I'm trying to solve ex. 4.6 in Fulton and Harris' book "Representation Theory". It asks about the Young diagram associated to the standard representation of $S_{n}$ and of its exterior powers. The one of the standard representation $V$ is the partition $\left( n-1, 1 \right)$, while for $\Lambda^{s}V$ is $\left( n-s , 1 , \ldots , 1 \right)$.

For the standard representation I found some hints, such as here, but for the second part I have no clue. The book advises to use Frobenius formula or the branching rule (known also as Pieri's formula)... I tried to think about something like induction on both $n$ and $s$, since the base of the induction for $s$ is given by the standard representation, but then I just stared at the sheet...

Thanks in advance for any hint!


Maybe I found an explanation, but I'm not sure it is ok!

  • $\begingroup$ Just a guess and far away from being a good one: Maybe character theory could help you? Since the characters of the exterior power of the standard representation are easy to calculate. $\endgroup$
    – BIS HD
    Nov 22, 2013 at 11:56

2 Answers 2


Here is a solution using Pieri's rule:

The representation $\wedge^s V$ has as basis vectors: $$ \{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n\}. $$

If we restrict this representation to $S_{n-1}$, then the representation on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n-1\}$ is just the representation of $S_{n-1}$ on $\wedge^s V_{n-1}$, where $V_{n-1}$ is the subspcace of $V$ consisting of vectors with the last coordinate equal to $0$.

On the other hand, the representation of $S_{n-1}$ on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_{s-1}}\wedge e_n\mid 1\leq i_1<\dotsb <i_{s-1}\leq n-1\}$ is isomorphic to the representation $\wedge^{s-1}V_{n-1}$ of $S_n$.

Therefore, by induction hypothesis, the restriction of $\wedge^s V$ to $S_{n-1}$ is the sum of the representation corresponding to $(n-s, 1^{s-1})$ and the representation corresponding to $(n-s-1, 1^s)$. It follows from Pieri's rule that $\wedge^s V$ is the representation corresponding to $(n-s, 1^s)$.

  • $\begingroup$ I guess this is not too different from your own answer. But I feel it's a little more direct. $\endgroup$ Nov 27, 2013 at 8:54
  • $\begingroup$ Thanks for the answer! Just two questions. What do you mean by "each $w \in S_{n}$ acts on basis vector $e_{i_{1}} \wedge \ldots \wedge e_{i_{s}}$ by $\pm 1$"? I see it if $w$ sends $\lbrace i_{1} \ldots i_{s} \rbrace$ in itself, but is it still true if the permutation does not send the set in itself? The second is just about the index. Since $\mathrm{dim}V_{n}=n-1$, shouldn't $n$ be replaced by $n-1$ and $n-1$ by $n-2$ when considering sets of index? $\endgroup$
    – Stefano
    Nov 27, 2013 at 15:18
  • $\begingroup$ @Stefano You are right - $w$ need not act by scalars on these basis vectors. Also these are basis vectors for $\wedge^s\mathbb C^n$, not $V$, so it seems this solution needs more work. $\endgroup$ Nov 29, 2013 at 4:24
  • $\begingroup$ I still haven't gotten around to fixing this. Making it Community Wiki. Maybe someone else can. $\endgroup$ Dec 7, 2013 at 2:50
  • 1
    $\begingroup$ What does it matter whether $w$ acts by scalars? It's not used anywhere in the argument. Not sure what you mean about basis vectors for $\Lambda^s \mathbb{C}^n$ instead of $V$. The argument seems fine to me if you just delete the line about acting by scalars. $\endgroup$ Apr 10, 2017 at 5:39

I am not sure that the proof I tried to sketch is ok. Any correction or hint is welcome!

I'll assume the following: given $S_{n}$ and $V$ it's standard representation, $\Lambda^{0}V , \ldots, \Lambda^{n-1}V$ are irreducible. Furthermore, I'll assume the Hook length formula and the consequence that the only representations of dimension lower than $n$ are given by the partitions $\left(n\right)$, $\left(1,\ldots,1\right)$ (both of dimension $1$), $\left(n-1,1\right)$, $\left(2, 1, \ldots, 1\right)$ (both of dimension $n-1$) (ex. 4.14 Fulton Harris). We'll also assume the branching rule (ex. 4.43, 4.44 Fulton Harris). Finally, we'll assume that it is known that the partition $\left(n\right)$ corresponds to the trivial representation (and hence, by dimension we are given that $\left(1, \ldots, 1\right)$ corresponds to the alternating one).

Notation: since things like $\left(1, \ldots , 1\right)$ can be ambiguous, I'll write $\left( 1, \ldots, 1 \right)_{n}$ to indicate it is a partition of $n$. I'll also write $V_{n}$ to indicate the standard representation of $S_{n}$

First, we check that the standard representation corresponds to $\left(n-1,1\right)$. Assume $n>1$. I'll identify the diagram with the representation as notation. We have $\mathbb{res}^{S_{n+1}}_{S_{n}}\left(n-1,1\right) = \left(n-1\right) \oplus \left(n-2,2\right)$ and $\mathbb{res}^{S_{n+1}}_{S_{n}}\left(2,1,\ldots,1\right)_{n+1} = \left(1,\ldots,1\right)_{n} \oplus \left(2,1,\ldots,1\right)_{n}$. It is easy yo see that the standard representation, when restricted, has a factor which is a trivial representation (we show it in the same way we decompose the representation $\mathbb{C}^{n}$ into the standard and the trivial one). So the standard representation must correspond to $\left(n-1,1\right)$. By dimension argument, $\Lambda^{n-2}V=\left(2,1,\ldots,1\right)_{n}$.

Now we'll proceed by induction both on $n$ and $s$, where $s$ is the index of the exterior power. For $n=1$ everything is trivial, so we have the basis of induction. Now we have to pass from $n$ to $n+1$. We have the base of induction on $s$ (for $s=0$ we have assumed known and for $s=1$ we have shown it). We have to pass from $s$ to $s+1$. Now $s \geq 1$, but since we already know $\Lambda^{n-2}V$ and $\Lambda^{n-1}V$, we can assume $s<n-3$. Now we'll use the fact that restriction and exterior power commute (in fact for a representation $M$, $\Lambda^{s} M = M^{\oplus r} / \left( I \cap M^{\oplus r}\right)$, where $I$ is the ideal in the tensor algebra of $M$ generated by the elements $m \otimes m$ and this operation commutes with the restriction). Now we have by branching rule $\mathrm{res}^{S_{n+1}}_{S_{n}}\left(n-s, 1, \ldots, 1 \right)_{n+1}=\left(n-s, 1 , \ldots, 1 \right)_{n} \oplus \left( n-s-1,1, \ldots, 1\right)_{n}$. By inductive hypothesis we know it is $\Lambda^{s+1}V_{n} \oplus \Lambda^{s}V_{n}$. Now we observe that because of the form of the diagrams we are considering (Fulton Harris call them hooks) and because of branching rule, no other irreducible representation of $S_{n+1}$ can restrict to this representation of $S_{n}$. Now we consider $\Lambda^{s+1}V_{n+1}$, which is known to be irreducible. We have $\mathrm{res}^{S_{n+1}}_{S_{n}}\Lambda^{s+1}V_{n+1}=\Lambda^{s+1}\mathrm{res}^{S_{n+1}}_{S_{n}}V_{n+1}=\Lambda^{s+1}\mathrm{res}^{S_{n+1}}_{S_{n}}\left(n,1\right)=\Lambda^{s+1}\left( \left(n\right) \oplus \left( n-1,1\right) \right)=\left(\Lambda^{s+1}\left(n-1,1\right)\otimes \Lambda^{0}\left(n\right)\right) \oplus \left(\Lambda^{s}\left(n-1,1\right)\otimes \Lambda^{1}\left(n\right)\right)=\Lambda^{s+1}V_{n} \oplus \Lambda^{s}V_{n}$.

So we are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.