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Character and cocharacter lattices of a split torus

Statement

Let T be a split torus over k, so that T≅Dk(M) for a free abelian group M of finite rank (Groups of multiplicative type and tori, Diagonalizable groups and their character modules). Then the character group X(T)=Hom⁡k(T,Gm) is canonically isomorphic to M, the cocharacter group X∗(T)=Hom⁡k(Gm,T) is canonically isomorphic to the dual lattice M∨=Hom⁡(M,Z), and the pairing X(T)×X∗(T)→Z, (χ,λ)↦⟨χ,λ⟩, defined by χ∘λ∈Hom⁡k(Gm,Gm)=Z, is a perfect Z-bilinear pairing identifying X∗(T) with X(T)∨. Both lattices are free of finite rank equal to dim⁡T, and the identifications and the pairing commute with extension of scalars. In particular X(T)⊗ZQ and X∗(T)⊗ZQ are Q-vector spaces in perfect duality.

Facts & Assumptions

Given: A field k, a free abelian group M of finite rank, and a split torus T with an isomorphism T≅Dk(M) over k.

[F1]

The diagonalizable group Dk(M)=Spec⁡k[M] has Dk(M)(R)=Hom⁡(M,R×) and Gm=Dk(Z), and characters are the homomorphisms X(G)=Hom⁡k-groups(G,Gm) (Diagonalizable groups and their character modules).

[F2]

For abelian groups M,N there are natural identifications X(Dk(M))=M, via m↦em, and Hom⁡k(Dk(M),Dk(N))=Hom⁡(N,M) (Split diagonalizable groups are dual to abelian groups).

[F3]

A torus is split when it is isomorphic over k to Gmr for some r≥0 (Groups of multiplicative type and tori). Its coordinate ring is k[t1,…,tr,(t1⋯tr)−1]; it has dimension r by A polynomial ring in n variables over a field has dimension n: localization cannot increase prime-chain length, and the chain (0)⊊(t1−1)⊊⋯⊊(t1−1,…,tr−1) survives this localization.

Proof

1.1F1F2F3algebra

By [F3] the split torus T is isomorphic over k to Gmr with r=dim⁡T, and Gmr=Dk(Zr) by [F1]; the given isomorphism T≅Dk(M) and [F2] therefore identify M with Zr. Applying [F2] to M and to Z gives X(T)=Hom⁡k(Dk(M),Dk(Z))=Hom⁡(Z,M)≅M and X∗(T)=Hom⁡k(Dk(Z),Dk(M))=Hom⁡(M,Z)=M∨. Both are free abelian of rank r=dim⁡T; the identifications are induced by the anti-equivalence and are compatible with any change of the splitting isomorphism, which only renames M by the induced automorphism.

2.1F1F2step 1.1algebra

For χ∈X(T) and λ∈X∗(T) the composite χ∘λ:Gm→Gm is a homomorphism, and [F1] identifies Hom⁡k(Gm,Gm)=Hom⁡k(Dk(Z),Dk(Z))=Hom⁡(Z,Z)=Z by [F2]; so ⟨χ,λ⟩:=χ∘λ is an integer and composition of homomorphisms is Z-bilinear. Under the identifications of step 1.1, an element of X(T) is an element m∈M and an element of X∗(T) is a homomorphism μ:M→Z, and the corresponding composite is μ(m): χ∘λ is evaluation of the character at the cocharacter. Therefore the map X∗(T)→X(T)∨, λ↦(χ↦⟨χ,λ⟩), is exactly the identity Hom⁡(M,Z)→Hom⁡(M,Z) and hence an isomorphism, so the pairing is perfect and identifies X∗(T) with X(T)∨.

3.1F2step 1.1step 2.1algebra∎

Let k′⊇k be a field extension. Base change of group algebras identifies Dk′(M)=Dk(M)k′, and [F2] applied over the field k′ gives X(Tk′)=X(Dk′(M))=M and X∗(Tk′)=M∨ with the same evaluation pairing, so the two identifications of step 1.1 commute with extension of scalars. For free lattices of finite rank the dual of a base change is the base change of the dual, so tensoring the perfect pairing of step 2.1 with Q exhibits (X(T)⊗Q)∨=X∗(T)⊗Q and gives a perfect Q-bilinear pairing of X(T)⊗Q with X∗(T)⊗Q.

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Sources