Absolute irreducibility of rational Specht modules #
Every endomorphism of a rational Specht module that commutes with the symmetric-group action is
scalar. Equivalently, the endomorphism algebra of S^mu is ℚ. This is the Schur-index-one
statement needed to pass from the rational Specht classification to irreducible complex
representations.
The proof uses the same column-antisymmetrizer calculation as James's submodule theorem. For a
tableau t, the antisymmetrizer b_t maps the whole permutation module onto the line spanned by
the polytabloid e_t. Moreover, e_t = b_t v for some v already in the Specht module. An
equivariant endomorphism therefore maps e_t to a scalar multiple of itself. Since the orbit of
e_t spans the Specht module, the endomorphism is that scalar everywhere.
Main results #
TauCeti.algebraMap_intertwiningMap_spechtSubrepresentation_bijective: the scalar map onto the equivariant endomorphisms of the diagram-indexed Specht representation is bijective.TauCeti.algebraMap_intertwiningMap_spechtModule_bijective: the partition-indexed version forspechtModule.TauCeti.spechtModuleEndAlgEquiv: theℚ[S_n]-linear endomorphism algebra ofS^muis isomorphic toℚ.
References #
- G. D. James, The Representation Theory of the Symmetric Groups, Chapter 4.
- B. E. Sagan, The Symmetric Group, 2nd ed. (2001), Section 2.4.
- Schur--Weyl roadmap, Layer 4, "Absolute irreducibility".
Every equivariant endomorphism of the diagram-indexed rational Specht representation is a
scalar. In the canonical algebra structure on intertwining endomorphisms, this says that the
algebra map from ℚ is bijective.
Every equivariant endomorphism of the partition-indexed rational Specht module is scalar.
The algebra of intertwining endomorphisms of a rational Specht module is ℚ.
Equations
Instances For
The algebra of ℚ[S_n]-linear endomorphisms of a rational Specht module is ℚ.