M Theory Lesson 173
$i \cdot j = k$
$j \cdot k = i$
$k \cdot i = j$
For example, if the objects were sides of a triangle, the operation might take distinct edges to the edge opposite their common vertex. Note that $k \cdot (j \cdot k) = k \cdot i = j$ so associativity implies that
$j \cdot j = i \cdot i = k \cdot k$
Rules of the form $k \cdot k \cdot k = i$ follow, so one need never encounter more than cubic terms. If the operation were also commutative, then $i \cdot i = k \cdot (j \cdot j) \cdot k = i \cdot k = j$. It follows that $j \cdot j = j$, and in fact all the objects are idempotent. But now there are just too many relations between these objects, so it might be more interesting to drop commutativity and/or associativity. Unfortunately, we can then no longer consider the simple example of an ordinary triangle.
Here is a picture of the three squares on the associahedron pair of pants. When a crossing is marked on each square, there is one path around the faces of the associahedron. By choosing crossings correctly, we can draw a trefoil knot. Note how the picture almost looks like two pieces of ribbon too. By mixing the shown edges with actual polytope edges, one can draw a trivalent ribbon vertex on both the front and back of the pair of pants. Now we can have fun dreaming up new quandle examples using this geometry, associated to the circulant mass operators.