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The full weight lattice need not integrate through a central quotient

Statement refuted

Assume the Axiom of Choice. Every dominant weight in the weight lattice of a compact semisimple group integrates to every compact group form with the given Lie algebra.

Facts & Assumptions

Given: The Axiom of Choice, the group $S U (2) = {g \in G L_{2} (C) : g^{*} g = I, \det g = 1}$ , the group $S O (3)$ with the surjective two-sheeted covering homomorphism $π : S U (2) \to S O (3)$ of kernel ${\pm I}$ (SU(2) to SO(3) as a covering homomorphism), the maximal torus $T = {diag (z, z^{- 1}) : ∣ z ∣ = 1}$ of $S U (2)$ and the element $- I = diag (- 1, - 1) \in T$ , the fundamental weight $ω_{1}$ of ${s l}_{2}$ (Fundamental weights, The special linear Lie algebra sl_2), and for each $m \geq 0$ the space $W_{m}$ of homogeneous polynomials of degree $m$ in the variables $x_{1}, x_{2}$ , with the action $(g \cdot p) (x) = p (g^{- 1} x)$ .

[A1]

The Axiom of Choice is assumed; it enters through the root-space and highest-weight theory used below (The Axiom of Choice).

[L1]

The covering $π$ is surjective with kernel ${\pm I}$ , so the fibers of $π$ are the two-element sets ${g, - g}$ and $S O (3) ≅ S U (2) / {\pm I}$ (SU(2) to SO(3) as a covering homomorphism).

[L2]

The polynomial action is a representation of $S U (2)$ : $(g h) \cdot p = p \circ (g h)^{- 1} = p \circ h^{- 1} \circ g^{- 1} = g \cdot (h \cdot p)$ , and the identity acts trivially; the differential at the identity makes $W_{m}$ the ${s l}_{2} (C)$ -module ${Sym}^{m} (C^{2})$ , which is irreducible of highest weight $m ω_{1}$ for the chosen positive system (Representations of Lie algebras, Symmetric powers as highest-weight modules).

[L3]

The torus element $diag (z, z^{- 1})$ acts on the monomial $x_{1}^{m - k} x_{2}^{k}$ by $z^{2 k - m}$ , so the weights of $W_{m}$ restricted to $T$ are the integers $m, m - 2, \dots, - m$ ; each of the weights $m ω_{1}, (m - 2) ω_{1}, \dots, - m ω_{1}$ lies in the weight lattice $Z ω_{1}$ (Weight and weight space, Fundamental weights).

Counterexample

technique · direct

1.1

The element $- I$ acts on a homogeneous polynomial of degree $m$ by $(- I) \cdot p (x) = p ((- I)^{- 1} x) = p (- x) = (- 1)^{m} p (x)$ , so the operator $ρ_{m} (- I)$ on $W_{m}$ is the scalar $(- 1)^{m}$ .

A1L2given

2.1

A homomorphism $ρ : S U (2) \to G L (V)$ factors as $\overset{ˉ}{ρ} \circ π$ for a homomorphism $\overset{ˉ}{ρ} : S O (3) \to G L (V)$ if and only if $ρ (- I) = {id}_{V}$ : if $ρ = \overset{ˉ}{ρ} \circ π$ then $π (- I) = I$ gives $ρ (- I) = id$ , while conversely $ρ (- g) = ρ (g) ρ (- I) = ρ (g)$ whenever $ρ (- I) = id$ , so $ρ$ is constant on the fibers ${g, - g}$ of the surjective $π$ from [L1] and descends uniquely to the quotient $S O (3) ≅ S U (2) / {\pm I}$ .

L1step 1.1

3.1

If $m$ is odd, then $ρ_{m} (- I) = - {id}_{W_{m}} \neq id$ by step 1.1, so by step 2.1 the representation $W_{m}$ of $S U (2)$ does not descend to $S O (3)$ ; but its highest weight $m ω_{1}$ is dominant integral and lies in the weight lattice by [L2] and [L3], so it is a dominant weight of the abstract weight lattice that does not integrate to the group form $S O (3)$ .

L2L3step 1.1step 2.1

4.1

If $m$ is even, then $ρ_{m} (- I) = id$ by step 1.1 and step 2.1 makes $W_{m}$ descend to $S O (3)$ ; in particular the highest-weight-one module $W_{1} = C^{2}$ integrates to $S U (2)$ but not to $S O (3)$ , whereas the even highest weights do descend.

step 1.1step 2.1step 3.1

5.1

The witness $(S U (2), S O (3), W_{1})$ shows that the dominant weight $ω_{1}$ of the weight lattice integrates to $S U (2)$ but not to the compact group form $S O (3)$ with the same Lie algebra, refuting the statement; the failed conclusion is that every dominant weight integrates to every group form.

step 3.1step 4.1∎

Depends on

Used by

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Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)