3
$\begingroup$

I have a question about the definition of the double shuffle lie algebra discussed in section 1.3 of Sarah Carr's thesis (see https://www.imj-prg.fr/theses/pdf/sarah_carr.pdf)

Recall the definition of the double shuffle lie algebra:
Consider two coproducts, coshuffle and costuffle, that are equipped in two noncommutative polynomial algebras respectively, $\mathbb{Q}⟨⟨x,y⟩⟩$ and $\mathbb{Q}⟨⟨y_i; 1\leq i < \infty⟩⟩$. These coproducts are defined on the generators and one can extend them through multiplication (concatenation): $$\begin{align} \Delta_⧢: \mathbb{Q}⟨⟨x,y⟩⟩ &\to \mathbb{Q}⟨⟨x,y⟩⟩ \otimes \mathbb{Q}⟨⟨x,y⟩⟩ \\ x &\mapsto x \otimes 1 + 1 \otimes x \\ y &\mapsto y \otimes 1 + 1 \otimes y \end{align}$$ $$\begin{align} \Delta_*: \mathbb{Q}⟨⟨y_i⟩⟩ &\to \mathbb{Q}⟨⟨y_i⟩⟩ \otimes \mathbb{Q}⟨⟨y_i⟩⟩ \\ y_i &\mapsto \sum_{m+n=i} y_m \otimes y_n \end{align}$$ Coshuffle can be considered as the dual of shuffle product of multiple zeta values, and costuffle as the dual of stuffle product of multiple zeta values.
Then the double shuffle lie algebra $\mathfrak{ds}$ is a vector subspace of $\mathbb{Q}⟨⟨x,y⟩⟩$ generated by elements $f$ such that $f$ are primitive for $\Delta_⧢$ and $\pi_y(f)$ are primitive for $\Delta_*$, where $\pi_y$ is a linear map that transforms $f$ to an element in $\mathbb{Q}⟨⟨y_i⟩⟩$ because costuffle is operated in $\mathbb{Q}⟨⟨y_i⟩⟩$. The definition of $\pi_y$ is $$\pi_y: \mathbb{Q}⟨⟨x,y⟩⟩ \to \mathbb{Q}⟨⟨y_i⟩⟩ \\ \tilde{\pi_y}(x^{k_1-1}yx^{k_2-1}yx^{k_3-1}y\cdots x^{k_n-1}yx^{k_{n+1}})= \begin{cases} 0, \; k_{n+1} \neq 0 \\ y_{k_1}y_{k_2}y_{k_3}\cdots y_{k_n}, \; k_{n+1} = 0 \end{cases} \\ \pi_y(f) = \tilde{\pi_y}(f) + \sum_{n\geq 2} (f|x^{n-1} y)\frac{(-1)^{n-1}}{n}y_1^n$$ My question is about this map $\pi_y$. The definition of $\tilde{\pi_y}$ is quite natural, but why do we need to add that summation term in the definition of $\pi_y$? Some other documentation tells me that the summation term comes from the extended double shuffle (EDS) relations of multiple zeta values, which is described in the first three sections of this paper. But I cannot figure out how EDS leads to the formula for $\pi_y$, nor can I find references that explain it. Very appreciated if anybody knows the references.

$\endgroup$

1 Answer 1

2
$\begingroup$

The double shuffle Lie algebra described relations among (formal) multiple zeta values, modulo products. However, the description in terms of primitivity with respect to the coshuffle and costuffle coproducts require us to assign finite values to divergent multiple zeta values.

For example, in weight three, we have two primitive elements with respect to the coshuffle coproduct:

$$x_0^2x_1 -2x_0x_1x_0 + x_1x_0^2$$ and $$x_1^2x_0 -2x_1x_0x_1 + x_0x_1^2$$ both of which involve divergent integrals.

One can show that, for the shuffle product, there is a unique way to associate a value to divergent integrals such that $\zeta(1)=0$ and the shuffle product holds for products of not-necessarily-convergent integrals $$\zeta(x_0^k)\zeta(w)=\text{Sum over shuffles of }x_0^k\text{ and }w.$$ This gives shuffle regularised MZVs.

Similarly, there is a unique way to associate a value to divergent sums such that $\zeta(1)=0$ and the stuffle product holds for products of not-necessarily-convergent sums, giving stuffle regularised MZVs.

However, the two regularisations do not agree: consider $\zeta(1,1)$. By the shuffle shuffle product, we get $$\zeta(1,1)=\frac{1}{2}\zeta(1)^2=0,$$ while the stuffle product tells us that $$\zeta(1,1)=\frac{1}{2}(\zeta(1)^2-\zeta(2))=-\frac{1}{2}\zeta(2).$$

As such, in order for the double shuffle Lie algebra to encapsulate all the double shuffle relations, we need to switch between the regularisations, which is where the extra term in $\pi_Y$ arises. One can compute directly that that additional term encodes the value of the stuffle regularised $\zeta(1,1,...,1)$. It is also easy enough to see that every divergent MZV can be written as a stuffle polynomial of convergent MZVs and $\zeta(1,1,...,1)$, so I'm going to hand-wave and say that this makes $\pi_Y$ sufficient to encode the change from shuffle regularisation to stuffle regularisation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .