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The S_4 character table by rim-hook recursion

Example

The Murnaghan–Nakayama rule (The Murnaghan–Nakayama character rule) computes the full ordinary character table of $S_{4}$ . With rows $λ = (4), (3, 1), (2, 2), (2, 1, 1), (1^{4})$ and columns the classes $(1^{4}), (2, 1, 1), (2, 2), (3, 1), (4)$ (of sizes $1, 6, 3, 8, 6$ ),

\begin{array}{crrrrr} χ^{(4)} & 1 & 1 & 1 & 1 & 1 \\ χ^{(3, 1)} & 3 & 1 & - 1 & 0 & - 1 \\ χ^{(2, 2)} & 2 & 0 & 2 & - 1 & 0 \\ χ^{(2, 1, 1)} & 3 & - 1 & - 1 & 0 & 1 \\ χ^{(1^{4})} & 1 & - 1 & 1 & 1 & - 1 \end{array}

(for example $χ^{(2, 2)}$ on the $4$ -cycle class is $0$ because $(2, 2)$ is not a hook (The value of an irreducible character on a full cycle is supported on hooks), and $χ^{(3, 1)} (3, 1) = 0$ , $χ^{(3, 1)} (4) = - 1$ by the same rule). Checking the first orthogonality relation (The first orthogonality relation for irreducible complex characters) row by row gives, for instance, $⟨ χ^{(3, 1)}, χ^{(2, 2)} ⟩ = \frac{1}{24} (1 \cdot 3 \cdot 2 + 6 \cdot 1 \cdot 0 + 3 \cdot (- 1) \cdot 2 + 8 \cdot 0 \cdot (- 1) + 6 \cdot (- 1) \cdot 0) = 0$ , and every row has norm $1$ .

Facts & Assumptions

Given: The partitions of $4$ , the conjugacy classes of $S_{4}$ indexed by cycle type with sizes $∣ C_{ρ} ∣$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ , If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ), and for every partition of an integer the character $χ^{λ}$ of the complex Specht module $S^{λ}$ (Column antisymmetrizers, polytabloids, and Specht modules).

[F1]

Murnaghan–Nakayama rule: for a cycle $c$ of length $r$ on a specified set of points fixed by $σ$ (a specified fixed point for length $1$ ), $χ^{λ} (σ c) = \sum_{λ / μ} (- 1)^{ht (λ / μ)} χ^{μ} (σ)$ , summed over the removable border strips of $λ$ of size $r$ , the empty sum being $0$ (The Murnaghan–Nakayama character rule).

[F2]

Hook lengths and dimensions: the hook lengths of $(3, 1), (2, 2), (2, 1, 1), (1^{4})$ are $4, 2, 1, 1$ ; $3, 2, 2, 1$ ; $4, 2, 1, 1$ ; $4, 3, 2, 1$ , and the dimensions are $f^{λ} = 4! / \prod h (x)$ , namely $1, 3, 2, 3, 1$ for $(4), (3, 1), (2, 2), (2, 1, 1), (1^{4})$ (Hook, arm, leg, and hook length of a box, The hook length formula).

[F3]

The five conjugacy classes of $S_{4}$ have cycle types $(1^{4}), (2, 1, 1), (2, 2), (3, 1), (4)$ and sizes $1, 6, 3, 8, 6$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ , If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ , Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

[F4]

Standard inner product $⟨ φ, ψ ⟩ = \frac{1}{∣ G ∣} \sum_{g \in G} φ (g) \overline{ψ (g)}$ (The standard inner product on $c f (G)$ ), and the first orthogonality relation $⟨ χ^{λ}, χ^{μ} ⟩ = δ_{λ μ}$ for irreducible characters (The first orthogonality relation for irreducible complex characters); the character table of a finite group has rows the irreducible characters and columns the classes (The character table of a finite group).

[F5]

For the empty partition $χ^{\emptyset}$ is the trivial character of $S_{0}$ , so $χ^{\emptyset} (1) = 1$ (Column antisymmetrizers, polytabloids, and Specht modules).

Verification

technique · direct

1.1F1F5algebra

Row $(4)$ : for each part $r$ of the cycle type the unique removable $r$ -strip of the row $(4)$ is horizontal and has height $0$ , leaving the row $(4 - r)$ , so iterating the rule [F1] gives $χ^{(4)} = 1$ on all five classes. Row $(1^{4})$ is the sign character: the unique $r$ -strip of the column is vertical of height $r - 1$ , so iterating [F1] gives $χ^{(1^{4})} (ρ) = \prod_{i} (- 1)^{r_{i} - 1}$ for the parts $r_{i}$ of the cycle type, that is the values $1, - 1, 1, 1, - 1$ on $(1^{4}), (2, 1, 1), (2, 2), (3, 1), (4)$ .

1.2F1F2F5algebra

Row $(3, 1)$ : by [F2] the dimension is $4! / (4 \cdot 2 \cdot 1 \cdot 1) = 3$ ; on $(2, 1, 1)$ the only $2$ -strip removes the domino ${(1, 2), (1, 3)}$ and leaves $(1, 1)$ with height $0$ , so the value is $χ^{(1, 1)} (1, 1) = 1$ ; on $(2, 2)$ the same strip gives $χ^{(1, 1)} (2) = - 1$ ; on $(3, 1)$ the hook lengths $4, 2, 1, 1$ contain no $3$ , so there is no $3$ -strip and the value is $0$ ; on $(4)$ the whole diagram $(3, 1)$ is the unique $4$ -strip, of height $1$ , so the value is $- χ^{\emptyset} = - 1$ . This gives the row $3, 1, - 1, 0, - 1$ .

1.3F1F2F5algebra

Row $(2, 2)$ : the two $2$ -strips are the bottom row to $(2)$ (height $0$ ) and the right column to $(1, 1)$ (height $1$ ), so on $(2, 1, 1)$ the value is $χ^{(2)} (1, 1) - χ^{(1, 1)} (1, 1) = 1 - 1 = 0$ , on $(2, 2)$ it is $χ^{(2)} (2) - χ^{(1, 1)} (2) = 1 - (- 1) = 2$ ; on $(3, 1)$ the unique $3$ -strip (leaving the single node $(1)$ ) has height $1$ and gives $- χ^{(1)} (1) = - 1$ ; on $(4)$ the hook lengths $3, 2, 2, 1$ contain no $4$ , so the value is $0$ ; the dimension is $2$ by [F2]. Row $(2, 1, 1)$ : the unique $2$ -strip removes the vertical domino ${(2, 1), (3, 1)}$ and leaves $(2)$ with height $1$ , so the values on $(2, 1, 1)$ and $(2, 2)$ are $- χ^{(2)} (1, 1) = - 1$ and $- χ^{(2)} (2) = - 1$ ; there is no $3$ -strip because the hook lengths are $4, 2, 1, 1$ ; the whole diagram is the unique $4$ -strip, of height $2$ , giving $χ^{\emptyset} = 1$ on $(4)$ ; the dimension is $3$ by [F2]. This gives the rows $2, 0, 2, - 1, 0$ and $3, - 1, - 1, 0, 1$ .

2.1step 1.1step 1.2step 1.3F3F4algebra∎

With the class sizes $1, 6, 3, 8, 6$ [F3] and the standard inner product [F4], the five rows have norms $\frac{1}{24} (1 + 6 + 3 + 8 + 6) = 1$ for $(4)$ and $(1^{4})$ , $\frac{1}{24} (9 + 6 + 3 + 0 + 6) = 1$ for $(3, 1)$ and $(2, 1, 1)$ , and $\frac{1}{24} (4 + 0 + 12 + 8 + 0) = 1$ for $(2, 2)$ ; pairwise orthogonality holds as in the displayed computation $⟨ χ^{(3, 1)}, χ^{(2, 2)} ⟩ = \frac{1}{24} (6 - 6) = 0$ , and likewise $⟨ χ^{(2, 2)}, χ^{(2, 1, 1)} ⟩ = \frac{1}{24} (1 \cdot 2 \cdot 3 + 6 \cdot 0 \cdot (- 1) + 3 \cdot 2 \cdot (- 1) + 8 \cdot (- 1) \cdot 0 + 6 \cdot 0 \cdot 1) = 0$ , the remaining pairs checking in the same way. Hence the five rows computed by the recursion are exactly the character table of $S_{4}$ displayed in the statement, with the rows orthonormal with respect to the class sizes of [F3].

Remarks

The $4$ -cycle column is the hook column. On the class $(4)$ only the hooks $(4), (3, 1), (2, 1, 1), (1^{4})$ can be nonzero and their values are $1, - 1, 1, - 1$ , while $χ^{(2, 2)} (4) = 0$ ; this is the S_4 instance of the full-cycle lemma The value of an irreducible character on a full cycle is supported on hooks.
Cross-check. The same array is obtained by the fixed-tabloid algorithm of Fixed-tabloid reconstruction of the character table (Fox–Mullineux), whose worked $S_{4}$ instance in the source recorded above produces the identical entries; the two routes use disjoint inputs, the recursion using hooks and the fixed-tabloid route using the Gram equation.

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Sources

G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notes in Mathematics 682, Springer 1978 (complete author-hosted scan, 161 pp.) (standard reference, not scraped)
William Woods, Representation theory of the symmetric groups (Cambridge Part III essay, 33 pp.) (standard reference, not scraped)
I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Oxford Mathematical Monographs, 1995 (complete author-hosted scan, 486 pp.) (standard reference, not scraped)