Induced and restricted representation

Consider the obvious inclusion $S_3 \subset S_4$ and let $V$ be the permutation representation of $S_3$.
 Compute the character of $Ind_{S_3}^{S_4} (V )$ (Induced representation) and decompose it as direct sum of irreducible representations; (b) Decompose $ Res_{S_3}^{S_4} \circ Ind_{S_3}^{S_4} (V )$ (Composition of restriction with the induced representation) as a direct sum of ireducible representations.

  • By permutation representation you mean the 3-dimensional one or the 6-dimensional one?

  • To be more precise with the names, is it the standard representation or the left-regular one?

  • Also, what is your background here? Do you know that irreducible representations of the symmetric group are indexed by partitions, and what the rules for induction and restriction are? Or do I have to do the computation "by hand"?


I assume that by "permutation representation" you mean $S_3$ acting on $\mathbb{C}^3$ by permuting the basis vectors.

The permutation representation on $S_n$ is the direct sum of the representationss indexed by the partitions $(n)$ and $(n-1,1)$. Now, there is a combinatorial rule for computing the induced representation: thinking of the partitions as corresponding to their Young diagrams, inducing the representation corresponding to a partition gives the sum over all partitions obtained by adding one box to the given partition. When you do this you get the direct sum \[ Ind^{S_{n+1}}_{S_n} ((n) \oplus (n−1,1))=(n+1) \oplus (n,1) \oplus (n,1) \oplus (n−1,2) \oplus (n−1,1,1). \] Similarly, restricting a representation of $S_{n+1}$ to $S_n$ we get the sum over all partitions obtained by removing one box from the given partition, so we get \[ Res_{S_n}^{S_{n+1}} Ind^{S_{n+1}}_{S_n} ((n) \oplus (n−1,1)) = (n) \oplus ((n) \oplus (n-1, 1)) \oplus ((n) \oplus (n-1, 1)) \\ \oplus ((n-1,1) \oplus (n-2, 2)) \oplus ((n-1,1) \oplus (n-2, 1, 1)). \] Note that, for $n=3$, $(n-2, 2)$ is not a partition and so that should be removed.

Note that these two rules are equivalent by Frobenius reciprocity, and both can be easily derived by definition of Specth module.

Let me know if this is satisfying or if the exercise was meant to be done with elementary tools.


Long answer (incomplete atm - will edit later):

Let's take as $S_4/S_3$ coset representatives the elements $g_i = (i,4)$ for $1 \leq i \leq 4$. The representation $Ind_{S_3}^{S_4}(V)$ acts on the space \[ W = \bigoplus_{i=1}^4 g_i V. \] Let us have $V = \langle e_1, e_2, e_3 \rangle$ with $\sigma(e_j) = e_{\sigma(j)}$ for $\sigma \in S_3$, and call $e_j^i$ the copy of $e_j$ in $g_i V$. Let $\sigma \in S_4$. We have $\sigma g_i = g_{\sigma(i)} \overline{\sigma}$ for some $\overline{\sigma} \in S_3$ (seen as an element of $S_4$ that fixes $4$. Indeed, $\sigma g_i(4) = \sigma(i)$ and $g_{\sigma(i)} \overline{\sigma}(4) = g_{\sigma(i)}(4) = \sigma(i)$, so the computation checks out. It follows that, is $\sigma$ has no fixed points, it permutes the subspaces and so the trace is $0$.

We want to compute the character now. We have $5$ conjugacy classes in $S_4$, given by the partitions $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, $(1,1,1,1)$. We immediately have that $\chi_W(4) = \chi_W(2,2) = 0$ as they do not have fixed points. Then, $\chi_W(3,1) = 0$ because it fixes one subspace (say $g_i V$), but then permutes the basis of that subspace $e_1^i, e_2^i, e_3^i$ keeping no fixed points. Next, $\chi_W(2,1,1) = 2$ as it fixes two subspaces, and one basis vector within each subspace. Finally, $\chi_W(1,1,1,1) = 12$ because the character of the identity always gives the dimension of the space.

We need the character table of $S_4$ now (see attachment - it can be computed by starting from the trivial representation and $\tilde{\rho}$ (the permutation rep minus the trivial), then tensoring with the sign, and then filling the rest by orthogonality relations.

We now want to decompose $Ind_{S_3}^{S_4}(V)$ in irreducibles. For sure, the sum of all the basis vectors is fixed by any permutation, so that gives an invariant subspace of dimension one, which corresponds to the partition $(4)$ (the trivial representation). Next, the space generated by the vectors $e^i_1 + e^i_2 + e^i_3$ for $1 \leq i \leq 4$, on which $S_4$ acts by permuting the upper indices $i$; this is the standard permutation representation, so, the trivial representation is a subrepresentation, and the orthogonal complement forms an irreducible representation associated to the partition $(3,1)$, (so, $\rho$).

Now, since $\chi_{Ind_{S_3}^{S_4}(V)}((123)) = 0$, and the only representation whose character has a minus sign in that column is $\phi$, we need to include at least one copy of that. So far, summing the characters of our three irreducibles gives the values $(6, 2, 0, 0, 0)$, so we need two more representations whose global character gives $(6,0,0,0,0)$. Looking at the values on $(12)(34)$, we immediately see that we cannot include another copy of $\phi$, as otherwise we would need two more $3$-dimensional representations to compensate and there's not enough space left. But if we can't include $\phi$, then we can't include the trivial or the sign either, because of the values on $(123)$. It is now immediate that we are left with a copy of $\tilde{\rho}$ and a copy of $\tilde{\rho} \otimes sgn$: the total dimension is $12$, so we have our decomposition. The characters check out. In the end,

\[ Ind^{S_{n+1}}_{S_n} (V)= 1_4 \oplus \tilde{\rho} \oplus \tilde{\rho} \oplus \phi \oplus (\tilde{\rho} \otimes sgn). \]
For the restriction, we can use Frobenius reciprocity. We have three irreducible representations of $S_3$, namely the trivial, the sign, and the permutation minus trivial (let us call them $1_3, sgn_3, \rho_3)$.

Now, using pretty much the same argument as above, $Ind_{S_3}^{S_4} 1_3$ gives the permutation representation of $S_4$, which decomposes as $1_4 \oplus \tilde{\rho}$.

Next, $Ind_{S_3}^{S_4} sgn_3$ gives the signed permutation representation of $S_4$ (the same, but tensoring with the sign), which decomposes as $sgn_4 \oplus (\tilde{\rho} \otimes sgn_4)$.

Finally, the induced of the standard representation $Ind_{S_3}^{S_4} \rho_3$ decomposes as the complement of the induced of the trivial in $Ind_{S_3}^{S_4} V$, so as $\tilde{\rho} \oplus \phi \oplus (\tilde{\rho} \otimes sgn_4)$.

Using the Frobenius reciprocity, $1_4$ restricts to $1_3$ (because $1_4$ appears once in the induced from $1_3$ and nowhere else, then $1_3$ appears with the same multiplicity in the restriction of $1_4$), $\tilde{\rho}$ restricts as $1_3 \oplus \rho$ (same reason - and the analogous when tensoring with the sign), and $\phi$ restricts as $\rho$ (again, same reason). Putting everything together, we have \[ Res_{S_n}^{S_{n+1}} Ind^{S_{n+1}}_{S_n} (V) = 1_3 \oplus (1_3 \oplus \rho) \oplus (1_3 \oplus \rho) \oplus (sgn_3 \oplus \rho) \oplus \rho.  \]

  • Jbuck Jbuck

    I would prefer it if you could do it with elementary tools , instead. You can use Frobenius reciprocity but I have no knowledge of Specht modules or Young diagrams. (And indeed the permutation representation is the one that pemrutes the basis vectors)

  • The Specth modules are just the irreducibles corresponding to a given partition, and Young diagrams are just a way to depict partitions. I'll try to redo the computation with elementary tools, however. Are you familiar with the hook-length formula?

  • Can you provide me a textbook or any reference material with the tools at your disposal?

  • Jbuck Jbuck
    0 Here's a rough draft of our lecture notes, see the pdf Group Representation Theory under the section Lecture Notes

  • I hope this is elementary enough, I don't know how to simplify it any further, bar finding the irreducibles by hand (which is not easy at all).

  • Jbuck Jbuck

    This should be fine, but could you please write down the sum decomposition of the induced explicitly?

  • Jbuck Jbuck

    Also, can you please elaborate a bit on the Frobenius reciprocity (show why the restrictions you mention are true by starting from the definition of the reciprocity?

  • Let me know if it's good now. The restrictions are like that because, if X appears n times in the induced from Y, then Y appears n times in the restricted from X, and I gave all the induces ones explicitly, so to find the multiplicities of the restricted ones you just need to count now many times they appear in total in the induced ones.

The answer is accepted.
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