Formula for calculating effect of Schur functor on character

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Statement

Suppose G is a group, K is a field (which we will assume for simplicity to have characteristic zero) and φ:GGL(d,K) is a linear representation of G over K with character χ. Suppose λ is a partition of a positive integer n and denote by Sλ the Schur functor corresponding to λ. We can consider a new representation Sλ(φ) and we denote its character by Sλ(χ). Our goal is to provide an explicit description of Sλ(χ).

Explicit description of degree of representation

The new degree is a polynomial in the old degree. The degree of the polynomial (not to be confused with degrees of representations) equals n. Explicitly, the polynomial is defined as follows. Denote by C(Sn) the set of conjugacy classes of the symmetric group Sn. Let ν:C(Sn)N be the function that outputs the number of cycles of a given conjugacy class. Denote by θλ the character of the irreducible representation of Sn corresponding to the partition λ of n. Then, the polynomial is:

1n!cC(Sn)θλ(c)|c|dν(c)

Explicit description of character of representation

The character of the representation is given as follows. Denote by C(Sn) the set of conjugacy classes of the symmetric group Sn. Let ν:C(Sn)N be the function that outputs the number of cycles of a given conjugacy class. Denote by θλ the character of the irreducible representation of Sn corresponding to the partition λ of n. Then, the character is:

Sλ(χ)(g)=1n!cC(Sn)θλ(c)|c|(parts in the cycle type of cχ(gsize of the part))

For instance, for a conjugacy class in S9 with cycle type 3+3+2+1, the product on the inside reads χ(g3)χ(g3)χ(g2)χ(g).

Description for small Schur functors

Size of set being partitioned Partition for Schur functor Name of functor Formula for computing degree if original representation has degree d Formula for computing character of the representation obtained after applying this functor in terms of the original character χ, on an element g Explanation for character formula
1 1 identity functor d χ(g) n=1, so the only summand corresponds to the partition 1 of 1, and is 11!(1)(1)χ(g1)=χ(g).
2 2 symmetric square d(d+1)/2 (χ(g)2+χ(g2))/2 n=2 and λ is the partition 2, so the summation is carried out over the partitions of 2. The representation θλ is the trivial representation. The two parts are:
partition 1 + 1 of 2, corresponding to conjugacy class of identity element: the summand is 12!(1)(1)χ(g)χ(g)=(χ(g))2/2.
partition 2 of 2, corresponding to conjugacy class of non-identity element: the summand is 12!(1)(1)χ(g2)=χ(g2)/2.
Adding, we get the result.
2 1 + 1 exterior square or alternating square d(d1)/2 (χ(g)2χ(g2))/2
3 3 symmetric cube d(d+1)(d+2)/6 (χ(g)3+3χ(g2)χ(g)+2χ(g3))/6
3 2 + 1 ? d(d+1)(d1)/3 (χ(g)3χ(g3))/3
3 1 + 1 + 1 exterior cube d(d1)(d2)/6 (χ(g)33χ(g2)χ(g)+2χ(g3))/6