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An invariant central character of the quaternion group with no linear extension

Counterexample

The nontrivial character $θ$ of the centre $Z (Q_{8}) = {\pm 1}$ of the quaternion group, $θ (- 1) = - 1$ , is invariant under $Q_{8}$ but has no linear extension to $Q_{8}$ : every linear character of $Q_{8}$ takes the value $1$ at $- 1$ , since $- 1 = [i, j]$ lies in the commutator subgroup. Thus invariance of a normal type does not by itself make the type extendible, which is why the little group method needs the split hypothesis or the projective correction.

Facts & Assumptions

Given: The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ of The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions and the character $θ$ of $Z (Q_{8}) = {\pm 1}$ with $θ (1) = 1$ , $θ (- 1) = - 1$ .

[F1]

$Q_{8}$ has $i^{2} = j^{2} = k^{2} = - 1$ , $i j = k$ , $j i = - k$ and $k = i j$ ; the element $- 1$ is central. (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions).

[F2]

A character of a normal subgroup $N ⊴ G$ is invariant when $^{g} θ = θ$ for all $g \in G$ , where $^{g} θ (n) = θ (g^{- 1} n g)$ , and then the inertia group is $G$ . (Inertia group and characters lying above a normal type).

[F3]

An extension of $θ$ to a subgroup $H$ with $N \leq H \leq G$ is a representation $\tilde{ρ} : H \to GL (S)$ on the space $S$ affording $θ$ with $\tilde{ρ} ∣_{N} = ρ$ ; at character level, an extension of $θ$ is a character $\tilde{θ}$ with ${Res}_{N}^{H} \tilde{θ} = θ$ . (An extension of a normal subgroup representation).

[F4]

An invariant irreducible representation extends to its inertia group if and only if its Clifford obstruction class is zero. (An invariant irreducible representation extends to its inertia group exactly when the Clifford obstruction vanishes).

[F5]

The faithful two-dimensional representation of $Q_{8}$ gives a projective representation of $Q_{8} / {\pm 1} ≅ C_{2} \times C_{2}$ whose factor set takes the values $α (y, x) = - 1 \neq 1 = α (x, y)$ and is not a coboundary, so that $H^{2} (Q_{8} / {\pm 1}, C^{\times})$ contains a nonzero class. (The quaternion group as a cocycle central extension of C2 x C2).

[A1]

A linear extension of the one-dimensional character $θ$ is a group homomorphism $λ : Q_{8} \to C^{\times}$ with $λ (- 1) = θ (- 1) = - 1$ , since a one-dimensional representation is a homomorphism and its character is itself.

Verification

technique · counterexample

1.1

The centre of $Q_{8}$ is $Z (Q_{8}) = {\pm 1}$ : the element $- 1$ is central by [F1], while $i$ , $j$ and $k$ are not central because $i j = k$ and $j i = - k \neq k$ , together with their cyclic analogues; since every element of $Q_{8}$ is one of $\pm 1, \pm i, \pm j, \pm k$ by [F1], these are all the central elements.

F1givenalgebra

1.2

Every linear character $λ : Q_{8} \to C^{\times}$ satisfies $λ (- 1) = 1$ : from the relations of [F1], $i j i^{- 1} j^{- 1} = (i j) (j i)^{- 1} = k \cdot (- k)^{- 1} = k \cdot k = k^{2} = - 1$ , so $- 1$ is a commutator, and multiplicativity gives $λ (i j i^{- 1} j^{- 1}) = λ (i) λ (j) λ (i)^{- 1} λ (j)^{- 1} = 1$ .

F1givenalgebra

2.1

The prescription $θ (1) = 1$ , $θ (- 1) = - 1$ defines a linear character of $Z (Q_{8})$ : products involving $1$ satisfy $θ (1 z) = θ (z 1) = θ (z) = θ (1) θ (z)$ , and the remaining product satisfies $θ ((- 1) (- 1)) = θ (1) = 1 = (- 1)^{2} = θ (- 1) θ (- 1)$ . These are all four pairs in $Z (Q_{8}) \times Z (Q_{8})$ , so the map is multiplicative. It acts on the nonzero one-dimensional space $C$ , which has no nonzero proper subspace, hence is irreducible.

F1step 1.1given

3.1

The character $θ$ is $Q_{8}$ -invariant: for $g \in Q_{8}$ and $z \in Z (Q_{8})$ one has $^{g} θ (z) = θ (g^{- 1} z g) = θ (z)$ because $z$ is central, so $^{g} θ = θ$ for every $g$ and the inertia group of $θ$ is all of $Q_{8}$ by [F2].

F2step 1.1step 2.1

4.1

Therefore $θ$ is invariant but does not extend to a linear character of $Q_{8}$ : a linear extension would be a homomorphism $λ : Q_{8} \to C^{\times}$ with $λ ∣_{Z (Q_{8})} = θ$ by [F3] and [A1], in particular $λ (- 1) = θ (- 1) = - 1$ , whereas step 1.2 forces $λ (- 1) = 1$ for every linear character. Hence no extension of $θ$ to $Q_{8}$ exists, and the invariance established in step 3.1 is not sufficient for extendibility.

A1F3step 3.1step 1.2algebra

5.1

By [F4] the failure of extension recorded in step 4.1 is exactly the statement that the Clifford obstruction class of $θ$ in $H^{2} (Q_{8} / {\pm 1}, C^{\times})$ is nonzero; this is a genuine obstruction, since [F5] exhibits a nonzero class in that same cohomology group, and it shows that the invariance hypothesis alone cannot replace the split hypothesis of the little group method. The character $θ$ is the central character of the faithful two-dimensional representation of $Q_{8}$ , so the example is exactly the nonsplit counterpart of the extendible invariant types.

F4F5step 4.1∎

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Sources

Britta Späth, Reduction theorems for some global-local conjectures — §1.B, printed pp. 3–5 (standard reference, not scraped)
Tammo tom Dieck, Representation Theory — Proposition (4.2.6) and Remark (4.2.7), printed p. 57 (standard reference, not scraped)