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%\journal{CIP Exercices-EVT-Hilbert}

\begin{document}


\title{About the antipode}


Let $(B,\mu,1_B,\Delta,\ep)$ be a bialgebra (the ground field will be taken as $k$), we define the convolution in $\End(B)$ for $f,g\in \End(B)$ and $x\in B$ as
\begin{equation}
	f*g(x)=\mu(f\otimes g)\Delta(x)
\end{equation}
one can check that (exercise 11.9 (2) of the tutorial) this is an algebra with the linear mapping
\begin{equation}
convunit\ :\	x\mapsto \ep(x)1_B
\end{equation}
as unit. 
\\
One can check easily that this mapping is the projector on the line $k.1_B$ and kernel
\begin{equation}
B^+=ker(\ep)=\{x\in B\ |\ \ep(x)=0\}\ .
\end{equation}
We will denote $I^+=Id_B-convunit$ the complementary projector (in fact, in my previous letter $\Delta^+=(I^+\otimes I^+)\circ \Delta$ and it can be proved that $\Delta^+$ is coassociative).\\

The antipode $S$, when it exists, is defined as the convolutional inverse of $Id_B$.\\
\textbf{But} $Id=convunit+I^+$ (of type $1+X$), and then if the sum 
\begin{equation}
	S=\sum_{n=0}^\infty (-1)^n (I^+)^{*n}
\end{equation}
converges (in some sense), it is the antipode.\\
\textbf{Here}, in the two cases of $\mathcal{U}$ and $\mathcal{G}$, the sequence of endomorphisms $(I^+)^{*n}$ is locally finite which means that, for all $x\in B$ it exists a number $N\in \N$ from which $(I^+)^{*n}(x)=0$. Formally 
\begin{equation}
	(\forall x\in B)(\exists N\in \N)(\forall n\geq N)((I^+)^{*n}(x)=0)
\end{equation}
then, it is easy to check that the sum
\begin{equation}
	S_n(x)=\sum_{m=0}^n (-1)^m (I^+)^{*m}(x)
\end{equation}
is stationnary from $N$. Taking as $S(x)$ the common value of the $S_n(x)$, one has then the antipode.\\
It is obvious that, if $x\in B^+$, $S(x)=\sum_{n=1}^\infty (-1)^n (I^+)^{*n}(x)$ as $(I^+)^{*0}(x)=convunit(x)=0$ and the explains formula (33) of the paper.

Now, if you have two bialgebras $(B_i,\mu_i,1_{B_i},\Delta_i,\ep_i),\ i=1,2$ with a morphism of bialgebras $\phi\ :\ B_1\ra B_2$, this is a morphism for the convolutions in the sense that, if $f_i,g_i\in \End(B_i)$ with 
$$
f_2\circ\phi=\phi\circ f_1\ ;\ g_2\circ\phi=\phi\circ g_1
$$
then $(f_2*_2g_2)\circ\phi=\phi\circ (f_1*_1g_1)$ (almost straigthforward). On the other hand, 
$$
I_2^+\phi=\phi\circ I_1^+
$$
and then, 
\begin{equation}
	S_2\circ\phi=(\sum_{n=0}^\infty (-1)^n (I_2^+)^{*n})\circ\phi=(\sum_{n=0}^\infty (-1)^n \phi\circ (I_1^+)^{*n})=\phi\circ(\sum_{n=0}^\infty (-1)^n (I_1^+)^{*n})=\phi\circ S_1
\end{equation}

\end{document}
