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All three Specht modules of $S_{3}$

Statement

For $S_{3}$ the partitions $(3), (2, 1), (1^{3})$ give Specht modules of dimensions $1, 2, 1$ respectively: trivial, the sum-zero subspace of the natural three-point permutation module with basis $v_{3} - v_{1}$ and $v_{2} - v_{1}$ , and sign. They are all the complex irreducibles.

Facts & Assumptions

Given: Work over $C$ with $n = 3$ and $S_{3} = Sym ({1, 2, 3})$ .

[F1]

A partition of $n$ is a finite weakly decreasing sequence of positive integers whose sum is $n$ (Partitions, English diagrams, and conjugation).

[F2]

The symmetric group $S_{n}$ is the group of permutations of ${1, \dots, n}$ , and a transposition exchanges two labels and fixes the rest (Partitions, English diagrams, and conjugation, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[F3]

A finite left $G$ -set $X$ gives the permutation representation with basis $e_{x}$ and action $g \cdot e_{x} = e_{g \cdot x}$ (The trivial representation, the regular representation, and permutation representations from finite $G$ -sets).

[F4]

The tabloids form a basis of the Young permutation module (Young subgroups, tabloids, and permutation modules).

[F5]

The action on tabloids is $σ \cdot {t} = {σ \cdot t}$ (Young subgroups, tabloids, and permutation modules).

[F6]

The Specht module is the complex span of all polytabloids of the given shape (Column antisymmetrizers, polytabloids, and Specht modules).

[F7]

In shape $(2, 1)$ , the vectors $a = v_{3} - v_{1}$ and $b = v_{2} - v_{1}$ form a basis of $S^{(2, 1)}$ (Polytabloids of shape $(2, 1)$ ).

[F8]

For $n \geq 1$ , $S^{(n)}$ is the one-dimensional trivial module and $S^{(1^{n})}$ is the one-dimensional sign module (The row and column Specht modules).

[F9]

A linear subspace contains zero (Linear subspace of a vector space).

[F10]

A linear subspace is closed under vector addition (Linear subspace of a vector space).

[F11]

A linear subspace is closed under scalar multiplication (Linear subspace of a vector space).

[F12]

A finite-dimensional representation is a finite-dimensional vector space with a group action by invertible linear maps (A finite-dimensional representation $ρ : G \to GL (V)$ over a field, and its degree).

[F13]

A subrepresentation is an invariant linear subspace (Subrepresentations, direct sums of representations, and irreducibility).

[F14]

The sign representation acts by $σ \cdot z = sgn (σ) z$ (The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup).

[F15]

Sign is computed by $sgn (σ) = (- 1)^{inv (σ)}$ on the standard finite ordinal; the Specht-module convention transports this sign along the order-preserving relabelling (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations, Column antisymmetrizers, polytabloids, and Specht modules).

[F16]

The standard polytabloids form a basis of each complex Specht module (Standard polytabloids form a basis of a complex Specht module).

[F17]

For every $n \geq 0$ , the Specht modules indexed by partitions of $n$ form a complete irredundant list of finite-dimensional irreducible complex $S_{n}$ -representations (Specht modules classify the complex irreducibles of $S_{n}$ ).

No form of the Axiom of Choice is used; every set needed here is explicitly finite.

Proof

technique · direct

1.1givenF1

By [F1], the partitions of $3$ are exactly $(3), (2, 1), (1^{3})$ .

1.2givenF8

By [F8], the Specht modules of shapes $(3)$ and $(1^{3})$ are respectively one-dimensional trivial and sign modules, so each has dimension $1$ .

1.3givenF3F4F5algebra

Write $v_{i}$ for the tabloid whose second row is ${i}$ ; these are the three distinct tabloid basis vectors by [F4]. Give $P = C^{({1, 2, 3})}$ the basis $w_{1}, w_{2}, w_{3}$ and natural action $σ \cdot w_{i} = w_{σ (i)}$ by [F3]. The basis map $Φ (v_{i}) = w_{i}$ is a linear isomorphism, and [F5] gives $Φ (σ \cdot v_{i}) = Φ (v_{σ (i)}) = w_{σ (i)} = σ \cdot Φ (v_{i})$ , so it is $S_{3}$ -equivariant.

1.4givenF2F5F7F15algebra

In the ordered basis $(a, b)$ , the left actions satisfy $(12) a = v_{3} - v_{2} = a - b$ , $(12) b = v_{1} - v_{2} = - b$ , $(23) a = v_{2} - v_{1} = b$ , and $(23) b = v_{3} - v_{1} = a$ , so their matrices (basis vectors as columns) are $ρ ((12)) = (\begin{array}{cc} 1 & 0 \\ - 1 & - 1 \end{array})$ and $ρ ((23)) = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ . After the order-preserving relabelling $1, 2, 3 \mapsto 0, 1, 2$ , the transpositions $(12), (23), (13)$ have respectively $1, 1, 3$ inversions, so every transposition has sign $- 1$ by [F15]; the relabelling preserves these counts. The six elements are $1, (12), (23), (13), (123), (132)$ , with $(13) = (12) (23) (12)$ , $(123) = (12) (23)$ , and $(132) = (23) (12)$ , so the two displayed actions determine the full $S_{3}$ -action.

2.1givenF3F6F7F9F10F11F13F16step 1.3algebra

The image of $S^{(2, 1)}$ under $Φ$ is precisely the coordinate-sum-zero subspace $P_{0} : = {x_{1} w_{1} + x_{2} w_{2} + x_{3} w_{3} : x_{1} + x_{2} + x_{3} = 0}$ , and it has basis $a = w_{3} - w_{1}$ , $b = w_{2} - w_{1}$ , hence dimension two.

Indeed, [F7] and step 1.3 send the basis of $S^{(2, 1)}$ to $a, b$ , each of coordinate sum zero. Conversely, for a vector with $x_{1} + x_{2} + x_{3} = 0$ we have $x_{1} = - x_{2} - x_{3}$ and $x_{1} w_{1} + x_{2} w_{2} + x_{3} w_{3} = x_{2} b + x_{3} a$ , proving equality with $P_{0}$ . The zero vector is in $P_{0}$ by [F9], while closure under addition and scalar multiplication follows from [F10,F11]. The permutation action preserves the coordinate sum by [F3], so $P_{0}$ is an invariant linear subspace, hence a subrepresentation by [F13]. The vectors $a, b$ are independent because the $w_{3}, w_{2}$ coefficients in $α a + β b = 0$ force $α = β = 0$ . To check the dimension by [F16] as well, the upper-left box of a standard $(2, 1)$ tableau must contain $1$ ; the remaining $2, 3$ may occupy the other two boxes in either order, and both orders satisfy the row and column inequalities. Thus the two standard tableaux give a two-element basis.

3.1givenF7F8F12F14step 1.4step 2.1

The trivial and sign modules are nonisomorphic, and $S^{(2, 1)}$ has dimension two rather than one.

Indeed, $(12)$ acts by $+ 1$ on the trivial module and by $- 1$ on the sign module by [F8,F14] and step 1.4. Their dimensions are $1$ , whereas [F7] and step 2.1 give dimension $2$ for $S^{(2, 1)}$ , so the latter is nonisomorphic to either one-dimensional module. All three are finite-dimensional complex representations by their displayed finite bases and [F12].

4.1givenF17step 1.1step 3.1

Applying [F17] at $n = 3$ shows that these three modules are all finite-dimensional complex irreducible $S_{3}$ -representations and form a complete irredundant list.

The theorem classifies the Specht modules indexed by partitions of $3$ ; step 1.1 lists exactly those partitions, so they are precisely the three modules just displayed. The explicit distinction in step 3.1 also verifies directly that the two one-dimensional models differ and that the third has dimension two. ∎

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Sources

Charlotte Chan, Representation Theory of Symmetric Groups, Example 3.14(a-b), printed p. 14; Theorem 4.4 and Corollary 4.5, printed pp. 16-17 (standard reference, not scraped)

All three Specht modules of S3S3​