I posed this question to [another user](https://mathoverflow.net/users/8320/domenico-fiorenza), but we weren't able neither to make a precise guess, nor to give a precise proof. Is it a "not-so-well" well-known fact?


Consider a connected non-finite CW complex $X$; denote its Poincare' series as $p\_X(T)=a\_0+a\_1T+\dots + a\_nT^n+\dots$


Since $X$ is connected, the series $p\_X(T)$ is an invertible element in $\mathbb Z[\![T]\!]$: what is the geometric meaning of its inverse series $q(T)$? Can it arise as the series of a space $Y$? Which is the relation between $X$ and $Y$?


**Edit**:
As it was pointed out in the comments, the question as it is stated makes no sense: whatever the series $p\_X(T)$, its inverse will have *negative* coefficients, so there is no way to interpret them as dimensions of homology objects associated to a wouldbe space $Y$.


Nevertheless, there are cases where utterly counterintuitive statements like "there is a [groupoid](http://ncatlab.org/nlab/show/groupoid+cardinality) whose measure is $\sum\_k \frac{1}{k!}$" have suitable meaning. So my question sounds more like 



> 
> is there a "groupoidal-trick" to turn the (multiplicative) inverse Poincare' series into something meaningful, by stretching a bit the meaning of the word "space"?
> 
> 
> 



The geometric meaning of an inverse Poincare' series
The short answer is "morally, $Y$ should be the loop space of $X$," and we'll see how far this answer gets towards justifying that. 




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First some words that are not about spaces. My go-to example of two power series, both of which are the Hilbert / Poincare series of something, whose product is $1$ is 


$$\left( \sum\_{k \ge 0} t^k \dim S^k(V) \right) \left( \sum\_{k \ge 0} (-1)^k t^k \dim \Lambda^k(V) \right) = 1$$


where $V$ is a finite-dimensional vector space. Explicitly, if $\dim V = d$ then $\dim S^k(V) = {d+k-1 \choose k}, \dim \Lambda^n(V) = {d \choose k}$, and we recover the power series identity


$$\frac{1}{(1 - t)^d} (1 - t)^d = 1.$$


Here we should think of the left factor as being the graded dimension of $S(V)$, where $V$ is regarded as being in degree $0$, and the right factor as being the graded dimension of $S(V[1])$, where the shift is the origin of the signs and the taking of exterior powers. 


The identity above reflects the fact that the symmetric and exterior algebras are Koszul dual. It can be categorified to the fact that a certain double complex refining the Koszul complex has homology concentrated in one bidegree, and can be used to write down a nice condition implying that a graded ring is Koszul; see Lemma 2.11.1 and Theorem 2.11.1 of Beilinson, Ginzburg, and Soergel's [Koszul Duality Patterns in Representation Theory](http://www.ams.org/journals/jams/1996-9-02/S0894-0347-96-00192-0/S0894-0347-96-00192-0.pdf) for details.




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Now some words about spaces. Below all cohomologies and homologies are with rational coefficients. 


Topologically our starting point is the intuition that fibrations


$$F \to E \to B$$


are like "twisted products" $E \cong B \times F$. One manifestation of this idea is that, under sufficiently nice hypotheses, $H\_{\bullet}(E)$ is something like $H\_{\bullet}(B) \otimes H\_{\bullet}(F)$ (maybe we want $B$ to be simply connected and the Serre spectral sequence to degenerate at the second page, or something like that). In particular, under sufficiently nice hypotheses, the Poincare series of $F, E, B$ should satisfy


$$p\_E(t) = p\_B(t) p\_F(t).$$


Now suppose $E \cong \ast$ is contractible, so $p\_E(t) = 1$ and the fibration above is (weakly homotopy equivalent to) the path space fibration


$$\Omega B \to PB \to B.$$


Then the identity $p\_E(t) = p\_B(t) p\_F(t)$ means that $p\_B(t)$ and $p\_{\Omega B}(t)$ are inverse power series. Of course by definition both must have non-negative integer coefficients so this is quite silly. Nevertheless the structure of the second page of the Serre spectral sequence is suggestive, and in particular something like the following can be deduced from Theorem 3.27 in McCleary's [A User's Guide to Spectral Sequences](http://www.cambridge.org/us/academic/subjects/mathematics/geometry-and-topology/users-guide-spectral-sequences-2nd-edition). 



> 
> **Theorem:** (Hypotheses). Then:
> 
> 
> 1) If $H^{\bullet}(B)$ is an exterior algebra on generators $x\_1, ..., x\_m$ of odd degrees $2r\_i - 1$, then $H^{\bullet}(\Omega B)$ is a polynomial algebra on generators $y\_1, ..., y\_m$ of even degrees $2r\_i - 2$. 
> 
> 
> 2) If $H^{\bullet}(\Omega B)$ is an exterior algebra on generators $x\_1, ..., x\_m$ of odd degrees $2r\_i - 1$, then $H^{\bullet}(B)$ is a polynomial algebra on generators $y\_1, ..., y\_m$ of even degrees $2r\_i$.
> 
> 
> 


In particular, the Poincare series are almost inverses except for the signs and the degree shift. For an example of the first case let $B$ be a product of odd-dimensional spheres (not including circles), and for an example of the second case let $B$ be the classifying space of a compact connected Lie group $G$, so that $\Omega B \cong G$. 




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Regarding the relationship between the first and second set of words, I have been told that there is a sense in which cochains $C^{\bullet}(B)$ on a space and chains $C\_{\bullet}(\Omega B)$ on its loop space are Koszul dual. Unfortunately I don't know a precise statement of this relationship. There should be a lot more to say here from the perspective of rational homotopy theory but I'm not the person to say it. 




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One more comment about in what sense $\Omega B$ can be understood as the inverse of $B$: note that when defined, [homotopy cardinality](http://ncatlab.org/nlab/show/groupoid+cardinality) of sufficiently connected spaces is multiplicative under fibrations by the long exact sequence of a fibration, and the homotopy cardinality of $\Omega B$ is the inverse of that of $B$. This generalizes the observation that if $B = BG$ is the classifying space of a finite group then it ought to have Euler characteristic $\frac{1}{|G|}$ in some sense. 


