How many finite-dimensional irreducible representations does 𝔰𝔩₂(ℂ) have?
AThree (the trivial, standard, and adjoint)
BFinitely many, one for each root of unity
CInfinitely many — one for each non-negative integer n, of dimension n+1
DTwo — the standard and its dual
For each n = 0, 1, 2, 3, …, there is exactly one irreducible representation V(n) of dimension n+1. V(0) is the trivial representation. V(1) is the standard 2-dimensional representation. V(2) is the 3-dimensional adjoint representation (≅ 𝔰𝔩₂ acting on itself via the Lie bracket). V(3) is 4-dimensional, and so on. This is the complete classification: every finite-dimensional irreducible representation is isomorphic to exactly one V(n).
Question 2 True / False
In the irreducible representation V(n) of 𝔰𝔩₂, the element h acts diagonalizably with eigenvalues n, n−2, n−4, …, −n.
TTrue
FFalse
Answer: True
V(n) has a basis {v_n, v_{n-2}, ..., v_{-n}} where h·v_m = m·v_m. The weights form the arithmetic sequence n, n−2, …, −n, decreasing by 2 at each step, giving n+1 weight spaces in total. Each weight space is 1-dimensional for irreducible representations of 𝔰𝔩₂. The element e maps v_m to a scalar multiple of v_{m+2} (raising operator) and f maps v_m to a scalar multiple of v_{m−2} (lowering operator).
Question 3 Short Answer
The Casimir element C = h² + 2ef + 2fe = h² + 2h + 4fe acts on the irreducible representation V(n) as a scalar. What scalar?
Think about your answer, then reveal below.
Model answer: C acts on V(n) as multiplication by n(n+2), equivalently n² + 2n.
The Casimir element lies in the center of the universal enveloping algebra U(𝔰𝔩₂). By Schur's lemma, it acts as a scalar on each irreducible representation. Computing on the highest weight vector v_n: h²·v_n = n²v_n, and 2h·v_n = 2nv_n, and 4fe·v_n = 0 (since e·v_n = 0 for the highest weight vector). So C·v_n = (n² + 2n)v_n = n(n+2)v_n. This scalar distinguishes non-isomorphic irreducibles.
Question 4 Multiple Choice
The standard representation V(1) of 𝔰𝔩₂ has dimension 2 with basis {v₁, v₋₁}. What is V(1) ⊗ V(1) as a direct sum of irreducibles?
AV(2)
BV(0) ⊕ V(2)
CV(1) ⊕ V(1)
DV(0) ⊕ V(1) ⊕ V(2)
V(1) ⊗ V(1) has dimension 4. The tensor product of representations with highest weights m and n decomposes by the Clebsch-Gordan formula: V(m) ⊗ V(n) ≅ V(m+n) ⊕ V(m+n−2) ⊕ ··· ⊕ V(|m−n|). For m = n = 1: V(1) ⊗ V(1) ≅ V(2) ⊕ V(0). Dimension check: 3 + 1 = 4. The V(2) component is the symmetric part (Sym²V(1)), and V(0) is the antisymmetric part (∧²V(1) ≅ det).
Question 5 Short Answer
Why is 𝔰𝔩₂ representation theory considered the 'template' for all semisimple Lie algebras?
Think about your answer, then reveal below.
Model answer: Every semisimple Lie algebra contains copies of 𝔰𝔩₂ as subalgebras (one for each simple root), and the weight space decomposition of any representation restricts to an 𝔰𝔩₂-representation along each root direction. The entire highest weight classification for semisimple Lie algebras is built by combining these 𝔰𝔩₂ analyses across all root directions.
The root space decomposition of a semisimple Lie algebra 𝔤 identifies a triple {eα, fα, hα} ≅ 𝔰𝔩₂ for each root α. The representation theory of 𝔤 is analyzed by restricting to these 𝔰𝔩₂-triples: the integrality of weights, the finite-dimensionality criterion, and the Weyl character formula all ultimately rest on the 𝔰𝔩₂ classification. This is why every Lie theory textbook begins with 𝔰𝔩₂.