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#1: Initial revision by user avatar celtschk‭ · 2025-01-04T07:19:24Z (4 days ago)
In a linear kei with more than one element (that is, $n>1$), for each line $a$ there exist lines $b\ne a$ such that $a\rhd b = a$, and moreover, for each such $b$, we also have $b\rhd a = b$.

In other words, for each line $a$, there exist lines that are orthogonal to it, and if $a$ is orthogonal to $b$, then $b$ is orthogonal to $a$.

That this is not a property of general keis can be seen from your example of point reflections. In point reflections, the only point that gets mapped to itself is the one you are reflecting on, that is for point reflections, $a\rhd b=a \implies a=b$.

However, there exist other keis with the property, for example, take the definition $a\rhd b = a$ for all $a$ and $b$ (I'll refer to it as the do-nothing kei, as we do nothing to $a$).

Another property, that builds on the previous, is that there is a maximal number of lines that are orthogonal to each other, that number is $n$. Written in term of the kei, any set $S$ of elements with the property that for any $a,b\in S$ we have $a\rhd b = a$ has at most $n$ elements. At the same time, the linear kei (for $n>1$) has infinitely many elements.

In addition, any such set can be extended to a set of $n$ elements with that property.

This is different from the do-nothing kei, where all elements are "orthogonal" to each other, and thus for infinitely many elements, there's no finite limit for such sets.

Note however that for $n=1$, the linear kei has only one element and is the same as the do-nothing kei with one element. Indeed, there is only one magma (and thus only one kei) with only one element.

Also, if you restrict a linear kei to a set of mutually orthogonal lines, you exactly get a finite do-nothing kei, for obvious reasons.