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Consider $V_{(n-1, 1)}$, the $n-1$ dimensional irreducible representation of $S_n$, i.e. the "standard" or "defining" representation. Is there a nice formula for how the $k$-th tensor power of $V_{(n-1, 1)}$ decomposes into irreps?

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$V = 1\oplus V_{(n-1,1)}$, so the problem boils down to figuring out how the tensor powers of $V_{(n-1,1)}$ decompose. In general, determining how tensor products decompose is hard (the Kronecker problem), but this case may be known. –  Amritanshu Prasad Oct 5 '12 at 6:12
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For symmetric powers there is recent paper by R.Stanley&K arxiv.org/abs/math/0002106 A Note on the Symmetric Powers of the Standard Representation of S_n In this paper, we prove that the dimension of the space spanned by the characters of the symmetric powers of the standard n-dimensional representation of the symmetric group S_n is asymptotic to n^2/2. This is proved by using generating functions to obtain formulas for upper and lower bounds, both asymptotic to n^2/2, for this dimension. In particular, for n>6, these characters do not span the full space of class functions on S_n. –  Alexander Chervov Oct 5 '12 at 9:05
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@Tobias Littlewood-Richardson coefficients are for $GL_n$ tensor products. $S_n$ is not so easy. –  David Speyer Oct 5 '12 at 11:24
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I think a combinatorial model for the case at hand is dealt with in emis.de/journals/SLC/wpapers/s54goupchau.html –  Vasu vineet Oct 5 '12 at 17:41
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Note that by Murnaghan's theorem, the multiplicity in V_(n-1,1)^()tensor k) of V_(n-a, lambda) for any partition lambda of a is going to be constant for n large enough. (Of course this means n large enough relative to a and k.) (The coefficients are called stable Kronecker coefficients, which are indeed not the same as Littlewood-Richardson coefficients.) –  JSE Oct 8 '12 at 4:10
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3 Answers 3

up vote 14 down vote accepted

For convenience consider the representation $Y=V_n\oplus V_{n-1,1}$ instead of $V_{n-1,1}$. Then the multiplicity of the representation of $S_n$ indexed by the partition $\lambda$ of $n$ in the $k$th tensor power of $Y$ equals the scalar product of the symmetric function $s_1^k$ (where $s_1=x_1+x_2+\cdots$ denotes a Schur function) with the plethysm $s_\lambda[1+h_1+h_2+h_3+\cdots]$, where $h_i$ is the complete symmetric function of degree $i$. This follows from the theory of inner plethysm; see Exercise 7.74 of Enumerative Combinatorics, volume 2. Since plethysm is in general intractable, I don't expect anything much simpler. This result does allow, however, these decompositions to be computed using Stembridge's Maple package SF for small values of $n$ and $k$.

Addendum. I used the method of Exercise 7.74 to get the analogous result for $V_{n-1,1}$. Namely, the multiplicity of the representation of $S_n$ indexed by the partition $\lambda$ of $n$ in the $k$th tensor power of $V_{n-1,1}$ equals the scalar product of $s_1^k$ with the symmetric function $(1-e_1+e_2-e_3+\cdots)\cdot s_\lambda[1+h_1+h_2+h_3+\cdots]$, where $e_i$ is an elementary symmetric function.

Addendum #2. A alternative formulation is the following. The multiplicity of the representation of $S_n$ indexed by the partition $\lambda$ of $n$ in the $k$th tensor power of $V_{n-1,1}$ equals the scalar product of $(s_1-1)^k$ with the symmetric function $s_\lambda[1+h_1+h_2+h_3+\cdots]$.

News flash! I said above that plethysm in in general intractable. Indeed, the Schur function expansion of $s_\lambda[1+h_1+h_2+\cdots]$ looks hopeless to me. However, taking the scalar product with $s_1^k$ results in a lot of simplification. I can show the following. The multiplicity of the representation of $S_n$ indexed by the partition $\lambda$ of $n$ in the $k$th tensor power of $V_n\oplus V_{n-1,1}$ equals the coefficient of $s_\lambda$ in the Schur function expansion of $(1+h_1+h_2+\cdots)\cdot \sum_{j=1}^k S(k,j)s_1^j$, where $S(k,j)$ is a Stirling number of the second kind. (After obtaining this result, I noticed that it is essentially the same as Corollary 2 of the Goupil-Chauve paper mentioned in Vasu Vineet's comment.) Since for fixed $j$ we have $S(k,j)=\frac{1}{j!}\sum_{i=1}^j (-1)^{j-i}{j\choose i}i^k$, we can get explicit formulas for the multiplicities for fixed $\lambda$ that don't involve Stirling numbers. For instance, when $\lambda=(3)$ the multiplicity is $\frac{1}{6}(3^k+3)$, for $\lambda=(2,1)$ it is $3^{k-1}$, and for $\lambda=(1,1,1)$ it is $\frac{1}{6}(3^k-3)$. In particular, the multiplicity for $\lambda = (1^n)$ (i.e., $n$ parts equal to 1) is $S(k,n)+S(k,n-1)$.

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You want the "partition algebras". Some references to get you started are:

MR1317365 (97b:82023) Jones, V. F. R. The Potts model and the symmetric group. Subfactors (Kyuzeso, 1993), 259--267, World Sci. Publ., River Edge, NJ, 1994.

MR1399030 (98g:05152) Martin, Paul . The structure of the partition algebras. J. Algebra 183 (1996), no. 2, 319--358.

MR2143201 (2006g:05228) Halverson, Tom ; Ram, Arun . Partition algebras. European J. Combin. 26 (2005), no. 6, 869--921.

The partition algebras are the endomorphism algebras of the tensor powers of, $V$, the natural representation of $S_n$. As has been mentioned in the comments this decomposes as the sum of the trivial representation and the representation you are interested in.

You can recover information about the representation you are interested in from the partition algebras. For example, instead of looking at all set partitions, you only consider set partitions with no singleton.

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The problem has been solved in the reference indicated in the comment by Vasu Vineet, namely: "Combinatorial Operators for Kronecker Powers of Representations of Sn" by Alain Goupil and Cedric Chauve. However, one cannot say that the formulas in Propositions 1 and 2 of this paper are "nice".

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