“…That is, (N θγk ) k∈{1,...,n} = n r=1 d s=1 N (rs) θk γ rs k∈{1,...,n} , where N (rs) θ uniquely solves (6). Furthermore, since e ı θ,· C d C n×d ⊆ P (θ), (19) implies that β, η C n×d = e ı θ,· C d β, e ı θ,· C d η = ι * θ aι θ (e ı θ,· C d γ + grad θ N θγ ), e ı θ,· C d η = a(grad θ N θγ + e ı θ,· C d γ), e ı θ,…”