Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Weyl character and dimension formulas for sl2

Example

For the finite-dimensional irreducible sl2-module V(n) with basis v0,…,vn of All irreducible finite-dimensional sl2 modules, put χn(z)=∑k=0nzn−2k=zn+zn−2+⋯+z−n(z∈C×). Then for z≠±1 χn(z)=zn+1−z−(n+1)z−z−1, and dim⁡V(n)=n+1.

Facts & Assumptions

Given: The module V(n) with its basis and h-eigenvalues n−2k (All irreducible finite-dimensional sl2 modules), the standard diagonal subalgebra Ch from The special linear Lie algebra sl_2, and the variable z∈C×. In this example we define the rank-one formal character by assigning the monomial zm to the h-eigenspace of eigenvalue m and summing with eigenspace multiplicities; this convention is not attributed to the weight-space definition.

[L1]

The h-eigenvalues on V(n) are n,n−2,…,−n, each with multiplicity one, and dim⁡V(n)=n+1 (All irreducible finite-dimensional sl2 modules, Finite-dimensional representations of sl2, The special linear Lie algebra sl_2).

Verification

technique · direct
1.1L1given

By [L1] the sum χn(z)=∑k=0nzn−2k is the sum of zm over the h-eigenvalues m of V(n), each counted with its multiplicity, so it is the rank-one formal character under the convention fixed in the given data.

1.2given

The telescoping identity (z−z−1)χn(z)=∑k=0n(zn−2k+1−zn−2k−1)=zn+1−z−(n+1) holds as an identity of Laurent polynomials.

2.1step 1.2

For z≠0 with z−z−1≠0, that is for z≠±1, division gives χn(z)=zn+1−z−(n+1)z−z−1, which is the displayed formula on the regular set.

2.2L1step 1.2

The identity of step 1.2 is the algebraic cancellation zn+1−z−(n+1)=(z−z−1)χn(z) in the Laurent polynomial ring; it exhibits χn as the quotient after cancelling the common factor z−z−1, and evaluating that Laurent polynomial at z=1 gives χn(1)=n+1, matching dim⁡V(n)=n+1 by [L1].

3.1step 2.1step 2.2∎

Hence the character identity on the regular set and the dimension formula both hold, as asserted.

Depends on

Used by

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