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Consider the standard permutation representation of $ S_n $. This representation decomposes as a trivial representation plus an irreducible representation of degree $ n-1 $, call it $ V $, sometimes called the "deleted permutation representation."

In this question $n-1$ dimensional permutation module for $S_n$ it proven that $ \Lambda^k(V) $ is irreducible for every $ k $.

Now suppose we restrict all this to $ A_n $. It is claimed in a comment here https://math.stackexchange.com/a/4557785/724711 that as an $ A_n $ representation $ \Lambda^k(V) $ is usually irreducible but there are some exceptions where it is reducible. However I have checked in GAP and I am not finding any exceptions. It seems that $ \Lambda^k(V) $ is irreducible as an $ A_n $ representation for all $ k $.

Could someone provide a reference for the correct fact about irreducibility of the exterior powers $ \Lambda^k(V) $ of the $ n-1 $ dimensional representation of $ A_n $? Either confirming my hunch or showing that the linked comment is correct.

Ian Gershon Teixeira
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    For a permutation representation that is $2k$-transitive, the corresponding exterior power $\Lambda^k$ is irreducible. – Steve D Jan 14 '25 at 02:10
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    Wait, maybe I misread what you're talking about. $\Lambda^2 V$ for $A_5$ has degree $6$, and is not irreducible. – Steve D Jan 14 '25 at 02:15
  • @SteveD ah yes I guess I was thinking more for $ n=4, n\geq 6 $ because then $ SO(n-1) $ is a simple Lie group and so (if we let $ V $ denote the natural module of $ SO(n-1) $) then $\Lambda^2(V)$ is the adjoint representation and is irreducible since $ SO(n-1) $ is simple for $ n=4, n \geq 6 $. As you point out the deleted permutation representation for $ n=5 $ is $ A_5 \hookrightarrow SO(4) $ and so $ \Lambda^2(V) $ is a degree 6 reducible representation since $ SO(4) $ is semisimple $ \Lambda^2(V) \cong \mathfrak{so}(4) \cong \mathfrak{so}(3) \oplus \mathfrak{so}(3) $ – Ian Gershon Teixeira Jan 14 '25 at 17:31

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Here is some info. Let $P$ be a permutation representation for the finite group $G$. We know we can write $P\cong I\oplus V$ for the trivial representation $I$. This shows $\Lambda^r P\cong \Lambda^r V \oplus\Lambda^{r-1} V$. This is also explained in Fulton-Harris, Proposition 3.12.

Assume $G$ is acting (2r)-transitively. This is equivalent to $|G|^{-1}\sum \chi^{2r}(g)=B_{2r}$, the $2r$th Bell number. Here $\chi$ is the permutation character. The proof of this equivalence is explained in exercises 11-13 of chapter 1 of Serre's Finite Groups: an Introduction.

Now inductively, these two facts -- together with the inductive definition of the Bell numbers -- show that if $\chi_r$ is the character associated to $\Lambda^r P$, then $\langle \chi_r,\chi_r\rangle=2$, and thus (again inductively) $\Lambda^rV$ is irreducible. It also shows that if $G$ is not $(2r+2)$-transitive, then $\langle \chi_{r+1},\chi_{r+1}\rangle>2$, so that $\Lambda^{r+1}V$ is reducible.

Applying this to the case of $A_n$ and the standard permutation action (which is $(n-2)$-transitive), we see that -- together with $\Lambda^n P$ being the sign representation -- $\Lambda^k V$ is irreducible if and only if $n\neq 2k+1$.

A little more character theory shows that $\Lambda^k V$ for $n=2k+1$ is the direct sum of two irreducible representations of $A_n$. This is somewhere in Isaacs's Character Theory book, but I don't have that handy to look it up.

Steve D
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    Ok thank you this is very helpful. I see now that when you start with $ A_n $ and there is a value $ k $ such that $ n=2k+1 $ then, for that particular value of $ k $, $ \Lambda^k(V) $ decomposes into two irreps, each of degree $ \frac{1}{2}$$ {2k}\choose{k} $ and related by the order 2 outer automorphism of $ A_n $. – Ian Gershon Teixeira Jan 14 '25 at 18:01
  • Side note: is the $ 2k \pm 1 $ a typo like you meant just $ 2k+1 $? because checking in GAP for $ A_9 $ it seems like when $ 9=2k-1 $ then $ k=5 $ but $ \Lambda^5(V) $ is still irreducible. And in general it seems like since $ V $ has degree $ n-1=2k $ then $ \Lambda^k(V) $ for $ k= \frac{n-1}{2} \pm 1 $ should be canonically isomorphic so either both irreducible or both reducible. – Ian Gershon Teixeira Jan 14 '25 at 18:05
  • @IanGershonTeixeira: Yes, that is a typo (I had it right the first time!). I mixed up odd/even when thinking about $\Lambda^kP$ and $\Lambda^kV$. Fixing it now. – Steve D Jan 14 '25 at 19:29