# Does Koszul duality between $Comm$ and $Lie$ imply the power series identity $\exp(\ln(1-z))-1 = -z$?

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?

• More generally, there is an inversion for plethysm in the ring of symmetric functions. Look at the original articles defining Koszul duality for operads. Sep 14 '18 at 7:01
• This unpublished (and probably silly) note from decades ago citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.8306 argues that the Goodwillie tower of the identity is inverse to the functor $\Omega^\infty \Sigma^\infty$, and that the relationship between them basically categorifies the relationship between $\ln(1+x)$ and $e^x-1$. Today we know that the derivatives of the identity are a topological version of the Lie operad, the Goodwillie derivatives of $\Sigma^\infty\Omega^\infty$ is the commutative (co)-operad, and they are Koszul dual to each other for a reason. Same idea, evolved. Sep 14 '18 at 12:25
• @GregoryArone Let's see... the Goodwillie derivatives of a functor $F$ form an operad if $F$ is a monad, right? Are you saying there's a general statement about the Koszul dual of the derivatives of a monad? Sep 17 '18 at 15:36
• I guess. If $F$ is a monad, then the derivatives of $F$ are also the derivatives of the identity functor on the category of algebras over $F$. Let $Alg_F$ be this category of algebras, and let $S_F$ be its stabilization (i.e. the spectra of $Alg_F$). Let $(\Sigma^\infty_F, \Omega^\infty_F)$ be the stabilization functor $Alg_F\to S_F$ and its right adjoint. Then $\Sigma^\infty_F \Omega^\infty_F$ is a comonad on $S_F$. The derivatives of $\Sigma^\infty_F \Omega^\infty_F$ form a cooperad, and it is the bar construction on the derivatives of $F$. Sep 17 '18 at 16:35

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 $\exp(\ln(1-x))-1=x$.

Yes. (Mathoverflow won't let me make this my total answer, so ...)

Koszul duality says that a certain chain complex of graded vector spaces is acyclic. Thus the alternating sum of the Poincare series gives $z$.

• Thanks! I don't have a very firm grasp on Koszul duality, so I wonder if you could spell out which statement of Koszul duality you're using and which complex it says is acyclic. I was trying to do something with the free commutative algebra / free Lie algebra functors, but it sounds like that was the wrong approach. Sep 14 '18 at 0:41
• That's about right, the complex is $Com \circ sLie$ (or also the opposite direction) in the category of symmetric sequences. The $s$ stands for a shift and twisting by sign. You can see Loday and Valette's book for details. Sep 14 '18 at 1:05
• Alternately, the operadic bar construction categorifies the Taylor expansion of inverse power series in the same way that the usual (reduced) bar construction categorifies the geometric series expansion. Sep 14 '18 at 1:11
• @PhilTosteson Thanks! I'm not familiar with how the usual bar construction categorifies the geometric series expansion. Could you explain? I think an explanation along those lines would make a great answer, actually. Especially because the one description of Koszul duality that I think I've started to grasp is in terms of the operadic bar construction. Sep 14 '18 at 2:06
• In particular, I'm still a bit confused because in the $Comm / Lie$ case it looks like I need to insert a sign on just one series, while in the $Ass$ case I need a sign on both copies of the series. So it's still not clear to me what the general statement even is. Sep 14 '18 at 2:12