Tensor product of quadratic forms (original) (raw)

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In mathematics, the tensor product of quadratic forms is most easily understood when one views the quadratic forms as quadratic spaces.[1] If R is a commutative ring where 2 is invertible, and if ( V 1 , q 1 ) {\displaystyle (V_{1},q_{1})} and ( V 2 , q 2 ) {\displaystyle (V_{2},q_{2})} are two quadratic spaces over R, then their tensor product ( V 1 ⊗ V 2 , q 1 ⊗ q 2 ) {\displaystyle (V_{1}\otimes V_{2},q_{1}\otimes q_{2})} is the quadratic space whose underlying _R_-module is the tensor product V 1 ⊗ V 2 {\displaystyle V_{1}\otimes V_{2}} of _R_-modules and whose quadratic form is the quadratic form associated to the tensor product of the bilinear forms associated to q 1 {\displaystyle q_{1}} and q 2 {\displaystyle q_{2}} .

In particular, the form q 1 ⊗ q 2 {\displaystyle q_{1}\otimes q_{2}} satisfies

( q 1 ⊗ q 2 ) ( v 1 ⊗ v 2 ) = q 1 ( v 1 ) q 2 ( v 2 ) ∀ v 1 ∈ V 1 , v 2 ∈ V 2 {\displaystyle (q_{1}\otimes q_{2})(v_{1}\otimes v_{2})=q_{1}(v_{1})q_{2}(v_{2})\quad \forall v_{1}\in V_{1},\ v_{2}\in V_{2}}

(which does uniquely characterize it however). It follows from this that if the quadratic forms are diagonalizable (which is always possible if 2 is invertible in R), i.e.,

q 1 ≅ ⟨ a 1 , . . . , a n ⟩ {\displaystyle q_{1}\cong \langle a_{1},...,a_{n}\rangle }

q 2 ≅ ⟨ b 1 , . . . , b m ⟩ {\displaystyle q_{2}\cong \langle b_{1},...,b_{m}\rangle }

then the tensor product has diagonalization

q 1 ⊗ q 2 ≅ ⟨ a 1 b 1 , a 1 b 2 , . . . a 1 b m , a 2 b 1 , . . . , a 2 b m , . . . , a n b 1 , . . . a n b m ⟩ . {\displaystyle q_{1}\otimes q_{2}\cong \langle a_{1}b_{1},a_{1}b_{2},...a_{1}b_{m},a_{2}b_{1},...,a_{2}b_{m},...,a_{n}b_{1},...a_{n}b_{m}\rangle .}

1. ^ Kitaoka, Yoshiyuki. "Tensor products of positive definite quadratic forms IV". Cambridge University Press. Retrieved February 12, 2024.