I am reading: https://www2.math.upenn.edu/~peal/polynomials/schur.htm where it says:

The Schur polynomial is defined as $$s_\lambda(x_1,...,x_n)=\prod_{1\le i<j\le n}(x_i-x_j)^{-1}\det(x_j^{\lambda_i+n-i})$$ And this can also be phrased as $$ s_\lambda(x_1,...,x_n)=\sum_{\sigma\in S_n}\sigma\left(\frac{x^\lambda}{\prod_{i<j}1-x_j/x_i}\right) $$ where $\sigma$ act by permuting the indices of the variables.

I am confused with the second expression of the Schur polynomial. For example, if we are given $n=2$ and $\lambda=(3,1)$, then we have $$ s_{(3,1)}(x_1,x_2)=\frac{1}{x_1-x_2}\det\begin{bmatrix}x_1^4&x_2^4\\x_1&x_2\end{bmatrix}=\frac{x_1^3x_2-x_2^4}{1-x_2/x_1} .$$ But this obviously does not agree with the second expression since the second term in the numerator of the last fraction above has partition type $(4)$ which is not $(3,1)$. What's wrong with my understanding? And how to prove the second expression is equal to the first expression?


1 Answer 1


The second expression gives $s_\lambda(x)=\sum_{w\in S_2}\dfrac{x_1^3x_2}{1-x_2/x_1}$, so $s_\lambda(x)=\dfrac{x_1^3x_2}{1-x_2/x_1}+\dfrac{x_2^3x_1}{1-x_1/x_2}=\dfrac{x_1^4x_2-x_1x_2^4}{x_1-x_2}$ which is what you got from the first formula.

To see the equivalence of the two definitions: Note that $$ \sum_{\sigma \in S_n} \sigma \bigg(\dfrac{x^\lambda}{\prod_{i<j} 1- x_j/x_i}\bigg)=\sum_{\sigma \in S_n} \sigma \bigg(\dfrac{x^\lambda x_1^{n-1}x_2^{n-2}...x_{n-1}}{\prod_{i<j} (x_i- x_j)} \bigg)$$.

Now $\Delta= \prod_{i<j} (x_i- x_j)$ is skew-symmetric, i.e, $w \Delta = \epsilon(w) \Delta$ for $w \in S_n$. Therefore, the right hand side of the above expression simplifies to $$ \Delta^{-1} \sum_{\sigma \in S_n} \epsilon (\sigma) (x^{\lambda+\rho}) $$ where $\rho=(n-1,n-2,...,1,0)$.

Now I leave you to check that the sum in the above expression is the determinant of the matrix $(x_j^{\lambda_i+n-i})$.

  • 1
    $\begingroup$ Thank you! I didn't realize that the $\sigma$ also permutes the denominator. $\endgroup$
    – Bach
    Commented Feb 8, 2021 at 17:40

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