# A "nice" Orthogonal Basis for Translation Invariant Symmetric Polynomials

It is going to be a rather long question, so I will first state it and then try to explain and motivate it.

Take $\Lambda_n$ as the graded ring of symmetric polynomials of a field $F$ in $n$ variables. We say $f\in \Lambda_n$ is translation invariant if for any $h\in F$ we have

$f(x_1+h, \cdots, x_n+h)=f(x_1, \cdots, x_n)$

Denote this graded subring of $\Lambda_n$ by $\Omega_n$. Suppose $F=\mathbb{Q}[q,t]$ (the field of rational functions on $q,t$) Define the scalar product on $\Lambda_n$ in the usual way for Macdonald polynomials, i.e. $\langle f,g\rangle_{q,t} = [f\bar{g}\Delta_{q,t}]_0$ where $[\bullet]_0$ means the constant term, and

$\Delta_{q,t}= \prod_{i\neq j}\frac{(x_ix_j^{-1};q)_\infty}{(tx_ix_j^{-1};q)_\infty}$, where $(a;q)\infty = \prod_{r=0}^\infty(1-aq^r)$.

Main Question: What is the basis for $\Omega_n$ such that if $f_\lambda$ and $f_\mu$ are in the basis then $\langle f_\lambda, f_\mu\rangle_{q,t}=0$ if $\lambda\neq \mu$. Is there any triangular property for this basis like there is for Macdonalds? i.e. something similar to $P_\lambda = m_\lambda +$ lower terms

Note: There is a retraction $\rho:\Lambda_n \to \Omega_n$, such that $\rho(f(x_1,\cdots, x_n))=f(x_1-x_\mathrm{avg},\cdots, x_n-x_\mathrm{avg})$ where $x_\mathrm{avg} = (x_1+\cdots+x_n)/n$. $\rho$ is surjective and acts identically on $\Omega_n$. So even if there is a triangular property probably it is in terms of something like $\rho(m_\lambda)$ instead of $m_\lambda$.

But the above question is not well-defined since basis is not unique. So we need a bit more restriction:

A Weaker Version: We know that for the case of $q=t^{\alpha}$ the MacDonald polynomials become Jack Polynomial. Also by a little work on Feigin et all paper (http://arxiv.org/abs/math/0112127), one can show that Jack polynomials with $\alpha = -(r-1)/(k+1)$ belong to $\Omega_n$ if $\lambda$ is $(k,r,n)$-admissible (i.e. $\lambda_i-\lambda_{i+k}\geq r$ for $1\leq i\leq n-k$). If we count the number of aforementioned Jack polynomials which are homogeneous of degree $d$ (i.e. they belong to $\Lambda_n^d$), we see that this number is less than $\dim \Omega_n^d$. Hence

What is the extension of Jack polynomials like mentioned above, which is an orthogonal basis for $\Omega_n$. And Furthermore can we generalize it to Macdonald polynomials?

Another Version Now motivated by the above, one can ask is there some $q,t$ such that the a subset of MacDonald polynomials $P_\lambda(q,t)$ is an orthogonal basis for $\Omega_n$? If not, then what $q,t$'s are better suited for the job (I know this is now exactly clear!) and what kinds of combinations of Macdonalds do we need to linear-span the whole space?

I hope I have made the question clear enough.

• This is an interesting question! I did some stuff on translation-invariant symmetric polynomials, perhaps that might be of interest? Discriminants, symmetrized graph monomials, and sums of squares, Experimental Mathematics 21 No. 4 (2012) 353–361 These are not orthogonal though, but I would like to see the relation with Macdonald polynomials if there is one. Jun 15, 2015 at 16:36
• Of course! Can you give me a few links? Thanks a lot. Jun 15, 2015 at 16:38
• Perhaps you can access it here: projecteuclid.org/euclid.em/1356038818 if your university access... otherwise, a preprint is on arxiv. Jun 15, 2015 at 16:47
• Thanks. Can you elaborate on the following fact asserted in your paper: "It is classically known that for any given number $n$ of vertices and $d$ edges, the linear span of the symmetrized graph monomials coming from all graphs with $n$ vertices and $d$ edges coincides with the linear space $\mathrm{PST}_{n,d}$ of all symmetric translation-invariant polynomials of degree $d$ in $n$ variables." Jun 15, 2015 at 17:11
• Actually now that you mention it, if one can construct an orthogonal basis (with Hall scalar product) for translation invariant symmetric polynomials, it shouldn't be hard to generalize it to what I want. But I don't know of such natural basis either. Jun 15, 2015 at 20:53