While studying the Drinfeld Double of a Hopf algebra, I came across two different definitions used for the multipliaction.
For a finite dimensional Hopf algebra $H$ (over a field $K$) we define $D(H)=H^{*cop}\otimes H$ (the Drinfeld double of H) as a Hopf algebra with the following structure: \begin{align*} &(\varphi \otimes g)(\psi \otimes h)= \psi_{(1)}(S^{-1}(g_{(3)}))\psi_{(3)}(g_{(1)})\varphi\psi_{(2)}\otimes g_{(2)}h && 1_{D(H)}=\varepsilon\otimes 1_H\\ & \Delta ( \varphi \otimes h)=(\varphi_{(2)}\otimes h_{(1)}) \otimes (\varphi_{(1)}\otimes h_{(2)}) && \varepsilon(\varphi \otimes h)=\varphi(1)\varepsilon(h) \\ & S(\varphi \otimes h)=(\varepsilon \otimes S(h))(S^{-1}(\varphi)\otimes 1). \end{align*}
In addition I came across the following definition of the multiplication: $$ (\varphi \otimes g)(\psi \otimes h)=\varphi(g_{(1)} \rightharpoonup \psi \leftharpoonup S^{-1}(g_{(3)}))\otimes g_{(2)}h .$$ The notation we use here is defined like this:
Let $A$ be an algebra. For $a \in A$, $\varphi \in A^*$ we define $a \rightharpoonup \varphi, \varphi \leftharpoonup a \in A^*$ with $(a \rightharpoonup \varphi)(b)=\varphi(ba)$ and $(\varphi \leftharpoonup a)(b)=\varphi(ab)$ for every $b \in A$.
My goal is to show that both multiplications are equal. This is my approach:
\begin{align*} (\varphi \otimes g)(\psi \otimes h) & = \psi_{(1)}(S^{-1}(g_{(3)}))\psi_{(3)}(g_{(1)})\varphi\psi_{(2)}\otimes g_{(2)}h \\ & = (\psi_{(1)} \leftharpoonup S^{-1}(g_{(3)}))(g_{(1)} \rightharpoonup \psi_{(3)})\varphi\psi_{(2)} \otimes g_{(2)}h \end{align*} On the other hand we have: \begin{align*} (\varphi \otimes g)(\psi \otimes h) & = \varphi(g_{(1)} \rightharpoonup \psi \leftharpoonup S^{-1}(g_{(3)}))\otimes g_{(2)}h \\ & = \varphi(g_{(1)} \rightharpoonup \psi_{(2)})\psi_{(1)}(S^{-1}(g_{(3)})) \otimes g_{(2)}h \\ & = \varphi\psi_{(2)}(g_{(1)})\psi_{(1)}(S^{-1}(g_{(3)})) \otimes g_{(2)}h \\ & = \psi_{(1)}(S^{-1}(g_{(3)})) \psi_{(2)}(g_{(1)}) \varphi \otimes g_{(2)}h \end{align*} As you can see, I am trying to show the equality by starting from both sides. But with my attempt at the bottom I am still missing a $\psi$ and have to make sure while adding it that it gets the right index. On the upper attempt I have to remove $\psi_{(2)}$ but do not see how it is done.
I got the answer: We have $g_{(1)}\rightharpoonup \psi_{(2)}=\psi_{(2)} \psi_{(3)}(g_{(1)})$ and use it here
\begin{align*} (\varphi \otimes g)(\psi \otimes h) & = \varphi(g_{(1)} \rightharpoonup \psi \leftharpoonup S^{-1}(g_{(3)}))\otimes g_{(2)}h \\ & = \varphi(g_{(1)} \rightharpoonup \psi_{(2)})\psi_{(1)}(S^{-1}(g_{(3)})) \otimes g_{(2)}h \\ & = \varphi \psi_{(2)} \psi_{(3)}(g_{(1)})\psi_{(1)}(S^{-1}(g_{(3)})) \otimes g_{(2)}h \end{align*} It is important to note that the last equation symbol of the following is simply wrong: \begin{align*} (\varphi \otimes g)(\psi \otimes h) & = \psi_{(1)}(S^{-1}(g_{(3)}))\psi_{(3)}(g_{(1)})\varphi\psi_{(2)}\otimes g_{(2)}h \\ & = (\psi_{(1)} \leftharpoonup S^{-1}(g_{(3)}))(g_{(1)} \rightharpoonup \psi_{(3)})\varphi\psi_{(2)} \otimes g_{(2)}h \end{align*}