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

Question

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?

Answer 1

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?

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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.