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Let $n$ be a positive integer and consider the ring $R$ of power series over $\mathbb{Q}$ in commuting variables $x_1,y_1,x_2,y_2,...$. Let the symmetric group $\mathfrak{S}$ of permutations of the positive integers (with finitely many non-fixed points, if you like) act diagonally: $w(x_i)=x_{w(i)}$, $w(y_i)=y_{w(i)}$. What is known about the invariant subring $R^\mathfrak{S}$? Of course, with just one variable set you get the symmetric functions, but I have not been able to find anything about this case. It seems like something that should have been studied a long ago.

What I have been able to figure out is that there is a natural vector space basis of elements analogous to the monomial symmetric functions, which lets one compute the bigraded Hilbert series as $$\prod_{(n,m)\in\mathbb{N}^2\setminus\{(0,0)\}} \frac{1}{1-x^ny^m}$$ (OEIS A054225). There is also an analogue of the power-sum symmetric functions; I have not yet tried constructing analogues of other well-known bases. The constructions of the bases extend easily to the case of $n$ variable sets.

Is this a thing that has been studied before? If so, I'd appreciate a pointer to the literature. There is plenty of material on the analogous diagonal (co)invariants for polynomials, but not (as far as I can tell) for the power series setting.

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  • $\begingroup$ In other words, you consider the symmetric functions of the vectors variables $v_1,v_2,...$ where $v_i=(x_i, y_i)$? $\endgroup$ Commented Nov 30 at 2:26
  • $\begingroup$ @Fedor Petrov I'd have to think about that. If it is helpful, the simplest nontrivial example is that in bidegree (1,1), the vector space of invariants has dimension 2. Here is what I would consider the "monomial basis": $$\left\{\sum_{i>0} x_iy_i, \quad \sum_{i,j>0,\ i\neq j} x_iy_j\right\}$$ $\endgroup$ Commented Nov 30 at 2:50
  • $\begingroup$ If you do not restrict the degree, the second guy is $(\sum x_i)(\sum y_i)-\sum x_iy_i$ is expressed via the invariant functions of the form $\sum g(v_i)$ $\endgroup$ Commented Nov 30 at 3:01
  • $\begingroup$ Yes, the pair $\{(\sum x_i)(\sum y_i),\ \sum x_iy_i\}$ is the analogue of the power-sum basis in bidegree (1,1). $\endgroup$ Commented Nov 30 at 5:10
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    $\begingroup$ These are also called multisymmetric functions. The theorem that multisymmetric power sums generate the ring of multisymmetric functions is due to Poisson and takes a couple sentences in his short article on the Poisson product formula for resultants, at the beginning of the 19th century. There is also earlier work in relation to elimination theory by Newton, MacLaurin if I remember well. $\endgroup$ Commented Nov 30 at 19:43

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Yes, these have been studied before. They were studied by MacMahon under the name "symmetric functions of several systems of quantities." Nowadays they are usually called MacMahon symmetric functions or vector symmetric functions. Here are some references:

P. A. MacMahon, VII. Memoir on Symmetric Functions of the Roots of Systems of Equations, Philosophical Transactions of the Royal Society of London. A 181 (1890), 481-536. (They are also discussed in MacMahon's book Combinatory Analysis.)

Ira M. Gessel, Enumerative applications of symmetric functions, Publ. I.R.M.A. Strasbourg, 1987, 229/S–08 Actes 17e Séminaire Lotharingien, 5-21.

G. C. Rota and J. A. Stein, Plethystic algebras and vector symmetric functions, Proc. Natl. Acad. Sci. USA 91 (1994), 13062–13066.

Mercedes H. Rosas, A combinatorial overview of the theory of MacMahon symmetric functions and a study of the Kronecker product of Schur functions, Ph.D. thesis, Brandeis University, 1999.

Mercedes H. Rosas, MacMahon Symmetric functions, the partition lattice, and Young subgroups, J. Combin. Theory Ser. A 96 (2001), 326–340.

Mercedes H. Rosas, Specializations of MacMahon symmetric functions and the polynomial algebra, Discrete Math. 246 (2002), 285–293.

Mercedes H. Rosas, Gian-Carlo Rota, and Joel Stein, A combinatorial overview of the Hopf algebra of MacMahon symmetric functions, Ann. Comb. 6 (2002), 195–207.

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  • $\begingroup$ Just what I needed. Thanks, Ira! $\endgroup$ Commented Nov 30 at 17:25

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