# How is the differential induced by $d_C$ on $\Omega C$ defined for $(C,d_C)$ is a dga coalgebra?

Again I am confused about something regarding the cobarconstruction of a dga coalgebra $(C,d_C)$. The cobar construction of $C$ is the dga algebra $(T(s^{-1}\bar{C}),d_1+d_2)$ where $d_2$ is induced by the reduced coproduct of $C$ and $d_1$ is induced by $d_C$. This is how I would try and define $d_1$:

(Below when I take tensor products of linear maps I use the Koszul sign convention.)

First of all we get a differential $1\otimes d_C$ on $s^{-1} \bar{C}=\mathbb{K}s^{-1}\otimes \bar{C}$. Then we get a differential $$\sum (1\otimes 1)\otimes...\otimes (1\otimes d_C)\otimes... \otimes (1\otimes 1)$$ on $(\mathbb{K}s^{-1}\otimes \bar{C})^{\otimes n}$ and taking the direct sum of all of these maps gives us a differential $d_1$ on all of $T(s^{-1}\bar{C})$.

Can someone tell me if this is the correct way to go about it?

The reason I started doubting myself is because I fail trying to prove that $d_1$ and $d_2$ anti-commute. In fact I get the feeling they commute.

Here is what I did. I want to check that $d_1d_2+d_2d_1$ is zero on $s^{-1}\bar{C}$ to begin with. We have $$d_1|_{s^{-1}\bar{C}}=1\otimes d_C:\mathbb{K}s^{-1}\otimes \bar{C}\to \mathbb{K}s^{-1}\otimes \bar{C},$$ $$d_1|_{(s^{-1}\bar{C})^{\otimes 2}}=1\otimes d_C\otimes 1\otimes 1+1\otimes 1\otimes 1\otimes d_C:(\mathbb{K}s^{-1}\otimes \bar{C})^{\otimes 2}\to (\mathbb{K}s^{-1}\otimes \bar{C})^{\otimes 2}$$ and $$d_2|_{s^{-1}\bar{C}}=(1\otimes \tau\otimes 1)\circ (\Delta_s\otimes \bar{\Delta}):\mathbb{K}s^{-1}\otimes \bar{C}\to (\mathbb{K}s^{-1}\otimes \bar{C})^{\otimes 2}$$ where $\Delta_s:s^{-1}\mapsto -s^{-1}\otimes s^{-1}$ and $\bar{\Delta}$ is the reduced coproduct in $C$ and $\tau$ is the signed switching map, in our case $\tau:\mathbb{K}s^{-1}\otimes \bar{C}\to \bar{C}\otimes \mathbb{K}s^{-1}$, $s^{-1}\otimes x\mapsto (-1)^{|x|}x\otimes s^{-1}$. Then this is what I get

$$d_2\circ d_1|_{s^{-1}\bar{C}}=(1\otimes \tau\otimes 1)\circ(\Delta_s\otimes \bar{\Delta})\circ(1\otimes d_C)=$$ $$(1\otimes \tau\otimes 1)\circ(\Delta_s\otimes (\bar{\Delta}\circ d_C))$$ Using that $d_C$ is a coderivation we get $$=(1\otimes \tau\otimes 1)\circ(\Delta_s\otimes ((1\otimes d_C+d_C\otimes 1)\circ\bar{\Delta}))=$$ $$(1\otimes \tau\otimes 1)\circ (1\otimes 1\otimes 1\otimes d_C+1\otimes 1\otimes d_C\otimes 1)\circ (\Delta_s\otimes \bar{\Delta})=$$ $$(1\otimes 1\otimes 1\otimes d_C+1\otimes d_C\otimes 1\otimes 1)\circ(1\otimes \tau\otimes 1)\circ(\Delta_s\otimes \bar{\Delta})=$$ $$d_1\circ d_2|_{s^{-1}\bar{C}}.$$

If someone could help me find my error either in how i define $d_1$ or if I defined it correctly in the computations above I would be very grateful!

Edit: I see now that I am missing signs from when maps "jump over each other". Maybe this fixes things..

I will use the notation $[c]$ for $s^{-1}c$, as usual, and, more generally, the bar notation for tensor monomials. The equation $d_1d_2+d_2d_1=0$ follows from the fact $d$ is a coderivation for $\Delta$. The suspension signs are what do the trick, so you must carefully follow them, using the Koszul sign rule.

Recall that $d_1[c] = -[dc]$ and that $d_2[c] = -(-1)^{|c_{(1)}|} [c_{(1)}\vert c_{(2)}]$, where I'm using Sweedler notation. That the differential $d$ of $C$ be a coderivation for $\Delta$ means that

$$\Delta d(c) = (d\otimes 1+1\otimes d)\Delta(c)$$

Writing this is Sweedler notation, you get that

$$\tag{1} (dc)_{(1)}\otimes (dc)_{(2)} = d(c_{(1)})\otimes c_{(2)} +(-1)^{|c_{(1)}|}c_{(1)}\otimes d(c_{(2)}).$$

Applying $s^{-1}\otimes s^{-1}$ on both sides, you obtain from $(1)$ the equation

$$\tag{2} (-1)^{|(dc)_{(1)}|}[(dc)_{(1)}\vert (dc)_{(2)}] = (-1)^{|d(c_{(1)})|}[d(c_{(1)})\vert c_{(2)}] +[c_{(1)}\vert d(c_{(2)})].$$

Now I will compute $d_1d_2$ and $d_2d_1$, and observe that your desired equation follows from $(2)$. To compute $d_1d_2$, I will use that $d_1$ is, by definition, extended as a derivation for the product (concatenation of bars) in $\Omega C$, so that (the degree of $[c]$ is $|c|-1$!)

$$d_1[c_1\vert c_2] = d_1[c_1]\cdot [c_2] +(-1)^{|c_1|-1} [c_1]\cdot d[c_2].$$

This gives that

\begin{align} d_1d_2[c]&= -(-1)^{|c_{(1)}|}d_1[c_{(1)}\vert c_{(2)}]\\ &= (-1)^{|c_{(1)}|}[d(c_{(1)})\vert c_{(2)}] + (-1)^{|c_{(1)}|+|c_{(1)}|-1}[c_{(1)}\vert d(c_{(2)})]\\ &= (-1)^{|c_{(1)}|}[d(c_{(1)})\vert c_{(2)}] -[c_{(1)}\vert d(c_{(2)})] \\ &= (-1)^{|d(c_{(1)})|+1}[d(c_{(1)})\vert c_{(2)}] -[c_{(1)}\vert d(c_{(2)})] \\ &= -((-1)^{|d(c_{(1)})|}[d(c_{(1)})\vert c_{(2)}] +[c_{(1)}\vert d(c_{(2)})]) \end{align}

On the other hand, we have that

\begin{align} d_2d_1[c] &= -d_2[dc] \\ &= (-1)^{|(dc)_{(1)}|} [(dc)_{(1)} \vert (dc)_{(2)} ] \end{align}

Equation $(2)$ now shows $d_1d_2+d_2d_1$ vanishes on $s^{-1}C$, which is what you wanted.

• Thanks! It now makes sense to me=) Sep 17 '18 at 16:55