Unital versions of the higher order peak algebras

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Problem: Reconstructed statement: let $q\in\mathbb C$ be a primitive $r$-th root of unity and let $\MR=\mathbb C\langle S_n,\bar S_n:n\ge1\rangle$, graded by $\deg S_n=\deg\bar S_n=n$. Define

$$F^\sharp=F*\sigma_1(A-q\bar A),\qquad \MR^\sharp=\operatorname{Im}(\sharp).$$

Put $S_n^+=S_n+\bar S_n$, $S_n^-=S_n-\bar S_n$. Color $0$ means $+$, color $1$ means $-$. The conjecture asserts that $\MR^\sharp$ is freely spanned by products

$$(S_{i_1}^{\varepsilon_1})^\sharp\cdots (S_{i_k}^{\varepsilon_k})^\sharp$$

where the colored composition $((i_1,\varepsilon_1),\dots,(i_k,\varepsilon_k))$ satisfies:

• if $r$ is odd: $i_j\not\equiv0\pmod r$ whenever $\varepsilon_j=0$, with no restriction for $\varepsilon_j=1$;
• if $r$ is even: additionally $i_j\not\equiv r/2\pmod r$ whenever $\varepsilon_j=1$.

The omitted hypothesis “$q$ primitive of order $r$” is forced by the surrounding section of the paper.

Result: The conjecture is true.

Let

$$\sigma(t)=\sum_{n\ge0}S_nt^n,\qquad \bar\sigma(t)=\sum_{n\ge0}\bar S_nt^n.$$

Write

$$U(t)=\sigma(t)^\sharp=\bar\sigma(qt)^{-1}\sigma(t)=1+\sum_{n\ge1}a_nt^n,$$ $$V(t)=\bar\sigma(t)^\sharp=\sigma(qt)^{-1}\bar\sigma(t)=1+\sum_{n\ge1}b_nt^n.$$

Then

$$(S_n^+)^\sharp=a_n+b_n,\qquad (S_n^-)^\sharp=a_n-b_n.$$

Modulo the subalgebra generated by $S_k^\pm$ with $k<n$,

$$(S_n^+)^\sharp\equiv (1-q^n)S_n^+,\qquad (S_n^-)^\sharp\equiv (1+q^n)S_n^-.$$

Thus all allowed generators have nonzero triangular leading terms, so their words are linearly independent.

It remains to show that the forbidden generators lie in the algebra generated by the allowed ones. Let $m=\operatorname{ord}(q^2)$, so $m=r$ if $r$ is odd and $m=r/2$ if $r$ is even. From the definitions,

$$V(q^{2k+1}t)U(q^{2k}t) =\sigma(q^{2k+2}t)^{-1}\sigma(q^{2k}t).$$

Multiplying for $k=m-1,\dots,0$ gives the telescoping identity

$$V(q^{2m-1}t)U(q^{2m-2}t)\cdots V(qt)U(t)=1.$$

Taking the coefficient of $t^n$, the linear part is

$$\sum_{k=0}^{m-1}\big(q^{2kn}a_n+q^{(2k+1)n}b_n\big) =\Big(\sum_{k=0}^{m-1}q^{2kn}\Big)(a_n+q^n b_n),$$

and all remaining terms involve only $a_i,b_i$ with $i<n$.

If a generator is forbidden, then $q^{2n}=1$, so the scalar sum equals $m\neq0$. Hence:

• if $q^n=1$, then $a_n+b_n=(S_n^+)^\sharp$ is expressed in lower-degree image generators;
• if $q^n=-1$, then $a_n-b_n=(S_n^-)^\sharp$ is expressed in lower-degree image generators.

Induction on $n$ shows every forbidden generator belongs to the subalgebra generated by the allowed ones. Therefore $\MR^\sharp$ is the free associative algebra on the allowed $(S_n^\pm)^\sharp$, and the asserted colored compositions give a basis.

The Hilbert series of a free associative algebra with generator series $P(t)$ is $1/(1-P(t))$. If $r$ is odd,

$$P(t)=\sum_{r\nmid n}t^n+\sum_{n\ge1}t^n,$$

so

$$H_r(t)=\frac{1-t^r}{1-2(t+t^2+\cdots+t^r)}.$$

If $r$ is even,

$$P(t)=\sum_{r\nmid n}t^n+\sum_{n\not\equiv r/2\!\!\pmod r}t^n,$$

so

$$H_r(t)=\frac{1-t^r}{1-2(t+t^2+\cdots+t^r)+t^{r/2}}.$$

Verification audit: the proof uses exactly the sharp transform and generators $S_n^\pm$ from the paper; the only implicit hypothesis added is the contextual one that $q$ is primitive of order $r$. The cases $r=1,2$ reduce to the identities already noted in the paper. The resulting basis and Hilbert series match Conjecture 6.1 exactly.

Citation: Conjecture source: Marcelo Aguiar, Jean-Christophe Novelli, Jean-Yves Thibon, “Unital versions of the higher order peak algebras,” arXiv:0810.4634, Conjecture 6.1. No prior published resolution is used here.