# Differential of the Twisted complex for algebraic operads

I have a question about the proof of lemma 6.4.12 in the book Algebraic Operads (Loday-Vallette) which I do not seem to be able to fully complete on my own. Hopefully, somebody here can point out what I am not seeing.

Let me sketch the situation. Given a cooperad $$(C,\Delta,\epsilon)$$ and a operad $$(P,\gamma,\eta)$$, we consider its free right P-module $$C \circ P$$. Let $$\alpha: C \longrightarrow P$$ be a twisting morphism (of degree $$-1$$). The map $$d^r_\alpha$$ is then defined as follows $$C \circ P \overset{\Delta_{(1)} \circ P}{\longrightarrow} (C \circ_{(1)} C) \circ P \overset{(C \circ_{(1)} \alpha) \circ P}{\longrightarrow} (C \circ_{(1)} P) \circ P \cong C \circ (P ; P\circ P) \overset{C \circ (P;\gamma)}{\longrightarrow} C \circ (P ; P) \cong C \circ P$$ In concrete terms, $$d^r_\alpha( c ; p_1,\ldots,p_n) = \sum_i \pm (c_{(1)}; p_1,\ldots, \alpha c_{(2)} \circ (p_i,\ldots,p_j), \ldots,p_n)$$ where $$\Delta_{(1)}(c) = c_{(1)} \otimes (i, c_{(2)})$$ denotes the sweedler notation.

The proof of the lemma then states the following $$[ d^r_\alpha , d^r_\alpha ] = d^r_{[\alpha,\alpha]}$$ which reduces to $$(d^r_\alpha)^2 = d^r_{\alpha * \alpha}$$ The book says the computation is similar to the case of twisted complex for associative (co)algebras. However, I think extra terms appear on the left-hand side, namely $$(c_{(11)}; p_1, \ldots, \alpha c_{(12)} \circ (p_i,\ldots,p_j), \ldots ,\alpha c_{(2)} \circ (p_k,\ldots,p_l), \ldots ,p_n)$$ and $$(c_{(11)}; p_1, \ldots, \alpha c_{(2)} \circ (p_i,\ldots,p_j), \ldots ,\alpha c_{(12)} \circ (p_k,\ldots,p_l), \ldots ,p_n)$$ Even using coassociativity, I still do not see how these terms cancel as they are not of the right shape? Perhaps I am overlooking something simple?

Since $$\alpha$$ is of degree $$-1$$, these terms come in pairs appearing with opposite signs that cancel each other. In other words, the (co)associativity for (co)operads has a sequential axiom and a parallel axiom, and these terms vanish because of the parallel axiom. A nice way to see that is to draw these as trees and remember that trees have levels helping you to remember the order of insertions, and that exchanging levels creates Koszul signs.
• Thank you for answering! I did indeed draw these using trees, but I do not see where the second half of the pair comes from in the calculation $(d^r_\alpha)^2$? We have that $(c_{(11)} \circ_i c_{(12)} ) \circ_j c_{(2)}$ appears exactly once for every $i$ and $j$ Sep 24, 2021 at 14:40