A C^{∗}-algebra A and a closed bi-ideal J of A such that \frac{A}{J} and J are liminal, but A is not liminal.

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Let H be an infinite dimensional Hilbert space and I_{H} be the identity operator on H. Then A = K(H) + CI_{H} isn’t liminal (otherwise, since identity representation K(H) + CI_{H} ⟶ B(H) is nonzero irreducible ((K(H) + CI_{H})′ = K(H)′ = CI_{H} (see CW 17)) we should have I_{H} ∈ K(H), a contradiction). But K(H) is liminal, since each nonzero irreducible representation of K(H) is unitarily equivalent to identity representation K(H) ⟶ B(H) (see [Mur, page 146]). Also \frac{A}{J} ≃ CI_{H} which is finite dimensional and so is liminal (Every finite dimensional C^{∗}-algebra B is liminal, since if (H_{1}, Ψ) is a nonzero irreducible representation of B, then for x ≠ 0 in H_{1}, Ψ(B)x is finite dimensional (for Ψ(B) is finite dimensional). If (u_{λ})_{λ} be any approximate unit for B, then (Ψ(u_{λ}))_{λ} strongly converges to I_{H} so x ∈ [Ψ(B)x] = Ψ(B)x. Hence Ψ(B)x is a nonzero (closed) subspace of H_{1} invariant for Ψ(B), so by irreducibility, Ψ(B)x = H_{1}. Therefore H_{1} is finite dimensional. Thus Ψ(B) ⊆ B(H_{1}) = K(H_{1})).

Ref.

[Mur] G.J. Murphy, C^{∗}-algebras and operator theory, Academic Press, 1990.