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The set of $n$-box Young diagrams classifies both conjugacy classes in $S_n$ and equivalence classes of irreducible representations of $S_n$. There is an outer automorphism of $S_6$, of order 2. This acts on the set of conjugacy classes of $S_6$ in an obvious way. It also acts on the set of equivalence classes of irreducible representations of $S_6$ in an obvious way.

So, we get two actions of $\mathbb{Z}_2$ on the set of 6-box Young diagrams---or in other words, two involutions on the set of 6-box Young diagrams.

Question 1: are these the same?

Question 2: is there some easy way to look at a 6-box Young diagram and see what Young diagram you get when you apply either of these involutions?

There is also a pictorially obvious involution on the set of $n$-box Young diagrams, which reflects them across the diagonal. There are 11 6-box Young diagrams, shown below, and the pictorially obvious involution sends the $i$th one to the $(11-i)$th one:

1.

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2.

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3.

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4.

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5.

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6.

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7.

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8.

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9.

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10.

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11.







It is easy to see that this pictorially obvious involution is neither of the other two involutions. Each column in the Young diagram corresponds to a cycle in the cycle decomposition of a permutation, so diagram 1 corresponds to the conjugacy class of the permutation

$$ (1 2 3 4 5 6) $$

which has $6!/6 = 120$ elements---this is the largest conjugacy class in $S_6$. Diagram 11 corresponds to the conjugacy class of the permutation

$$ (1) (2) (3) (4) (5) (6) $$

that is, the conjugacy class of the identity, with just one element---this is the smallest conjugacy class. So, the outer automorphism of $S_6$ can't switch these.

Similarly, the irreducible representation corresponding to diagram 1 corresponds to the trivial representation of $S_6$, while diagram 11 corresponds to the sign representation. Both these are 1-dimensional, but the outer automorphism of $S_6$ doesn't switch these: the trivial representation has to get mapped to the trivial representation.

Some extra hints:

  1. The conjugacy classes of $S_6$ have cardinalities

$$ 1, 15, 15, 40, 45, 90, 90, 120, 120, 144 $$

which is consistent with

$$1 + 15 + 15 + 40 + 40 + 45 + 90 + 90 + 120 + 120 + 144 = 720 $$

Just for fun, I conjecture that all the pairs of equal cardinality get switched by the outer automorphism. Clearly the cardinalities put strong limitations on how the outer automorphism can act, since a conjugacy class has to be mapped to an equal-sized conjugacy class.

  1. The irreducible representations of $S_6$ have dimensions

$$ 1, 1, 5, 5, 5, 5, 9, 9, 10, 10, 16 $$

which is consistent with

$$ 1^2 + 1^2 + 5^2 + 5^2 + 5^2 + 5^2 + 9^2 + 9^2 + 10^2 + 10^2 + 16^2 = 720 $$

Here we know that not all pairs of equal dimension get switched by the outer automorphism, since the trivial representation is mapped to itself (and therefore so is the sign representation). What happens to the 4 representations of dimension 5?

I apologize for being too lazy to match up the cardinalities and dimensions to the Young diagrams.

  1. For a concrete description of the outer automorphism of $S_6$, see this chart drawn by Greg Egan:

enter image description here

Any permutation of $S =\{1,2,3,4,5,6\}$ induces a permutation of the 15 'synthemes' shown in blue ovals in this chart, and thus a permutation of the 6 'synthematic totals' shown as colored bands at the edge of the chart. (The duads, shown in blue ovals, are the 15 2-element subsets of $S$. Each syntheme is a way of partitioning $S$ into three duads, and each synthematic total is a choice of 5 synthemes that contain all 15 duads.)

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  • $\begingroup$ I feel as though the identification of Young diagrams with irreducible representations of the symmetric group is only defined up to the row/column symmetry anyways. $\endgroup$ – Sam Hopkins Aug 21 '15 at 4:26
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    $\begingroup$ For representations, you have to pick a convention about whether columns correspond to symmetrization or antisymmetrization; I'm using the usual convention, which is antisymmetrization. There's also a convention involved with conjugacy classes: I'm using the convention that cycles correspond to columns. If the answer to Question 1 is "no, you idiot, because you picked incompatible conventions", we can just change one convention. $\endgroup$ – John Baez Aug 21 '15 at 4:29
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    $\begingroup$ Section 2.1 of my paper on labelling the character tables of symmetric groups is relevant: please see ma.rhul.ac.uk/~uvah099/Maths/labels.pdf. $\endgroup$ – Mark Wildon Aug 21 '15 at 11:44
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    $\begingroup$ In a similar spirit as the question: there are three involutions on the set of Specht and dual Specht modules for $S_n$. (1) conjugation: $S^\lambda \mapsto S^{\lambda'}$; (2) twisting by the sign representation: $S^\lambda \mapsto S^{\lambda} \otimes \mathrm{sgn}_{S_n}$; (3) duality: $S^\lambda \mapsto (S^{\lambda})^\star$. Doing any two of these three is the same as doing the third; in characteristic zero (3) is the identity. $\endgroup$ – Mark Wildon Aug 21 '15 at 14:34
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Let $t$ be an outer automorphism of $S_6$. It's known that $t$ sends a $6$-cycle to an element of cycle type $(3,2)$. This forces three pairs of conjugacy classes to be swapped:

$$ (6) \leftrightarrow (3,2), \ (3,3) \leftrightarrow (3),\ (2,2,2) \leftrightarrow (2). $$

The only other classes sharing a common size are $(4)$ and $(4,2)$ but these turn out to be fixed. (I've included a proof below.) From the character table and the equation $\chi^t(g^t) = \chi(g)$ one then gets the action on characters:

$$ (5,1) \leftrightarrow (2,2,2), \ (2,1,1,1,1) \leftrightarrow (3,3), \ (4,1,1) \leftrightarrow (3,1,1,1). $$

Incidentally, this gives a nice way to construct the 'exceptional' reflection representations labelled by $(3,3)$ and $(2,2,2)$: just apply $t$ to the natural permutation module, and take the non-trivial constituent.

As expected from Brauer's Permutation Lemma, the actions of $t$ on conjugacy classes and characters have the same cycle type.

Construction of $t$. Let $H \cong S_5$ be the transitive subgroup of $S_6$ given by the action of $S_5$ on its $6$ Sylow $5$-subgroups. By identifying $S_6/H$ with $\{1,\ldots,6\}$ we can define $t$ to be the map $S_6 \rightarrow \mathrm{Sym}(S_6/H)$. Since $(1,2,3)(4,5) \in S_5$ acts as a $6$-cycle on the Sylow $5$-subgroups of $S_5$, $H$ contains a $6$-cycle. Hence a $6$-cycle acting on $S_6/H$ has a fixed point, so must have cycle type $(3,2)$. Similarly $H$ contains a $4$-cycle, so $t$ sends $4$-cycles to $4$-cycles.

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    $\begingroup$ Quick proof that $(4)$ and $(4,2)$ are fixed by $t$: One of them is in the kernel of the unique nontrivial map $S_6 \to \mathbb{Z}/2$, and the other isn't. (The unique map is taking the sign of the permutation.) $\endgroup$ – David E Speyer Aug 21 '15 at 20:23
  • $\begingroup$ Are Irr and Conj isomorphic as Z/2-sets ? $\endgroup$ – Alexander Chervov Aug 24 '17 at 6:43
  • $\begingroup$ Yes, since the outer automorphism acts with cycle type $(2^3,1^5)$ on both. This also follows from Brauer's Permutation Lemma. $\endgroup$ – Mark Wildon Aug 24 '17 at 23:22

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