# Representation ring of the symmetric group $S_n$ in the limit as $n \to \infty$

Let $$S_n$$ denote the symmetric group on $$n$$-letters and let $$\mathrm{Rep}(S_n)$$ denote its representation ring. For every $$n$$ restriction along the inclusion $$S_{n-1} \to S_n$$ induces a ring homomorphism $$\mathrm{Res}^{S_n}_{S_{n-1}} :\mathrm{Rep}(S_n) \to \mathrm{Rep}(S_{n-1 })$$. Considering these together we get a diagram of the following form $$\dots \to \mathrm{Rep}(S_n) \to \mathrm{Rep}(S_{n-1 }) \to \dots \to \mathrm{Rep}(S_2) \to \mathrm{Rep}(S_1) \simeq \mathbb{Z}$$

Question: Is there a generators and relations description of the limit $$\underset{n }{\varprojlim} \,\mathrm{Rep}(S_n)$$ as ring?

Edit: Will Sawin has pointed out in the comments that the limit, must have uncountably many generators in any presentation. Let me slightly modify the question in the hopes of making it more answerable.

Lets consider instead of $$\mathrm{Rep}(S_n)$$ its completion with respect to the augmentation ideal $$\widehat{\mathrm{Rep}}(S_n):=\varprojlim_m \mathrm{Rep}(S_n)/I^m$$ where $$I:= \ker(\dim : \mathrm{Rep}(S_n) \to \mathbb{Z})$$. We still have a tower as above and we can ask the following variation.

Question: Is there an explicit countable presentation of the limit $$\underset{n }{\varprojlim} \,\widehat{\mathrm{Rep}}(S_n)$$ in some suitable category of complete topological rings?

• Do you not just get the famous “ring of symmetric functions” in the limit? Dec 12, 2021 at 14:48
• @SamHopkins I believe this is the case if $S_n$ is replaced with polynomial representations of $GL_n$ (by Schur Weyl duality). I suspect the answer for symmetric groups (if it exists) should be quite different. Dec 12, 2021 at 14:49
• Given that this ring is uncountable (it's isomorphic, as a group, to $\prod_{n=1}^{\infty} \mathbb Z$) we would need uncountably many generators. Dec 12, 2021 at 15:08
• Dec 12, 2021 at 16:40
• Not quite what you ask but: Character theory roughly says that Rep(G) is dual to the center of the group ring of G. On that side there is the Farahat-Higman ring, which is a sort of limit of the centers of the group algebras of S_n, and it has a pretty nice presentation.
– Nate
Dec 15, 2021 at 16:27

It follows from Hooks generate the representation ring of the symmetric group by Ivan Marin that $$\operatorname{Rep}(S_n)$$ is generated, and thus the inverse limit is topologically generated, by the elements $$x_k = \sum_{i=0}^k (-1)^i \binom{n+i -2}{i}\wedge^{k-i} (\operatorname{std} )$$

So the ring is some kind of completion of $$\mathbb Z[x_1,x_2,\dots ]$$.

Moreover, the augmentation ideal of this ring is profinite, since the augmentation ideal of $$\operatorname{Rep}(S_n)$$ modulo any power of the augmentation ideal of $$\operatorname{Rep}(S_n)$$ is finite.

So the simplest guess is that it is obtained from $$\mathbb Z[x_1, x_2,\dots]$$ by the inverse limit over all ideals which are finite-index subgroups of the "augmentation ideal" $$(x_1,x_2,\dots)$$.

However, this is not correct, because the sign representation is an element of the inverse limit which squares to $$1$$ but is nontrivial modulo $$I^2$$ for all $$n$$. However, the completion of $$\mathbb Z[x_1,x_2,\dots]$$ along all ideals which are finite-index subgroups of $$(x_1,x_2,\dots)$$ does not contain a nontrivial element which squares to $$1$$.

A first step to understand this completion would be to understand it modulo $$I^2$$. In other words, what is the inverse limit over $$n$$ of $$I_n /I_n^2$$, where $$I_n$$ is the augmentation ideal in the representation ring of $$\operatorname{Rep}(S_n)$$? This is some kind of countably-topologically-generated profinite abelian group, but which one?

As an alternate, more well-behaved ring, I propose to restrict attention to representations whose norms have at most polynomial growth in $$n$$, where the norm of a class $$\alpha$$ is $$\min\left\{ \dim V + \dim W \mid V, W \textrm{ representations of }S_n, [V]-[W]=\alpha \right\} .$$ I expect that this restriction of the inverse limit is a polynomial ring in the $$x_k$$ plus one more generator, the sign representation, that squares to one, but I haven't checked this.

• I guess the norm of $V - W$ should be $\lvert\dim(V) - \dim(W)\rvert$? Dec 12, 2021 at 16:56
• @LSpice Sorry, I meant an implicit minimization, now made explicit. And you're right that the absolute values were unnecessary. Dec 12, 2021 at 17:32
• That paper by Ivan Marin is great! It doesn't quite answer my original question but it did shift my view on this. Thanks for sharing it! Dec 14, 2021 at 7:46