Arithmetic product of symmetric functions: why is it integral?

For every commutative ring $A$, let $\mathbf{Symm}_A$ be the ring of symmetric functions over $A$. Let $\mathbf{Symm}$ without a subscript denote $\mathbf{Symm}_{\mathbb{Z}}$.

We can define a bilinear map $\boxdot : \mathbf{Symm}_{\mathbb{Q}} \times \mathbf{Symm}_{\mathbb{Q}} \to \mathbf{Symm}_{\mathbb{Q}}$ by setting

$p_{\lambda} \boxdot p_{\mu} = \prod\limits_{i\geq 1,\ j\geq 1} p_{\operatorname*{lcm}\left(\lambda_i,\mu_j\right)}^{\gcd\left(\lambda_i,\mu_j\right)}$

for any two partitions $\lambda = \left(\lambda_1,\lambda_2,\lambda_3,...\right)$ and $\mu = \left(\mu_1,\mu_2,\mu_3,...\right)$. Here, we are writing $\boxdot$ as an infix operator (that is, $a\boxdot b$ means $\boxdot\left(a,b\right)$), and $p_\nu$ means the $\nu$-power sum symmetric function.

This bilinear map $\boxdot$ is associative. I call it the "arithmetic product", as it boils down to the arithmetic product of species viewed through the cycle index series.

Now, species theory can be used to show that $\boxdot$ restricts to a well-defined map $\mathbf{Symm} \times \mathbf{Symm} \to \mathbf{Symm}$ (that is, the restriction of $\boxdot$ to $\mathbf{Symm} \times \mathbf{Symm}$ has its image in $\mathbf{Symm}$). My question is: Can this be proven more elementarily? Is there a good way to describe this map on an actual basis of $\mathbf{Symm}$ rather than on the power-sum symmetric functions? Is there a more direct combinatorial or even representation-theoretical significance of this map?

(This is somewhat similar to MO question #120924, where another operation on $\mathbf{Symm}$ is constructed on the power sums first and then happens to be integral for weird reasons.)

• By "restricts to a well-defined map" you mean "this map, defined on a basis, is a ring homomorphism"? – Allen Knutson Jul 30 '13 at 10:33
• No, he means that the map is defined over the integers even though its definition makes sense only over the rationals (since the power sums only form a $\mathbf Q$-basis for the ring of symmetric functions). – Dan Petersen Jul 30 '13 at 11:30
• Dan is right; I've now edited the post to make this clearer. – darij grinberg Jul 30 '13 at 13:14
• I am not sure what you mean by an elementary proof in this context. Arithmetic product of species is a nice and elementary combinatorial operation. Could you be a bit more specific? – Vladimir Dotsenko Jul 30 '13 at 15:07
• How this product expresses on the monomial, elementary, and Schur bases of symmetric functions? – Samuele Giraudo Jul 30 '13 at 20:34

About the question "Is there a more direct combinatorial or even representation-theoretical significance of this map?" Probably you know that the arithmetic product is an operation between representations of the symmetric group, because the species is a theory of representations of the symmetric group that gives concrete set theoretical constructions of operations, without the use of induced representations. The ordinary species correspond to permutation-representations and the tensor species to vector representations (representations on the general linear group). In the classical language, the arithmetic product of two represntations $\rho$ and $\tau$ of $S_m$ and $S_n$ respectively is given as the induced representation: $$\rho\boxdot\tau=\mathrm{Ind}_{S_m\times S_n}^{S_{m.n}}\rho\otimes\tau.$$ But this seems to me less concrete than the direct recipe, given in the language of species, of the vector space where the symmetric group $S_U$ acts naturally, $$(R\boxdot T)[U]=\bigoplus_{(\pi,\sigma)}R[\pi]\otimes T[\sigma]$$ The arithmetic product of two homogeneous symmetric functions in terms of the monomial s.f. is as follows $h_m\boxdot h_n=\sum M_{\lambda} m_{\lambda}$ where $M_{\lambda}$ is the number of $m\times n$ matrices with $\lambda_i$ i-es (up to row and column permutations). This is because: $$Ch(E_m\boxdot E_m)(x)=\sum_{\lambda}|(E_m\boxdot E_n)[m.n]/S_{\lambda_1}\times S_{\lambda_2}\times\dots| \, m_{\lambda}(x)$$ $Ch$ being the Frobenius character. An analogous result is obtained for the arithmetic product of elementary symmetric functions, taking the arithmetic product of the sign representations $\Lambda_m$ and $\Lambda_n$. Maybe this can be helpful for expressing their arithmetic product in terms of themselves.

• Very nice, and welcome to MathOverflow! I only knew the arithmetic product of species from your paper. I reconstructed the arithmetic product on symmetric functions from it, but I was not aware that people had considered the arithmetic product of representations of symmetric groups and found such a nice formula for it. This answers the question very well. – darij grinberg Aug 15 '13 at 9:25
• Thanks for the welcoming. I should have put it in our paper with Manuel. In a talk that I gave at UQAM some years ago I formulated the same questions about expansions of symmetric functions that you have done, but never wrote that. I would be very interested in receiving updates of your current work on these problems! – Miguel Mendez Aug 16 '13 at 15:00
• Thanks once again. The integrality statement is now in Exercise 4.33 of Vic Reiner's and my Hopf notes web.mit.edu/~darij/www/algebra/HopfComb-sols.pdf (and the induction formula is in its solution -- although I had to rule out the case when $n$ or $m$ is $0$ first because we don't currently treat induced characters over noninjective morphisms). Incidentally, do you happen to know how to prove (or disprove) the very similar question mathoverflow.net/questions/182083 ? – darij grinberg Dec 16 '14 at 0:49

Here is a way of looking at the arithmetic product of species that generalizes to a product on representations of symmetric groups corresponding to the arithmetic product of symmetric functions.

First let me recall some important operations on species. First is the Cartesian product. If $F$ and $G$ are species then the Cartesian product $F\times G$ is defined by $(F\times G)[A] = F[A]\times G[A]$ for any finite set $A$, where $\times$ on the right is the usual Cartesian product of sets. The Cartesian product of species corresponds to the internal (or Kronecker or inner) product of symmetric functions and to the (inner) tensor product of symmetric group representations. I'll need a partial generalization to multisort species. Suppose that $F(X)$ is a one-sort species and $G(X,Y)$ is a two-sort species. Then $F(X)\times_X G(X,Y)$ is the two-sort species whose value at the pair of finite sets $(A,B)$ is the Cartesian product $F[A]\times G[A,B]$. Next we have the operation of setting a variable to 1". If $F$ is a one-sort species homogeneous of degree $n$ then $F(1)$ or $F(X)|_{X=1}$ is the set of orbits of $F$-structures on $[n]=\{1,2,\dots,n\}$. More generally, if $F(X,Y)$ is a two-sort species which is homogeneous of degree $n$ in $Y$, then $F(X,1)$ or $F(X,Y)_{Y=1}$ is the one-sort species whose value at a set $A$ is the set of orbits of $F[A,[n]]$ under the action of the symmetric group on $[n]$. The operation of setting $X=1$ corresponds to setting all power sums $p_i[x]$ equal to 1 in a symmetric function or to extracting the multiplicity of the trivial representation from a symmetric group representation.

Finally if $F$ and $G$ are one-sort species then their scalar product $\langle F,G\rangle$ is just $(F\times G)_{X=1}$, and this is extended in a natural way to multisort species, so for example if $F(Y)$ is a one-sort species and $G(X,Y)$ is a two-sort species then $\langle F(Y), G(X,Y) \rangle_Y$ is a one-sort species. (The scalar product of species corresponds to the scalar product of symmetric functions.)

Now we can construct the arithmetic product of species in a way that generalizes to an arithmetic product of symmetric group representations. We start off with a certain three-sort species $P(X,Y,Z)$ defined as follows: $P[A,B,C]$ is the set of bijections from $A$ to $B\times C$. We can think of these objects as bipartite graphs with vertex bipartition $(B,C)$ in which the edges are labeled by the elements of $A$, or (more intuitively but less precisely) rectangular arrangements of the elements of $A$ in which the rows are indexed by $B$ and the columns by $C$.

Then for two species $F(X)$ and $G(X)$ the arithmetic product $F(X)\boxdot G(X)$ is just $\langle P(X,Y,Z), F(Y)G(Z)\rangle_{Y,Z}$. (If you have the right picture in mind, this is clear, but I'm not going to try to prove it.) This definition extends easily to arbitrary representations: we replace $P(X,Y,Z)$, $F(Y)$, and $G(Z)$ with the corresponding representations and replace the scalar product with an appropriate tensor product and extraction of trivial representations.

While we're on the subject of arithmetic products of species, I'll mention something that I've wondered about (and that anyone interested in arithmetic products of species might find interesting to think about). Roughly speaking, we get the usual composition of species by symmetrizing powers of a species, and this operation on species corresponds to the usual composition or plethysm of symmetric functions. If we use the Cartesian product of species instead of the usual product, we get the inner plethysm of species, corresponding to the inner plethysm of symmetric functions. Both the usual (or "outer") and inner plethysm correspond to lambda ring structures on symmetric functions. We can similarly define a composition of species corresponding to the arithmetic product, called "exponential composition". (This was studied by my former student Ji Li, in "Prime graphs and exponential composition of species", J. Combin. Theory Ser. A 115 (2008), 1374-1401, based on earlier work by E. M. Palmer and R. W. Robinson.) Unfortunately exponential composition of symmetric functions does not give a lambda ring, and the formulas seem to me quite mysterious. Is there some sort of generalization of a lambda ring, or some other way of looking at these formulas, that can help to explain them?

• Thank you, this is very interesting; I have yet to fully understand it (my knowledge of species is cursory and does not include anything beyound one-sort). Am I seeing it right that if we are able to expand the $P\left(X,Y,Z\right)$ integrally in the tensor products of Schur functions, we will automatically get an algebraic proof of the integrality of the arithmetic product? – darij grinberg Aug 13 '13 at 20:08

In response to Samuele Giraudo's (very natural) question in the comments: I've added the arithmetic product as a method to Sage patch #14775 ( http://trac.sagemath.org/ticket/14775#comment:24 ). This allows to easily see how it behaves on standard bases of the symmetric functions. Unfortunately, I don't see much of a pattern:

sage: Sym = SymmetricFunctions(QQ)
sage: Sym.inject_shorthands()
/home/darij/sage-5.11.beta3/local/lib/python2.7/site-packages/sage/combinat/sf/sf.py:1192: RuntimeWarning: redefining global value e
inject_variable(shorthand, getattr(self, shorthand)())
sage: s.arithmetic_product(s[1,1])
s[2, 1, 1] + s[3, 1]
sage: s.arithmetic_product(s)
s[1, 1, 1, 1] + 2*s[2, 2] + s
sage: s.arithmetic_product(s[1,1])
s[1, 1, 1, 1, 1, 1] + 2*s[2, 2, 1, 1] + s[3, 2, 1] + 2*s[3, 3] + s[4, 1, 1] + s[5, 1]
sage: s.arithmetic_product(s)
s[2, 1, 1, 1, 1] + 2*s[2, 2, 2] + s[3, 1, 1, 1] + s[3, 2, 1] + 2*s[4, 2] + s
sage: s[2,1].arithmetic_product(s[1,1])
s[2, 1, 1, 1, 1] + 2*s[2, 2, 1, 1] + s[3, 1, 1, 1] + 3*s[3, 2, 1] + s[3, 3] + 2*s[4, 1, 1] + s[4, 2] + s[5, 1]
sage: s[2,1].arithmetic_product(s)
s[2, 1, 1, 1, 1] + s[2, 2, 1, 1] + s[2, 2, 2] + 2*s[3, 1, 1, 1] + 3*s[3, 2, 1] + s[4, 1, 1] + 2*s[4, 2] + s[5, 1]
sage: s[1,1,1].arithmetic_product(s[1,1])
s[2, 2, 2] + 2*s[3, 1, 1, 1] + s[3, 2, 1] + s[4, 1, 1] + s[4, 2]
sage: s[1,1,1].arithmetic_product(s)
s[2, 2, 1, 1] + s[3, 1, 1, 1] + s[3, 2, 1] + s[3, 3] + 2*s[4, 1, 1]
sage: h.arithmetic_product(h[1,1])
h[1, 1, 1, 1] - 2*h[2, 1, 1] + 2*h[2, 2]
sage: h.arithmetic_product(h)
h[1, 1, 1, 1] - 3*h[2, 1, 1] + 3*h[2, 2]
sage: e.arithmetic_product(e[1,1])
2*e[2, 1, 1] - 2*e[2, 2]
sage: e.arithmetic_product(e)
e[2, 1, 1] - e[2, 2]
sage: e[1,1].arithmetic_product(e[1,1])
e[1, 1, 1, 1]
sage: s[1,1].arithmetic_product(s[1,1])
s[2, 1, 1] + s[3, 1]
sage: e[2,1].arithmetic_product(e)
e[2, 1, 1, 1, 1] - e[2, 2, 1, 1]
sage: e[2,1].arithmetic_product(e[1,1])
2*e[2, 1, 1, 1, 1] - 2*e[2, 2, 1, 1]
sage: e[2,1].arithmetic_product(e[2,1])
e[2, 1, 1, 1, 1, 1, 1, 1] - 4*e[2, 2, 2, 1, 1, 1] + 4*e[2, 2, 2, 2, 1]

If there is anything hopeful here, then it's probably the e's, but their simplicity doesn't persist:

sage: e[3,1].arithmetic_product(e)
e[2, 2, 1, 1, 1, 1] - 2*e[2, 2, 2, 1, 1] + 2*e[2, 2, 2, 2] + e[3, 2, 1, 1, 1] - 4*e[3, 2, 2, 1] + e[3, 3, 1, 1] + e[3, 3, 2] + e[4, 1, 1, 1, 1] - 3*e[4, 2, 1, 1] + 2*e[4, 2, 2] - e[5, 1, 1, 1] + 2*e[5, 2, 1] + e[6, 1, 1] - 2*e[6, 2]