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This question was inspired by the question posed by John Baez here: and Neil Strickland's answer to that question.

Let $X$ be a CW complex. If $X$ is finite, there are no problems defining its Euler characteristic whatsoever. However, when $X$ is infinite, there are at least two different ways which give similar but not identical results:

If the number $c(d)$ of cells of $X$ of dimension $d$ is finite for each $d$, one can form a generating function $f(t)=\sum_d c(d)t^d$ and define $\chi(X)=\lim_{t\to -1^+}f(t)$ (assuming the limit makes sense and exists).

If $X$ has a finite $d$-fold cover $\tilde X$ which is homotopy equivalent to a finite CW complex, then we can set $\chi(X)=\frac{1}{d}\sum_{i=1}^\infty (-1)^i\dim H^i(\tilde X,k)$ where $k$ is a field. This is the Wall characteristic (a.k.a. the rational Euler characteristic) of $X$ and it is a homotopy invariant and does not depend on $k$.

Now, if we try to calculate the characteristic of $\mathbb{R} P^\infty$ with its standard cell decomposition (one cell in each dimension from $0$ to $\infty$), then both definitions agree: the first one gives the generating function $\frac{1}{1-t}$, which gives $\frac{1}{2}$, when evaluated at $t=-1$; the second one also gives $\frac{1}{2}$ using the double cover $pt\cong S^\infty\to \mathbb{R} P^\infty$.

However, if we take a finite group $G$ and try to compute the characteristic of the Milnor construction of $BG$, then the first definition will give $\frac{1}{1+\# G}$, since the number of simplices of dimension $d$ is $(\# G)^d$, while the second one gives $\frac{1}{\# G}$.

I would like to ask: is there a conceptual reason the two definitions agree in the first case and almost (but not quite) agree in the second? This is, of course, a rather vague question, so here is a slightly more specific version: if the number of cells of $X$ of each dimension is finite, then is there a way to deduce the Wall characteristic of $X$ from the generating series for those numbers, assuming the Wall characteristic exists?

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The first definition must not be homotopy invariant (be dependent on the cell structure) as you have illustrated for $G$ the group of two elements (the Milnor construction of $BG$ gives $\chi(BG)=1/3$ while the standard cell decomposition give $\chi(BG)=1/2$). – Mark Grant Jun 11 '11 at 7:27
Mark -- indeed, it is not homotopy invariant. But still, the fact that it almost manages to give the right answer for $BG$, $G$ a finite group, seems somewhat surprising (to me). – algori Jun 11 '11 at 7:35
Number of d-simplices of BG for reduced bar-construction is $(|G|-1)^d$ so it gives the right answer... – Grigory M Jun 11 '11 at 7:59
do you need k there to be of characteristic zero, or would any field serve? – hugo Jun 11 '11 at 14:13
Topologically (in the usual way of realizing simplicial sets) there is a cell for each nondegenerate simplex, so what Grigori says makes perfect sense to me. – Tom Goodwillie Jun 11 '11 at 18:18

I hope a topologist provides an answer. I'll make some general comments about sums of series.

Something to keep in mind when working with divergent series is that they do diverge. In particular, divergent sums are not associative. (This is similar to the fact that conditionally convergent sums are not commutative, as is standardly explained in freshman calculus.) Perhaps the most famous manifestation of the non-associativity is the "proof" that $0=1$ by writing $0 = \sum 0 = \sum (1-1) = \dots$. But a better example, because it's more illustrative, is the proof that $\frac12 = \frac13$: as you point out in the case of $\mathbb R P^\infty$, one should predict that $1 - 1 + 1 - 1 + \dots = \frac12$, because if you translate the sum by one step and then add the translated series to the original series in columns, all terms cancel except the first one. But if you instead consider $1 - 1 + 0 + 1 - 1 + 0 + \dots$, you see that you need three shifts the sum to $1$. Put another way, the Abel sum of $1 - 1 + 0 + 1 - 1 + 0 + \dots$ is $\sum t^{3n} - \sum t^{3n+1} = \frac1{1-t^3} - \frac t{1-t^3} = \frac1{1+t+t^2} \to \frac13$.

I think you're finding some similar behavior in the case of the Milnor $\mathrm BG$. Of course, the "correct" Euler characteristic is $1/\chi(G)$. What I think needs to happen is that the "new" simplices when you construct $\mathrm B G$ should have the even spacing in the sum, the same as they have in the reduced bar construction. But the degenerate simplices in Milnor's cell complex should be associated in such a way as to cancel out like in $(1-1) + (1-1) + \dots$.

Of course, this is a bit ad hoc. A possibility is that the dimensions of the homology groups do have the correct Abel sum? That would be the answer if you believe me that only the nondegenerate simplices should contribute. But I haven't thought whether this is correct on the nose.

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