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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 $C_n^{\perp}=J^{2^n}$...)

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  • $\begingroup$ I never remember how this is defined, so I keep my copy of Sweedler's book on my desk :) $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '10 at 18:44
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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.

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    $\begingroup$ You could just delete the question. $\endgroup$ – Martin Brandenburg Dec 21 '10 at 18:33
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    $\begingroup$ I don't like deleting anything with non-trivial math in it... $\endgroup$ – darij grinberg Dec 21 '10 at 19:54

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