Corresponding space to a irreducible part of a given representation of $S_n$

Question

My setting is the following:

I have an action of $S_n$ on some $\mathbb{C}$-vector space, $V$, by permuting a special basis:

$$<a^1_{1},\dots,a^1_{n!},a^2_{1},\dots,a^2_{n!},\dots,a^k_{n!},b_1,\dots,b_r> $$

where the orbit of $a^i_{1}$ is the set of all ${a^i_j}$ for $j=1,\dots,n!$ and $S_n$ acts on the rest of the basis by some unknown permutation of elements, but its orbit has not $n!$ elements.

There I have the subspace such that every element of $S_n$ acts by its sign, with the notation of my reference the subspace associated with $[1,\dots,1]$ (lets say the alternated space of $S_n$, also $V_{S_n}$). This can be written as

$$V_{S_n}=<\sum_{\sigma\in S_n}(-1)^{sign(\sigma)}\sigma(a^1_1),\dots,\sum_{\sigma\in S_n}(-1)^{sign(\sigma)}\sigma(a^k_1)> $$

Then I have the action of $S_{n-1}$ as a subgroup of $S_n$, formed by the elements that fix the first entry. My problem is that I know that $V_{S_n}$ will give me, a space that is alternated for $S_{n-1}$. This is because for each $j$ the set of $a^j_i$ for $i=1,\dots,n!$ contains $n$ orbits for the subgroup $S_{n-1}$, and I can perform a similar computation than the one to find $V_{S_n}$. But in general it is not the whole space such that $S_{n-1}$ acts by its sign, i.e. the alternated space for $S_{n-1}$ could have more elements from the $b_i$.

I have been told to look for representation theory (I am reading James Kerber, The Representation Theory of the Symmetric Group) and I have concluded that I want to find the subspace associated to $[2,1,\dots,1]$, as defined in page 36 of that book, to find the rest of the alternated space that does not come from $V_{S_n}$.

My question is the following:

How do I find the subspace associated to $[2,1,\dots,1]$?

To find $[2,1,\dots,1]$ I need to induce some representations but the induced representation is an action over a different space, how can I find my subspace related to $[2,1,\dots,1]$ if I am changing the space to find $[2,1,\dots,1]$? Am I getting something wrong?

Answer 1

Since the question has changed to include a particular representation, I'll add a separate answer.

The regular representation of $S_n$ is an $n!$-dimensional vector space with basis $\{v_\sigma:\sigma\in S_n\}$ a collection of linearly independent vectors with $\sigma\cdot v_{\tau}=v_{\sigma\tau}$ the group action, for $\sigma,\tau\in S_n$. In the representation you give, the orbit condition means that it splits up into $k$ copies of the regular representation along with one copy of the unknown $\langle b_1,\dots, b_r\rangle$ representation. $$ V = R\oplus R\oplus\dots\oplus R\oplus\langle b_1,\dots, b_r\rangle$$ That is, each $\langle a_1^j,\dots,a_{n!}^j\rangle$ is an invariant subspace isomorphic to $R$. Let $B=\langle b_1,\dots, b_r\rangle$.

The representation $R$ is known to contain each irreducible representation of $S_n$ with multiplicity given by the dimension of that irreducible representation. For example, the trivial and sign/alternating representations are each contained with multiplicity $1$ since they are both $1$-dimensional.

Restricting $V$ to $S_{n-1}$ involves seeing how each of the irreducible representations that exist in $R$ and $B$ split, using the branching rules.

If all we care about are the alternating representations, then $[2,1,\dots,1]$ and $[1,1,\dots,1]$ (partitions of $n$) are the only partitions that restrict to $[1,1,\dots,1]$ (partition of $n-1$). The irreducible representation for $[2,1,\dots,1]$ is known to be $(n-1)$-dimensional, so $R$ contains a $(n-1)^2$-dimensional invariant subspace corresponding to this, which when restricted to $S_{n-1}$ contains a $(n-1)$-dimensional alternating invariant subspace. So each $R$ contributes $1+(n-1)=n$ dimensions of alternating $S_{n-1}$ representations, $n-1$ dimensions of which did not come from the alternating $S_n$ representation.

Without knowing what $B$ is, there is not much to say. It could have any number of $[1,1,\dots,1]$ and $[2,1,\dots,1]$ components already.

That said, we can use the character-based projection formulas. The $S_n$ alternating part of $R$ is spanned by $\sum_{\sigma \in S_n} (-1)^\sigma v_\sigma$, and the $S_{n-1}$ alternating part is projected onto by $$\pi_{\mathrm{alt}}(w) = \frac{1}{(n-1)!}\sum_{\sigma\in S_{n-1}} (-1)^{\sigma} \sigma\cdot w$$ It is the same formula for $B$.

If you want to locate the alternating part that came from the $[2,1,\dots,1]$ part itself, you can use the projection formula for it then intersect the two subspaces. To describe the character, let $\operatorname{tr}\sigma$ denote the number of fixed points for the permutation $\sigma$ (i.e., $\operatorname{tr}\sigma=\#\{i:\sigma(i)=i\}$). The $[2,1,\dots,1]$ representation is the tensor product of the "reduced standard representation" with the alternating representation; I'll denote this by $-\mathrm{std}$. $$\chi_{-\mathrm{std}}(\sigma) = (-1)^\sigma(\operatorname{tr}(\sigma)-1)$$

Then, for a representation $\phi$ of $S_n$, the projector is $$\pi_{-\mathrm{std}}(w) = \frac{n-1}{n!}\sum_{\sigma\in S_n} (-1)^\sigma(\operatorname{tr}(\sigma)-1)\phi_\sigma(w)$$ (note that $(-1)^\sigma=(-1)^{\sigma^{-1}}$ and $\operatorname{tr}(\sigma)=\operatorname{tr}(\sigma^{-1})$).

The composition $\pi_{\mathrm{alt}}\circ\pi_{-\mathrm{std}}$ gives the $S_{n-1}$ alternating part that was not the $S_n$ alternating part.

Answer 2

To give some notation: let's have $v_1,\dots,v_n\in V_{S_n}$ be a basis where $S_n$ acts by permutation: for all $\sigma\in S_n$ and $1\leq i\leq n$, then $\sigma$ acts on $v_i$ by $\sigma\cdot v_i= v_{\sigma(i)}$. This is the "natural permutation representation."

This representation splits into two subrepresentations, the one corresponding to the partition $[n]$ and the one corresponding to the partition $[n-1,1]$. These are conjugate to the partitions you give. The $[n]$ partition corresponds to the trivial representation, and it is spanned by $\sum_{i=1}^n v_i$. The $[n-1,1]$ partition is spanned by an orthogonal complement of the trivial representation, a subspace characterized by vectors $\sum_{i=1}^nc_i v_i$ with $\sum_{i=1}^n c_i=0$; the subspace is spanned by the set $$\{v_j-\frac{1}{n}\sum_{i=1}^nv_i:1\leq j\leq n\}.$$ While this is not a basis, removing any vector from this set gives an independent set.

If you modify the action of $S_n$ on $V_{S_n}$ so that, in addition, the vector is multiplied by the sign (so that $\sigma\cdot v_i = (-1)^\sigma v_{\sigma(i)}$, where $(-1)^\sigma$ is the sign of $\sigma$), then you do get the subrepresentations corresponding to the partitions you mentioned. The representation still splits along the same subspaces given above.

The effect on a decomposition from restricting a representation to a subgroup (like $S_{n-1}\subset S_n$) is that the subrepresentions might no longer be irreducible, and they might split further. The branching rules for the symmetric group are that you take a partition and find all the ways of subtracting $1$ so that the partition remains valid (in nonincreasing order). The $[n]$ partition has only one way, so it gives $[n-1]$, the trivial representation of $S_{n-1}$. This is spanned by the same vector as before. The $[n-1,1]$ partition has two ways, giving $[n-2,1]$ and $[n-1,0]=[n-1]$, so the second subspace splits into two irreducible representations of $S_{n-1}$, with the second being an additional trivial subrepresentation.

By staring at the basis for a little while, I noticed that $v_1-\frac{1}{n}\sum_{i=1}^nv_i$ is left invariant by $S_{n-1}$. This is the additional trivial subrepresentation. Again, orthogonal to this must be the $[n-2,1]$ subrepresentation. Here are spanning sets: \begin{align*} A_1 &= \{\sum_{i=1}^n v_i\}\\ A_2 &= \{v_1-\frac{1}{n} \sum_{i=1}^n v_i\}\\ A_3 &= \{v_j-\frac{1}{n-1}\sum_{i=2}^n v_i:2\leq j\leq n\} \end{align*} where $A_1\cup A_2$ together spans the two-dimensional space for the trivial $S_{n-1}$ subrepresentation of $V_{S_n}$ and $A_3$ spans the $[n-2,1]$ subrepresentation. Just like the case of repeated eigenvalues, that two-dimensional subspace should be taken as a two-dimensional eigenspace; it only splits into two subspaces somewhat canonically from the noncanonical choice of how $S_{n-1}$ is a subgroup of $S_n$ (there are many ways!)

Again, if we do the modified representation where we multiply by the sign, to match the partitions you give in your question, these are still the correct spanning sets.

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A general computational way to locate a subrepresentation for a representation of a finite group is to calculate characters and then use those calculate a projection matrix. If $V$ is an irreducible representation of a finite group $G$ and $\chi_V(g)=\operatorname{tr}(\rho_g)$ is the character function, where $\rho_g:V\to V$ the action of $g\in G$ on $V$, then for $W$ another representation of $G$ with $\psi_g$ the action of $g\in G$ on $W$ we have a projector $\pi_V:W\to W$ whose image is the $V$ part of $W$. It can be given by $$\pi_V(w) = \frac{\dim V}{|G|} \sum_{g\in G} \chi_V(g^{-1})\psi_g(w).$$

For example, the trivial representation $T$ of $S_{n-1}$ has the character function $\chi_T(g)=1$ for all $g\in S_{n-1}$. The projector then is $$\pi_T(v_i) = \frac{1}{(n-1)!}\sum_{\sigma\in S_{n-1}}v_{\sigma(i)}.$$ Notice what this does to a vector is average components $2$ through $n$. That is, $$\pi_T(\sum_{i=1}^n c_iv_i) = c_1v_1 + \frac{1}{n-1}\left(\sum_{i=2}^n c_i\right)\left(\sum_{i=2}^n v_i\right).$$ So, another basis for the $S_{n-1}$ trivial part of $V_{S_n}$ is $$\{v_1,\sum_{i=2}^n v_i\}.$$