Laurent subtraction factors a character into counterterm and finite characters

For homogeneous $h$ of degree $n>0$, every term $h_{(1)}\otimes h_{(2)}$ in $\tilde{\mathrm{\Delta }}(h)$ has both factors of positive degree strictly less than $n$, because $H$ is connected and graded by [F2]. There are finitely many terms by [F1], so induction on $n$ defines $B_{\varphi }(h)$ and ${\varphi }_{-}(h)$ without circularity; extend linearly and set ${\varphi }_{-}(1)=1$.

Let $x,y$ be positive-degree homogeneous elements with total degree $d$, and assume inductively that ${\varphi }_{-}(uv)={\varphi }_{-}(u){\varphi }_{-}(v)$ whenever $\mathrm{deg}u+\mathrm{deg}v<d$. Expand $\mathrm{\Delta }(xy)=\mathrm{\Delta }(x)\mathrm{\Delta }(y)$. In $({\varphi }_{-}\ast \varphi )(x)({\varphi }_{-}\ast \varphi )(y)$, every summand except the unique pair $(x\otimes 1)(y\otimes 1)$ has left-factor product degree less than $d$; for those terms the induction hypothesis factors ${\varphi }_{-}(x_{(1)}{y}_{(1)})$, while the character property factors $\varphi (x_{(2)}{y}_{(2)})$. Thus these lower-degree summands agree term-by-term with those in $({\varphi }_{-}\ast \varphi )(xy)$. The only remaining terms are the top pair ${\varphi }_{-}(x){\varphi }_{-}(y)$ in the product and ${\varphi }_{-}(xy)$ in the convolution of $xy$. Since $B_{\varphi }(xy)=({\varphi }_{-}\ast \varphi )(xy)-{\varphi }_{-}(xy)$ and $({\varphi }_{-}\ast \varphi )(x)={\varphi }_{-}(x)+B_{\varphi }(x)$, subtracting the top pair gives $B_{\varphi }(xy)=({\varphi }_{-}\ast \varphi )(x)({\varphi }_{-}\ast \varphi )(y)-{\varphi }_{-}(x){\varphi }_{-}(y)={\varphi }_{-}(x)B_{\varphi }(y)+B_{\varphi }(x){\varphi }_{-}(y)+B_{\varphi }(x)B_{\varphi }(y)\mathrm{.}$ The equality uses commutativity of $A$.

If either $x$ or $y$ has degree zero, multiplicativity follows from ${\varphi }_{-}(1)=1$. Otherwise write $X=B_{\varphi }(x)$ and $Y=B_{\varphi }(y)$. By the recursion and the identity in [F4], ${\varphi }_{-}(xy)=-R{B}_{\varphi }(xy)=R(R(X)Y+XR(Y)-XY)=R(X)R(Y)={\varphi }_{-}(x){\varphi }_{-}(y)\mathrm{.}$ Thus degree induction proves that ${\varphi }_{-}$ is a character on all of $H$.

For positive-degree $h$, expanding its full coproduct gives $({\varphi }_{-}\ast \varphi )(h)={\varphi }_{-}(h)+B_{\varphi }(h)=-R{B}_{\varphi }(h)+B_{\varphi }(h)=(1-R)B_{\varphi }(h)\in A_{+}$; also $({\varphi }_{-}\ast \varphi )(1)=1\in A_{+}$. For $x,y\in H$, the character properties already proved and multiplicativity of $\mathrm{\Delta }$ give $({\varphi }_{-}\ast \varphi )(xy)=\sum {\varphi }_{-}(x_{(1)}){\varphi }_{-}(y_{(1)})\varphi (x_{(2)})\varphi (y_{(2)})=({\varphi }_{-}\ast \varphi )(x)({\varphi }_{-}\ast \varphi )(y),$ where commutativity of $A$ permits the middle factors to be regrouped. Thus ${\varphi }_{+}={\varphi }_{-}\ast \varphi$ is a character with the stated range.

For $h\in H$, multiplicativity and unitality of ${\varphi }_{-}$ and the antipode identities in [F2] give $(({\varphi }_{-}\circ S)\ast {\varphi }_{-})(h)=\sum {\varphi }_{-}(S(h_{(1)})){\varphi }_{-}(h_{(2)})={\varphi }_{-}(\sum S(h_{(1)})h_{(2)})={\varphi }_{-}(u\epsilon (h))=u_A\epsilon (h),$ and likewise $({\varphi }_{-}\ast ({\varphi }_{-}\circ S))(h)=\sum {\varphi }_{-}(h_{(1)}){\varphi }_{-}(S(h_{(2)}))={\varphi }_{-}(\sum h_{(1)}S(h_{(2)}))=u_A\epsilon (h)\mathrm{.}$ Hence ${\varphi }_{-}^{-1}={\varphi }_{-}\circ S$. From ${\varphi }_{+}={\varphi }_{-}\ast \varphi$, associativity and cancellation by this inverse give $\varphi ={\varphi }_{-}^{-1}\ast {\varphi }_{+}$.