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Richard Stanley
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The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

Note added 1/10/14: Originally I inadvertently wrote $1/(1-p_1-p_2-p_3-\cdots)$ instead of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$. This has been corrected. It seems thatThe "incorrect" symmetric function $1/(1-p_1-p_2-p_3-\cdots)$ is also has a Schur-positiveinteresting. It is equal to $$ \frac{\sum_{n\geq 0}h_n}{1-\sum_{n\geq 1}(n-1)h_n}. $$ See for instance (evenhttp://www.math.miami.edu/~wachs/papers/chrom.pdf and let $h$-positive) expansion$t\to 1$. Does it have some natural representation-theoretic significance?

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

Note added 1/10/14: Originally I inadvertently wrote $1/(1-p_1-p_2-p_3-\cdots)$ instead of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$. This has been corrected. It seems that $1/(1-p_1-p_2-p_3-\cdots)$ also has a Schur-positive (even $h$-positive) expansion. Does it have some natural representation-theoretic significance?

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

Note added 1/10/14: Originally I inadvertently wrote $1/(1-p_1-p_2-p_3-\cdots)$ instead of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$. This has been corrected. The "incorrect" symmetric function $1/(1-p_1-p_2-p_3-\cdots)$ is also interesting. It is equal to $$ \frac{\sum_{n\geq 0}h_n}{1-\sum_{n\geq 1}(n-1)h_n}. $$ See for instance http://www.math.miami.edu/~wachs/papers/chrom.pdf and let $t\to 1$.

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Richard Stanley
  • 50.8k
  • 14
  • 155
  • 279

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1-p_2-p_3-\cdots)$$1/(1-p_1)(1-p_2)(1-p_3)\cdots$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

Note added 1/10/14: Originally I inadvertently wrote $1/(1-p_1-p_2-p_3-\cdots)$ instead of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$. This has been corrected. It seems that $1/(1-p_1-p_2-p_3-\cdots)$ also has a Schur-positive (even $h$-positive) expansion. Does it have some natural representation-theoretic significance?

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1-p_2-p_3-\cdots)$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.

Note added 1/10/14: Originally I inadvertently wrote $1/(1-p_1-p_2-p_3-\cdots)$ instead of $1/(1-p_1)(1-p_2)(1-p_3)\cdots$. This has been corrected. It seems that $1/(1-p_1-p_2-p_3-\cdots)$ also has a Schur-positive (even $h$-positive) expansion. Does it have some natural representation-theoretic significance?

Source Link
Richard Stanley
  • 50.8k
  • 14
  • 155
  • 279

The multiplicity of the irreducible character of $S_n$ indexed by the partition $\lambda$ of $n$ in the action of $S_n$ on itself by conjugation is the coefficient of the Schur function $s_\lambda$ in the Schur function expansion of $1/(1-p_1-p_2-p_3-\cdots)$, where $p_i$ is a power sum symmetric function. (This is an exercise in Chapter 7 of Enumerative Combinatorics, vol. 2.) This makes it very easy to compute the decomposition of the conjugacy action using Stembridge's SF package for Maple. For instance, I computed on my laptop the entire decomposition for $n=20$ in a couple of seconds.