# Stuck showing that the group ring $\mathbb{C}[G]$ is a bi-algebra

I am learning Hopf algebras via the following online notes on geometric representation theory,

https://www.maths.ed.ac.uk/~djordan/QGpublic.pdf

Exercise 1.7 of the Hopf Algebra section asks to show that $$\mathbb{C}[G]$$ with coproduct $$\Delta(g) = g \otimes g$$ and counit $$\epsilon(g) = \delta_{e,g}$$ gives a bi-algebra structure, where $$G$$ is a finite group. Using definition 1.6 this amounts to showing that both $$\Delta$$ and $$\epsilon$$ are algebra morphisms.

I am ok with everything except showing that $$\epsilon$$ is an algebra morphism, that is, showing that the following square commutes

$$\require{AMScd} \begin{CD} \mathbb{C}[G] @<{\mu}<< \mathbb{C}[G] \otimes \mathbb{C}[G];\\ @V{\epsilon}VV @VV{\epsilon \otimes \epsilon}V \\ \mathbb{C} @<{\cdot}<< \mathbb{C} \otimes \mathbb{C}; \end{CD}$$

as if $$r_1 \otimes r_2 \in \mathbb{C}[G] \otimes \mathbb{C}[G]$$ then surely $$\epsilon \circ \mu(r_1\otimes r_2) = \epsilon(r_1r_2) = {(r_1r_2)}_e$$, i.e. the $$[e]$$ coefficient of $$r_1r_2 \in \mathbb{C}[G]$$ whereas $$(\cdot \circ \epsilon \otimes \epsilon)(r_1 \otimes r_2) = (r_1)_e \cdot (r_2)_e \in \mathbb{C}$$, and these are not the same thing.

What am I misunderstanding? Thanks!

• Are you sure the statements are correct? It seems that these notes are still at an incomplete level. What you quoted as Exercise 1.7 appears now as Exercise 1.8 and the sentence ends with a comma. Commented Jan 5, 2022 at 18:40
• I thought the counit sends each $g \in G$ to $1 \in \mathbb{C}$ Commented Jan 5, 2022 at 18:44
• @WhatsUp Oh sorry that's a typo on my end, I mean Exercise 1.8. I am unsure whether the notes are complete or if statement is even true, but I only learnt what a bi-algebra was around 2 hours ago, so it would be bold of me to assume it is wrong and I am correct haha Commented Jan 5, 2022 at 18:44
• @leibnewtz Perhaps, this would make the square commute, however I am interpreting $\delta_{e,g}$ to be a kronecker-delta-esque operation, do you think the author meant $\epsilon(g) = 1$ instead? Commented Jan 5, 2022 at 18:48
• @Dylan I don't see how that notation could be interpreted as anything else. I think that's the issue here; the counit is indeed given by $g \mapsto 1$ extended linearly to $\mathbb{C}[G]$. Commented Jan 5, 2022 at 18:50

In the usual definition of the coalgebra structure on $$\mathbb{C}[G]$$, the counit is defined as $$\epsilon(g) = 1$$ for all $$g\in G\subset \mathbb{C}[G]$$, i.e. it sends all the group elements to $$1$$, then extended linearly. Thus $$\epsilon(r_1 r_2)$$ should sum up the coefficient of all basis elements of $$\mathbb{C}[G]$$, instead of just taking the $$[e]$$ coefficient.