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It is a basic fact about the symmetric group $S_n$ that its irreducible representations are indexed by partitions of $n$.

My question is, can the association between partitions and irreps be specified without having to explicitly construct the irreps? In other words, is there for each partition $\lambda$ some relatively simple property that is possessed by the Specht module $S^\lambda$ but none of the other irreps of $S_n$?

In particular I was imagining that an answer might come from Jucys-Murphy type theory but I am not familiar enough with the material to know how to (or if it is possible to) use it to characterize $S^\lambda$ without having to give a bunch of background information.

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    $\begingroup$ $S^\lambda$ appears as a direct summand of the permutation module on the Young subgroup with the same partition, with multiplicity one, and all other constituents correspond to higher partitions. For example, the trivial Specht module $S^{(n)}$ is obtained by inducing the trivial module from $S_n$ to itself, while the sign module, i.e. the Specht module $S^{(1^n)}$, appears in the regular representation (induction of the trivial module from the trivial subgroup). Does it help? $\endgroup$ – Daniel Juteau Apr 8 '15 at 22:10
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    $\begingroup$ It is also the representation of $S_n$ on the top cohomology of the Springer fiber associated to a nilpotent element with Jordan type corresponding to the same partition. $\endgroup$ – Daniel Juteau Apr 8 '15 at 22:11
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    $\begingroup$ Using Jucys-Murphy elements there is a natural bijection between Specht modules and partitions that is uniquely determined up to taking conjugates. The best way to see this is using the approach of A. Okounkov and A. Vershik, (A new approach to representation theory of symmetric groups)[link.springer.com/article/10.1007%2FBF02433451], Selecta Math. (N.S.), 2 (1996), 581–605. $\endgroup$ – Andrew Apr 9 '15 at 2:04
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    $\begingroup$ $S^\lambda$ is also the unique irreducible of $\mathfrak S_n$ whose restriction to $\mathfrak S_\lambda$ contains the trivial module and whose restriction to $\mathfrak S_{\lambda^*}$ contains the sign module. $\endgroup$ – Daniel Juteau Apr 9 '15 at 11:48
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    $\begingroup$ The one about it being "the intersection of the induced trivial and induced sign" is actually very nice IMO (it is obvious how to check it using characters, and it gives information about the structure of the irrep in a very easy to understand way), you should put it as an answer. $\endgroup$ – Sam Clearman Apr 9 '15 at 15:20
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The characterization using Jucys-Murphy elements is this: let $z$ be a formal variable, and $X_i$ the Jucys-Murphy elements. The product $(z-X_1)\cdots (z-X_n)$ is central in the group algebra of the symmetric group, so it acts on any irreducible representation by a polynomial with scalar coefficients. On the Specht module $V_\lambda$, the polynomial it acts by the product of $z+i-j$ where $(i,j)$ are the boxes of the Young diagram. This is one of the consequences of the Okounkov-Vershik theory as mentioned by Steven and Andrew above.

EDIT: By way of normalization, note that in $S_2$ this is $z^2-z\cdot (12)$, so it acts on the trivial rep by $z(z-1)$ and on the sign by $z(z+1)$. The difference comes from the fact that the box $(1,2)$ is in the diagram of the trivial, and $(2,1)$ is in the diagram of the sign. So, if somewhere you flipped this convention (which is very easy), you must change $z+i-j$ to $z-i+j$.

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  • $\begingroup$ I guess this is what I was asking for, thanks very much! $\endgroup$ – Sam Clearman Apr 9 '15 at 15:21
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    $\begingroup$ It's probably worth mentioning that $\prod (z-X_i)$ has an explicit formula that doens't mention Jucys-Murphy elements: The coefficient of $z^k$ is $(-1)^{n-k}$ times the sum of all permutations that have $k$ cycles. $\endgroup$ – DES-SupportsMonicaAndTransfolk Apr 9 '15 at 17:16
  • $\begingroup$ @DavidSpeyer You mentioning that was definitely worthwhile (I had no idea that that was the case). $\endgroup$ – Ben Webster Apr 9 '15 at 20:23
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Schur-Weyl duality tells you that under the commuting actions of $GL(V)$ and $S_n$ on $V^{\otimes n}$, it is a direct sum $\bigoplus_{\lambda} S^{\lambda} \otimes V_\lambda$ where $S^\lambda$ is some irreducible representation of $S_n$ and $V_\lambda$ is some irreducible representation of $GL(V)$ and where the $\lambda$ range over some indexing set. Furthermore, the correspondence is a bijection, i.e., $S^\lambda \cong S^\mu$ if and only if $\lambda = \mu$ if and only if $V_\lambda \cong V_\mu$.

So you can transfer the question to $GL(V)$, and for that group there is a more satisfactory answer: $V_\lambda$ is the unique irreducible representation which has a eigenvector for the subgroup of upper-triangular matrices whose generalized eigenvalue is given by $\lambda$, i.e., the eigenvalue is a function $b \mapsto b^\lambda$ where $b^\lambda = \prod_i x_i^{\lambda_i}$ and $x_1, \dots, x_n$ are the diagonal entries of $b$.

This clarifies certain combinatorial operations on partitions (like addition, which corresponds to the fact that the eigenvalue of a tensor product of eigenvectors is a sum) that are used in combinatorial statements like Murnaghan's stability theorem.

Alternatively, you can use Jucys-Murphy elements to use eigenvalues to label representations if you don't want to introduce $GL(V)$.

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  • $\begingroup$ Thank you for this answer. Can you give me any additional guidance on how exactly you would use Jucys-Murphy elements' associated eigenvalues to label representations? $\endgroup$ – Sam Clearman Apr 9 '15 at 3:17
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    $\begingroup$ That comes from the theorem in Okounkov-Vershik arxiv.org/abs/math/0503040 (see Section 5) that the spectra of the J-M elements on an irreducible representation correspond to the contents of the boxes in the Young diagram. But I would think this falls under "explicit" rather than "non-explicit". $\endgroup$ – Steven Sam Apr 9 '15 at 5:37
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I'm not sure what makes a characterization good or bad for you. Here is the fastest way I know to describe $S^{\lambda}$.

Let $\lambda$ be a partition of $n$. Let $M^{\lambda}$ be the permutation representation of $S_n$ on the set partitions of $\{1,2,\ldots, n \}$ into sets of size $\lambda_1$, $\lambda_2$, ..., $\lambda_r$. Let $\epsilon$ be the sign representation of $S_n$. Then $\mathrm{Hom}_{S_n}(M^{\lambda}, M^{\lambda^T} \otimes \epsilon)$ is one dimensional. (Here $\lambda^T$ is the transpose, or conjugate, partition.) Let $\alpha$ be a non-zero $S_n$-equivariant Hom from $M^{\lambda}$ to $M^{\lambda^T} \otimes \epsilon$. The Specht module $S^{\lambda}$ is the image of $\alpha$.

I wrote up some notes constructing $S^{\lambda}$ from this perspective.

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As Sam told me to do so, I convert my comment to an answer, although it is essentially the same as David Speyer's.

The Specht module $S^\lambda$ is the unique simple module of $\mathfrak S_n$ whose restriction to the $\mathfrak S_\lambda$ contains the trivial and whose restriction to $\mathfrak S_{\lambda^*}$ contains the sign (where the star denotes conjugation of partitions, and with the usual notation for Young subgroups).

To see this, one just needs to know that if $S^\lambda$ occurs in the permutation module $M^\mu$, then $\lambda \geq \mu$ (for the dominance order), that tensoring with sign induces the transposition of partitions, and that the transposition is an anti-automorphism of the poset of partitions. The books by James, or James and Kerber, for example, can be used as a reference for those facts.

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    $\begingroup$ To elaborate on the relation between Daniel's answer and mine, $M^{\lambda}$ is the induction of $1$ from the Young subgroup, and $M^{\lambda^T} \otimes \epsilon$ is the induction of the sign rep. So I am talking about induction, Daniel is talking about restriction and they are related by Frobenius reciprocity. I like my framing better because it sounds like a construction, whereas Daniel's only sounds like a characterization. $\endgroup$ – DES-SupportsMonicaAndTransfolk Apr 9 '15 at 23:06
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    $\begingroup$ That's what you get when you ask for a "non-explicit characterization"! :) $\endgroup$ – Daniel Juteau Apr 9 '15 at 23:36
  • $\begingroup$ Fair enough! :-) $\endgroup$ – DES-SupportsMonicaAndTransfolk Apr 9 '15 at 23:42

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