Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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=0nzn2k=zn+zn2++zn(zC×). Then for z±1 χn(z)=zn+1z(n+1)zz1, and dimV(n)=n+1.

Facts & Assumptions

Given: The module V(n) with its basis and h-eigenvalues n2k (All irreducible finite-dimensional sl2 modules), the standard diagonal subalgebra Ch from The special linear Lie algebra sl_2, and the variable zC×. 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,n2,,n, each with multiplicity one, and dimV(n)=n+1 (All irreducible finite-dimensional sl2 modules, Finite-dimensional representations of sl_2, The special linear Lie algebra sl_2).

Verification

technique · direct
1.1

By [L1] the sum χn(z)=k=0nzn2k 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.

L1given
1.2

The telescoping identity (zz1)χn(z)=k=0n(zn2k+1zn2k1)=zn+1z(n+1) holds as an identity of Laurent polynomials.

given
2.1

For z0 with zz10, that is for z±1, division gives χn(z)=zn+1z(n+1)zz1, which is the displayed formula on the regular set.

step 1.2
2.2

The identity of step 1.2 is the algebraic cancellation zn+1z(n+1)=(zz1)χn(z) in the Laurent polynomial ring; it exhibits χn as the quotient after cancelling the common factor zz1, and evaluating that Laurent polynomial at z=1 gives χn(1)=n+1, matching dimV(n)=n+1 by [L1].

L1step 1.2
3.1

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

step 2.1step 2.2

Depends on

Used by

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