Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I posed this question to another user, 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 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"?

share|improve this question
What exactly do you mean by «Poincaré series of X»? –  Mariano Suárez-Alvarez Oct 13 '13 at 21:21
If you mean the Poincaré series associated to the Betti numbers, any series can be realized by a space. –  Fernando Muro Oct 13 '13 at 21:23
With positive coefficients, though. And inverses tend to have negative ones. –  Mariano Suárez-Alvarez Oct 13 '13 at 21:23
The problem is exactly this, even if I did not explain it really well. Let me edit the OP. –  tetrapharmakon Oct 13 '13 at 21:35
Maybe this question also sounds dumb, but I'm particularly led to think (I don't rememeber who said it) that beautiful simmetries are (almost) always meaningful; if something is one inch far from being a "homomorphism"... then I want to know if it can effectively become one. –  tetrapharmakon Oct 13 '13 at 21:47
show 2 more comments

2 Answers

up vote 2 down vote accepted

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.

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 for details.

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.

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 $\Omega B \cong G$.

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.

One more comment about in what sense $\Omega B$ can be understood as the inverse of $B$: note that when defined, homotopy 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.

share|improve this answer
Beautiful! Thank you for your time. –  tetrapharmakon Nov 17 '13 at 19:09
add comment

Let me propose a fanta-mathematical story (sorry, I guess it's not that polite to answer my own question, especially with a long post not cointaining a single precise statement): it is a completely different path that that proposed by Domenico (thank you!), and unfortunately I tend to be incredibly verbose.

First of all, I think it is reasonable to expect that sending a space to its series is either a monoid homomorphism or some kind of functor.

In the first case, forget for a moment about the problem of the definition of this functor. I think it is reasonable to expect that the well known adjunction $G\colon {\bf Mon\leftrightarrows Grp}\colon U$ extends (internalizes?) to a 2-adjunction $$ \underline G\colon {\bf Mon(Cat) \leftrightarrows Grp(Cat)}\colon \underline U $$ sending a monoidal category into its 2-Grothendieck group. Google points me to Joyal's "Traced Monoidal Cats", but I didn't follow him in the white rabbit hole yet.

If this is true, there would be a single way to extend a monoidal functor $F$ to a functor between categorical groups.

Again if it is true, it is reasonable to expect that the Poincare' series $p_X(T)^{-1}$ corresponds to the image under the extended functor $\underline G({\bf Top})\to \mathbb Z[\![T]\!]$ of the formal inverse "$-X$" of the object $X$. However this naive approach is plenty of apparently unsolvable problems. Just to mention a couple of them:

(Fix an universe which is implicitly mentioned everywhere things have to be a set). The class of objects in $\bf Top$ is not a cancellative monoid, which seems to be the minimal hypotesis ensuring that the unit of the adjunction $\underline G\dashv \underline U$ is monic, which is reasonable if we want to figure out the Grothendieck 2-group as an "extension" of the starting monoidal category. We could ask $h\bf Top$ instead of $\bf Top$ to be cancellative; again it is not true, and it is also a subtle problem to find out why. Can I egt rid of the cancellation property, still obtaining a sensible adjunction "a la Grothendieck"?

Is it right to ask $P\colon X\mapsto p_X(T)$ to be a functor? What is the category structure on $\mathbb Z[[T]]$ if objects are elements of the ring, and the ring product is the tensor functor? Instead, it's maybe better to ask $p$ to be a bare genus: this is more compatible with the decategorifications I can do in such a setting (homology -> polynomial -> Euler Characteristic). This would reduce the question to a 1-categorical Grothendieck-group issue. Does this approach seem sensible to you?

Is the aforementioned adjunction really meaningful? Is the categorical group / crossed module associated to the category $\bf Top$ "something interesting"?

As you maybe noticed, I'm a good question-finder, but a really bad problem solver.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.