# Help with an expression of the Schur polynomial

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

The Schur polynomial is defined as $$s_\lambda(x_1,...,x_n)=\prod_{1\le i And this can also be phrased as $$s_\lambda(x_1,...,x_n)=\sum_{\sigma\in S_n}\sigma\left(\frac{x^\lambda}{\prod_{i 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?

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 $$\sum_{\sigma \in S_n} \sigma \bigg(\dfrac{x^\lambda}{\prod_{i.
Now $$\Delta= \prod_{i 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 $$(x_j^{\lambda_i+n-i})$$.
• Thank you! I didn't realize that the $\sigma$ also permutes the denominator.