Grouplike elements is a group in a Hopf algebra

Question

I have been looking at this question and two questions arose for me:

(1) A simpler one: Why is $(g \otimes g)(h \otimes h)=(g h \otimes g h)$? My guess: $$ (g \otimes g)(h \otimes h) = g^2(1 \otimes 1)(1 \otimes 1) h^2 = g^2(1 \otimes 1) h^2 = g h \otimes g h \text{ (and also $= g^2 \otimes h^2$)} \,. $$

(2) How can I see that the following definition of grouplike elements is equivalent to the usual one?

Consider the free tensor algebra $T(V) := \bigoplus_{k=0}^\infty V^{\otimes k}$ where $\dim(V) = d < \infty$ for simplicity and $V^{\otimes 0} := \mathbb{R}$. Let $I$ be an $m$-letter word in the letters $\{ 1, \dotsc, d \}$, put $I = \{i_1 i_2 \dotso i_m\}$. Call $e_I := e_{i_1} \otimes \dotsb \otimes e_{i_m} \in V^{\otimes m}$ and $e^*_I := e^*_{i_1} \otimes \dotsb \otimes e^*_{i_m}\in (V^*)^{\otimes m}$. Take some $\mathbf{a} \in T(V)$, which will be of the form $\mathbf{a} = (a_0, a_1, \dotsc)$ for some $a_k \in V^{\otimes k}$. Define the action of $e^*_I$ on $\mathbf{a}$ through:

\begin{align*} e^*_I(\mathbf{a}) &:= \langle (0, 0, \dotsc, 0, \overbrace{e_I}^{\text{$m$-th entry}}, 0, \dotsc), (a_0, a_1, \dotsc, a_{m-1}, a_m, a_{m+1}, \dotsc) \rangle_{T(V)} \\ &:= \langle e_I, a_m \rangle_{V^{\otimes m}} \\ &= \biggl\langle e_I, \sum_{J \in \{1, \dotsc, d\}^m} \alpha_J e_J \biggr\rangle_{V^{\otimes m}} \\ & = \alpha_I, \end{align*} where $\sum_{J \in \{1,\ldots,d\}^m} \alpha_J e_J$ is the decomposition of $a_m$ in the canonical basis of $V^{\otimes m}$ given by $\{ e_J \}_{J \in \{1, \dotsc, d\}^m}$. The book I'm reading defines grouplike elements $\mathbf{a} \in T(E)$ so that, for any two word $I$ and $J$ of given lengths $m$ and $n$, the following identity holds $$ e^*_I(\mathbf{a}) e^*_J(\mathbf{a}) = e^*_{I ⧢ J}(\mathbf{a}) \,, $$ where ⧢ denotes the shuffle product: $$ e_I^* ⧢ e_J^* := \sum_{\sigma \in \operatorname{Shuffles}(m, n)} e_{\left( k_{\sigma^{-1}(1)}, \dotsc, k_{\sigma^{-1}(m+n)} \right)}^* $$

It would seem natural to me that my book's definition of grouplike elements should coincide with the usual definition of grouplike elements for a Hopf algebra, which is that $\Delta(x) = x \otimes x$ where $\Delta$ denotes comultiplication. For $T(V)$, a Hopf algebra structure can be imposed by defining comultiplication through the shuffle product:

\begin{align*} \Delta( e_{i_1} \otimes \dotsb \otimes e_{i_m} ) &= \Delta(e_{i_1}) \otimes \dotsb \otimes \Delta(e_{i_m}) \\ &= \sum_{p = 0}^m (e_{i_1} \otimes \dotsb \otimes e_{i_p}) ⧢ (e_{i_{p+1}} \otimes \dotsb \otimes e_{i_m}) \\ &= \sum_{p=0}^m \sum_{\sigma \in \operatorname{Sh}(p, m-p)} \Bigl( e_{i_{\sigma^{-1}(1)}} \otimes \dotsb \otimes e_{i_{\sigma^{-1}(p)}} \Bigr) \boxtimes \Bigl( e_{i_{\sigma^{-1}(p+1)}} \otimes \dotsb \otimes e_{i_{\sigma^{-1}(m)}} \Bigr) \,. \end{align*} extending by linearity of $\Delta$ to all elements of $T(V)$. I drew this from Wikipedia but fixed the $\sigma$’s which should be $\sigma^{-1}$’s in the last line, and for some reason, their definition of the shuffle product uses $\boxtimes$ instead of $\otimes$.

Answer 1

(1) none of your calculations are right here. If $H$ is a Hopf algebra, the multiplication on $H\otimes H$ is defined by (linear extension of) $(x\otimes y)(v \otimes w) = xv \otimes yw$ which is why the given identity holds.

For (2), I think the key thing is to prove that for any $I$ and $J$ you have $e_I^* ⧢ e_J^* = \mu \circ (e_I^* \otimes e_J^*) \circ \Delta$ (where $\mu$ is the multiplication and $\Delta$ the comultiplication). I didn't check this but it feels right. Everything is linear, so it's enough to verify it on basis elements $e_K$. It would follow from the identity that if $\mathbf{a}$ is grouplike in the traditional sense then it is grouplike in the new sense. Conversely if $\mathbf{a}$ is grouplike in the new sense, $e_I(\mathbf{a}_{(1)})e_J(\mathbf{a}_{(2)}) = e_I^*(\mathbf{a})e_J^*(\mathbf{a})$ (Sweedler notation) for any $I$ and $J$ so the coefficient of $e_I\otimes e_J$ in $\Delta(\mathbf{a})$ is the same as that in $\mathbf{a} \otimes \mathbf{a}$.