# Macdonald polynomials

macdonald_polynomials

We follow Chapter 6 of Macdonald's book [{macdonald98}].

## Derivation

Let

$z_\lambda(q, t) := z_\lambda \prod_{i = 1 : l(\lambda)} {1 - q^{\lambda_i} \over 1 - t^{\lambda_i}}$

and define inner product

$\langle p_\lambda, p_\mu \rangle = \langle p_\lambda, p_\mu \rangle_{q, t} = \delta_{\lambda \mu} z_\lambda(q, t).$

Then the $$\braket{\cdot, \cdot}_{q, q}$$ is the same as the one defined in Cauchy-Littlewood identities. Let $$\Lambda_F = \Lambda \otimes F$$ where $$F = \rtnl(q, t)$$ Also define

$\Pi(x, y; q, t) = \prod_{i j} {(t x_i y_j; q)_\infty \over (x_i y_j; q)_\infty}$

then

$\Pi(x, y; q, t) = \sum_\lambda z_\lambda(q, t)^{-1} p_\lambda(x) p_\lambda(y) \qquad (2.6)$

and that

$\braket{u_\lambda, u_\mu} = \delta_{\lambda \mu} \Leftrightarrow \sum_\lambda u_\lambda(x) u_\lambda(y) = \Pi(x, y; q, t). \qquad (2.7)$

Define $$g_n$$ by

$\prod_i {(t x_i y; q)_\infty \over (x_i y; q)_\infty} = \sum_n g_n(x; q, t) y^n. \qquad (2.8)$

Then since $$p_\lambda(y_1) = y_1^n$$

$\sum_n g_n(x; q, t) y_1^n = \Pi (x, (y_1, 0, 0, ... ); q, t) = \sum_\lambda z_\lambda(q, t)^{-1} p_\lambda(x) y_1^{|\lambda|}$

so

$g_n(x; q, t) = \sum_{\lambda \vdash n} z_\lambda(q, t)^{- 1} p_\lambda (x)$

Also by (2.8),

$\Pi(x, y; q, t) = \prod_j \sum_n g_n(x; q, t) y_j^n = \sum_\lambda g_\lambda(x; q, t) m_\lambda(y)$

hence $$g_\lambda$$ and $$m_\lambda$$ are basis dual to each other. For $$u, v \in F$$, let $$\omega_{u, v} : \Lambda_F \to \Lambda_F$$ be defined by

$\omega_{u, v} (p_r) = (-1)^{r - 1} {1 - u^r \over 1 - v^r} p_r$

hence

1. $$\omega_{q, t} p_\lambda = \epsilon_\lambda p_\lambda \prod_i {1 - q^{\lambda_i} \over 1 - t^{\lambda_i}} p_\lambda = \epsilon_\lambda z_\lambda^{-1} z_\lambda(q, t) p_\lambda$$.
2. $$\omega_{u, v} = \omega_{v, u}^{-1}$$ and $$\omega_{u, u} = \omega$$.
3. $$\omega_{u, v}$$ is self-adjoint: $$\braket{\omega_{u, v} f, g} = \braket{f, \omega_{u, v} g}$$, as this can be verified when $$f = p_\lambda$$ and $$g = p_\mu$$.
4. $$\braket{\omega_{t, q} f, g}_{q, t} = \braket{\omega f, g}_{q, q}$$, which can also be verified when $$f = p_\lambda$$ and $$g = p_\mu$$: $\braket{\omega_{q, t}^{-1} p_\lambda, p_\mu}_{q, t} = \epsilon_\lambda z_\lambda z_{\lambda}(q, t)^{-1} \braket{p_\lambda, p_\mu}_{q, t} = \epsilon_\lambda z_\lambda \delta_{\lambda \mu} = \braket{\omega p_\lambda, p_\mu}_{q, q}.$

Hence by (2.9), and (2.14') in Ch 1 of [{macdonald98}]

$\omega_{q, t} g_n = \sum_{\lambda \vdash n} \omega_{q, t} z_\lambda(q, t)^{-1} p_\lambda = \sum_{\lambda \vdash n} \epsilon_\lambda z_\lambda^{-1} p_\lambda = e_n$

Moreover

\begin{align} \omega_{q, t, x} \Pi(x, y; q, t) &= \omega_{q, t, y} \Pi(x, y; q, t) = \sum_\lambda z_\lambda(q, t)^{-1} \omega_{q, t} p_\lambda(x) p_\lambda(y) \\ &= \sum_\lambda z_\lambda^{-1} \epsilon_\lambda p_\lambda(x) p_\lambda(y) = \Pi'(x, y) \end{align}

## Definition of the Macdonald polynomials

Def. $$P_\lambda$$ is defined as the unique symmetric function $$P_\lambda \in \Lambda_F$$ such that

1. $$P_\lambda = \sum_{\mu \le \lambda} u_{\lambda \mu} m_\mu$$ where for some $$u$$ such that $$u_{\lambda \lambda} = 1$$.
2. $$\braket{P_\lambda, P_\mu} = 0$$ if $$\lambda \neq \mu$$.

In particular

\begin{align} P_{1^r} = e_r \qquad (4.8) \\ P_{(r)} = {(q; q)_r \over (t; q)_r} g_r. (4.9)\\ Q_{(n)} = g_n \qquad (5.5) \end{align}

Now let

\begin{align} b_\lambda = \braket{P_\lambda, P_\lambda}^{-1},\qquad (4.11) \\ Q_\lambda = b_\lambda P_\lambda \qquad (4.12) \end{align}

So that $$P$$ and $$Q$$ are duals of each other. Hence by (2.7)

$\sum_\lambda P_\lambda(x) Q_\lambda(y) = \Pi(x, y) \qquad (4.13)$

## Special cases

1. $$q = t$$: $$P$$ and $$Q$$ are Schur functions
2. $$q = 0$$: $$P$$ and $$Q$$ are Hall-Littlewood polynomials
3. $$q = t^\alpha$$ and $$t \to 1$$: Jack polynomials
4. $$t = 0$$: q-Whittaker functions
5. $$t = 1$$: $$P_\lambda = m_\lambda$$
6. $$q = 1$$: $$P_\lambda = e_{\lambda'}$$.

## Duality

\begin{align} \omega_{q, t} P_\lambda(x; q, t) &= Q_{\lambda'}(x; t, q) \qquad (5.1) \\ \omega_{q, t} Q_\lambda(x; q, t) &= P_{\lambda'} (x; t, q). \end{align}

By (4.11), (4.12) and (5.1)

$b_\lambda(q, t) b_{\lambda'}(t, q) = 1 \qquad (5.3)$

By (5.1) and (4.13)

$\sum_\lambda P_\lambda(x; q, t) P_{\lambda'} (y; t, q) = \sum_\lambda Q_\lambda(x; q, t) Q_{\lambda'}(x; t, q) = \Pi'(x, y)$

## Pieri Formulas

Define

\begin{align} T_{q, x_i} f(x_{1 : n}) &= f(x_{1 : i - 1}, q x_i, x_{i + 1 : n}) \\ A_I (x; t) &= t^{r(r - 1) / 2} \prod_{i \in I, j \notin I} {t x_i - x_j \over x_i - x_j} \\ D_n^r &= \sum_I A_I(x; t) \prod_{i \in I} T_{q, x_i} \end{align}

Define a homomorphism $$u_\mu$$ by $$u_\mu(x_i) = q^{\mu_i} t^{n - i}$$, then

$D_n^r P_\lambda = u_\lambda(e_r) P_\lambda \qquad (6.1)$

hence

$\sum_I A_I(\prod_{i \in I} T_{q, x_i}) P_\lambda = u_\lambda(e_r) P_\lambda \qquad (6.2)$

where the first sum is over $$r$$-element subsets $$I$$ of $$[n]$$ and

$A_I = a_\delta^{-1} (\prod_{i \in I} T_{t, x_i}) a_\delta = \prod_{1 \le i < j \le n} {x_i t^{\ind_I(i)} - x_j t^{\ind_I(j)} \over x_i - x_j},$

When $$r = 1$$, (6.2) becomes

$\sum_{k = 1 : n} (\prod_{j \neq k} {x_j - x_k t \over x_j - x_k}) P_\lambda(x_{1 : k - 1}, q x_k, x_{k + 1 : n}) = (\sum_{k = 1 : n} q^{\lambda_k} t^{n - k}) P_\lambda(x)$

We skip some derivation and show the main points:

Define the armlength and leglenth

$a_\lambda(s) = \lambda_i - j, \qquad a_\lambda'(s) = j - 1, \qquad l_\lambda(s) = \lambda'_j- i, \qquad l'_\lambda(s) = i - 1 \qquad (6.14)$

and

$b_\lambda(s) = s \in \lambda ? {1 - q^{a_\lambda(s)} t^{l_\lambda(s) + 1} \over 1 - q^{a_\lambda(s) + 1} t^{l_\lambda(s)}} : 1 \qquad (6.20)$

then

$b_\lambda = \prod_{s \in \lambda} b_\lambda(s)$

Let $$C_{\lambda / \mu}$$ ($$R_{\lambda / \mu}$$) be the union of the columns (rows) intersecting $$\lambda - \mu$$.

Claim. (Pieri's rule) Define

\begin{align} \phi_{\lambda / \mu} &= \prod_{s \in C_{\lambda / \mu}} {b_\lambda(s) \over b_\mu(s)}\\ \psi_{\lambda / \mu} &= \prod_{s \in R_{\lambda / \mu} - C_{\lambda / \mu}} {b_\mu(s) \over b_\lambda(s)}\\ \phi'_{\lambda / \mu} &= \prod_{s \in R_{\lambda / \mu}} {b_\mu(s) \over b_\lambda(s)}\\ \psi'_{\lambda / \mu} &= \prod_{s \in C_{\lambda / \mu} - R_{\lambda / \mu}} {b_\lambda(s) \over b_\mu(s)} \end{align}

then

\begin{align} P_\mu g_r &= \sum_\lambda \phi_{\lambda / \mu} P_\lambda \qquad (i)\\ Q_\mu g_r &= \sum_\lambda \psi_{\lambda / \mu} Q_\lambda \\ Q_\mu e_r &= \sum_\lambda \phi'_{\lambda / \mu} Q_\lambda \\ P_\mu e_r &= \sum_\lambda \psi'_{\lambda / \mu} P_\lambda \end{align}

where the sum is over horizontal (vertical) $$r$$-strips in the first and second (third and fourth) identities.

Define $$\phi_T$$

$\phi_T = \prod_{i = 1 : r} \phi_{\sh T^i / \sh T^{i - 1}}$

then by (i) of Pieri rule we have

$P_\mu g_nu = \sum_\eta (\sum_{T : \sh T = \eta - \mu, \ty T = \nu} \phi_T) P_\eta. \qquad (i')$

## Skew Macdonald polynomials

Let

$f^\lambda_{\mu \nu} = \braket{Q_\lambda, P_\mu P_\nu} \qquad (7.1)$

so that

$P_\mu P_\nu = \sum_\lambda f^\lambda_{\mu \nu} P_\lambda$

Applying $$\omega_{q, t}$$ to both sides and noting the duality (5.1) one has

$Q_{\mu'} (t, q) Q_{\nu'} (t, q) = \sum_\lambda f^\lambda_{\mu, \nu}(q, t) Q_{\lambda'} (t, q)$

i.e.

$Q_\mu Q_\nu = \sum_\lambda f^{\lambda'}_{\mu' \nu'}(t, q) Q_\lambda$

Define

\begin{align} &Q_{\lambda / \mu} = \sum_\nu f^\lambda_{\mu \nu} Q_\nu \\ &\Leftrightarrow \braket{Q_{\lambda / \mu}, P_\nu} = \braket{Q_\lambda, P_\mu P_\nu} \\ &\Leftrightarrow \braket{Q_{\lambda / \mu}, f} = \braket{Q_\lambda, P_\mu f} \end{align}

Define $$P_{\lambda / \mu}$$ by

\begin{align} &\braket{P_{\lambda / \mu}, Q_\nu} = \braket{P_\lambda, Q_\mu Q_\nu}\\ &\Leftrightarrow \braket{P_{\lambda / \mu}, f} = \braket{P_\lambda, Q_\mu f} = \braket{b_\lambda^{-1} Q_\lambda, b_\mu P_\mu f} = b_\lambda^{-1} b_\mu \braket{Q_{\lambda / \mu}, f} \end{align}

so

$P_{\lambda / \mu} = b_\mu b_\lambda^{-1} Q_{\lambda / \mu}$

Claim.

1. $$Q_\lambda(x, z) = \sum_\mu Q_{\lambda / \mu} (x) Q_\mu (z)$$ (7.9)
2. $$P_\lambda(x, z) = \sum_\mu P_{\lambda / \mu}(x) P_\mu(z)$$.

Proof.

$\sum_\lambda Q_{\lambda / \mu} (x) P_\lambda(y) = \sum_{\lambda, \nu} f^\lambda_{\mu \nu} Q_\nu(x) P_\lambda(y) = \sum_\nu P_\mu(y) P_\nu(y) Q_\nu(x) = P_\mu(y) \Pi(x, y)$

so

$\sum_{\lambda, \mu} Q_{\lambda / \mu}(x) P_\lambda(y) Q_\mu(z) = \sum_\mu P_\mu(y) \Pi(x, y) Q_\mu(z) = \Pi((x, z), y) = \sum_\lambda Q_\lambda(x, z) P_\lambda(y)$

And the proof the $$P$$ is similar. $$\square$$

Now we can obtain the branching identity for the Macdonald polynomials:

Claim.

1. $$Q_{\lambda / \mu} (x) = \sum_{T : \sh T = \lambda - \mu} \phi_T(q, t) x^T$$ (7.13)
2. $$P_{\lambda / \mu} (x) = \sum_{T : \sh T = \lambda - \mu} \psi_T(q, t) x^T$$

Proof. Since $$g$$ and $$m$$ are dual to each other, by (i') of the Pieri rule

\begin{align} Q_{\lambda / \mu} = \sum_\nu \braket{Q_{\lambda / \mu}, g_\nu} m_\nu = \sum_\nu \braket{Q_\lambda, P_\mu g_\nu} m_\nu \\ = \sum_\nu \braket{Q_\lambda, \sum_\eta (\sum_{T: \sh T = \eta - \mu, \ty T = \nu} \phi_T) P_\eta} m_\nu\\ = \sum_\nu (\sum_{T : \sh T = \lambda - \mu, \ty T = \nu} \phi_T) m_\nu = \sum_{T: \sh T = \lambda - \mu} \phi_T x^T \end{align}

Similar method for $$P$$. $$\square$$

## References

• [macdonald98] Symmetric Functions and Hall Polynomials, , 1998.