# Young diagram for exterior powers of standard representation of $S_{n}$

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!

Edit

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

• 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. – BIS HD Nov 22 '13 at 11:56

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 $(n-s, 1^s)$.

• I guess this is not too different from your own answer. But I feel it's a little more direct. – Amritanshu Prasad Nov 27 '13 at 8:54
• 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? – Stefano Nov 27 '13 at 15:18
• @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. – Amritanshu Prasad Nov 29 '13 at 4:24
• I still haven't gotten around to fixing this. Making it Community Wiki. Maybe someone else can. – Amritanshu Prasad Dec 7 '13 at 2:50
• 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. – J Swanson Apr 10 '17 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.