(This was originally posted here, https://math.stackexchange.com/questions/4687466/cauchy-identity-for-schur-functions, and I am reposting it here as it seems to be more appropriate.)

PRELIMINARY. The Cauchy identity for Schur polynomials reads $$ \sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\prod_{i,j=1}^n\frac 1{1-x_iy_j}, $$ where $s_\lambda$ are the Schur polynomials and the sum on the left-hand side runs over all partitions (of length $\leq n$).

Denoting $p_k(x_1,...,x_n)=x_1^k+...+ x_n ^k$, we also have the identity $$ \exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)}{k}y^k\right) = \prod_{i=1}^n \frac 1{1-x_iy}, $$ which implies that we can rewrite Cauchy identity as $$ \sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)p_k(y_1,...,y_n)}{k}\right).\qquad (\star) $$

THE QUESTION. Let now $p_1,p_2,...$ be an infinite set of independent variables, and let $S_\lambda(p_1,p_2...)$ be the Schur functions, namely the expression of Schur polynomials in the power-sum polynomials. Can the identity $$ \sum_{\lambda}S_\lambda(p_1,p_2,...)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\frac{p_k}{k}(y_1^k+...+y_n^k)\right)\qquad (\star\star) $$ be deduced directly from $(\star)$ by an $n\to\infty$ limit argument?

I guess I am missing something (trivial!) about rigorously inferring identities in the ring of symmetric functions (which is the inverse limit as $n\to\infty$ of the rings of symmetric polynomials in $n$ variables) from identities in the rings of symmetric polynomials in $n$ variables.

Any help is greatly appreciated!


2 Answers 2


Here is one possible approach to make the aruement rigorous:

Define the two series on the left and the right as Taylor series over the monomials $$ y_1^{e_1} \cdot ... \cdot y_n^{e_n}; $$ in particular, let $$[y_1^{e_1} \cdot ... \cdot y_n^{e_n}]f(y_1,...,y_n) $$ denote the coefficient of $ y_1^{e_1} \cdot ... \cdot y_n^{e_n} $ in $f(y_1,...,y_n)$. It is important to note that the coefficients are in $\mathbb{C}[[x_1,x_2,...]]$, so that we have that the entire domain of discussion is contained within $$ (\mathbb{C}[[x_1,x_2,...]])[[y_1,...,y_n]]. $$

Also let $$ p_{k,n} := p_k(x_1,,...,x_n) $$ so that $$ \lim_{n \to \infty} p_{k,n} = p_k. $$

Now you have that the identity $(\star)$ is equivalent to $$ \sum_{\lambda}s_{\lambda,n}(p_1,...,p_n)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\frac{p_{k,n}p_k(y_1,...,y_n)}{k}\right),\qquad (\star\star\star) $$ where $s_{\lambda,n}(p_1,...,p_n)$ is defined anlogously to $p_{k,n}$.

Now define a norm, $||\_||_y$, on $(\mathbb{C}[[x_1,x_2,...]])[[y_1,...,y_n]]$ as $$ ||f||_y : = \max_{e \in \mathbb{N}^n} \left|\left|\frac{[y_1^{e_1} \cdot ... \cdot y_n^{e_n}]f}{A(e_1\cdot...\cdot e_n,e_1\cdot...\cdot e_n)}\right|\right|_x, $$ where we define the norm $||\_||_x$ on $\mathbb{C}[[x_1,x_2,...]]$ as $$ ||g||_x : = \max_{n \in \mathbb{N}}\max_{e \in \mathbb{N}^n} \left|\left|\frac{[x_1^{e_1} \cdot ... \cdot x_n^{e_n}]g}{A(e_1\cdot...\cdot e_n,e_1\cdot...\cdot e_n)}\right|\right|_\mathbb{C}, $$ where $A(x,y)$ is the Ackermann function (see [1] below).

The main idea is that all of the coeffcients of $s_{\lambda,n}$ and $p_{k,n}$ are recursively compuatable and the Ackermann function therefore grows far faster than any of those coefficients (since the Ackermann function is not primitive recursively computable). It is easy to prove this: simply open up a combinatorics book (e.g., Volume II Chapter 7 of Stanley's book, see [2] below) and notice that the deifnitions of $s_{\lambda,n}$ and $p_{k,n}$ can be written as a computer program using only recursion and for-loops. Furthermore, sums, products, compositions, etc ... and all of the standard combinatorial operations on primitive recursive generating functions are themselves primitive recursive generating functions.

Therefore, we have that $(\star\star)$ is in the space of functions that converges under this norm and, furthermore, that the limit of $(\star\star\star)$ as $n \to \infty$ is equal to $(\star\star)$ by a straightforward calculation, since we forced "deeper" coefficients involving $x_n$ to have smaller contribution to the norm. In particular, I mean that that the limit of LHS $(\star\star\star)$ as $n \to \infty$ is equal to LHS $(\star\star)$ and likewise for RHS.

There might be a more "elementary" arguement that doesn't involve the Ackermann function, but it will likely involve more messy details invloving $\ell^p$ norms that I, as a computer scientist, would find less simple than invoking elementary computability theory (the result is from ~100 years ago!).

[1] Ackermann, W., Zum Hilbertschen Aufbau der reellen Zahlen., Math. Ann. 99, 118-133 (1928). ZBL54.0056.06.

[2] Stanley, Richard P., Enumerative combinatorics. Volume 2., Cambridge Studies in Advanced Mathematics. 62. Cambridge: Cambridge University Press. xii, 585 p. (2001). ZBL0978.05002.

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    $\begingroup$ Whoa, that's one meta trick. I'd be careful not to overuse it, as I'm not sure if it can be applied several times in succession. But anyway, it is overkill: The OP needs a topology, not a norm :) $\endgroup$ May 10, 2023 at 4:31
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    $\begingroup$ I think you mean to say "it could be overkill". In order to confidently say that a topology suffices, you must provide a proof that uses a topology that is not a norm (or metric) and it must be shorter/simpler than my proof. That is totally possible, but I am not going to spend days pulling out my hair to figure that out. Mathematical beauty leads people astray just as often as it guides people :) $\endgroup$ May 10, 2023 at 9:10

Yes. Here is the strategy that I typically use to derive symmetric-functions identities from symmetric-polynomials identities. It might not be the most general strategy, but it has so far been sufficient for myself.

Let $\mathbf{k}$ be the base ring. Consider the $\mathbf{k}$-algebra $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ of formal power series in infinitely many commuting indeterminates $x_{1} ,x_{2},x_{3},\ldots$. If $\mathfrak{m}$ is a monomial and $f\in\mathbf{k} \left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ is a power series, then $\left[ \mathfrak{m}\right] f$ shall denote the coefficient of $\mathfrak{m}$ in $f$.

We equip $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ with a topology, which is the product topology coming from viewing $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ as an infinite direct product of copies of $\mathbf{k}$ (indexed by monomials). Explicitly, this topology can be easily described in terms of limits of nets (see Stijn Vermeeren's Sequences and nets in topology for an introduction to nets): A net $\left( f_{s}\right) _{s\in S}$ of power series $f_{s}\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ (where $S$ is a directed set) converges to a power series $f\in\mathbf{k} \left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ if and only if for each monomial $\mathfrak{m}$, the net $\left( \left[ \mathfrak{m}\right] f_{s}\right) _{s\in S}$ converges to $\left[ \mathfrak{m}\right] f$ in the discrete topology on $\mathbf{k}$ (that is, $\left[ \mathfrak{m}\right] f_{s}=\left[ \mathfrak{m}\right] f$ for all sufficiently high $s$, where the specific meaning of "sufficiently high" depends on $\mathfrak{m}$).

This topology turns $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3} ,\ldots\right] \right] $ into a topological $\mathbf{k}$-algebra. It is easy to see that any $f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3} ,\ldots\right] \right] $ satisfies \begin{equation} f=\lim\limits_{n\rightarrow\infty}f\left( x_{1},x_{2},\ldots,x_{n} ,0,0,0,\ldots\right) \label{eq.darij1.1} \tag{1} \end{equation} (since each monomial $\mathfrak{m}$ uses only finitely many indeterminates, and thus remains unchanged when we replace all the higher indeterminates by $0$). If $f$ is a symmetric function, then we conventionally write $f\left( x_{1},x_{2},\ldots,x_{n}\right) $ for $f\left( x_{1},x_{2},\ldots ,x_{n},0,0,0,\ldots\right) $, so this equality becomes \begin{align*} f=\lim\limits_{n\rightarrow\infty}f\left( x_{1},x_{2},\ldots,x_{n}\right) . \end{align*}

Infinite sums $\sum_{s\in S}f_{s}$ of summable families $\left( f_{s}\right) _{s\in S}$ are easily defined using the above topology. Summability of a family $\left( f_{s}\right) _{s\in S}$ means that the net $\left( \sum_{s\in T}f_{s}\right) _{T\subseteq S\text{ finite}}$ converges, or, equivalently, that for each monomial $\mathfrak{m}$, only finitely many $s\in T$ satisfy $\left[ \mathfrak{m}\right] f_{s}\neq0$. In particular, if $\left( f_{s}\right) _{s\in S}$ is a family of homogeneous power series, and if it contains only finitely many degree-$n$ entries for each given $n\in\mathbb{N}$, then it is summable. Thus, sums like $\sum_{\lambda \in\operatorname*{Par}}s_{\lambda}$ and $\sum_{k\geq1}p_{k}$ are well-defined.

Now, it is easy to see that if $\left( f_{s}\right) _{s\in S}$ and $\left( g_{t}\right) _{t\in T}$ are two summable families of power series satisfying \begin{align*} \sum_{s\in S}f_{s}\left( x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\right) =\sum_{t\in T}g_{t}\left( x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\right) \end{align*} for each $n\in\mathbb{N}$, then \begin{equation} \sum_{s\in S}f_{s}=\sum_{t\in T}g_{t}. \label{eq.darij1.2} \tag{2} \end{equation} Indeed, this follows by comparing the coefficient of each monomial $\mathfrak{m}$ (noticing that $\mathfrak{m}$, once again, contains only the indeterminates $x_{1},x_{2},\ldots,x_{n}$ for some $n\in\mathbb{N}$). Note that I am not sure whether limits commute with summable sums in general, but fortunately this is not needed to prove \eqref{eq.darij1.2}.

This almost suffices to derive your equality ($\star\star$) from your ($\star $) (upon taking $\mathbf{k}=\mathbb{Q}\left[ p_{1},p_{2},p_{3},\ldots\right] $), except that we need one more ingredient: The exponential. The definition of the exponential is well-known: If $\mathbf{k}$ is a $\mathbb{Q}$-algebra, and if $f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ is a power series with constant term $0$, then the exponential $\exp f$ of $f$ is defined by \begin{align*} \exp f=\sum_{k\in\mathbb{N}}\dfrac{f^{k}}{k!}, \end{align*} which is easily seen to be a sum of a summable family. Thus, we obtain a map $\exp$ \begin{align*} & \text{from }\left\{ \text{power series }f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \text{ with constant term }0\right\} \\ & \text{to }\left\{ \text{power series }f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \text{ with constant term }1\right\} . \end{align*} This map $\exp$ is furthermore continuous (because for each monomial $\mathfrak{m}$, we can compute $\left[ \mathfrak{m}\right] \left( \exp f\right) $ from knowing only the finitely many coefficients $\left[ \mathfrak{n}\right] f$ for the finitely many monomials $\mathfrak{n}$ that divide $\mathfrak{m}$). As a consequence, it commutes with taking limits and with taking sums of summable families. The substitution homomorphism \begin{align*} \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] &\to \mathbf{k}\left[ \left[ x_{1},x_{2},\ldots,x_{n}\right] \right] , \\ f &\mapsto f\left(x_1,x_2,\ldots,x_n,0,0,0,\ldots\right) \end{align*} is also continuous for each $n \in \mathbb{N}$, and commutes with $\exp$. The rest is substitution.


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