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, and if \( {\displaystyle (V_{1},q_{1})} \) and \( {\displaystyle (V_{2},q_{2})} \) are two quadratic spaces over R, then their tensor product\( {\displaystyle (V_{1}\otimes V_{2},q_{1}\otimes q_{2})} \) is the quadratic space whose underlying R-module is the tensor product \( {\displaystyle V_{1}\otimes V_{2}} \) of R-modules and whose quadratic form is the quadratic form associated to the tensor product of the bilinear forms associated to \( q_{1} \) and \( q_{2}. \)
In particular, the form \( {\displaystyle q_{1}\otimes q_{2}} \) satisfies
\( {\displaystyle (q_{1}\otimes q_{2})(v_{1}\otimes v_{2})=q_{1}(v_{1})q_{2}(v_{2})\quad \forall v_{1}\in V_{1},\ v_{2}\in V_{2}} \)
(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.,
\( q_{1}\cong \langle a_{1},...,a_{n}\rangle \)
\( q_{2}\cong \langle b_{1},...,b_{m}\rangle \)
then the tensor product has diagonalization
\( {\displaystyle q_{1}\otimes q_{2}\cong \langle a_{1}b_{1},a_{1}b_{2},...a_{1}b_{m},a_{2}b_{1},...,a_{2}b_{m},...,a_{n}b_{1},...a_{n}b_{m}\rangle .} \)
Undergraduate Texts in Mathematics
Graduate Studies in Mathematics
Hellenica World - Scientific Library
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