Why are $321,213,132$ in cyclic order

Question

Could anyone help me understand this concept, if we have $1,2,3$ why are $$123,231,213$$ Considered in cyclic order and $$321,213,132$$ Considered in acyclic order? I am asking this in regards to the Levi-Civita symbol, many thanks in advance

Answer 1

We take $123$ to be cyclic (essentially by definition) because the numbers are all in order. If you look at this as a cycle around a circle, then the other sequences that start at different points and continue in the same direction are "cyclic" as well. These are $231$ and $312$, found by taking the first number and putting it at the end. I.e. going around the circle we find $1\to 2\to 3\to 1\to 2$, and the cyclic permutations are in this progression. The anti-cyclic permutations are found by going the other way around the circle: $321,213,$ and $132$, found in a similar way.

In general this isn't good enough. For example, $1234$ is cyclic, and the "rotations" still are just like before, but there are permutations like $1324$ that don't correspond visually to a cycle at all. So what do we do? Well it turns out that we can look at the number of transpositions it takes to get to the sequence from the ordered one. A transposition is the name for swapping two elements. For example:

$$123\text{ is 0 transpositions away from 123}$$ $$213, 321,\text{ and } 132\text{ are 1 transposition away from 123}$$ $$231 \text{ and } 312 \text{ are 2 transpositions away from 123}$$

Note that the anti-cyclic permutations are all $1$ away, while the cyclic ones are $0$ or $2$ away. This generalizes quite nicely: a cyclic permutation is one that is an even number of permutations away from the ordered (i.e identity) permutation, while the anti-cyclic permutations are an odd number of transpositions away.