3

I'm beginning the study of coalgebras and the sigma notation using the book called Hopf Algebras of M.E. Sweedler. I'm doing the exercises and I don't know if this ideas are realy clear for me and if what I do is realy what I need do here.

Let $C$ a $\mathbb{k}$-vectorial space $\Delta$ the comultiplication ($\Delta:C \rightarrow C\otimes C $) and $\varepsilon$ the counity ($\varepsilon: C\rightarrow \mathbb{k}$

Exercise: verify the following identity. For any $c\in C$ $$\Sigma_{(c)}\varepsilon(c_{(2)})\otimes\Delta(c_{(1)})=\Delta(c)$$

Here I'm using the Sigma Notation, $\Delta(c)=\Sigma_{(c)}c_{(1)}\otimes c_{(2)}$.

What I did:

We have that each $\varepsilon(c_{(2)})$ is one scalar, say $k_{(2)}$. Furthermore $\Delta(c_{(2)})=\Sigma_{(c_{(2)})}c_{(2)_{(1)}}\otimes c_{(2)_{(2)}}$.Therefore, $$ \Sigma_{(c)}\varepsilon(c_{(2)})\otimes\Delta(c_{(1)})=\Sigma_{(c)}\varepsilon(c_{(2)})\otimes\Sigma_{(c_{(2)})}c_{(2)_{(1)}}\otimes c_{(2)_{(2)}}=\Sigma_{(c)}k_{(2)}\otimes\Sigma_{(c_{(2)})}c_{(2)_{(1)}}\otimes c_{(2)_{(2)}}=^*\Sigma_{(c)}k_{(2)}\otimes\Sigma_{(c_{(2)})}c_{(2)_{(1)}}\otimes c_{(2)_{(2)}}=\Sigma_{(c)}\Sigma_{(c_{(2)})}c_{(2)_{(1)}}\otimes c_{(2)_{(2)}}=\Sigma_{(c)}c_{(1)}\otimes c_{(2)}=\Delta(c) $$

But, after $=^*$ I don't know if each equality makes sense. If makes, why?

Felipe
  • 1,539

1 Answers1

4

We have $$\sum{\epsilon(c_{(1)})\otimes c_{(2)}}=\sum{c_{(1)}\otimes \epsilon(c_{(2)})}=c$$ (This is an axiom.) Since scalar multiplication is commutative, we also have $$\sum{\epsilon(c_{(2)})\otimes c_{(1)}}=c$$ (Really the identity is $\sum{\epsilon(c_{(2)})c_{(1)}}=c$). Thus $$\sum{\epsilon(c_{(2)})\otimes \Delta(c_{(1)})}=\sum{1\otimes \Delta(\epsilon(c_{(2)})c_{(1)})}=1\otimes \Delta(c)$$ (the first equality is true since scalars can be pulled in and out of terms of a tensor product) and the ever-confusing abuse of notation has us calling this $\Delta(c)$.

Matt Samuel
  • 59,287
  • I have some questions about your answer: 1 - Why you can say that $\Sigma\varepsilon(c_{(1)})\otimes c_{(2)} = c =\Sigma(c_{(1)})\otimes \varepsilon(c_{(2)})$? What I have is, using the axiom of coalgebras about the counity, that $\Sigma\varepsilon(c_{(2)})c_{(1)}=c$, but how arrive in your affirmation (changing the "multiplication" to a tensor symbol). 2 - the first equality after "Thus" is ok, I understand that property of a tensor product, but it's ok say that $1\otimes \Delta(c)=\Delta(c)$? – Felipe Sep 09 '15 at 13:51
  • 3
    @Felipe in general it's not okay to replace multiplication with tensor product, but when applying a counit there's an implied contraction of the tensor through the scalar multiplication map. I've never looked at a text on this subject by the way, so you may have to adjust this to be acceptable. The general idea should be correct though. – Matt Samuel Sep 09 '15 at 14:06