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



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> 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 $\forall n\ge 1 : \varepsilon|\_{C(n)}=0$ 
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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 ? 



Difference between two definitions of graded coalgebra
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 $n>0$.


