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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?

a196884
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  • In other words, you want sufficient conditions for $(ab)(\bar{b}\bar{a})=(a\bar{a})(b\bar{b})$. – J.G. Jun 06 '22 at 15:23
  • In those terms I'm asking if $ab\ne 0$ implies $(ab)(\bar{b}\bar{a})=(a\bar{a})(b\bar{b})$ – a196884 Jun 06 '22 at 15:51
  • As a curiosity you might check a calculator I put online almost ten years ago addressing this very topic. – John Wayland Bales Jun 30 '23 at 22:42
  • If I got it correctly, we still have $|z^n|=|z|^n$ in every step of the Cayley–Dickson construction. This is because we can write $z=a+v$ where $a$ is a scalar (a real number) and $v$ is a vector (purely imaginary); we have $av=va$ and $v^2=-|v|^2$. This allows us to conclude that the norms of powers of $a+v$ behave just like norms of powers of a complex number. – Jianing Song May 26 '25 at 10:26

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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

$$1=n(x)n(y)<n(xy)= n(-e_4+e_{15}) = 2,$$

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

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

This follows from Theorem 8.3 of this article.

pregunton
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