3
$\begingroup$

Background

Let $V_n$ be the $\mathbb{C}$-module spanned by the set of derangements (permutations with no fixed points) inside the group ring of $S_n$. We make $V_n$ into a $\mathbb{C}S_n$-module with $S_n$ acting on $V_n$ by conjugation extended by linearity. That is

\begin{align} V_n &:= \{\sum\limits_i c_i \sigma_i \,| \, \sigma_i \in S_n, \, \sigma_i \,\text{has no 1-cycles}, \, c_i \in \mathbb{C} \} \\ g\cdot \sigma_i &:= g \sigma_i g^{-1}, \quad g \in S_n \end{align}

$V_n$ is the direct sum of $\mathbb{C}S_n$-invariant submodules where each submodule is spanned by derangements of the same cycle type:

\begin{align} V_n = \bigoplus_{\mu} V_{\mu} \end{align}

where we denote cycle type as an integer partition $\mu = (\mu_1,\dots, \mu_k )$. So for example

\begin{align} V_4 \cong V_{(4)} \oplus V_{(2,2)} \end{align}

because the only derangements when $n=4$ have cycle type $(4)$ or $(2,2)$. If we choose some derangement $x_{\mu} \in S_n$ where $x_{\mu}$ has cycle type $\mu$ and define

\begin{align} H := \text{Stab}_{S_n}(x_{\mu}) = \{ g \, | \, g x_{\mu} g^{-1} = x_{\mu}, \, g \in S_n \} \end{align}

then $V_{\mu}$ is acted on by the representation of $S_n$ that is induced from the trivial $H$-action by conjugation on the $\mathbb{C}H$-module that is spanned by the single element $x_{\mu}$. Explicitly, let

\begin{align} W(x_{\mu}) &:= \{c x_{\mu} \, |\, c \in \mathbb{C} \} \\ \rho &: H \rightarrow \text{End}(W(x_{\mu})) \\ \rho(h)\cdot w &:= h w h^{-1} \\ &= w \quad (\text{by definition of the Stabilizer}) \end{align}

then

\begin{align} \text{Ind}_{H}^{S_n} \rho : S_n \rightarrow \text{End}(V_{\mu}) \end{align}

is the induced representation describing the $S_n$-action on $V_{\mu}$. I don't know if this is tedious or unnecessary, but writing things this way buys us:

By Frobenius reciprocity, the multiplicity of an irreducible representation of $S_n$ labelled by the integer partition $\lambda = ({\lambda_1, \dots , \lambda_j})$ in a decomposition of $V_{\mu}$ is given by

\begin{align} m^{\lambda} = \frac{1}{|H|} \sum\limits_{h\in H} \chi_{S_n}^{\lambda}(h) \end{align}

And so because characters of $S_n$ are straightforward to compute, we know how $V_{\mu}$ and thus $V_n$ decompose into irreducibles.

The Question

It's known that given any $n$-fold tensor product of a vector space $M$

\begin{align} M^{\otimes n} := M \otimes \dots \otimes M, \quad (n \, \text{times}) \end{align}

one can choose a basis for each $S_n$-invariant irreducible submodule of $M^{\otimes n}$ such that there is a basis element for each standard Young Tableaux with $n$ boxes and all of these basis elements are mutually orthogonal in the sense that if $e_i$ is the element of the group ring of $S_n$ whose image is the span of the basis element in question, then $e_i e_j = \delta_{ij}$, where the multiplication is the normal group ring multiplication (see here, or here).

For clarity: one example of a non-orthongal list of group algebra elements whose image would be a non-orthogonal basis for $M^{\otimes n}$ would be the set of Young Symmetrizers.

However, applying the cited methods for obtaining orthogonal bases to $V_n$ will not produce a fully orthogonal basis, but it will produce an orthogonal basis for each submodule $V_{\mu}$. The reason for this is that same irreducible submodule indexed by a partition $\lambda$ can appear in say $V_{\mu}$ and $V_{\nu}$ $(\mu \neq \nu)$ and there is no reason why the basis for $V_{\mu}^{\lambda}$ should be orthogonal to $V_{\nu}^{\lambda}$.

For example, I find that (remember subscripts label cycle type of derangements, superscripts label irreducible modules)

\begin{align} V_{(4)} &\cong V_{(4)}^{(4)} \oplus V_{(4)}^{(2,2)} \oplus V_{(4)}^{(2,1,1)} \\ V_{(2,2)} &\cong V_{(2,2)}^{(4)} \oplus V_{(2,2)}^{(2,2)} \end{align}

and so the 6 basis elements of $V_{(4)}$ will be mutually orthogonal and the 3 basis elements of $V_{(2,2)}$ will be mutually orthogonal, but the basis elements of $V_{(4)}^{(4)}$ will not be orthogonal to those of $V_{(2,2)}^{(4)}$ using the procedures cited above.

So my question is: what is a canonical way to ensure that submodules corresponding to derangements of different cycle types are orthogonal to each other? Is such a procedure possible/obvious?

In general this is possibly an unreasonable thing to ask, but maybe the structure of derangements and cycle types makes this a tractable problem?

Or maybe at least this particular setup of irreducible representations of the $\mathbb{C}S_n$-module spanned by derangements ($S_n$ action by conjugation) is known under some other name, which could lead me to resources and ideas?

Attempt at a solution

I have wondered if the solution lies in working with non-standard tableaux, or higher $m$ such that $n < m$, but honestly I don't really see a preliminary "proof of concept" in these ideas and haven't gone further. Of course I could just apply Gram-Schmidt, but I'm wondering if there's something "canonical" that I don't know about.

$\endgroup$
  • $\begingroup$ I'm lost from the beginning, $V_n $ is not a group, right? So how is the product defined in this group ring? $\endgroup$ – Arnaud Mortier Jan 13 '18 at 12:34
  • 2
    $\begingroup$ Do you mean the $\mathbb{C}$-module spanned by the derangements inside the group ring of $S_{n},$ which becomes a $\mathbb{C}S_{n}$-module under the conjugation action you describe? $\endgroup$ – Geoff Robinson Jan 13 '18 at 14:34
  • $\begingroup$ Yes, precisely what Geoff Robinson has written. Sorry I'll edit it to make this correction. $\endgroup$ – Jonathan Rayner Jan 13 '18 at 14:38

Your Answer

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

Browse other questions tagged or ask your own question.