# Example: $n=3$

As an example, let’s calculate the Hall-Littlewood polynomials whose partition $\mu$ has size $n=3$. That is, we wish to find $\widetilde{H}_{(3)}$, $\widetilde{H}_{(2,1)}$, and $\widetilde{H}_{(1,1,1)}$. In particular let’s try to find their explicit expansions in terms of Schur functions.

To work out this example, I took out my new Microsoft Surface and its magnetic tablet pen, and by the time I was done with the entire calculation, it looked like this:

It’s a long slog that is better done by computer, so I’ll just provide a summary of the calculation with some of the main details. Recall that we are working with the conditions:

- $\left\langle \widetilde{H}_\mu(X;t),s_\lambda\right\rangle=0$ for any $\lambda<\mu$ in dominance order, and $\langle\widetilde{H}_\mu,s_\mu\rangle=1$
- $\left\langle \widetilde{H}_\mu(X;t),\widetilde{H}_{\lambda}[(1-t)X;t]\right\rangle=0$ whenever $\lambda\neq \mu$.

We can start by finding $\widetilde{H}_{(3)}$, which cannot have any $s_{(2,1)}$’s or $s_{(1,1,1)}$’s in its expansion by condition (1) above. So $$\widetilde{H}_{(3)}=a\cdot s_{(3)}$$ for some $a\in \mathbb{Q}(t)$. In fact, the diagonal condition in (1) says that $a=1$, so we have $$\widetilde{H}_{(3)}=s_{(3)}.$$

To find $\widetilde{H}_{(2,1)}$, we first use condition (1) to find that we can express it as $$\widetilde{H}_{(2,1)}=s_{(2,1)}+b\cdot s_{(3)}$$ for some $b\in \mathbb{Q}(t)$. Furthermore, by condition (2), it must be orthogonal to $\widetilde{H}_{(3)}[(1-t)X;t]$. So we now need to understand $\widetilde{H}_\mu[(1-t)X;t]$. This is equal to $s_{(3)}[(1-t)X]$, and so it suffices to understand what the substitution $X\mapsto (1-t)X$ does to Schur functions.

To make this substitution, we first express them in terms of the power sum symmetric functions and then make the substitution on each. The character table for $S_3$ is:

$$\begin{array}{c|ccc}

& [(1)(2)(3)] & [(12)(3)] & [(123)] \\\hline

\chi^{(3)} & 1 & 1 & 1 \\

\chi^{(1,1,1)} & 1 & -1 & 1 \\

\chi^{(2,1)} & 2 & 0 & -1

\end{array}

$$

Using the formula $s_\lambda=\sum \frac{1}{z_\mu} \chi^\lambda(\mu)p_\mu$ where $z_\mu=\prod i^\alpha_i\alpha_i!$ and $\alpha_i$ is the number of times $i$ occurs in $\mu$, we have

- $s_{(3)}=\frac{p_{(1,1,1)}}{6}+\frac{p_{(2,1)}}{2}+\frac{p_{(3)}}{3}$
- $s_{(1,1,1)}=\frac{p_{(1,1,1)}}{6}-\frac{p_{(2,1)}}{2}+\frac{p_{(3)}}{3}$
- $s_{(2,1)}=\frac{p_{(1,1,1)}}{3}-\frac{p_{(3)}}{3}$

Notice that plugging in $x_1,-tx_1,x_2,-tx_2,\ldots$ into a power sum symmetric function $p_{k}$ yields $p_k-t^kp_k=(1-t^k)p_k$, so we have $$p_\mu[(1-t)X]=p_\mu\cdot \prod (1-t^{\mu_i}).$$ Thus

$$s_{(3)}[(1-t)X]=(1-t)^3\frac{p_{(1,1,1)}}{6}+(1-t^2)(1-t)\frac{p_{(2,1)}}{2}+(1-t^3)\frac{p_{(3)}}{3},$$ and we can use the equations above to convert this back into a sum of Schur functions:

\[\begin{eqnarray*}

s_{(3)}[(1-t)X] &=& s_{(3)}-t(s_{(3)}+s_{(2,1)})+t^2(s_{(1,1,1)}+s_{(2,1)})-t^3s_{(1,1,1)} \\

&=& (1-t)s_{(3)}+(t^2-t)s_{(2,1)}+(t^2-t^3)s_{(1,1,1)}

\end{eqnarray*}\]

We can now finish the computation of $\widetilde{H}_{(2,1)}(X;t)$. The orthogonality relation says that

$$\langle s_{(2,1)}+b\cdot s_{(3)}, s_{3}[(1-t)X]\rangle=0.$$ By the expansion of $s_{(3)}[(1-t)X]$ above and using the orthogonality of the Schur functions, the left hand side becomes $(t^2-t)+b(1-t)$. Thus $b(1-t)=t-t^2$, and so $b=t$. It follows that $$\widetilde{H}_{(2,1)}(X;t)=s_{(2,1)}+ts_{(3)}.$$

Finally, a similar calculation shows that $$\widetilde{H}_{(1,1,1)}(X;t)=s_{(1,1,1)}+(t+t^2)s_{(2,1)}+t^3s_{(3)}.$$

Indeed, the degree of these polynomials $\widetilde{H}_\mu(X;t)$ as a polynomial in $t$ is equal to $n(\mu)$ in each case above.

# The Cocharge Formula

A similar method can be used to compute the characters of the $W$-modules $H^\ast(\mathcal{B}_u)$ for any Lie type. In type A, however, there is a much simpler formula for the Hall-Littlewood polynomials. It is based on a combinatorial statistic on semistandard Young tableaux called *cocharge*. We write $\DeclareMathOperator{\cc}{cc}\cc(T)$ to denote the cocharge of a tableau $T$.

Let $\widetilde{K_{\lambda\mu}}(t)$ denote the coefficient of $s_\lambda$ in $\widetilde{H}_\mu(X;t)$, so in other words $$\widetilde{H}_\mu(X;t)=\sum_{\lambda} \widetilde{K_{\lambda\mu}}(t) s_\lambda.$$ Since the Hall-Littlewood polynomials are the Frobenius characteristic of a graded $S_n$-module, we know that $\widetilde{K_{\lambda\mu}}(t)$ is always a polynomial in $t$ with positive integer coefficients. In fact we have $$\widetilde{K_{\lambda\mu}}(t)=\sum_{T\in SSYT(\lambda,\mu)}t^{\cc(T)}$$ where $SSYT(\lambda,\mu)$ denotes the set of all semistandard Young tableaux of shape $\lambda$ and content $\mu$. That is, $T$ ranges over all fillings of the diagram of $\lambda$ that are weakly increasing along rows and strictly increasing up columns, and use exactly $\mu_i$ copies of the entry $i$ for each $i$.

To define the cocharge of such a filling, we define a cocharge statistic on words having partition content, and then simply define $\cc(T)$ to be the cocharge of its reading word. Given a word $w=w_1\cdots w_n$ of content $\mu$, we decompose it into standard subwords $w^{(1)},\ldots,w^{(\mu_1)}$ as follows. Search from the right until you find a $1$, then continue until finding a $2$, and so on, possibly wrapping around to the beginning if you reach the left end of the word, and stopping when a copy of the largest letter in the word is found. Then these letters form the subword $w^{(1)}$. We then remove $w^{(1)}$ and repeat the process on the remaining word to find $w^{(2)}$ and so on.

Now, for a standard subword $w^{(k)}$, let $a_i$ be the number of times we wrapped around the left edge of $w$ in our search before arriving at the entry $i$ in $w^{(k)}$. We define $$\cc(w^{(k)})=\sum_i i-1-a_i,$$ and define $$\cc(w)=\cc(w^{(1)})+\cdots+\cc(w^{(\mu_1)}).$$

For instance, if $$w=13224121353,$$ then our standard subwords are $w^{(1)}=34215$, $w^{(2)}=213$, and $w^{(3)}=123$. The cocharge of $w^{(1)}$ is the sum of the subscripts in the labeling $$3_2 4_2 2_1 1_0 5_2,$$ since we had to wrap around once to find the $4$ and a total of twice to find the $5$. So $\cc(w^{(1)})=7$, $\cc(w^{(2)})=2$, and $\cc(w^{(3)})=0$. Hence $\cc(w)=7+2=9$.

Finally, the cocharge of a tableau is the cocharge of its reading word. So to compute, say, $\widetilde{K}_{(1,1,1),(2,1)}(t)$, we consider the two semistandard fillings of shape $(2,1)$ and content $(1,1,1)$:

Their reading words are $213$ and $312$, which have cocharge $2$ and $1$ respectively. Hence $$\widetilde{K}_{(1,1,1),(2,1)}(t)=t+t^2.$$ Notice that this is indeed the coefficient of $s_{(2,1)}$ in $\widetilde{H}_{(1,1,1)}(X;t)$ that we found above by orthogonality!

Finally, it also follows immediately from this formula that the degree of $\widetilde{H}_\mu(X;t)$ is $n(\mu)$. Indeed, it is the cocharge of the superstandard filling

We know it is maximal, since the cocharge of a filling with content $\mu$ is maximized when the number of times we wrap around the left edge to find each standard subword is minimized. Since there is no wrapping around at all in the reading word of the superstandard filling, it follows that $n(\mu)$ is the degree of $\widetilde{H}_\mu(X;t)$.