
# Hall-Littlewood Polynomials

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We know the schur basis, and many more, for the ring of symmetric functions over a field $$F$$. The next step of generalization is consider the field $$F(t)$$, and twist a little bit the inner product. In contrast with Macdonald polynomials, we can give a closed expression for Hall-Littlewood polynomials

### Definition and first properties

First we need the following $$t$$ - analogues

$[k]_t := \dfrac{1-t^k}{1-t}= 1+t+t^2+\cdots+t^{k-1}$

$[k]_t! := [k]_t[k-1]_t\cdots[1]_t$

Then the Hall-Littlewood polynomial $$P_{\lambda}(x;t)$$ in n variables is given by the following formula

$P_{\lambda}(x;t) = \dfrac{1}{\prod_{i\geq 0}[\alpha_i]_t!} \sum_{w\in Sn}w\left(x^{\lambda}\dfrac{\Pi_{i<j}(1-tx_j/x_i)}{\Pi_{i<j}(1-x_j/x_i)}\right)$

Where $$\lambda = (1^{\alpha_1},2^{\lambda2},\cdots)$$ and $$\alpha_0$$ is such that $$\sum_{i\geq 0} \alpha_i = n$$

Note that when $$t=0$$ the denominator $$\Pi_{i\geq 0}[\alpha_i]_t!$$ goes away and we get precisely the Weyl's character formula for the schur functions, so

$P_{\lambda}(x,0) = s_{\lambda}(x)$

at $$t=1$$ the products inside cancel and we get the usual monomial funcitons

$P_{\lambda}(x,1) = m_{\lambda}(x)$

The Hall-Littlewood polynomials will form a basis, then we can expand schur in this new basis. The "Kostka-Foulkes polynomials" $$K_{\lambda\mu}(t)$$ are defined by

$s_{\lambda}(x) = \sum_{\mu} K_{\lambda\mu}(t) P_{\lambda}(x;t)$

They don't deserve the name polynomials yet, because so far we just know that they are rational functions in t. But we will see why they're actual polynomials.

### Definition with raising operators

Define the Jing Operators as t deformations of the Bersntein operator in the following way

$S^t_m f = [u^m]f[X+(t-1)u^{-1}]\Omega[uX]$

and their modified version

$\tilde{S}^t_m f = [u^m]f[X-u^{-1}]\Omega[(1-t)uX]$

which are related by

$\tilde{S}^t_m = \Pi_{(1-t)}S^t_m\Pi^{-1}_{(1-t)}$

where $$\Pi_{(1-t)}$$ is the operator with the plethystic substituion $$f\to f[X(1-t)]$$, and $$\Pi^{-1}_{(1-t)}$$ is its inverse, namely $$f\to f[X/(1-t)]$$

Analogously to the schur functions now defined the transformed Hall-Littlewood polynomials as

$H_{\mu}(x;t) = S^t_{\mu_1}S^t_{\mu_2}\cdots S^t_{\mu_l}(1)$

And if we set $$Q_{\mu}(x;t) = H_{\mu}((1-t)X;t)$$ we get

$Q_{\mu}(z;t) = \tilde{S}^t_{\mu_1}\tilde{S}^t_{\mu_2}\cdots \tilde{S}^t_{\mu_l}(1)$

Recall that the Bernstein operators added one part to a partition. This new operators behave in a more complicated way, but of similar spirit

Theorem: Jing Operators

If $$m\geq \mu_1 \gamma$$ and $$\lambda\geq \mu$$ then

$S^t_m s_\lambda \in \mathbb{Z}[t] \{ s_{\gamma} : \gamma \geq (m,\mu) \}$

Moreover, $$s_{(m,\lambda)}$$ appears with coefficient 1

The last part is saying something similar to the previous situation, we will get the schur function with an additional part m added, but the theorem is saying that we get also polynomial combinations of other schur functions.

By repeated use of the theorem we can conclude that

$H_{\mu}(x;t) = \sum_{\lambda\geq \mu} C_{\lambda \mu}(t) s_{\lambda}(x)$

where $$C_{\lambda \mu}(t)$$ are polynomials with $$C_{\mu \mu}(t) = 1$$

That means that we have upper unitriangularity with respect to the schur basis.

We have analogous statements for Q (although with different proof!)

Theorem: Modified Jing Operators

If $$m\geq \mu_1 \gamma$$ then

$\tilde{S}^t_m s_\lambda \in \mathbb{Z}[t] \{ s_{\gamma} : \gamma \leq (m,\lambda) \}$

Moreover, $$s_{(m,\lambda)}$$ appears with coefficient $$1-t^{\alpha}$$ where $$\alpha$$ is the multiplicity of m as a part of $$(m,\lambda)$$

Again by repeated use of the theorem we can conclude that

$Q_{\mu}(x;t) = \sum_{\lambda\leq \mu} B_{\lambda \mu}(t) s_{\lambda}(x)$

where $$B_{\lambda \mu}(t)$$ are polynomials with $$B_{\mu \mu}(t) = (1-t)^{l(\mu)}\prod_{i\geq 0}[\alpha_i]_t!$$

Which means that we have lower triangularity (but with a messier diagonal elements) with respect to the schur basis.

The operator $$\Pi_{(1-t)}$$ is self adjoint for the inner product, i.e. we have

$\langle f,g[(1-t)X] \rangle = \langle f[(1-t)X],g \rangle$

By the opposite triangularities of $$H$$ and $$Q=H[(1-t)X]$$ we have that if $$\langle H_\mu, H_\upsilon[(1-t)X] \rangle \neq 0$$ then $$\mu \leq \upsilon$$. Passing the $$(1-t)$$ to the other side, we obtain the opposite conclusion $$\mu \geq \upsilon$$ and hence $$\mu = \upsilon$$. Which implies the following claim

The transformed Hall-Littlewood polynomials are orthogonal with respect to the inner product $$\langle f,g[(1-t)X]\rangle$$ and their self inner products are given by

$\langle H_{\mu},H_{\mu}[(1-t)X] \rangle=(1-t)^{l(\mu)}\prod_{i\geq 0}[\alpha_i]_t$

### Now everything fits smoothly

Really. First, from the definition of  $$Q$$ one can get the following formula by induction

$Q_{\lambda}(x;t)=(1-t)^{l(\lambda)}[n-l(\lambda)]_t! \sum_{w\in S_n} w\left(x^{\lambda} \dfrac{\Pi_{i<j}(1-tx_j/x_i)}{\Pi_{i<j}(1-x_j/x_i)} \right)$

The relation with the original Hall-Littlewood polynomials is

$P_{\lambda}(x;t) = \dfrac{ Q_{\lambda}(x;t)}{(1-t)^{l(\lambda)}\prod_{i\geq 0}[\alpha_i]_t!}$

Note that the denominator is precisely the self inner product of the $$H$$ in the inner product $$\langle f,g[(1-t)X]\rangle$$. Classically something a bit different is defined

$\langle f,g\rangle_t = \langle f,g[X/(1-t)]\rangle$

In this product, the basis $$\{P_{\lambda}\}$$ and $$\{Q_{\lambda}\}$$ are orthogonal and furthermore, they are dual! So recall that we defined the Kostka - Foulkes polynomial as

$s_{\lambda}(x) = \sum_{\mu} K_{\lambda\mu}(t) P_{\lambda}(x;t)$

By taking inner products, and using the duality just mentioned we arrive at

$K_{\lambda\mu}(t) = \langle s_{\lambda},Q_{\mu}\rangle_t = \langle s_{\lambda},H_{\mu}\rangle$

But that last coefficient is equal to our previously defined polynomials $$C_{\lambda \mu}(t)$$, showing that the Kostka-Foulkes polynomials are in fact polynomials.

### Positivity of Kostka-Foulkes polynomials

It turns out that they are not just integer polynomials, but their coefficients are positive. It may not sound very interesting to show that a quantity is positive, but usually the question is implicitly asking for a interpretation. There are many different approaches here, all far from trivial. Let's briefly review them.

### Representation theory

The work of Hotta, Lusztig, and Springer showed deep connections with representation theory. I cannot say more than a few words: They relate the Kostka-Foulkes polynomials, and a variation of them, called ''cocharge '' Kostka-Foulkes polynomials to some hardcore math where the keywords are ''Unipotent Characters, local intersection homology, Springer fiber and perverse sheaves''.

The important point is that they found a ring, the cohomology ring of the Springer fiber, whose Frobenius series is given by the cocharge transformed Hall-Littlewood polynomials, implying they expand schur positively.

### Combinatorics of Tableaux

Lascoux and Schutzenberger proved the following simple and elegant formula, that gives a concrete meaning to each coefficient

$K_{\lambda\mu}(t) = \sum_T t^{c(T)}$

the sum is over all SSYT of shape $$\lambda$$ and content $$\mu$$. The new definition is the ''charge'' $$c(T)$$ which is easier to define in terms of cocharge $$cc(T)$$ which is an invariant characterized by

1. Cocharge is invariant under jeu-de-taquin slides
2. Suppose the shape of $$T$$ is disconnected, say $$T = X \cup Y$$ with $$X$$ above and left of $$Y$$, and no entry of $$X$$ is equal to 1. Then $$S = Y \cup X$$, obtained by swapping, has $$cc(S) = |X| - cc(X)$$
3. If $$T$$ is a single row, then $$cc(T) = 0$$

And then $$c(T) = n(\mu) - cc(T)$$. The existence of such an invariant requires proof. There is a process to compute the cocharge called ''catabolism''.

### Alternative description using tableaux

Kirillov and Reshetikhin gave the following formula

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where the sum is over all $$(\lambda,\mu)$$ - admissible configurations $$\upsilon$$.

Complicated as it seems, this expression has clearly positive coefficients. The origin of this formula is from a technique in mathematical physics known as ''Bethe ansatz'', which is used to produced highest weight vectors for some tensor products. The theorem is relating $$K_{\lambda\mu}(t)$$ with the enumeration of highesst weight vectors in $$V_{\mu_1}\otimes\cdots\otimes V_{\mu_r}$$ by a quantum number. For more info, stay tuned, probably Anne has something to say about in class.

### Commutative Algebra

This may be the less technical. Garsia and Procesi simplified the first proof by giving a down to earth interpretation of the cohomology ring of the springer fiber $$R_{\mu}$$. Now the action happens inside the polynomial ring $$C[x] = C[x_1,x_2,\cdots,x_n]$$. And

$R_{\mu} = C[x]/I_{\mu}$

For an ideal with a relatively explicit description. They manage to give generators, and finally they proof with more elementary methods that the frobenius series is the cocharge invariant

$F_{R_{\mu}}(x;t) = t^{n(\mu)}H_{\mu}(x;t^{-1}) = \sum_{\lambda} \tilde{K}_{\lambda\mu}(t)s_{\lambda}$

where $$\tilde{K}_{\lambda\mu}(t) = t^{n(\mu)}K_{\lambda\mu}(t^{-1})$$ is the cocharge Kostka-Foulkes poylnomial.