I am reading the appendix of Fulton and Harris pg. 459 and am trying to understand the following setup. Suppose $\lambda : \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_k \geq 0$ is a partition of a positive integer $d$ into at most $k$ parts. Suppose $P$ is a symmetric polynomial of degree $d$ in $k$ variables. Write

$$\omega_\lambda(P) = [\Delta \cdot P]_l$$

where $\Delta$ is the discriminant $\prod_{1 \leq i < j \leq k} (x_i - x_j)$ and $[\Delta\cdot P]_l$ denotes the coefficient of the monomial $X^l = x_1^{l_1}x_2^{l_2} \ldots x_k^{l_k}$ in $\Delta\cdot P$. The tuple $l = (l_1,\ldots,l_k)$ has entries

$$l_1 = \lambda_1 + k-1, \hspace{2mm} l_2 =\lambda_2 + k -2,\hspace{2mm}, \ldots \hspace{2mm}, l_k = \lambda_k.$$

Now Fulton and Harris claim the identity

$$P = \sum_{\lambda \hspace{2mm} \text{a partition of $d$ into at most $k$ parts}} \omega_\lambda(P) s_\lambda$$

where $s_\lambda$ is the Schur polynomial

$$s_\lambda = \frac{ \left|\begin{array}{cccc}x_1^{l_1} & x_2^{l_1} &\ldots & x_k^{l_1} \\ x_1^{l_2} & x_2^{l_2} &\ldots & x_k^{l_2} \\ &&\vdots\\ x_1^{l_k} & x_2^{l_k} & \ldots & x_k^{l_k}\end{array}\right|}{\Delta}.$$

Why should this be the case? They claim it is easy to see but I have been starring at this for several days now and don't understand how this claim comes. Am I missing something or is this another proof in Fulton and Harris that they just glance over?



1 Answer 1


As the Wikipedia article you link to states, the Schur polynomials of degree $d$ in $k$ variables form a basis of the space of symmetric polynomials of degree $d$ in $k$ variables. Thus we can write $P=\sum_\lambda c_\lambda s_\lambda$. Then multiplying by $\Delta$ yields $\Delta\cdot P=\sum_\lambda c_\lambda \Delta\cdot s_\lambda=\sum_\lambda c_\lambda d_\lambda$, where $d_\lambda$ is the determinant in your last equation. For every $\lambda$, only the corresponding determinant contains the corresponding monomial $X^l$, and the coefficient of $X^l$ in it is $+1$, so $[\Delta\cdot P]_l=c_\lambda$. (This would be clearer if the connection between $\lambda$ and $l$ were reflected in the notation.)

  • $\begingroup$ Right. I guess I know now where I was confused as all the $\lambda's$ in that sum got me confused. I guess it must be the case that given a $\lambda$ and its corresponding determinant $d_\lambda$, for no other $\lambda'$ does the monomial $X^l$ appear in $d_{\lambda'}$. $\endgroup$
    – user38268
    Commented Oct 18, 2012 at 11:45

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