I have seen the coradical filtration of a coalgebra $C$ defined as follows:


$C\_0 = \text{sum of all simple subcoalgebras of }C$;


for any $n\geq 1$, let $C\_n$ be $\Delta^{-1}\left(C\otimes C\_{n-1}+C\_{n-1}\otimes C\right)$.


Now I am trying to prove that this really is a coalgebra filtration, and something seems not to work. Is it possible that $C\_n$ should rather be $\Delta^{-1}\left(C\otimes C\_0+C\_{n-1}\otimes C\right)$ ? Or is this secretly the same?


(*Motivation:* I am working with the dual algebra $C^{\ast}$, at least in the case of $C$ finite-dimensional. In this case, if $J$ denotes the Jacobson radical of $C^{\ast}$, then we have $C\_n^{\perp}=J^{2^n}$ if $C\_n$ is defined as $\Delta^{-1}\left(C\otimes C\_{n-1}+C\_{n-1}\otimes C\right)$, we have $C\_n^{\perp}=J^{n+1}$ if $C\_n$ is defined as $\Delta^{-1}\left(C\otimes C\_1+C\_{n-1}\otimes C\right)$. Now, if my computations are right, $\left(C\_0,C\_1,C\_2,...\right)$ is a coalgebra filtration if and only if $\left(C^{\ast},C\_0^{\perp},C\_1^{\perp},C\_2^{\perp},...\right)$ is an algebra filtration. The latter holds for $C\_n^{\perp}=J^{n+1}$ but not for $C\_n^{\perp}=J^{2^n}$...)



How is the coradical filtration defined?
I am voting to close this. $C\_n$ should be defined as $\Delta^{-1}\left(C\otimes C\_0+C\_{n-1}\otimes C\right)$. My wrong definition was due to my bad memory. Sorry for spamming.


