# The Cali Garmo

does Math

## Schur Functions

By Cali G , Published on Tue 14 January 2020
Category: math / symmetric functions

## Background

Let's talk about Schur functions: a nice basis for the ring of symmetric functions. Recall that a Young tableau with shape $\lambda$, is a Young diagaram with shape $\lambda$ where each box is filled in with some number.

Definitions: A weak composition $\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_k, \ldots)$ of $n \in \N$ is a set of numbers such that $\alpha_i \in \Z_{\geq 0}$, $\alpha_{j} = 0$ for $j > k$ and $\sum \alpha_i = n$. A Young tableau $T_{\alpha, \lambda}$ with content $\alpha$ and shape $\lambda$ is a Young diagram of shape $\lambda$ whose boxes are filled in with $\alpha$.

As an example, let $\alpha = (1,2,1)$ be a weak composition of $4$ and $\lambda = (2,2)$ be a partition of $4$. To construct the (only) Young tableau with content $(1,2,1)$ and shape $(2,2)$ we first draw the Young diagram of shape $(2,2)$ and then we fill in the boxes with the number $\left\{1,2,2,3\right\}$ coming from $(1,2,1) = 1^12^23^1$ in such a way that the rows are weakly increasing and the columns are strictly increasing. The only tableau that works is the following:

Definitions: For each partition $\lambda$ we can associate what is called a Schur function. The Schur function of $\lambda \vdash n$ is given by: where $\alpha$ is a weak composition of $n$.

As an example, let us find $s_{(2,2)}$. In particular, we will first only look at weak compositions of $4$ which are contained in the set: Since $\lambda = (2,2)$ all of our tableau must have shape: In fact, we end up having the following tableau for each of the following weak partitions above:

This implies that we have the following (in order) for the Schur function:

Notice that if we continue with all possible weak compositions then we would have a symmetric function. For example, we would have $x_1^2x_3^2$ and $x_2^2x_3^2$ by using the weak compositions $(2,0,2)$ and $(0,2,2)$ respectively.

Notice that the coefficients of the monomials are given by the Kostka numbers.

## Relationship with other bases

### Jacobi-Trudi identity

The first relationship we will look at is with complete homogeneous symmetric functions $\left\{h_\lambda\right\}$. We call this identity the Jacobi-Trudi identity: where $h_0 = 1$ and $h_k = 0$ whenever $k < 0$.

This is best seen in an example. Let $\lambda = (4,3,3,1)$. Then $\ell(\lambda) = 4$ since there are $4$ entries in $\lambda$ and we construct the $4 \times 4$ matrix as defined by above:

Next we take the determinant of this matrix.

The key thing to notice here is that our partition $\lambda = (4,3,3,1)$ appears on the diagonal and then we just add $1$ each time we go to the right and subtract $1$ each time we go to the left.

### Pieri Rule

There are two other relationships with other bases that we will discuss next. In particular, we will look at what happens when we multiple a Schur function with either a complete homogeneous symmetric function or a elementary symmetric function. These two rules are known as the Pieri rules. Before we define the following two sets for ease of notation: Here, horizontal strip means that each column in the Young diagram has at most one block and similarly vertical strip means that each row in the Young diagram has at most one block.

Then our two Pieri rules are:

Let's look at an example for the Pieri rules. Suppose that $\lambda = (2,1)$ and $k = 3$. Recall that the Young diagram for $\lambda$ is the following:

We first look at $s_\lambda h_k$. Since we want to find all $\gamma$ such that:

• $\lambda$ is contained in $\gamma$ ($\lambda \subseteq \gamma$),
• we only add $k = 3$ new blocks ($\abs{\gamma / \lambda} = k$), and
• no column as more than one new block added ($\gamma / \lambda$ is a horizontal strip)

This gives us exactly four different possible Young diagrams (the green boxes are the new boxes): By the Pieri rule we have:

Now let's look at $s_\lambda e_k$. Recall that we want our $\gamma$ to be almost exactly the same as in $s_\lambda h_k$ but instead of horizontal strips we want vertical strips. So no row should get more than one new block added. This also (by chance) gives us exactly four new tableau: By the Pieri rule we have:

### Littlewood-Richardson rule

In the Pieri rule we multiplied a Schur function by one of the other symmetric function bases. What happens when we multiply it by another Schur function? This is given by the Littlewood-Richardson rule. Let $\lambda$ and $\beta$ be two partitions. Then the Littlewood-Richardson rule states: where $c_{\lambda, \beta}^\gamma$ is the number of semi-standard Young tableau of (skew) shape $\gamma / \beta$ with weight $\lambda$ such that the reading word $\sigma_1 \ldots \sigma_k$ of $\gamma/\beta$ has the property that any initial segment has at least as many $i$'s as $i+1$'s where the reading word is read from right to left and bottom to top. Such a tableau is often called a Littlewood-Richardson tableau.

Let's look at an example. Let $\lambda = (2,1)$ and let $\beta = (1)$. Then $\gamma$ are going to be all partitions of $\abs{\lambda} + \abs{\beta} = 3 + 1 = 4$. This gives us the following possible Young diagrams:

We have five different possible diagrams. In particular we have: so it only remains to find each of the coefficients.

Let's first look at $\gamma_1 = (4)$. Recall that this it the number of semi-standard Young tableau of skew shape $\gamma_1 / \beta = (4)/(1)$ whose weight is $\lambda = (2,1) = 1^22$ such that the reading word has a particular property. There is only one way we can fill in this diagram: Recall that reading word is read from right to left and bottom to top, so we have as a reading word $211$. But notice that the initial section $2$ has one $2$ and zero $1$s which means this is not a valid tableau! So $c_{(2,1),(1)}^{(4)} = 0$.

Next we look at $\gamma_2 = (3,1)$. In this case, there are two different ways we can fill in the skew diagram $(3,1)/(1)$ in order to get a (skew) semi-standard Young tableau: The reading word (right to left, bottom to top) of the left tableau is: $112$ and the reading word of the right tableau is $211$. Just as in the previous example, $211$ is not a valid reading word(!), but $112$ is. Notice that the initial section $1$ is ok, $11$ is also ok, and $112$ is ok since when we hit the first $2$ we already have two $1$s. Therefore $c_{(2,1),(1)}^{(3,1)} = 1$.

Continuing, we look at $\gamma_3 = (2,2)$. In this case, there is only one way to fill in the skew diagram $(2,2)/(1)$ to get a (skew) semi-standard Young tableau: The reading word for this tableau is $121$ which is a valid reading word since the initial section $1$ is ok, $12$ is ok since there is one $1$ and one $2$, and $121$ is ok since we are adding another $1$. Therefore, $c_{(2,1),(1)}^{(2,2)} = 1$.

Let's next look at $\gamma_4 = (2,1,1)$. As in the $\gamma_3$ case, there is only one way to fill in the skew diagram $(2,1,1)/(1)$ to get a (skew) semi-standard Young tableau: This gives us a reading word of $112$ which is valid. Therefore, $c_{(2,1),(1)}^{(2,1,1)} = 1$.

Finally, let's consider $\gamma_5 = (1,1,1,1)$. Notice that there is no way to fill in the (skew) diagram $(1,1,1,1)/(1)$ with two $1$s since we would have to place one $1$ above another and the columns must be strictly increasing. Therefore, $c_{(2,1),(1)}^{(1,1,1,1)}= 0$.

This implies (finally) that

Although this computation takes a while to compute by hand, you can use sagemath (which uses Anders Buch's Littlewood-Richardson Calculator) in order to calculate these numbers quickly and efficiently without all of the above work. Documentation for this calculator on sagemath are found here.