The Specht ideal is irreducible #
For a Young tableau t of shape μ, the left ideal ℚ[Sₙ] c_t generated by the Young
symmetrizer c_t = a_t b_t carries an irreducible representation of Sₙ
(TauCeti.YoungTableau.isIrreducible_spechtIdealRep). This is the Specht module in its ideal
presentation, and the theorem is the one all of its representation theory rests on.
Everything the proof needs is already in place. The sandwich calculation of
TauCeti.RepresentationTheory.Symmetric.Factorization says that c_t x c_t is always a rational
multiple of c_t, and Maschke's theorem says that ℚ[Sₙ] is semisimple. Take a left ideal
W ≤ ℚ[Sₙ] c_t and split into the two cases the sandwich calculation allows.
- If
c_t w = 0for everyw ∈ W, thenWis square-zero: an element ofWisa c_tby hypothesis, so a product of two elements ofWisa (c_t w) = 0. A square-zero left ideal of a semisimple ring vanishes (TauCeti.Submodule.eq_bot_of_forall_mul_eq_zero), soW = ⊥. - Otherwise
c_t w ≠ 0for somew = a c_t ∈ W, andc_t w = c_t a c_tisκ • c_twithκ ≠ 0. It lies inW, hence so doesc_t, hence so does the whole ideal it generates, andWis everything.
So ℚ[Sₙ] c_t is an atom of the lattice of left ideals, which is exactly simplicity of the
module it carries; the transfer to the language of representations is
TauCeti.Representation.isIrreducible_ofModule'_iff.
Running the first case on W = ℚ[Sₙ] c_t itself, which is nonzero, shows that the sandwich is
not identically zero: some c_t a c_t is a nonzero multiple of c_t
(TauCeti.YoungTableau.exists_ne_zero_eq_smul_youngSymmetrizer_mul_mul). Whether the scalar
attached to a = 1 is itself nonzero -- that is, whether c_t * c_t = (n! / f^μ) • c_t with the
stated nonzero scalar -- does not follow from the sandwich calculation alone (over M₂(ℚ) the
element e₁₂ sandwiches to a line and squares to zero), and is not proved here.
The statement is made at the Representation level. Its FDRep mirror is
Simple (spechtIdealFDRep t), which follows from FDRep.simple_iff_isIrreducible and the
irreducibility theorem below.
On the one-row and the one-column shape the Specht ideal is a line, by the dimension count of
TauCeti.RepresentationTheory.Symmetric.Specht.Ideal.Extremes, so on those shapes this theorem is
visible without any of the above; there is nothing extra to say about them here. The
identification of this representation with the span of the polytabloids inside the Young
permutation module, and the classification saying that these exhaust the irreducibles of Sₙ and
are pairwise non-isomorphic across shapes, are separate targets and are not claimed here.
Main results #
TauCeti.YoungTableau.exists_ne_zero_eq_smul_youngSymmetrizer_mul_mul: the sandwichc_t x c_tis a nonzero multiple ofc_tfor somex.TauCeti.YoungTableau.isAtom_spechtIdeal:ℚ[Sₙ] c_tis an atom of the lattice of left ideals.TauCeti.YoungTableau.isSimpleModule_spechtIdeal: it is a simpleℚ[Sₙ]-module.TauCeti.YoungTableau.isIrreducible_spechtIdealRep: it is an irreducible representation ofSₙ.
References #
- W. Fulton, Young Tableaux, Section 7.2, Theorem 5.
- W. Fulton and J. Harris, Representation Theory: A First Course (1991), Theorem 4.3.
- Schur--Weyl roadmap, Layer 4, "irreducibility".
The sharpened form of TauCeti.YoungTableau.exists_eq_smul_youngSymmetrizer_mul_mul: for some
element of the group algebra the scalar it sandwiches the Young symmetrizer by is nonzero.
The Specht ideal is an atom of the lattice of left ideals of ℚ[Sₙ]: it is nonzero, and a
left ideal strictly inside it is zero.
The Specht ideal is a simple ℚ[Sₙ]-module.
The Specht module is irreducible, in its presentation as the left ideal ℚ[Sₙ] c_t.