slash fi—I see where this is done in but when I suppress the corresponding lines, display style equations don’t get the WordPress latex affix and suffix appended to them— or better yet, to provide support for listings environment?


About aquazorcarson

math PhD at Stanford, studying probability
This entry was posted in algebraic combinatorics and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s