The Cali Garmo

does Math

Schur Functions

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


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 1s 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 1s. 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 1s 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.