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 $\Omega_n$. But actually if we limit oursleves to $\Omega_n^k$ we can see that these Jack polynomials are not enough to generate the whole $\Omega_n$. So
Alternation 1: What is the extension of Jack polynomials like mentioned above, which generates the whole $\Omega_n$ while respecting orthogonality. 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 $P_\lambda(q,t)$ actually generates the whole graded ring $\Omega_n$? 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 generate the whole space?
I hope I have made the question clear enough.