ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Character lattices of SU(2) and SO(3)

Example

Assume the Axiom of Choice. For type $A_{1}$ with fundamental weight $ω$ one has $P = Z ω$ and $Q = 2 Z ω$ . The character lattice of the maximal torus of $S U (2) = {g \in U (2) : \det g = 1}$ is $P$ , while that of the maximal torus of $S O (3) ≅ S U (2) / {\pm I}$ is $Q$ ; consequently precisely the even $S U (2)$ highest weights descend to $S O (3)$ .

Facts & Assumptions

Given: Assume the Axiom of Choice; $S U (2)$ with maximal torus $T_{S U (2)} = {diag (z, z^{- 1}) : ∣ z ∣ = 1}$ and the standard central quotient identification $S O (3) ≅ S U (2) / {\pm I}$ .

[L1]

For a compact connected semisimple group $Q \subseteq X^{*} (T) \subseteq P$ , the simply connected form has character lattice $P$ and the adjoint form has character lattice $Q$ ; central quotients correspond to intermediate lattices (Root and weight lattice sandwich, Central quotients and intermediate character lattices).

[L2]

For type $A_{1}$ , $Q = Z α = 2 Z ω$ and $P = Z ω$ , and characters of $T_{S U (2)}$ are the maps $diag (z, z^{- 1}) \mapsto z^{n}$ with weight $n ω$ (Characters are the integral weights). [L1]

Verification

technique · direct

1.1

$S U (2)$ is simply connected and semisimple with root system $A_{1}$ , so $X^{*} (T_{S U (2)}) = P = Z ω$ by [L1]; the characters are $diag (z, z^{- 1}) \mapsto z^{n}$ , with weight $n ω$ .

L1L2

2.1

The kernel of $S U (2) \to S O (3)$ is ${\pm I}$ . Since $- I = diag (- 1, - 1)$ corresponds to $z = - 1$ in $T_{S U (2)}$ , the character of weight $n ω$ takes the value $(- 1)^{n}$ there, so it is trivial on the kernel exactly when $n$ is even.

L2step 1.1

3.1

By [L1] the intermediate lattice of $S O (3)$ is $X^{*} (T_{S O (3)}) = {characters trivial on {\pm I}} = 2 Z ω = Q$ , and the adjoint-form computation of [L1] gives the same answer.

L1step 2.1

4.1

Hence a dominant $S U (2)$ highest weight $n ω$ descends to $S O (3)$ exactly when $n$ is even, i.e. exactly when $n ω \in Q$ ; this is the explicit form of the finite central quotient obstruction.

L1step 2.1step 3.1∎

Depends on

Used by

SU(2) and SO(3) share roots but are not isomorphic Counterexample

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Brian Conrad and Aaron Landesman, Compact Lie Groups (standard reference, not scraped)