We know that $0 \subseteq V_n \subseteq \mathbb{C}^n \cong \mathbb{1}_n \oplus V_n$ is a composition series for the natural $\mathbb{C}[S_n]$-module $\mathbb{C}^n$ for all $n \geq 2$. Now we have compatible inclusions $S_n \subseteq S_{n+1}$, $\mathbb{C}^n \subseteq \mathbb{C}^{n+1}$, $V_n \subseteq V_{n+1}$, which gives $$ 0 \subseteq \bigcup_n V_n \subseteq \bigcup_n\mathbb{C}^n =: \mathbb{C}^\infty $$ as $S_\infty := \bigcup_n S_n$-modules. In fact, it can be easily argued that $\bigcup_n V_n$ — the vectors whose coefficients sum to $0$ — is the only proper submodule of $\mathbb{C}^\infty$. Hence the length of $\mathbb{C}^\infty$ agrees with the length of $\mathbb{C}^n$ for $n$ big enough.

More generally, for $k$ fixed, the length of $(\mathbb{C}^n)^{\otimes k} \cong (\mathbb{1}_n \oplus V_n)^{\otimes k} \cong \bigoplus_{0 \leq i \leq k} \binom{k}{i} V_n^{\otimes i}$ stabilises for $n \geq 2k$ (and the multiplicities are given in terms of the stable Kronecker coefficients). On the other hand, Sam and Snowden's paper Stability patterns in representation theory as referred to by this answer to Specht modules for symmetric group $S_{\infty}$ proves that $(\mathbb{C}^\infty)^{\otimes k}$ has finite length as well. In fact, I believe an elementary argument shows that here the length of the infinite module is bounded above by the length of the finite one: if $0 \subsetneq M_1 \subsetneq \dots \subsetneq M_l = (\mathbb{C}^n)^{\otimes k}$, we can pick $l$ vectors $v_i \in M_i \setminus M_{i-1}$ witnessing the strictness of inclusions; then by picking $N$ big enough to accommodate all the $l$ vectors, we have $$0 \subsetneq \mathbb{C}[S_N] \cdot v_1 \subsetneq \mathbb{C}[S_N] \cdot v_1 + \mathbb{C}[S_N] \cdot v_2 \subsetneq \dotsb \subsetneq \sum_{1 \leq i \leq l} \mathbb{C}[S_n] \cdot v_i \subseteq (\mathbb{C}^N)^{\otimes k}$$ which shows that $l \leq \operatorname{length} (\mathbb{C}^N)^{\otimes k}$. My question is: do we have equality?

For $k = 2$, I worked out two composition series

which have inductive limits in $(\mathbb{C}^\infty)^{\otimes k}$, as I found generators in $\mathbb{C}^4 \otimes \mathbb{C}^4$ for each subspace: for example, if we write $\mathbb{C}^n = \bigoplus_i \mathbb{C} e_i$,

  • $U_n = \mathbb{C}[S_n] \cdot (e_1 e_2 + e_2 e_3 + e_3 e_4 + e_4 e_1)$ where $e_i e_j := \frac{1}{2} (e_i \otimes e_j + e_j \otimes e_i)$ for all $n \geq 4$;
  • $S^2 V_n = \mathbb{C}[S_n] \cdot (e_1 e_1 - e_1 e_2 + e_2 e_2)$ — viewing $e_i \otimes e_j$ as an edge $i \to j$, this space consists precisely of the weighted undirected graphs in $n$ vertices (possibly with loops) where at each vertex the edge weights sum to $0$;
  • $V_n \otimes \mathbb{C}^n = \mathbb C [S_n] \cdot (e_1 \otimes e_1 - e_1 \otimes e_2)$ consists precisely of weighted directed graphs where at each vertex, the outgoing edge weights sum to $0$.

Therefore, the infinite length and the finite length agree for $k = 2$ as well.

Of course this method of ‘finding a chain and hoping that it lifts to a chain for $S_\infty$’ is very elementary and ad hoc. Are there any machineries I can use?

  • $\begingroup$ $\newcommand\C{\mathbb C}$Just to have it written, $V_n = \ker\bigl(\sum : \C^n \to \C\bigr)$ is the reflection representation of $S_n$, right? $\endgroup$
    – LSpice
    Nov 19, 2022 at 19:10
  • 3
    $\begingroup$ @LSpice that's right. I'm very much a novice, so for me this is yet another new mysterious name after the 'standard representation'. $\endgroup$ Nov 19, 2022 at 19:13
  • 1
    $\begingroup$ Since you ask for a machine, I give an algorithm (including sage code!) for calculating tail multiplicities in the Sam-Snowden category here: arxiv.org/abs/1808.07803 $\endgroup$ Nov 23, 2022 at 12:50

1 Answer 1


To make sure I have understood correctly, I have repeated part of what you wrote in the next paragraph.

The aim is to to compare the lengths of the representation $(\mathbb{C}^{\infty})^{\otimes k}$ of $S_\infty$ with the related representation $(\mathbb{C}^n)^{\otimes k}$ of $S_n$ (where $n$ is sufficiently large). You have observed that if we take a filtration $$ 0 \subsetneq \mathbb{C}S_\infty \cdot v_1 \subsetneq \mathbb{C}S_\infty \cdot v_2 \subsetneq \cdots \subsetneq \mathbb{C} S_\infty \cdot v_l = (\mathbb{C}^\infty)^{\otimes k}, $$ then since each $v_r$ is in $\mathbb{C}^\infty = \varinjlim \mathbb{C}^n$ (or $\bigcup_n \mathbb{C}^n$, if you prefer), then in particular each $v_r$ is contained in $\mathbb{C}^{n_r}$ for some $n_r$. Letting $n$ be the maximum of $n_1, n_2, \ldots, n_l$, we find that each $v_r$ is an element of $\mathbb{C}^n$. Moreover, since $\mathbb{C}S_\infty \cdot v_r \subsetneq \mathbb{C}S_\infty \cdot v_{r+1}$, we have $v_{r+1} \notin \mathbb{C}S_\infty \cdot v_r$, so certainly $v_{r+1} \notin \mathbb{C}S_n \cdot v_r$, and hence $\mathbb{C}S_n \cdot v_r \subsetneq \mathbb{C}S_n \cdot v_{r+1}$. The conclusion is that for large $n$, $(\mathbb{C}^n)^{\otimes k}$ has length at least as long as in the case where $n=\infty$.

Here is a quick way to see that the lengths are the same. The length of a module $M$ can be found by passing to the Grothendieck group, and expressing $[M]$ as a linear combination of (the classes in the Grothendieck group of) simple modules; the length is simply the sum of the coefficients. Both $(\mathbb{C}^n)^{\otimes k}$ and $(\mathbb{C}^\infty)^{\otimes k}$ are obtained from the trivial representation (of $S_n$ and $S_\infty$ respectively) by tensoring $k$ times with $\mathbb{C}^n$ and $\mathbb{C}^\infty$ respectively. The tensor product multiplicities for representations of the finite symmetric groups $S_n$ are the Kronecker coefficients, while for $S_\infty$ are the stable Kronecker coefficients. Because the reduced Kronecker coefficients are stable limits (meaning the limit of an eventually constant sequence) of Kronecker coefficients, the tensor product multiplicities in the case of $S_\infty$ are just the stable limits of those in the case of $S_n$, so also the lengths (which are the sum of these multiplicities) agree.

It would be reasonable to ask whether there is a proof of this fact that (like the proof you gave of one inequality) does not rely on a detailed understanding of $\mathrm{Rep}(S_\infty)$. I don't know such a proof offhand, and here is a different example to illustrate why a very general lifting argument might not suffice.

Let $A_n = \mathrm{Mat}_n(\mathbb{C}) \oplus \mathrm{Mat}_1(\mathbb{C})$, which we may view as $(n+1) \times (n+1)$ matrices of complex numbers such that the final row and column are zero except at their intersection (the bottom-right entry). That is, $A_n$ consits of block-diagonal matrices where the blocks have size $n$ and $1$: $$ \begin{bmatrix} \mathbf{M} & 0 \\ 0 & N \end{bmatrix} $$ (here $\mathbf{M}$ is an $n \times n$ matrix and $N$ is a complex number). Expressing $A_n$ in terms of $(n+1) \times (n+1)$ matrices this way gives us an obvious inclusion: $$ A_n \subseteq \mathrm{Mat}_{n+1}(\mathbb{C}) \subseteq \mathrm{Mat}_{n+1}(\mathbb{C}) \oplus \mathrm{Mat}_1(\mathbb{C}) = A_{n+1}. $$ On one hand it is clear that $\mathbb{C}^{n+1} = \mathbb{C}^n \oplus \mathbb{C}^1$ is a length two module over $A_n$. On the other hand, we have $\varinjlim \mathbb{C}^{n+1} = \mathbb{C}^\infty$ and $\varinjlim A_n = \mathrm{Mat}_\infty(\mathbb{C})$, which acts irreducibly on $\mathbb{C}^\infty$. So in the limiting case, the length is only one, and the length in the finite case is strictly larger than in the infinite case.


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