
# Background on Symmetric Functions

### General Background

Here we will be giving a general background on the ring of symmetric functions. We start by letting $$n$$ be an integer. A partition $$\lambda$$ of $$n$$, which is written as $$\lambda \vdash n$$ is a weakly decreasing sequence with values in $$\mathbb{Z}_{\geq 0}$$ whose sum is $$n$$. A weak composition $$\alpha$$ of $$n$$ is a sequence with values in $$\mathbb{Z}_{\geq 0}$$ whose sum is $$n$$. We will let $$x^{\alpha} = x_1^{\alpha_1} x_2^{\alpha_2} \cdots$$ and we will consider the formal power series $$\Lambda = \mathbb{Q}[[x_1, x_2, \ldots]]$$ which are invariant under $$S_{+\infty}$$, the group of permutations of $$\{1, 2, 3, \ldots\}$$, where $$\sigma \in S_{+\infty}$$ acts by $$\sigma x_i = x_{\sigma(i)}$$.  The ring (algebra) $$\Lambda$$ is known as the ring (algebra) of symmetric functions. We can consider this as the direct limit of polynomials in $$k$$ variables that are invariant under $$S_k$$, the symmetric group on $$k$$ letters, where $$\sigma \in S_k$$ acts by $$\sigma x_i = x_{\sigma(i)}$$.

Partitions can be represented as diagrams where we have $$\lambda_i$$ boxes left-aligned where the top row corresponds to $$\lambda_1$$. This diagram is called a Young diagram or Ferrers' diagram in English convention. Another notation often seen is for the bottom row to correspond to $$\lambda_1$$ and this is known as French convention. Partitions have a natural partial ordering given by containment of the Young diagrams, or more formally, we have $$\lambda \subseteq \mu$$ if $$\lambda_i \leq \mu_i$$ for all $$i$$. However the partial ordering we will be predominantly using is known as dominance order, where $$\lambda \leq \mu$$ if $$\sum_{j=1}^i \lambda_j \leq \sum_{j=1}^i \mu_j$$ for all $$i$$.

We also need to define a few more things for partitions. We denote $$|\lambda| = n$$ as the size of $$\lambda$$ and $$\ell(\lambda)$$ is number of non-zero entries of $$\lambda$$ which we call the length. Let $$m_i(\lambda)$$ be the number of parts of $$i$$ in $$\lambda$$, and if the partition is clear, we will simply write this as $$m_i$$.

### Monomial basis

It is straightforward to see that the following is a basis for $$\Lambda$$: $m_{\lambda} = \sum_{\alpha} x^{\alpha}$ where we sum over all distinct permutations $$\alpha$$ of $$\lambda$$. Now there is a natural grading on polynomials given by the degree, and since $$m_{\lambda}$$ is homogeneous of degree $$n = |\lambda|$$. Therefore this is a graded basis of $$\Lambda$$, which means that for $$m_{\lambda} \in \Lambda^n$$ and $$m_{\mu} \in \Lambda^m$$, we have $$m_{\lambda} m_{\mu} \in \Lambda^{n+m}$$.

### Elementary basis

We now define a new basis, the elementary basis, by $e_k = \sum_{i_1 < i_2 < \cdots < i_k} x_{i_1} x_{i_2} \cdots x_{i_k} \\ e_{\lambda} = e_{\lambda_1} e_{\lambda_2} \cdots.$ We note that $$e_k = m_{1^k}$$. From this definition, it is not clear that this forms a basis. However we can show that the transition map is upper unitriangular with respect to dominance order, and so this really is a basis.

### (Complete) Homogeneous basis

Our next basis, the (complete) homogeneous basis, is defined similar to the elementary basis: $h_k = \sum_{i_1 \leq i_2 \leq \cdots \leq i_k} x_{i_1} x_{i_2} \cdots x_{i_k} \\ h_{\lambda} = h_{\lambda_1} h_{\lambda_2} \cdots.$ We note that $$h_k = \sum_{\mu \vdash k} m_{\mu}$$. However the transition map to the monomial symmetric functions is not as nice as above, so we have to define an algebra morphism $$\omega \colon \Lambda \to \Lambda$$ by $$\omega(e_k) = h_k$$. Using generating functions for the elementary and homogeneous basis, one can show that $$\omega(h_k) = e_k$$ and hence $$\omega$$ is an involution. Therefore, since the elementary symmetric functions are a basis, so are the homogeneous symmetric functions.

### Powersum Basis

Our last natural choice of basis is the powersum basis and is given by

$p_k = \sum_i x_i^k \\ p_{\lambda} = p_{\lambda_1} p_{\lambda_2} \cdots.$

We note that $$p_k = m_{k}$$. While the transition map is upper triangular with respect to dominance order, it is not unitriangular. We can use the powersum basis to define an inner product on $$\Lambda$$ by

$\langle p_{\lambda}, p_{\mu} \rangle = \delta_{\lambda\mu} z_{\lambda}$

where $$z_{\lambda} = m_1! 1^{\lambda_1} m_2! 2^{\lambda_2} \cdots$$ (which is the size of the conjugacy class of cycle type $$\lambda$$ in $$S_k$$). So this does form a basis, but in a way, it is not a particularly nice basis.

A side remark, the inner product defined above is known as the Hall inner product and implies that

$\langle m_{\lambda}, h_{\mu} \rangle = \delta_{\lambda\mu}.$

### Schur functions

Since we will be working in an infinite number of variables, we will be using the definition using semistandard Young tableaux. We start by considering a partition $$\lambda$$, and now we will fill $$\lambda$$ with entries in $$\{1, 2, \ldots\}$$ such that the following conditions hold:

1. it is weakly increasing across rows,
2. it is strictly increasing down columns.

Such a filling $$T$$ is called a semistandard Young tableau. We define the weight of $$T$$ as the weak composition $$\alpha = \mathrm{wt}(T)$$ by $$\alpha_i$$ is the number of boxes filled with an $$i$$. We denote the set of all semistandard (Young) tableaux of shape $$\lambda$$ by $$SSYT(\lambda)$$ and for a fixed weight by $$SSYT(\lambda, \alpha)$$.

Next we will denote $$x^T = x^{\mathrm{wt}(T)} = x_1^{\alpha_1} x_2^{\alpha_2} \cdots$$, and so we can define the Schur function by $s_{\lambda} = \sum_{T \in SSYT(\lambda)} x^T.$ We want to show that this is a symmetric function, and we must extend the action of $$S_{+\infty}$$ to semistandard tableaux. We do so by defining the action of the simple generator $$s_i$$ by finding all partial rows that looks like $$(i, \ldots, i, i+1, \ldots, i+1)$$ where any $$i$$ does not have an $$i+1$$ below it and the any $$i+1$$ does not have an $$i$$ above it. Let $$r, s$$ denote the number of $$i, i+1$$ respectively, and $$s_i$$ acts by interchanging the number of $$i, i+1$$ occurring in each partial row. This is a valid action of $$S_{+\infty}$$ on $$SSYT(\lambda)$$, and so the Schur functions are symmetric functions.

Schur functions exhibit many remarkable properties. The first is that this is a basis, which can be seen by $s_{\lambda} = \sum_{\mu \vdash |\lambda|} K_{\lambda \mu} m_{\mu}$ where $$K_{\lambda \mu} = |SSYT(\lambda, \mu)|$$ and is known as a Kostka number. It is easy to see that $$K_{\lambda \mu} = 0$$ if $$\lambda \not\leq \mu$$ and $$K_{\lambda \lambda} = 1$$. The next is that from the RSK (a bijection from SSYT to integer matrices) and dual RSK algorithms, which hold many interesting properties within themselves, one can deduce that $$\langle s_{\lambda}, s_{\mu} \rangle = \delta_{\lambda \mu}$$ and $$\omega s_{\lambda} = s_{\lambda^t}$$ where $$\lambda^t$$ is the transpose partition (reflect the Young diagram along the main diagonal).

Schur functions​

There is another basis, called the schur functions, with strong connections to combinatorics and representation theory.  We give several equivalent definitions:

Tableaux
We first give a purely combinatorial definition

$s_{\lambda} = \displaystyle \sum_{T} x^{cont(T)}$

where the sum is over all Semi-Standard Young Tableaux of shape $$\lambda$$ and $$cont(T)$$ is the monomial representing its content. An example in 3 variables is

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

corresponding to the tableaux

$$[11,2], [11,3], [12,3], [13,2], [12,2], [13,3], [22,3], [23,3]$$

Here $$[ab,c]$$ its just a one line notation for the more familiar

$$a b$$

$$c$$

Note in the example that the schur function is indeed symmetric, a fact which is not direct from its definition. In general, the symmetry of schur functions requires a proof, not hard, and once we know it, we can write

$s_\lambda= \sum_\mu K_{\lambda\mu}m_\mu.\$

with the following natural interpretation $$K_{\lambda\mu}$$ is the number of Semi-Standard Young Tableaux (SSYT for short) with shape  $$\lambda$$ and content $$\mu$$. A careful analysis reveals that $$K_{\lambda\mu}$$ is nonzero if and only if $$\lambda \geq \mu$$. Even more $$K_{\lambda\lambda} = 1$$, with the only tableaux being the one with all 1's in the first row, 2's in the second, and so on. So in fact we have

$s_\lambda= \sum_{\lambda\geq\mu} K_{\lambda\mu}m_\mu.\$

Which means that the transition matrix will be lower unitriangular with respect to any order extending the dominance order. This shows that the schur functions are an integral basis for the ring of symmetric functions.

Remark: This is stronger that triangularity since being smaller in the linear order doesn't guarantees being smaller in dominance order. It is triangular, but several other positions are forced to be zero too.

One advantage of the combinatorial definition is that it makes sense also to define schur functions indexed by skew shapes.

Ratio of determinants

There is also a purely algebraic definition of $$s_{\lambda}$$

For any integer vector $$(l_1,l_2,\cdots, l_k)$$ define $$a_{(l_1,l_2,\cdots, l_k) } (x_1, x_2, \dots , x_k)$$ as the determinant of​

$\left[ \begin{matrix} x_1^{l_1} & x_2^{l_1} & \dots & x_k^{l_1} \\ x_1^{l_2} & x_2^{l_2} & \dots & x_k^{l_2} \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{l_k} & x_2^{l_k} & \dots & x_k^{l_k} \end{matrix} \right]$

Set $$\delta = (n-1,n-2,\cdots,1,0)$$ and note that $$a_{\delta}$$ is the Vandermonde determinant $$\v(x)$$. We have the alternative expression for the schur functions

$$s_{\lambda} = \dfrac{a_{\lambda+\delta}}{a_{\delta}}$$

Remark: This makes sense even if $$\lambda$$ is not a partition. In its useful to define $$s_{\upsilon}$$ this way for any integer vector. If nonzero, $$s_{\upsilon}$$ will be equal (up to sign) to $$s_{\lambda}$$ for a unique partition.

One common application, that we will use later, is the following: to find the coefficient of $$s_{\lambda}$$ in a symmetric function $$f(x)$$ we can just compute the coefficient of $$x^{\lambda+\delta}$$ in $$v(x)f(x)$$. The hook length formula can be proven this way (See http://en.wikipedia.org/wiki/Hook_length_formula)

More determinants

From the first definition we were able to expand schur in terms of the monomial basis. What about the other bases? The answer for the homogeneous (and elementary) basis is given by the Jacobi-Trudi determinant

$s_{\lambda} = \det(h_{\lambda_i-i+j})_{i,j=1}^n$

Applying the involution $$\omega$$ we get

$s_{\lambda} = \det(e_{\lambda'_i-i+j})_{i,j=1}^n$

Relation with the representation theory of the symmetric group

The remarkable connection between schur functions and the characters of the irreducible representations of the symmetric group is given by the following magical formula

$\chi^{\lambda}(w)=\langle s_{\lambda}, p_{\tau(w)}\rangle$

where $$\chi^{\lambda}(w)=$$ is the value of the character of the irreducible representation $$V_{\lambda}$$ in the element $$w$$ and $$p_{\tau(w)}$$ is the power symmetric function of the partition $$\tau(w)$$ associated to the cycle decomposition of $$w$$. For example, if $$w=(154)(238)(6)(79)$$ then $$\tau(w) = (3,3,2,1)$$.

Since $$e=(1)(2)\cdots(n)$$ in cycle notation, $$\tau(e) = 1+1+\cdots+1=1^{(n)}$$. Then the formula says

$\dim V_\lambda =\chi^\lambda(e) = \langle s_\lambda, p_{1^{(n)}}\rangle$

Considering the expansion of schur functions in terms on monomial symmetric functions using the Kostka numbers

$$s_\lambda= \sum_\mu K_{\lambda\mu}m_\mu.\$$

the inner product with $$p_{1^{(n)}} = h_{1^{(n)}}$$ is $$K_{\lambda 1^{(n)}}$$, because $$\langle m_{\mu},h_{\nu} \rangle = \delta_{\mu\nu}$$. Note that $$K_{\lambda 1^{(n)}}$$ is equal to $$f^{\lambda}$$, the number of Standard Young Tableaux of shape $$\lambda$$. Hence

$$\dim V_\lambda = f^{\lambda}$$

and

$$\langle s_\lambda, p_{1^{(n)}}\rangle = f^{\lambda}$$

which will be useful later

The magical formula is equivalent to

$s_{\lambda} = \dfrac{1}{n!} \sum_{w\in S_n} \chi^{\lambda}(w)p_{\tau(w)}$

This gives a conceptual proof of the identity $$\omega(s_{\lambda}) = s_{\lambda'}$$, by comparing the coefficients and taking into account that $$\chi^{\lambda}(w) = \textrm{sgn}(w) \chi^{\lambda'}(w)$$ because tensoring by the sign representation gives the irreducible representation for the conjugate partition.

Frobenius Series

The expansion in terms of the power symmetric functions suggest we define the following map The Frobenius Characteristic map F takes class functions on the symmetric group to symmetric function by sending $$\chi^{\lambda} \to s_{\lambda}$$ and extending by linearity. An important fact is that F is an isometry with respect to the inner products.

Remark F does not commute with multiplication

The formula

$$s_\lambda= \sum_\mu K_{\lambda\mu}m_\mu.\$$

is equivalent to

$$h_\mu= \sum_\mu K_{\lambda\mu}s_\lambda.\$$

And this also comes from representation theory. There is a module $$M^{\mu}$$ decomposition in $$V^{\lambda}$$ is given by the multiplicities $$K_{\lambda\mu}$$, and the above equation is simply the Frobenius translation. (See Sagan)

Let $$A = \bigoplus_r A_r$$ be a graded $$S_n$$ module, with each $$A_r$$ finite dimensional, define the Frobenius Series of A as

$F_A(z;t) = \sum_r t^r F\textrm{char}(A_r)$

where $$F\textrm{char}(A_r)$$ is the image of $$\textrm{char} (A_r)$$ under the Frobenius map.

And similarly if $$A = \bigoplus_{r,s} A_{r,s}$$ is a doubly graded $$S_n$$ module, with each $$A_{r,s}$$ finite dimensional, define the Frobenius Series of A as

$F_A(z;q,t) = \sum_r t^rq^s F\textrm{char}(A_r)$

It is clear that the Frobenius series expand positively in terms of schur functions, because the coefficients of schur come from the multiplicities (obviously nonnegative) of the irreducibles on each graded piece. The proof of the positivity conjecture for Macdonald Polynomials consist of finding a module whose Frobenius series is the desired symmetric function.

Characters of the General Linear group

Another way of thinking about schur functions is as the characters of the irreducible representations of $$GL(n)$$. Lets go through a simple example

The first nontrivial representation of $$GL(2)$$ that comes into mind is itself, which comes from the natural action on $$k^2$$, call this representation $$V$$. The character is just the trace, which is, as function of the eigenvalues, equal to $$x_1+x_2$$. What happens if we tensor $$V\otimes V$$? The character gets raised by two and we have the identity
$$(x_1+x_2)^2 = x_1^2+x_2^2 + 2x_1x_2 = x_1^2+x_2^2 + x_1x_2 + x_1x_2 = s_{2}(x_1,x_2) + s_{1,1}(x_1,x_2)$$
Since decomposing the characters gives the information to decompose the representation, the above identity says that $$V\otimes V$$ decomposes into two irreducibles one corresponding to the partition $$(2)$$ and other to the partition $$(1,1)$$. This are the symmetric and antisymmetric part, respectively.

More generally, consider $$GL(n)$$ and its defining representation $$V$$ given by the natural action on $$k^n$$. If we want to decompose $$V^{\otimes m}$$ into irreducibles we will need to write $$(x_1+x_2+\cdots+x_n)^m$$ in term of schur. The remarkable formula, in the crossroads between symmetric functions, representation theory and combinatorics is

$(x_1+x_2+\cdots+x_k)^n = \sum_{\lambda\vdash n} s_\lambda f^\lambda.$

Which is the expansion of $$p_{1^{(n)}}$$ in terms of schur functions using the coefficients given by the inner product, because $$\langle s_{\mu},s_{\nu}\rangle = \delta_{\mu\nu}$$ and $$\langle s_\lambda, p_{1^{(n)}}\rangle = f^{\lambda}$$. The above equality can be proven also checking the coefficients of each monomial at both sides and using the Robinson–Schensted–Knuth correspondence. For a more detailed analysis of the decomposition of  $$V^{\otimes m}$$ see Schur–Weyl duality.

In this context we can express the schur functions by using Weyl's Character Formula

$s_{\lambda}(x) = \sum_{w\in S_n} w\left(\dfrac{x^{\lambda}}{\Pi_{i<j}(1-x_j/x_i)}\right)$

which is equivalent to the ratio of determinants.

Raising operators

Check the Plethystic notation section. For a set of variables $$x = (x_1,x_2,\cdots)$$ define

$$\Omega(X) = \prod_{i}\dfrac{1}{1-x_i}$$

Now define the ''Bernstein Operators'' $$S^0_m$$ on symmetric functions as

$$S^0_m(f) = [u^m]f(X-u^{-1})\Omega(uX)$$

In words this means we have some variables x, and we add one more variable u, so $$uX = (ux_1,ux_2,\cdots)$$, and $$f(X-u^{-1})$$ can be thought as adding the variable $$-u^{-1}$$ to the set. So $$f(X-u^{-1})\Omega(uX)$$ is an expression containing the variables x and u. $$[u^m]$$ takes coefficient of $$u^m$$ in the big mess

The following theorem is fundamental, not by itself, but because a lot of the theory is developed by deforming this operator, and while the proofs are different for the other cases, there is a pattern in the proofs. so lets describe this one to get a flavor of what's going on

Theorem The Bernstein operators add a part to the indexing of the Schur function, that is, for $$m \geq \lambda1$$ we have $$S^0_m s_\lambda = s_{m,\lambda}$$

Sketch of proof

We need the following ingredients. First by partial fraction expansion

$$\Omega(uX) = \prod_{i}\dfrac{1}{1-ux_i} = \sum_i \dfrac{1}{1-ux_i}\prod_{j\neq i} \dfrac{1}{1-x_j/x_i}$$

The second ingredient is Weyl's character formula

$$s_{\lambda}(x) = \sum_{w\in S_n} w\left(\dfrac{x^{\lambda}}{\prod_{i<j}(1-x_j/x_i)}\right)$$

By expanding its easy to check that for a polynomial f

$$[u^m]f(u^{-1})\dfrac{1}{1-uz} = z^mf(z)$$

Now we're ready to stir the elements. First lets mix the first with definition, so for any f

$$[u^m]f(X-u^{-1})\Omega(uX) = [u^m]f(X-u^{-1})\sum_i \dfrac{1}{1-ux_i}\prod_{j\neq i} \dfrac{1}{1-x_j/x_i}$$

$$= \sum_i [u^m]f(x-u^{-1}) \dfrac{1}{1-ux_i}\prod_{j\neq i} \dfrac{1}{1-x_j/x_i}$$

because $$[u^m]$$ is an operator, it distributes over sum. Furthermore, considering $$f(x-u^{-1})$$ as a polynomial in $$u^{-1}$$ we can use the third ingredient, with $$x_i$$ playing the role of $$z$$ to get

$$\sum_i x_i^mf(X-x_i)\prod_{j\neq i} \dfrac{1}{1-x_j/x_i} = \sum_i x_i^m\prod_{j\neq i} \dfrac{f(X-x_i)}{1-x_j/x_i}$$

By the virtues of the plethystic substitution $$f(X-x_i)$$ is the same thing as evaluating in the other variables, i.e., taking $$x_i$$ out. Now specialize to the schur function and consider the second ingredient, Weyl's formula, then it follows easily by induction, because note the factor

$$x_i^m\prod_{j\neq i} \dfrac{s_{\lambda}(X-x_i)}{1-x_j/x_i}$$

is the same as for the terms in the formula for $$s_{(m,\lambda)}$$ when the permutation sends i to the first position.

This gives our final definition of schur functions

$s_\lambda = S^0_{\lambda_1}S^0_{\lambda_2}\cdots S^0_{\lambda_l}(1)$

Final Remarks

The schur functions are a basis for the symmetric functions with the following properties

1. Lower unitriangularity with respect to monomials $$s_\lambda= \sum_\mu K_{\lambda\mu}m_\mu.\$$

2. Orthogonality  $$\langle s_{\mu},s_{\nu}\rangle = \delta_{\mu\nu}$$

The Kostka numbers $$K_{\lambda\mu}$$ have two interpretations, a combinatorial and an algebraic one. These properties are important to keep in mind while generalizing with one or two parameters.

### Hall-Littlewood symmetric functions

Now that we've looked at the classical basis for $$\Lambda$$, we are going to look at a $$t$$-deformation which specializes to the monomial and the Schur bases (plus two additional bases we haven't discussed here). We do this deformation considering a different inner product $\langle p_{\lambda}, p_{\mu} \rangle_{(t)} = \delta_{\lambda \mu} z_{\lambda} \prod_{i=1}^{\ell(\lambda)} (1 - t^{\lambda_i})^{-1}.$  Now we define the Hall-Littlewood symmetric functions by the following two properties:

1. (Triangularity) $$P_{\lambda}(t) = m_{\lambda} + LOT$$ where LOT stands for lower order terms with respect to dominance order.
2. (Orthogonality) $$\langle P_{\lambda}(t), P_{\mu}(t) \rangle_{(t)} = 0$$ if $$\lambda \neq \mu$$.

There is a well-known formula for the Hall-Littlewood polynomials for $$n$$ variables and partitions such that $$|\lambda| \leq n$$ by

$P_{\lambda}(x_1, \ldots, x_n; t) = v_{\lambda}^{-1}(t) \sum_{w \in S_n} w\left( x^{\lambda} \prod_{i < j} \frac{x_i - t x_j}{x_i - x_j} \right)$

where

$v_{\lambda}(t) = \prod_{i \geq 0} \prod_{j=1}^{m_i} \frac{1 - t^j}{1 -t}$

is just to normalize the coefficient of $$x^{\lambda}$$ to $$1$$.

Now if we take $$t = 0$$, we get $$P_{\lambda}(0) = s_{\lambda}$$ by the inner product specializing to the Hall inner product, and if we take $$t = 1$$, we can see that

$P_{\lambda}(1) = \sum_{w \in S_n} w( x^{\lambda} ) = m_{\lambda}.$

As above, we take the limit as $$n \to \infty$$ to get an element in $$\Lambda$$.

### Contributors

• Travis Scrimshaw (UC Davis)
• Colombian Power