For the binomial Hopf algebra, what is the group of grouplike elements of the dual algebra?

Question

The linear space of finite polynomials over a field $\mathbb{K}$ of zero characteristic has a structure of a graded connected bialgebra (meaning the zeroth subspace is isomorphic to the base field):

$\newcommand{\bra}{\langle}$

$$\Omega(\infty)=\bigoplus_{k=0}^{\infty}\langle x^{k}\rangle$$

with multiplication being the regular multiplication of polynomials, comultiplication $\Delta 1 = 1 \otimes 1$, $\Delta x^n = \sum_{k=0}^{n}{n \choose k}x^{k}\otimes x^{n-k}$, and counit $\varepsilon(x^{n})=\delta_{n 0}$ , where $\delta_{ij}$ is the Kronecker delta. The antipode $S$ is given by $S(x^{n})=(-1)^{n}x^{n}$. The dual Hopf algebra to the Hopf algebra of polynomials is the Hopf algebra of exponential formal power series

$$\bra f|:\bra f| x^{n}= \sum_{k=0}^{\infty} \frac{(\frac{d}{dx})^k}{k!}f_{k}x^{n}\Bigg|_{x=0}$$

, with multiplication being the regular exponential formal power series multiplication, counit $\varepsilon^{*}$ being the value of the zeroth term of the series, comultiplication being defined by $$\Delta^{*} \langle f| = \langle f| \nabla$$ and antipode $S^{*}$ being $$S^*(\langle f|)= \langle f| S$$

which amounts to flipping the sign of the odd coefficients of $f$.

Grouplike elements of a coalgebra are such elements $g$ that $\Delta g = g \otimes g$ and $\varepsilon(g)=1$. There is a well-known theorem that grouplike elements of a Hopf algebra form a group, and the inverse of an element $g$ of this group is given by $S(g)$. So, as far as I can understand, the grouplike elements of $\Omega^{*}(\infty)$ must be such invertible formal power series $f(t)$ that $f(t)f(-t)=1$. At the very least, the group should contain all the exponents, and, my guess, all generating functions of the form $e^{tf(t)}$ (those that define sequences of binomial type). But when I try to derive the general formula, I stumble upon a problem: invertible elements are such that $\Delta^{*}\langle f| = \langle f| \otimes \langle f|$. Multiplication on the dual algebra is adjoint to comultiplication on coalgebra, so, if $\Delta^{*}\bra f|= \bra f|\otimes \bra f|$,

$$(\bra f| \otimes \bra f|)\Delta x^{n}=\bra f^2 |x^n = \sum_{k=0}^{n}{n \choose k}f_{k}f_{n-k}$$ At the other hand, $$\Delta^{*}\bra f|\Delta x^{n}:= \bra f|\nabla \Delta x^{n}=\bra f|\sum_{k=0}^{n}{n \choose k}x^{n}=2^{n}\bra f|x^{n}=2^{n}f_{n}$$ So, apparently, $$f_{n}=\frac{1}{2^{n}}\sum_{k=0}^{n}{n \choose k}f_{k}f_{n-k}$$

which we can rewrite as recurrence for $f_{0}=1$ and $f_{1}=\alpha$: $$f_{n}=\frac{1}{2^{n}-2}\sum_{k=1}^{n-1}{n \choose k}f_{k}f_{n-k}$$

This recurrence is satisfied by exponents and (so it seems) only exponents, which is somewhat disappointing given the fact that it should contain the uniparametric group of exponents as a proper subgroup. Indeed, consider the generating function $$e^{t f(t)}=1+ tf(t)+ \frac{t^2}{2}f(t)^2 + \cdots$$ such that $f(t)$ is even. Then $$e^{tf(t)}\cdot e^{-tf(-t)}=e^{0}=1$$ and such series has the first term equal to 1. What am I missing?

Answer 1

So, as far as I can understand, the grouplike elements of $\Omega^{*}(\infty)$ must be such invertible formal power series $f(t)$ that $f(t)f(-t)=1$.

This is incorrect. The comultiplication sends a formal power series $f(t)$ to the formal power series $f(t + s)$ in two variables, in the tensor product. So the grouplike condition $\Delta f = f \otimes f$ actually says that $f(t + s) = f(t) f(s)$ (and $\varepsilon(f) = 1$ means $f(0) = 1$).

Over a field of characteristic zero this implies that $f(t) = \exp(ct)$ where $c \in K$. Interestingly, in positive characteristic $p$ there are more solutions, given by

$$f(t) = \exp \left( \sum_{k \ge 0} c_k t^{p^k} \right).$$

Let's trace through this really explicitly. I find your notation a little confusing so I'll adopt different notation. An element of the dual Hopf algebra is a linear functional $F : K[x] \to K$ on polynomials, which can be identified with its coefficients $f_k = F(x^k)$. The multiplication is dual to the comultiplication, which gives

$$FG(x^n) = \sum_{k=0}^n {n \choose k} F(x^k) G(x^{n-k})$$

and hence that the $n^{th}$ coefficient of the product $FG$ is the binomial convolution $\sum_{k=0}^n {n \choose k} f_k g_{n-k}$ of the coefficients of $F$ and $G$; this reproduces the usual multiplication of exponential formal power series $\sum \frac{f_k}{k!} t^k$, as you say. Now the comultiplication is dual to the multiplication, which gives

$$\Delta(F)(x^n \otimes x^m) = F(x^{n+m}) = f_{n+m}$$

meaning that $\Delta(F)$, as an exponential formal power series in two variables $t, s$, is

$$\begin{eqnarray*} \Delta(F) &=& \sum_{n, m \ge 0} f_{n+m} \frac{t^n}{n!} \frac{s^m}{m!} \\ &=& \sum_{n \ge 0} f_n \left( \sum_{k=0}^n \frac{t^k}{k!} \frac{s^{n-k}}{(n-k)!} \right) \\ &=& \sum_{n \ge 0} \frac{f_n}{n!} \left( \sum_{k=0}^n {n \choose k} t^k s^{n-k} \right) \\ &=& \sum_{n \ge 0} f_n \frac{(t+s)^n}{n!} \end{eqnarray*}$$

as promised (here we're allowing ourselves to work in characteristic zero for simplicity; this argument doesn't actually require any divisions and can be adapted to work over an arbitrary commutative ring but it becomes notationally more annoying).

So the grouplike condition says that $f(t + s) = f(t) f(s)$ and $f(0) = 1$, as mentioned above. In characteristic zero, formally differentiating with respect to $s$ gives that $f(t)$ satisfies $f'(t) = f(t) f'(0)$, and using a formal version of existence and uniqueness of ODEs there is a unique solution to this ODE with initial condition $f(0) = 1$, namely the expected $f(t) = \exp( f'(0) t)$.

I can also describe the argument in positive characteristic if you're interested.