# Associated graded of a filtration of a tensor product

I'm trying to understand a part of the PhD thesis of Kenji Lefèvre-Hasegawa (e.g. available here). My question is about the proof of Lemma 1.3.2.3b stating:

Remarquons que nous avons un isomorphisme de complexes $\bigoplus_{i\geq 1}Gr_i(\Omega BA)\to \Omega \overline{T^c}V$ qui identifie à la composante $Gr_i(\Omega BA), i\geq 1$, à la somme des $S^{-1} V^{\otimes i_1}\otimes\dots\otimes S^{-1} V^{\otimes i_k}\subset (S^{-1} \overline{T^c}V)^{\otimes k}$, où $k\geq 1$ et où $i_1+\dots+i_k=i$.

Here $V=SA$, where $S$ is the shift functor. $A$ is a dg algebra and $B$ and $\Omega$ are the bar and cobar resolution, respectively, i.e. $BA=\bigoplus_{i\geq 1} V^{\otimes i}$ (with the coproduct $\Delta: V^{\otimes i}\to \bigoplus_{p+q=i} V^{\otimes p}\otimes V^{\otimes q}$ and the filtration $BA_{[i]}=ker\Delta^{(i)}$) and $\Omega BA=\bigoplus_{i\geq 1}S^{-1}BA^{\otimes i}$ with the filtration induced by that of $BA$.

Can someone give more details?

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The key idea is very simple - the bar differential has a part coming from the differential on $A$, and the remaining part, encoding the product of $A$. The first part preserves the number of tensor factors, the remaining part consists of terms with fewer tensor factors. Therefore, for the filtration by the number of tensor factors, the corresponding graded object will just remember the differential on $A$, that is the structure of a chain complex, not the product. It remains to notice that for obvious reasons the number of tensor factors is precisely what is counted by the filtration by kernels of iterated deconcatenations... Does it help?
If I understand you correctly, your answer is on the associated graded of $BA$, whose parts are $(SA)^{\otimes i}$. That I did understand, what I am missing is how this leads to the associated graded parts of $\Omega BA$. –  Julian Kuelshammer Aug 30 '12 at 15:56
Well, the induced filtration on the cobar-bar construction is naturally by the total number of tensor factors, $i_1+\ldots+i_k$ in the notation of your post, so I am puzzled by what worries you. $BA$ is a cofree coalgebra, so the differential of its cobar complex has a part coming from the differential of $BA$, and you seem to agree that after passing to the graded object this becomes the differential on $\overline{T^c(sA)}$ obtained by extending the differential of $A$, and a part coming from the (deconcatenation) coproduct on $BA$, it preserves the number of tensor factors and is unchanged. –  Vladimir Dotsenko Aug 30 '12 at 16:10