Fact 3: Given G 1 , G 2 , T G 1 ⊗ T G 2 ∘ ◻ = T G 1 ⊕ G 2 {\displaystyle {\mathcal {G}}_{1},\ {\mathcal {G}}_{2},\ \ {\mathcal {T}}_{{\mathcal {G}}_{1}}\otimes {\mathcal {T}}_{{\mathcal {G}}_{2}}\circ \Box ={\mathcal {T}}_{{\mathcal {G}}_{1}\oplus {\mathcal {G}}_{2}}} under the canonical isomorphism.