# Explicit formula of Macdonald polynomials

macdonald_polynomials special_functions

Recall q-Pochhammer notation $$(a)_n$$.

We derive the explicit formula of the Macdonald polynomials based on Pieri's rule. Let

$f(u) = {(t u)_\infty \over (q u)_\infty}.$

We denote

\begin{align} G(n, k) &:= {(q^n t^{k + 1})_\infty \over (q^{n + 1} t^k)_\infty} = f(q^n t^k) \\ g(n, k) &:= {(q t^k)_n \over (t^{k + 1})_n} \end{align}

Then

${g(n_1, k) \over g(n_2, k)} = {G(n_1, k) \over G(n_2, k)} = {f(q^{n_1} t^k) \over f(q^{n_2} t^k)}.$

Recall the branching rule of the Macdonald polynomials.

We start from

\begin{align} \psi_{\lambda / \mu} &= \prod_{1 \le i \le j \le \ell(\mu)} {f(q^{\mu_i - \mu_j} t^{j - i}) f(q^{\lambda_i - \lambda_{j + 1}} t^{j - i}) \over f(q^{\lambda_i - \mu_j} t^{j - i}) f(q^{\mu_i - \lambda_{j + 1}} t^{j - i})} \\ &= \prod_{1 \le i \le j \le \ell(\mu)} {g(\lambda_i - \lambda_j, j - i) g(\mu_i - \mu_{j + 1}, j - i) \over g(\lambda_i - \mu_j, j - i) g(\mu_i - \lambda_{j + 1}, j - i)}. \qquad (1) \end{align}

This first equality can be found on page 342 of [{macdonald98}].

Furthermore, note for $$n_1 \ge n_2$$

${g(n_1, k) \over g(n_2, k)} = {(q^{n_2 + 1} t^k)_{n_1 - n_2} \over (q^{n_2} t^{k + 1})_{n_1 - n_2}}.$

Therefore for a semistandard Young tableaux $$T \leftrightarrow (\lambda^k_j)_{1 \le j \le k \le \ell}$$ (recall correspondence between Young tableaux and Gelfand-Tsetlin patterns),

$\psi_T = \prod_{1 \le i \le j \le k - 1 \le \ell - 1} {(q^{\lambda^{k - 1}_i - \lambda^{k - 1}_j} t^{j - i + 1})_{\lambda^k_i - \lambda^{k - 1}_i} (q^{\lambda^{k - 1}_i - \lambda^k_{j + 1} + 1} t^{j - i})_{\lambda^k_i - \lambda^{k - 1}_i} \over (q^{\lambda^{k - 1}_i - \lambda^{k - 1}_j + 1} t^{j - i})_{\lambda^k_i - \lambda^{k - 1}_i} (q^{\lambda^{k - 1}_i - \lambda^k_{j + 1}} t^{j - i + 1})_{\lambda^k_i - \lambda^{k - 1}_i}}.$

And for $$\lambda = \lambda_{1 : \ell}$$ the Macdonald polynomial

$P_\lambda(x) = \sum_{T:\sh T = \lambda} \psi_T x^T.$

Similarly one can write down the formula for

$Q_\lambda (x) = \sum_{T: \sh T = \lambda} \phi_T x^T.$

In this case again on page 342 of [{macdonald98}]:

\begin{align} \phi_{\lambda / \mu} &= \prod_{1 \le i \le j \le \ell(\lambda)} {f(q^{\lambda_i - \lambda_j} t^{j - i}) f(q^{\mu_i - \mu_{j + 1}} t^{j - i}) \over f(q^{\lambda_i - \mu_j} t^{j - i}) f(q^{\mu_i - \lambda_{j + 1}} t^{j - i})} \\ &= \prod_{1 \le i \le j \le \ell(\lambda)} {g(\lambda_i - \lambda_j, j - i) g(\mu_i - \mu_{j + 1}, j - i) \over g(\lambda_i - \mu_j, j - i) g(\mu_i - \lambda_{j + 1}, j - i)}. \qquad (2) \end{align}

And thus

$\phi_T = \prod_{1 \le i \le j \le k \le \ell} {(q^{\lambda^k_i - \lambda^k_j} t^{j - i + 1})_{\lambda^k_j - \lambda^{k - 1}_j} (q^{\lambda^k_i - \lambda^k_{j + 1} + 1} t^{j - i})_{\lambda^k_{j + 1} - \lambda^{k - 1}_{j + 1}} \over (q^{\lambda^k_i - \lambda^k_j + 1} t^{j - i})_{\lambda^k_j - \lambda^{k - 1}_j} (q^{\lambda^k_i - \lambda^k_{j + 1}} t^{j - i + 1})_{\lambda^k_{j + 1} -\lambda^{k - 1}_{j + 1}}}.$

## References

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