Brian Bi

Remark: I had some trouble understanding why the expression given for $$\theta_\lambda$$ in the proof of Theorem 5.15.3 follows, as Etingof says, from Theorem 5.14.3. In fact, it follows from Theorem 5.14.3 and the following well-known result: $\Delta(x) = \sum_{\sigma \in S_N} \sgn(\sigma) \, x^{\sigma(\rho)}$ This identity is not difficult to prove. See here.

Next, Etingof explains why $$\mu = \lambda + \rho - \sigma(\rho)$$ is always a larger partition (in the lexicographic ordering introduced in section 5.13) than $$\lambda$$ except when $$\sigma$$ is the identity. It's not immediately clear to me why his statement about the $$e_i - e_j$$ vectors (with $$i < j$$) is true, but it's clear enough to me that the former claim is true. Since $$\lambda_1 = \max_i(\lambda_i)$$, any partition with any part greater than $$\lambda_1$$ will be larger than $$\lambda$$. But this happens whenever $$\sigma$$ doesn't fix the first element, as the first element of $$\rho - \sigma(\rho)$$ will then be positive, as it is $$N - 1$$ minus some smaller integer. If $$\sigma$$ fixes its first element but not its second, then by the same logic, $$\mu$$ will compare larger than $$\lambda$$ when a part with value $$\lambda_1$$ is removed from both of them; but that implies $$\mu > \lambda$$ when we compare them in their entirety. If $$\sigma$$ fixes the first two elements but not the third, then $$\mu$$ will compare greater than $$\lambda$$ when we compare them with a part of value $$\lambda_1$$ and a part of value $$\lambda_2$$ is removed from both; which likewise implies $$\mu > \lambda$$ when they are compared in their entirety. Continuing this logic, we get $$\mu > \lambda$$ whenever $$\sigma$$ is not the identity.

In the expression for $$S(x, y)$$, note that $$H_m(x)$$ has been written out explicitly for the sake of the algebraic manipulations that follow. Thus, the intimidating-looking expression really just follows from the definition of $$\theta_\lambda$$ and the preceding results after rearranging some factors.

To prove Corollary 5.15.4, we use the fact that multiplying any row of a matrix by a constant factor also multiplies the determinant by the same factor, thus: $\det \left[\frac{1}{1 - x_i y_j}\right]_{ij} = \left(\prod_i x_i^{-1}\right) \det \left[\frac{x_i}{1 - x_i y_j}\right]_{ij} = \left(\prod_i x_i^{-1}\right) \det \left[\frac{1}{z_i - y_j}\right]_{ij}$ The remaining algebraic manipulations after applying Lemma 5.15.3 are straightforward. $$N$$ factors of each $$x_i^{-1}$$ will be pulled out from the denominator and $$N-1$$ from the numerator, so all such factors cancel.

Finally, let us consider the obvious leap from Corollary 5.15.4 to the conclusion that the coefficient of $$x^{\lambda + \rho} y^{\lambda + \rho}$$ is 1. The coefficient will be obtained by summing over the coefficient of said monomial in $$\prod_j (1 - x_j y_{\sigma(j)})^{-1}$$ as $$\sigma$$ ranges over $$S_N$$. Let us write out the infinite series in each factor explicitly: $\prod_j (1 - x_j y_{\sigma(j)})^{-1} = (1 + x_1 y_{\sigma(1)} + x_1^2 y_{\sigma(1)}^2 + \ldots) \times (1 + x_2 y_{\sigma(2)} + x_2^2 y_{\sigma(2)}^2 + \ldots) \times \ldots \times (1 + x_N y_{\sigma(N)} + x_N^2 y_{\sigma(N)}^2 + \ldots)$ The only way to get a monomial of the form $$x^{\lambda + \rho} y^q$$ is to take the term with $$x$$-degree $$\lambda_1 + \rho_1$$ from the first factor, the term with $$x$$-degree $$\lambda_2 + \rho_2$$ from the second factor, and so on. The product of all such terms is $x^{\lambda + \rho} \prod_i y_{\sigma(i)}^{\lambda_i + \rho_i}$ Thus, in order for this to equal $$x^{\lambda + \rho} y^{\lambda + \rho}$$ we must have $$\lambda_i + \rho_i = \lambda_{\sigma(i)} + \rho_{\sigma(i)}$$ for all $$i \in \{1, \ldots, N\}$$. If $$\sigma$$ satisfies this property, the coefficient of $$x^{\lambda + \rho} y^{\lambda + \rho}$$ is 1; otherwise it is 0. But note that whenever $$i > \sigma(i)$$, we have $$\lambda_i \ge \lambda_{\sigma(i)}$$ and $$\rho_i > \rho_{\sigma(i)}$$, so that $$\lambda_i + \rho_i > \lambda_{\sigma(i)} + \rho_{\sigma(i)}$$. There will always be such $$i$$ for any $$\sigma$$ other than the identity. When $$\sigma$$ is the identity, the required property evidently holds. Thus, just as Etingof says, the coefficient of $$x^{\lambda + \rho} y^{\lambda + \rho}$$ is 1 for $$\sigma = 1$$ and 0 for all other $$\sigma$$.