I am trying to show that $d²=0$ where $d$ is the derivation on $T(s^{-1}\bar{C})$ induced by the map $s^{-1}\bar{C}\to T(s^{-1}\bar{C})$ defined by $$s^{-1}x\mapsto -\sum (-1)^{|x_{(1)}|}s^{-1}x_{(1)}\otimes s^{-1}x_{(2)}$$ where $\bar{\Delta}(x)=\sum x_{(1)}\otimes x_{(2)}$.

I tried doing this by considering an element $s^{-1}c_1\otimes...\otimes s^{-1}c_k\in T(s^{-1}\bar{C})$. Because $d$ is linear and the elemets of $T(s^{-1}\bar{C})$ are sums of elements of this form this should suffice. Leibniz rule tells us that $$d(s^{-1}c_1\otimes...\otimes s^{-1}c_k)=\\ (\sum 1\otimes...\otimes 1 \otimes d\otimes 1\otimes ...\otimes 1)(s^{-1}c_1\otimes...\otimes s^{-1}c_k).$$ By writing out the formulas and keeping track of all the signs I managed to prove that when we apply $d$ twice, the term we get from first applying $d$ on the $i$'th position and then applying $d$ on the $j$'th position cancels with the term we get from first applying $d$ on the $j$'th position and then applying $d$ on the $i$'th position as long as $j\neq i,i+ 1$. However, I got stuck with the terms we get from first applying $d$ on the $i$'th position and then applying $d$ on the $(i+1)$'th position. I know coassociativity of $C$ should imply that these terms cancel with the terms we get by applying $d$ on the $i$'th position twice but I cannot prove it.

In the book Algebraic Operads by Loday and Vallette they write out a proof of it (page 44) but I do not understand it. They only consider $d^2$ applied to an element of the form $s^{-1}x$. Then I think that they identify terms of $$(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x):=\sum x_{(1)}\otimes x_{(2)}\otimes x_{(3)}$$ with terms of $$(1\otimes \bar{\Delta})\circ \bar{\Delta}(x):=\sum x_{(1)}'\otimes x_{(2)}'\otimes x_{(3)}'$$ and show how they cancel given the signs they get in $d²$. This is what I don't understand because coassociativity of $C$ only tells us $(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x)=(1\otimes \bar{\Delta})\circ \bar{\Delta}(x)$ but $$(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x)=\sum x_{(1)}\otimes x_{(2)}\otimes x_{(3)}$$ need not contain the same terms as $$(1\otimes \bar{\Delta})\circ \bar{\Delta}(x)=\sum x_{(1)}'\otimes x_{(2)}'\otimes x_{(3)}'.$$

Maybe someone who is familiar with the book and/or the theory can help bring clarity here=)


You are missing many signs from desuspension, which you did write in your first sentence, but didn't afterwards.

First, recall that since the differentail, call it $d$, induced on $\Omega C$, obtained from the comultiplication, is a derivation, and since $\Omega C$ is generated in degree $1$ by $V=s^{-1}\overline C$, it suffices you prove the claim on these generators. Now the differential is given as follows: take $s^{-1}c$ a generator, desuspend, and comultiply to get $c_{(1)}\otimes c_{(2)}$, and then multiply by $s^{-1}\otimes s^{-1}$ to get (from the Koszul rule and the interchange of $c_{(1)}$ and $s^{-1}$)

$$d(s^{-1}c) = -(-1)^{|c_{(1)}|}s^{-1}c_{(1)}\otimes s^{-1}c_{(2)}.$$

If you apply $d$ again, you will get two terms, one is

$$ (-1)^{|c_{(1)}|+|c_{(1)(1)}|}s^{-1}c_{(1)(1)}\otimes s^{-1}c_{(1)(2)}\otimes s^{-1}c_{(2)},$$

where the sign $|c_{(1)(1)}|$ comes from the same rule as above. The other is

$$ (-1)^{|c_{(1)}|+|c_{(2)(1)}|+|c_{(1)}|-1}s^{-1}c_{(1)}\otimes s^{-1}c_{(2)(1)}\otimes s^{-1}c_{(2)(2)},$$

where $|c_{(1)}|-1$ comes from $d$ going over $s^{-1}c_{(1)}$, so the overall sign is just $|c_{(2)(1)}|-1$.

To get the global sign of the first term, note that $|c_{(1)}|= |c_{(1)(1)}| +|c_{(1)(2)}|$, so the sign is $|c_{(1)(2)}|$. Writing everything in Sweedler notation, you get

$$(-1)^{|c_{(2)}|}c_{(1)}\otimes c_{(2)}\otimes c_{(3)} + (-1)^{|c_{(2)}|-1}c_{(1)}\otimes c_{(2)}\otimes c_{(3)} = 0$$

  • $\begingroup$ Thank you for your answer. I am still confused however, because Sweedlers notation implies there's a sum right, and coassociativity tells us that the two sums with out any signs are equal. But the terms in the sums need not equal so $c_{(2)}$ in the first term of your last expression is not necessarily the same as $c_{(2)}$ in the second term. Sweedler's notation still confuses me a lot so I tried writing it all out with sums. I'm sorry about the horrible expressions in my next comment but if you can make sense of it, maybe you can help me get it to equal zero. $\endgroup$
    – budwarrior
    Aug 27 '18 at 7:43
  • $\begingroup$ If $\bar{\Delta}(c)=\sum_i a^i_1\otimes a^i_2$, $\bar{\Delta}(a^i_1)=\sum_j b^{ij}_1\otimes b^{ij}_2$ and $\bar{\Delta}(a^i_2)=\sum_j d^{ij}_1\otimes d^{ij}_2$ i get $$d\circ d(s^{-1}c)=\sum_{ij}(-1)^{|a_1^i|+|b_1^{ij}|}s^{-1}b_1^{ij}\otimes s^{-1}b_2^{ij}\otimes s^{-1}a_2^{i}+$$$$ \sum_{ij}(-1)^{|d_1^{ij}|-1}s^{-1}a_1^i\otimes s^{-1}d_1^{ij}\otimes s^{-1}d_2^{ij}.$$ $\endgroup$
    – budwarrior
    Aug 27 '18 at 7:51
  • $\begingroup$ Coassociativity tells us $\sum_{ij}b_1^{ij}\otimes b_2^{ij}\otimes a_2^i=\sum_{ij}a_1^i\otimes d_1^{ij}\otimes d_2^{ij}$ and we also know $|a_1^i|+|a_2^i|=|c|$ and $|b_1^{ij}|+|b_2^{ij}|=|a_1^i|$ and $|d_1^{ij}|+|d_2^{ij}|=|a_2^i|$ but I cannot put all of this together to get $0$. $\endgroup$
    – budwarrior
    Aug 27 '18 at 7:51
  • $\begingroup$ @budwarrior All you need to observe is that $|a_1^i|+|b_1^{ij}|$ (which is what I denoted $|c_{(1)}|+|c_{(1)(1)}|$) is congruent to $|b_2^{ij}|$ modulo $2$ (which is what I denoted $c_{(1)(2)}$), and your second sign is $|d_1^{ij}|-1$ which by coassociativity matches with $|b_2^{ij}|$ if you group terms by their homogenous degree. $\endgroup$
    – Pedro Tamaroff
    Aug 28 '18 at 23:14
  • $\begingroup$ Thanks a lot for your answer! Now I get it=) $\endgroup$
    – budwarrior
    Aug 29 '18 at 4:26

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