To a symmetric sequence $V\_\bullet$ of vector spaces, associate the generating function $F\_V(z) = \sum\_n \frac{\dim(V\_n)}{n!} z^n$. Then


$$F\_{Comm\_\ast}(z) = \exp(z)-1 \qquad F\_{Lie}(z) = \ln(1-z)$$


where $Comm\_\ast$ is the reduced commutative operad and $Lie$ is the Lie operad. Notice that


1. On the one hand, these power series are inverse up to a sign.
2. On the other hand, these operads are Koszul dual (and perhaps the sign corresponds to the shift that appears in Koszul duality?).


Similarly,


$$F\_{Ass\_\ast}(z) = \frac{z}{1-z}$$


where $Ass\_\ast$ is the reduced associative operad. On the one hand, this power series is its own inverse up to a sign. On the other hand, this operad is Koszul self-dual.


**Question:** Is this a coincidence? Or is there some deeper connection between (1) and (2)? More concretely, is it the case (under certain conditions, perhaps) that Koszul dual operads have inverse generating functions, up to some sign?



Does Koszul duality between $Comm$ and $Lie$ imply the power series identity $\exp(\ln(1-z))-1 = -z$?
Let me flesh out the answer a little. The general statement is given by Theorem 7.5.1 in the book *Algebraic Operads* by Loday and Vallette.


First a definition. Let $P = P(E,R)$ be a quadratic operad, with generators $E$ (f.gen. s.t. $E(0) = 0$) and $R \subset E \circ E$ quadratic relations. Let $P^{(r)}(n)$ be the subspace of operations of weight $r$, where $E$ is of weight $1$. There is a generating series, aka Hilbert-Poincaré series:
$$f^P(x,y) = \sum\_{r \ge 0, n \ge 1} \frac{\dim P^{(r)}(n)}{n!} y^r x^n.$$


The theorem states that if $P$ is Koszul, with dual $P^!$, then there is a functional equation:
$$f^{P^!}(f^P(x,y),-y) = x.$$


Remark: as Nicholas Kuhn explained, this equality follows from the acyclicity of the Koszul complex, the product $P^¡ \circ P$ with the Koszul differential.


$\newcommand{\Com}{\mathsf{Com}}\newcommand{\Lie}{\mathsf{Lie}}$
Apply this to $P = \Lie$, $P^! = \Com$. It's well-known that $\Com(n) = \Com^{(n-1)}(n)$ is of dimension $1$ for $n \ge 1$, while $\Lie(n) = \Lie^{(n-1)}(n)$ is of dimension $(n-1)!$ for $n \ge 1$. So in particular you get
\begin{align}
f^\Com(x,1) & = \sum\_{n \ge 1} \frac{x^n}{n!} = \exp(x) - 1,\\
f^\Lie(x,-1) & = \sum\_{n \ge 1} \frac{(-1)^{n-1} x^n}{n} = \ln(1+x)
\end{align}


Apply the functional equation to $y = -1$ and you get $\exp(\ln(1-x))-1=x$.


