I am reading "Geometric versus Non-Geometric Rough Paths" by Martin Hairer and David Kelly. I don't know much about Hopf algebras but $T(V)$ is the one example I'm comfortable with so far and I'm trying to understand it's dual properly.
They look at the (truncated) tensor algebra $T(V)$ as a Hopf Algebra, by giving it the shuffle product (here denoted $\varpi$ since SE doesn't have \Sha) as the product and the deconcatenation product $\Delta$ as the coproduct.
They then say that we can define a dual Hopf Algebra built up out of the dual of the $T(V)^* \cong T(V^*)$ (I believe this is true crucially only when we truncate things, else see this question here). The dual Hopf algebra has product given by the ordinary tensor product $\otimes$, and a coproduct $\delta$ which they don't define, other than "it is dual" to the deconcatenation product.
What is $\delta$ explicitly? Are there any digestible references, specifically focusing on $T(V)$ rather than anything too obtuse/abstract? I found Reutenauer too abstract for me to find what I wanted.
I searched for a while and didn't find any explicit constructions, funnily enough, I got some interesting information from ChatGPT, which told me the dual of:
$(T(V), \varpi, \Delta)$ is the Hopf Algebra $(S(V^*), \otimes, \Delta)$, where $S$ is the symmetric tensor powers. Which I found rather interesting but have no idea whether this is correct. This is slightly agreeing with Hairer and Kelly in terms of $\varpi$ swapping to $\otimes$ and would explain why they didn't say much about $\delta$, if it's just the same as $\Delta$.
It also told me $(T(V), \otimes, \Delta)$ is dual to $(T(V^*), \otimes, \otimes)$ which seems a bit fishier. Id be curious to know if either of these are true.
The (graded) dual Hopf algebra to the shuffle algebra is $(T(V^*), \otimes, \delta)$ where $\delta$ is dual to the shuffle product in the following sense: let's say that $\dim V=d$ and pick a basis $\{e_1,\dotsc,e_d\}$. Consider the dual basis $\{e^i,\dotsc,e^d\}$ and the induced duality pairing between $V$ and $V^*$, given by $\langle e^i,e_j\rangle=e^i(e_j)=\mathbf{1}_{i=j}$. Extend it to a pairing between $T(V)$ and $T(V^*)$ by $$\langle e^{i_1,\dotsc,i_n},e_{j_1,\dotsc,j_m}\rangle:=\mathbf{1}_{n=m}\prod_{k=1}^n\langle e^{i_k},e_{j_k}\rangle.$$
In order to avoid any confusion I'll use the symbol $\dot\otimes$ to denote the tensor product external to $T(V)$ (and $T(V^*)$) used in the definition of the coproduct. Then, $\delta$ is determined by
$$\langle\delta e^{i_1,\dotsc,i_n},x\mathbin{\dot\otimes}y\rangle=\langle e^{i_1,\dotsc,i_n},x\mathbin{\varpi}y\rangle$$ for all $x,y\in T(V)$. From this one can get an expression, although is not very useful:
$$\delta e^{i_1,\dotsc,i_n}=\sum_{\substack{x,y\in T(V)\\\langle e^{i_1,\dotsc,i_n},x\mathbin{\varpi}y\rangle\neq 0}}x\mathbin{\dot\otimes}y.$$ Another expression, which you can check as an exercise, and that has a more combinatorial flavor is: $$\delta e^{i_1,\dotsc,i_n}=\sum_{I\subset[n]}e^{I}\mathbin{\dot\otimes}e^{[n]\setminus I}$$ where the notation $e^I$ should be self-explanatory.
Hint: observe that $(T(V^*),\otimes)$ is the free associate algebra over the set $\{e^1,\dotsc,e^d\}$ and use the fact that $\delta$ must be an algebra morphism $T(V^*)\to T(V^*)\mathbin{\dot\otimes} T(V^*)$ where the codomain has the usual product $(a\mathbin{\dot\otimes}b)\otimes(a'\mathbin{\dot\otimes}b'):=(a\otimes b)\mathbin{\dot\otimes}(a'\otimes b')$. This property plus its image on the generating set ($\delta e^i=e^i\mathbin{\dot\otimes} 1+1\mathbin{\dot\otimes}e^i$) can also be taken as the definition of $\delta$. In fact, this is one way of defining it in a basis-independent way; that is, by requiring that the subspace $V^*$ is primitive.
PS: Forget ChatGPT!