# “C choose k” where C is topological space

One day I read a generating function in a paper. For any "sufficietly nice topological space", $C$: $$\sum_{l \geq 0 } q^{2l}\chi(\mathrm{Sym}^l[C]) = (1 - q^2)^{-\chi(C)} = \sum_{l \geq 0} \binom{l + \chi(C)-1}{l} q^{2l}$$ First of all, I'm not sure what "sufficiently nice" means here. I'm guessing any CW complex will do. I wonder what's an example of a space where this formula doesn't work.

This formula suggests $$\chi(\mathrm{Sym}^l[C]) = \binom{l -1+ \chi(C)}{l} = \chi \binom{l-1 + C }{l}$$ where the right hand side is some "categorification".
Removing the $\chi$'s, is there some sense in which $$\mathrm{Sym}^l[C] = \binom{l -1+ C }{l}$$ is rigorous?

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There are some papers by Schanuel and, I think, Leinster where ideas like this are explored for a restricted class of C. I don't remember their titles, but you can start here: math.ucr.edu/home/baez/counting – Qiaochu Yuan Dec 29 '10 at 12:40

I presume you are asking is whether one can make sense of $\binom{X}{k}$ as a space in such a way that it relates to the formula you are asking about.

First let $l = 1$. For a space $X$ define $\binom{X}{k}$ to be the configuration space of subsets of X having cardinality $k$. Then $\binom{X}{1} = X$. In this case $\text{Symm}^1(X) = X$ and we have agreement $$\binom{X}{1} = X = \text{Symm}^1(X) .$$

Now consider the case $l=2$, and let {1} be the one element set
Then as sets there is an evident bijection $$\binom{X \amalg \text{{1}} }{2} = \binom{X}{2} \amalg (X\times \text{{1}})$$ As a set, the right side is the same thing as $\text{Symm}^2(X) = X\times_{Z_2} X =$ the orbits of the cyclic group of order two acting on $X\times X$ by permutation. To see this, note that $X\times X$ has two kinds of isotropy: one coming from the diagonal copy of $X$ (with trivial action) and the other being it's complement which is $X\times X - X$ with free action having quotient $\binom{X}{2}$. With respect to this identification, your formula makes sense. However, these spaces have different topology when $X$ isn't discrete.

A similar observation works in the $l >2$ case.

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The papers I was thinking of are actually by Propp:

(although I think these papers are outdated and more is known these days.) One has to restrict to a certain class of particularly nice spaces and modify the definition of Euler characteristic.

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Exactly... but what is this "nice"? – john mangual Dec 29 '10 at 18:08