# Norm of the Sedenions

Let the Cayley-Dickson doubling of the octonions be called the sedenions. The sedenions are not a division algebra, because they contain zero divisors. The presence of zero divisors means that the norm of a sedenion $$x$$, denoted $$n(x)=x\bar{x}$$, is not multiplicative: if $$a,b$$ are such that $$a\cdot b=0,$$ then $$n(ab)=0\ne n(a)n(b)$$. My question is: if $$a,b$$ are sedenions such that $$a\cdot b\ne0$$, is the norm multiplicative? If not, is it submultiplicative?

• In other words, you want sufficient conditions for $(ab)(\bar{b}\bar{a})=(a\bar{a})(b\bar{b})$.
– J.G.
Jun 6, 2022 at 15:23
• In those terms I'm asking if $ab\ne 0$ implies $(ab)(\bar{b}\bar{a})=(a\bar{a})(b\bar{b})$ Jun 6, 2022 at 15:51
• As a curiosity you might check a calculator I put online almost ten years ago addressing this very topic. Jun 30, 2023 at 22:42

The answer to both questions is no. For example, if we take $$x=\frac{e_1-e_{10}}{\sqrt2}$$ and $$y=\frac{e_5+e_{14}}{\sqrt2}$$, using the multiplication table in Wikipedia one has

$$1=n(x)n(y)

so the norm fails to be (sub)multiplicative even when zero divisors are ignored.

What is always true is that $$0 \le n(xy) \le 2 n(x)n(y)$$ for any sedenions $$x$$ and $$y$$. More generally, for $$k \ge 4$$ the norm of the $$2^k$$-dimensional Cayley-Dickson algebra over $$\mathbb{R}$$ satisfies the sharp bound

$$0 \le n(xy) \le 2^{k-3} n(x)n(y).$$