Let $\Omega^{*}_{\text{poly}}\: : \: sSet\to dg_{\geq 0}Comm_{+}$ be the polynomial De Rahm functor on simplicial sets, where the codomain is the category of commutative differential graded algebras over a field of charachteristic zero. Let $G$ be a group, what is $\Omega^{*}_{\text{poly}}(BG)$? Its cohomology? Do you have some references for that?
1 Answer
Let $\mathbb{F}$ be a field of characteristic zero.
For any simplicial set $Y$ denote by $RY$ its topological realization then you have an isomorphism of graded algebras: $$H^*(\Omega_{poly}^*(Y,\mathbb{F}))\cong H^*_{Sing}(RY,\mathbb{F}).$$ Any book about rational homotopy theory will be a good reference:
A. K. Bousfield and V. K. A. M. Gugenheim, On $\mathrm{PL}$ de Rham theory and rational homotopy type, Mem. Amer. Math. Soc. 8 (1976), no. 179
Griffiths, P.; Morgan, J. (1981), Rational homotopy theory and differential forms, Progress in Mathematics, 16, Birkhäuser
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Edit: let me give you a conceptual proof
1) you have a first adjunction $$\Omega^*_{Poly}:SSets^{op}\leftrightarrows cdga_{\mathbb{F}}:\mathfrak{R}$$ this is a Quillen adjunction we get on homotopy classes: $$[Y,\mathfrak{R}A^*]_{SSets}\cong [A^*,\Omega^*_{Poly}(Y)]_{cdga_{\mathbb{F}}}.$$
2) let $\Lambda_{\mathbb{F}}(x_n)$ be the free graded commutative algebra on a generator $x_n$ of degree $n$ we have: $$[\Lambda_{\mathbb{F}}(x_n),B^*]_{cdga_{\mathbb{F}}}\cong H^n(B^*)$$ putting 1)+2) together we have that $H^n(\Omega^*_{poly}(Y))\cong [Y,\mathfrak{R}\Lambda_{\mathbb{F}}(x_n)]_{SSets}$.
3) We use the fact that $\mathfrak{R}B^*$ is fibrant (it is always a Kan simplicial set) and the Quillen equivalence between SSets and Top (given by the realization and the singular set of a topologica space) to get that $$[Y,\mathfrak{R}B^*]_{Ssets}\cong [RY,R\mathfrak{R}B^*]_{Top}$$ this gives us: $H^n(\Omega_{Poly}^*(Y))\cong [RY,R\mathfrak{R}\Lambda(x_n)].$
4) The last step is the identification of $R\mathfrak{R}\Lambda(x_n)$ together with the Eilenberg-MacLane space $K(\mathbb{Q},n)$.
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1$\begingroup$ Thanks, it seems to me that if the simplicial set is the nerve of some group, then you get the group cohomology with coefficients in the field of charachteristic zero. I'm wrong? $\endgroup$– CepuOct 10, 2015 at 9:10
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$\begingroup$ Exactly, you are right, you get group cohomology. $\endgroup$– David COct 10, 2015 at 9:11