If $\Delta(c) = \Delta^{\mathrm{op}}(c)$, then under which permutations is $c_{(1)} \otimes \dotsb \otimes c_{(n)}$ invariant?

Question

Let $(C, \Delta)$ be a coalgebra and $c\in C$ an element with $\Delta(c) = \Delta^{\mathrm{op}}(c)$.

For certain permutations $\sigma \in S_n$, we will have that

$$c_{(\sigma(1))} \otimes \dots \otimes c_{(\sigma(n))} = c_{(1)} \otimes \dots \otimes c_{(n)}$$

where the notation is Hopf–Sweedler notation. Is it possible to characterise/describe the permutations $\sigma \in S_n$ for which this identity holds (on the purely formal level, so you don’t need to take into account equalities that occur due to coincidences in the coalgebra itself – so say that $C$ is free as a coalgebra).

Answer 1

The precise set of permutations depends on $c$. For example, if $c = 0$ then every permutation does the job; but in general, not every permutation will work. So the question should probably be understood as ‘which permutations always work’.

Let us call an element $c$ of $C$ with $Δ(c) = Δ^{\mathrm{op}}(c)$ cocommutative. That $c$ is cocommutative means that $c_{(1)} ⊗ c_{(2)} = c_{(2)} ⊗ c_{(1)}$.

I assume that we are working with coalgebras over a field. By the fundamental theorem of coalgebras, every element of $C$ is contained in a finite-dimensional subcoalgebra. We may therefore assume that $C$ is finite-dimensional. The coalgebra $C$ is then isomorphic to $A^*$ for some algebra $A$, so we may assume that $C = A^*$. The comultiplication of $C$ is then the dual of the multiplication of $A$, in the sense that $$ c(xy) = c_{(1)}(x) \, c_{(2)}(y) \quad \text{for all $x, y ∈ A$.} $$ More generally, the iterated coproduct $c_{(1)} ⊗ \dotsb ⊗ c_{(n)}$ is implicitly defined via the formula $$ c(x_1 \dotsm x_n) = c_{(1)}(x_1) \,\dotsm\, c_{(n)}(x_n) \quad \text{for all $x_i ∈ A$.} $$ That $c$ is cocommutative is now equivalent to the condition $$ c(xy) = c(yx) \quad \text{for all $x, y ∈ A$.} $$ We can thus rephrase our question as follows:

Question. Let $A$ be an algebra and let $c ∈ A^*$ with $c(xy) = c(yx)$ for all $x, y ∈ A$. For which permutations $σ$ do we have $$ c(x_1 \dotsm x_n) = c(x_{σ(1)} \dotsm x_{σ(n)}) \quad \text{for all $x_i ∈ A$?} \tag{$\ast$} $$

This problem has a well-known special case: if $A$ is the matrix of $(n × n)$-matrices then we have $\mathrm{tr}(AB) = \mathrm{tr}(BA)$ for any two matrices $A$ and $B$. So for which permutations $σ$ do we have $\mathrm{tr}(A_1 \dotsm A_n) = \mathrm{tr}(A_{σ(1)} \dotsm A_{σ(n)})$? It is a common false belief that this holds for all permutations (see https://mathoverflow.net/a/23481/200357), but the case of $n = 2$ only generalizes to cyclic permutations, i.e., $$ \mathrm{tr}(A_1 \dotsm A_n) = \mathrm{tr}(A_2 \dotsm A_n A_1) = \mathrm{tr}(A_3 \dotsm A_n A_1 A_2) = \dotsb = \mathrm{tr}(A_n A_1 \dotsm A_{n-1}) \,. $$

The same principle also works for the general question:

Observation. Every cyclic permutation satisfies the condition $(\ast)$.

The cyclic permutations are in fact the only permutations that always work. To prove this, it suffices to consider the above special case of matrices and the trace.

Claim. Let $σ ∈ \mathrm{S}_n$ be a non-cyclic permutation. Then there exist $(n × n)$-matrices $A_1, \dotsc, A_n$ such that $$ \mathrm{tr}(A_1 \dotsm A_n) ≠ \mathrm{tr}(A_{σ(1)} \dotsm A_{σ(n)}) \,. $$

To prove this claim we consider the standard basis matrices $E_{ij}$ (with $E_{ij} E_{kl} = δ_{jk} E_{il}$) and set $$ A_1 = E_{12} \,, \quad A_2 = E_{23} \,, \quad \dotsc \,, \quad A_{n-1} = E_{n-1, n} \,, \quad A_n = E_{n 1} \,. $$ We have $$ \mathrm{tr}(A_1 \dotsm A_n) = \mathrm{tr}(E_{11}) = 1 \,. $$ But for every non-cyclic permutation $σ$ we have $$ A_{σ(1)} \dotsm A_{σ(n)} = 0 \,, $$ because at some point we have $A_{σ(i)} A_{σ(i+1)} = 0$ since the indices don’t match up.