Tensor product of quadratic forms

In mathematics, the tensor product of quadratic forms is most easily understood when one views the quadratic forms as quadratic spaces. If R is a commutative ring where 2 is invertible (that is, R has characteristic [math]\displaystyle{ \text{char}(R) \neq 2 }[/math]), and if [math]\displaystyle{ (V_1, q_1) }[/math] and [math]\displaystyle{ (V_2,q_2) }[/math] are two quadratic spaces over R, then their tensor product [math]\displaystyle{ (V_1 \otimes V_2, q_1 \otimes q_2) }[/math] is the quadratic space whose underlying R-module is the tensor product [math]\displaystyle{ V_1 \otimes V_2 }[/math] of R-modules and whose quadratic form is the quadratic form associated to the tensor product of the bilinear forms associated to [math]\displaystyle{ q_1 }[/math] and [math]\displaystyle{ q_2 }[/math].

In particular, the form [math]\displaystyle{ q_1 \otimes q_2 }[/math] satisfies

[math]\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 }[/math]

(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.,

[math]\displaystyle{ q_1 \cong \langle a_1, ... , a_n \rangle }[/math]

[math]\displaystyle{ q_2 \cong \langle b_1, ... , b_m \rangle }[/math]

then the tensor product has diagonalization

[math]\displaystyle{ q_1 \otimes q_2 \cong \langle a_1b_1, a_1b_2, ... a_1b_m, a_2b_1, ... , a_2b_m , ... , a_nb_1, ... a_nb_m \rangle. }[/math]