# 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. – F. C. 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• This unpublished (and probably silly) note from decades ago citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.8306 argues that the Goodwillie tower of the identity is inverse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. – Gregory Arone 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? – Tim Campion 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 $F$. – Gregory Arone 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. – Tim Campion 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. – Phil Tosteson 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. – Phil Tosteson 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. – Tim Campion 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. – Tim Campion Sep 14 '18 at 2:12