Norm of product

Norm of a product

Normed division algebras require that the norm of the product is equal to the product of the norms. Lagrange's identity exhibits this equality. Due to Hurwitz theorem, it admits this interpretation only for algebras isomorphic to the real numbers, complex numbers, quaternions and octonions. If divisors of zero are allowed, many other algebraic structures in $\mathbb {R} ^{n}$ are possible , . One approach has been presented in the context of a deformed Lorentz metric. This latter proposal is based on a transformation stemming from the product operation and magnitude definition in hyperbolic scator algebra . The product identity used as a starting point here, is a consequence of the $\left\Vert \mathbf {ab} \right\Vert =\left\Vert \mathbf {a} \right\Vert \left\Vert \mathbf {b} \right\Vert$ equality for scator algebras.

The fourth order identity gives Lagrange's identity.

sixth order identity

The non trivial identities for real numbers obtained to sixth order

series expansion of the product identity

$\prod _{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=\prod _{i=1}^{n}\left(1-a_{i}^{2}\right)\prod _{i=1}^{n}\left(1-b_{i}^{2}\right)$

are

$\sum _{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)\right]+\sum _{i<j<k}^{n}\left[a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right]=\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right)$

and its counterpart, obtained by interchanging the variables $a$ and $b$.

Expand the product identity in series up to sixth order. The LHS is

$\prod _{i=1}^{n}\left(1-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)=1+\sum _{i=1}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)+\sum _{i<j}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)\left(-a_{j}^{2}-b_{j}^{2}+a_{j}^{2}b_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(-a_{i}^{2}-b_{i}^{2}+a_{i}^{2}b_{i}^{2}\right)\left(-a_{j}^{2}-b_{j}^{2}+a_{j}^{2}b_{j}^{2}\right)\left(-a_{k}^{2}-b_{k}^{2}+a_{k}^{2}b_{k}^{2}\right)+{\mathcal {O}}^{7+}.$

Consider only the sixth order terms ${\mathcal {O}}^{6}\left({\textrm {LHS}}\right)=-\sum _{i<j}^{n}\left[a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)\right]-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}+b_{i}^{2}b_{j}^{2}b_{k}^{2}-\sum _{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)-\sum _{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)$

The RHS of the product identity is similarly expanded in series up to sixth order

$\prod _{i=1}^{n}\left(1-a_{i}^{2}\right)\prod _{i=1}^{n}\left(1-b_{i}^{2}\right)=\left(1-\sum _{i=1}^{n}a_{i}^{2}+\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}+{\mathcal {O}}^{7+}\right)\left(1-\sum _{i=1}^{n}b_{i}^{2}+\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}-\sum _{i<j<k}^{n}b_{i}^{2}b_{j}^{2}b_{k}^{2}+{\mathcal {O}}^{7+}\right),$

and only sixth order terms retained

${\mathcal {O}}^{6}\left({\textrm {RHS}}\right)=-\sum _{i<j<k}^{n}a_{i}^{2}a_{j}^{2}a_{k}^{2}-\sum _{i<j<k}^{n}b_{i}^{2}b_{j}^{2}b_{k}^{2}-\left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\right)-\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right).$

These two results are equated for equal powers of $a^{n}b^{m}$. The terms $a^{6}$ and $b^{6}$ give trivial identities whereas the terms involving $a^{4}b^{2}$ and $a^{2}b^{4}$ give the non trivial sixth order identities

$\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\left(b_{i}^{2}+b_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(a_{i}^{2}a_{j}^{2}b_{k}^{2}+a_{i}^{2}b_{j}^{2}a_{k}^{2}+b_{i}^{2}a_{j}^{2}a_{k}^{2}\right)=\left(\sum _{i=1}^{n}b_{i}^{2}\right)\left(\sum _{i<j}^{n}a_{i}^{2}a_{j}^{2}\right)$

and

$\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\left(a_{i}^{2}+a_{j}^{2}\right)+\sum _{i<j<k}^{n}\left(a_{i}^{2}b_{j}^{2}b_{k}^{2}+b_{i}^{2}a_{j}^{2}b_{k}^{2}+b_{i}^{2}b_{j}^{2}a_{k}^{2}\right)=\left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i<j}^{n}b_{i}^{2}b_{j}^{2}\right).$