Derivative of Schur function

Question

In his answer to https://mathoverflow.net/questions/129854, R. Stanley says that the partial derivative (over the relevant x[i]) of the Schur function of a partition lambda of n equals the sum the Schur functions over the trimmed partions of lambda:

"Let $s_λ$ denote a Schur function and $p1=s1=x1+x2+\dots$. Then, $\frac{\partial\lambda}{\partial p_1}=\sum_\mu s_\mu$, where the $μ$'s are obtained by removing a single box from $λ$."

If I work this out for $\lambda = \{3,1\}$, I find for the derivative $12 X[3]+ 108 X[2,1]+ 60 X[1,1,1]$ (substituting $X[i,j,k]$ for terms $x[1]^i*x[2]^j*x[3]^k$) while the trimmed partitions $\{2,1\}$ and $\{3\}$ produce $4 X[3]+ 24 X[2,1] + 12 X[1,1,1]$.

What am I missing? Can anyone provide an example of what is intended?

Answer 1

My understanding of Stanley's answer on MathOverflow is that:

• The schur functions are taken to be generating functions in infinitely many variables;
• In order to differentiate wrt $p_1$, one must first write $s_\lambda$ as a generating function in the power sum symmetric polynomials as $s_\lambda=f(p_1,p_2,\cdots)$.

Let's do a simpler example, say $\lambda=(2,1)$. The semistandard tableaux of shape $\lambda$ will come in a number of possible forms; we split them up into types for all naturals with $a<b<c$:

$$\color{blue}{ \begin{matrix} a & b \\ b\end{matrix} \qquad \begin{matrix} a & a \\ b\end{matrix}} \qquad \color{green}{ \begin{matrix} a & b \\ c\end{matrix} \qquad \begin{matrix} a & c \\ b\end{matrix}}$$

Thus, the schur function is $s_{(2,1)}=\color{blue}{m_{(2,1)}}+\color{green}{2m_{(1,1,1)}}$. Since $p_1^3=p_3+3m_{(2,1)}+6m_{(1,1,1)}$, we may write $s_{(2,1)}=(p_1^3-p_3)/3$, hence $\partial s_{(2,1)}/\partial p_1=p_1^2=p_2+2m_{(1,1)}$. Now the semistandard tableaux of shape $\color{purple}{(2)}$ and $\color{teal}{(1,1)}$ are all of the form (demarcated with a $|$)

$$\left. \color{purple}{\begin{matrix}a \\ b\end{matrix}} \quad \right| \quad \color{Teal}{\begin{matrix}a& b\end{matrix} \qquad \begin{matrix}a & a\end{matrix}} $$

Hence $\color{purple}{s_{(2)}=m_{(1,1)}}$ and $\color{teal}{s_{(1,1)}=m_{(1,1)}+p_2}$. Therefore

$$\frac{\partial s_{(2,1)}}{\partial p_1}=p_2+2m_{(1,1)}=s_{(1,1)}+s_{(2)},$$

exactly as claimed.

This should hold for schur functions of finitely many variables; simply apply the "evaluate $x_l=0$ for all $l>n$" ring homomorphism to the relations $s_\lambda = f(p_1,p_2,\cdots)$ and $\partial f/\partial p_1=\sum s_\mu$.

---

However, I am getting a missing or extraneous term in my own computation involving $s_{(3,1)}$. The three semistandard tableaux of shape $(3,1)$ are

$$\begin{matrix}1 & 1 \\ 2 \\ 3\end{matrix} \qquad \begin{matrix}1 & 2 \\ 2 \\ 3\end{matrix} \qquad \begin{matrix}1 & 3 \\ 2 \\ 3\end{matrix}$$

hence

$$s_{(3,1)}(x_1,x_2,x_3)=x_1^2x_2x_3+x_1x_2^2x_3+x_1x_2x_3^2=x_1x_2x_3(x_1+x_2+x_3).$$

Consulting the NG formulae and the recursions they yield, this should be equal to

$$\frac{p_1^3-3p_1p_2+2p_3}{6}p_1$$

hence

$$\frac{\partial s_{(3,1)}}{\partial p_1}=\frac{4p_1^3-6p_1p_2+2p_3}{6}=x_1^2x_2+x_1x_2^2+x_1^2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2+4x_1x_2x_3.$$

The only semistandard tableau of shape $(3)$ is $\begin{matrix}1\\2\\3\end{matrix}$, hence $s_{(3)}(x_1,x_2,x_3)=x_1x_2x_3$.

The semistandard tableaux of shape $(2,1)$ they are

$$\begin{matrix} 1 & 1 \\ 2\end{matrix} \qquad \begin{matrix} 1 & 1 \\ 3\end{matrix} \qquad \begin{matrix} 1 & 2 \\ 2\end{matrix} \qquad \begin{matrix} 1 & 2 \\ 3\end{matrix} \qquad \begin{matrix} 1 & 3 \\ 2\end{matrix} \qquad \begin{matrix} 1 & 3 \\ 3\end{matrix} \qquad \begin{matrix} 2 & 2 \\ 3\end{matrix} \qquad \begin{matrix} 2 & 3 \\ 3\end{matrix} $$

hence

$$s_{(2,1)}(x_1,x_2,x_3)=x_1^2x_2+x_1x_2^2+x_1^2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2+2x_1x_2x_3.$$

But this would indicate that $\partial s_{(3,1)}/\partial p_1=s_{(2,1)}+s_{(3)}\color{red}{+x_1x_2x_3}$, so I must have made a mistake somewhere. Perhaps someone else can see where my error is.

Answer 2

Stanley uses a cryptic description whose interpretation is easy for the initiated, but not for others. Remember that the ring of symmetric functions is equipped with an inner product for which the Schur functions form an orthonormal basis (you can take that as its definition, although there are also other descriptions). Now the operation written $\frac\partial{\partial p_1}$ is just the adjoint (in the sense of linear mappings) of the operation of multiplication by $p_1=s_1=\sum_i x_i$. Since that multiplication applied to $s_\lambda$ gives the sum of all $s_\mu$ where the diagram of$~\mu$ can be obtained by adding a square to that of$~\lambda$, its adjoint applied to $s_\lambda$ gives the sum of all $s_\mu$ where the diagram of$~\mu$ can be obtained by removing a square from that of$~\lambda$.

A way to view this operation as applied to any symmetric function $f$ is to single out one of the variables in which $f$ is a symmetric function (since it is symmetric it doesn't matter which one), call that variable $y$ and the others $x_1,x_2,\ldots$; observing that $f$ is certainly symmetric in the $x_1,x_2,\ldots$, write it as a polynomial in $y$ with as coefficients symmetric functions in the $x_1,x_2,\ldots$, and then take the coefficient of the first power $y^1$ of$~y$ as outcome (we've used up one variable, but given the infinite supply this makes no difference; this is the Hilbert hotel in operation). In practice one doesn't need infinitely many variables: one can write $f$ as symmetric polynomial in variables $x_1,\ldots,x_n$ where $n$ is at least $\deg f$, and choose $y=x_n$ as "ultimus inter pares" which avoids renumbering (the degree conveniently drops by $1$, so $x_n$ is expendable).

If one takes the description of $s_\lambda(x_1,\ldots,x_n)$ in terms of semistandard tableaux, one can partition the set of its terms according to the set of squares containing an entry$~n$ in the tableau. The parts of this partition of terms each define a product $s_\mu(x_1,\ldots,x_{n-1})x_n^d$, where $d$ is the number of squares in the fixed set containing an entry$~n$, and the diagram of $\mu$ consists of the remaining squares. It follows that the coefficient of $x_n^d$ is the sum of all such $s_\mu(x_1,\ldots,x_{n-1})$, where the diagram of$~\mu$ is obtained from that of$~\lambda$ by removing a horizontal strip of size$~d$. In particular, our operator maps $s_\lambda$ to the sum of all $s_\mu$ whose diagram is obtained from that of $\lambda$ by removing one corner.

To see that the operator coincides with formal differentiation with respect to $p_1$ when applied to a polynomial in $p_1,p_2,\ldots$, interpreted as power sum symmetric functions, one can either prove that it is a derviation which vanishes on $p_2,p_3,\ldots$ and sends $p_1\mapsto1$, or one can directly apply our description to a product $p_1^mf$ where $f$ is a factor on which the operator vanishes (a monomial in $p_2,p_3,\ldots$), using $p_1(x_1,\ldots,x_n)=p_1(x_1,\ldots,x_{n-1})+x_n$ and expansion of $(p_1(x_1,\ldots,x_{n-1})+x_n)^m$, giving as result of the operation $mp_1^{m-1}f$.