Difference between two definitions of graded coalgebra

Question

I'm reading articles about applications of Hopf algebras in physics. In one of them there is following definition of graded coalgebra:

A coalgebra $C$ is called graded if $C=\bigoplus\limits_{n\ge 0}C(n)$ as vector space and $\forall n\ge 0 : \ \Delta (C(n))\subset \bigoplus_\limits{i=0}^{n}C(i)\otimes C(n-i)$ and $\forall n\ge 1 : \varepsilon|_{C(n)}=0$

where $\Delta$ is a coproduct and $\varepsilon$ is a counit.

In the other article one can find definition in which it is assumed only that

$\forall n\ge 0 : \ \Delta (C(n))\subset \sum_\limits{i=0}^{n}C(i)\otimes C(n-i) $

Are these definitions equivalent ?

Answer 1

Yes, these are equivalent. This is easier to think about in terms of the dual $C^\vee=\prod C(n)^\vee$, which is an algebra; we will think of $C(i)^\vee$ as sitting inside $C^\vee$ as those sequences $(x_n)\in C^\vee$ such that $x_i=0$ unless $i=n$. The condition $\Delta(C(n))\subset\sum C(i)\otimes C(n-i)$ implies that $xy\in C(i+j)^\vee$ if $x\in C(i)^\vee$ and $y\in C(j)^\vee$, and furthermore that if $x=(x_n)\in C^\vee$ and $y=(y_n)=C^\vee$, then $xy=(\sum_{i+j=n} x_i y_j)$. The condition $\epsilon(C(n))=0$ for $n>0$ which we wish to prove is equivalent to $1\in C(0)^\vee$, where $1$ is the unit of the algebra $C^\vee$.

To prove this write $1=(a_n)$; we wish to show $a_n=0$ for $n>0$. We must have $1\cdot a_0=a_0$, which implies $a_na_0=0$ for all $n>0$, and similarly $a_0\cdot 1=a_0$ implies $a_0a_n=0$ for all $n>0$. Looking at the degree $1$ part of $1=1\cdot 1$, we find that $a_1=a_0a_1+a_1a_0=0$. But then looking at the degree $2$ part, we have $a_2=a_0a_2+a_1^2+a_2a_0=0$. Continuing by induction, we find that $a_n=0$ for all $n>0$.