Define trace-like function (e.g. cyclically invariant) involving operators without built-in meaning

Question

I am interested in defining a function, say $\mathrm{Tra}$, that imply operators without build-in meaning, say it involves products with the $\cdot$ (Esc . Esc), not assumed commuatative.

I want to mimic a trace for it, the main property being cyclic invariance, which I expect to use to simplify expressions. Now I face a syntax problem: the last two operators being the same, as is in general, when one wants to impose some invariance.

esq[A_CenterDot] := RotateLeft@A (*rotates to the CenterDot-product*)
Tra[A_Plus] :=Tra[#] & /@ A (*make Tra additive*)
Tra[A_CenterDot] := Tra[esq[A]] 

Second line I've learnt from this answer by Jules Lamers, mutatis mutandis.

With last command I was expecting $ \mathrm{Tra}(x\cdot y\cdot y\cdot x\cdot y\cdot x) $ to yield $ \mathrm{Tra}(x\cdot x \cdot y\cdot y\cdot x\cdot y)$ (with only one left shift, but see below the aimed lexicographic order).

But this of course gives a loop instead. In general,

• how does one correct this?

• if possible: how to deal with situations when "the two last arrows of the commutative diagram are the same operation"?

• is there an way out to defining relations at the level of the algebra just before evaluating this "trace"? What is meant is: assuming that $\mathrm{Tra}: A \to \mathbb k$, for any $n$, here there is a commuting pentagon (joining the first and last $A^n$ at the same node)

$A^n \stackrel{left-shift}{\to} A^n \stackrel{\cdot}{\to} A \stackrel{Tra}{\to} \mathbb k\stackrel{Tra}{\leftarrow } A \stackrel{\cdot}{\leftarrow} A^n $

Edit: After the comment(s): I wish the routine to stop, say, after the lexicographic-like order is reached. All x before y and the highest power of x showing up first, independently of what happens thereafter. So, inside the Tra-operator:

• $ x \cdot y\cdot x \cdot x \to x\cdot x \cdot x \cdot y $ (cyclicity allows here to group all the x together)
• $ y \cdot y\cdot y \cdot x \to x\cdot y \cdot y\cdot y $
• $ x\cdot y \cdot x\cdot y\cdot y \cdot x \cdot x \to x \cdot x \cdot x\cdot y \cdot x\cdot y\cdot y $ (this cannot be tested with OrderedQ, will almost always false)

Output example is then:

$\mathrm{Tra}[ y \cdot y\cdot y \cdot x + x\cdot y \cdot y\cdot y ] \to 2\mathrm{Tra}[x\cdot y \cdot y\cdot y]$

since the second line above makes Tra additive, but need not be. It suffices that $\mathrm{Tra}[ y \cdot y\cdot y \cdot x]$ is replaced by $ \mathrm{Tra}[x\cdot y \cdot y\cdot y]$ in order to simplify.

The issue with the lexicographic for the non-commutative product order is that the test in the given answer will be always false, unless there is no interlaced y's and x's, but that's the exception.

Answer 1

Perhaps something like:

ClearAll[esq, Tra]

esq[A_CenterDot] :=  NestWhile[RotateLeft, A, Not @* OrderedQ, 1, Length[A] - 1]

Tra[Y_Plus] := Tra[#] & /@ Y

Tra[A_CenterDot] := esq @ A

Examples:

Tra[x·y·y·x·y·x]

 x·x·y·y·x·y

Tra[x·y·y·y]

 x·y·y·y

 Tra[y·x·y·y·y]

  x·y·y·y·y

Tra[y·y·y·x +  x·y·y·y]

 2 x·y·y·y