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Let $G$ be a discrete group and let $BG \simeq K(G,1)$ be its classifying space. Let $H$ be a topological group with classifying space $BH$.

  • In case $H$ is also discrete, it was pointed out in the (comments to one) answer of this question that $hom(G,H)$ is in bijection with $[BG,BH]_0$.

Is there anything along the lines of the above statements that can be said in case $H$ is a nonabelian topological group?

Let $A \to C$ be a (topological) crossed module. This also has a classifying space $B(A \to C)$ - the classifying space of the associated $2$-group.

What is the relationship between the first group cohomology of $G$ with coefficients in $A \to C$, i.e. $H^1(G, A \to C)$ and $[BG, B(A \to C)]_0$.

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Dear @Ulrich Pennig: For the case of $H$ a topological abelian group, do you mean to replace $\operatorname{hom}(G,H)$ with its set of path components? Also, do you have a reference for that result? Thanks. –  Ricardo Andrade Oct 21 '13 at 10:16
    
@RicardoAndrade: I have to check what I really meant in the topological abelian case :-) and deleted the statement for now. Sorry for causing confusion. –  Ulrich Pennig Oct 21 '13 at 11:50
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With regard to the first question, the answer is much more complicated, even for compact Lie groups; see the answers to this question. –  Danny Ruberman Oct 21 '13 at 14:08
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In some sense, the first question is addressed in a number of my papers. For instance, one can get information about the relationship between Hom($\pi_1 M^g$, U(n)) and Map($M^g$, BU(n)), where $M^g$ is a Riemann surface, using Yang-Mills theory. For sufficiently large n, these spaces have the same homotopy groups above dimension zero. General results on the low-dimensional difference between the homotopy groups of Hom(G, U(n)) and Map(BG, BU(n)) appear in my paper with Baird (arXiv:1206.3341), which also addresses the case of general linear representations. –  Dan Ramras Oct 21 '13 at 18:16
    
@DanRamras: Thanks! I will look into it. –  Ulrich Pennig Oct 22 '13 at 5:23

1 Answer 1

Let $\mathbf{H}$ be an $\infty$-topos (in your case, you want to take the $\infty$-topos of spaces, $\mathcal{S}$). There is an equivalence of $\infty$-categories $$ \Omega : \operatorname{Grp}(\mathbf{H}) \to \mathbf{H}^{\ast/}_{\geq 1}, \quad \mathbf{B} : \mathbf{H}^{\ast/}_{\geq 1} \to \operatorname{Grp}(\mathbf{H}) $$ between the $\infty$-categories of $\infty$-group objects in $\mathbf{H}$ (with $\infty$-group homomorphisms between them) and that of pointed, connected objects in $\mathbf{H}$ (where maps should respect the basepoints), mediated by the usual looping and delooping functors —here $\Omega$ is left adjoint to $\mathbf{B}$. This is all in Lurie's Higher Topos Theory, but I recommend the exposition in Nikolaus, Schreiber and Stevenson's Principal $\infty$-bundles - General Theory.

Discrete groups are $0$-truncated objects in $\operatorname{Grp}(\mathcal{S})$, and the equivalence above implies an equivalence of mapping spaces: $$ B : \operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H) \to \operatorname{Map}_{\mathcal{S}^{\ast/}_{\geq 1}}(\mathbf{B}G, \mathbf{B}H) $$ The left hand side here is homotopy equivalent to the set of group homomorphisms from $G$ to $H$, and taking $\pi_0$ gets you the desired bijection: $\hom(G, H) \cong [\mathbf{B}G, \mathbf{B}H]_0$.

If $H$ is not discrete, $\mathbf{B}H$ is no longer an Eilenberg-Maclane space: rather, $\pi_{i+1}(\mathbf{B}H, \ast) = \pi_i (H, e)$.

  • If what you really want is $K(H,1)$, you can give $H$ the discrete topology —call it $H^\delta$. Then you have $\mathbf{B}H^\delta \simeq K(H,1)$, and a bijection $$\hom(G, H) \cong \pi_0\operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H^\delta) \cong [K(G, 1), K(H, 1)]_0$$ Forgetting the topology on $H$ is no big deal because $G$ is discrete.

  • For the classifying space $\mathbf{B}H$ things are more complicated. The most you (or at least I) can say is that there is a bijection $$\pi_0\operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H) \cong [\mathbf{B}G, \mathbf{B}H]_0$$


As for your second question, the first group cohomology group with values in the 2-group associated to the crossed module $A \to C$ is given by the set of homotopy classes of (unbased) maps from $\mathbf{B}G$ to $\mathbf{B}(A \to C)$: $$ H^1(G, A \to C) = [ \mathbf{B}G, \mathbf{B}(A \to C) ] $$ (see the nLab pages on cohomology and group cohomology).

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I think that the statement $\operatorname{hom}(G,H) = [BG,BH]_0$ when $G$, $H$ are discrete groups is significantly simpler than the equivalence of homotopy categories between pointed connected spaces and loop spaces (or group objects in the quasi-category of spaces). –  Ricardo Andrade Oct 21 '13 at 14:48
    
@Ricardo: I was just using the discrete case for illustration purposes. I'm sure there are proofs that don't require that much technology. –  Alberto García-Raboso Oct 21 '13 at 14:59
    
@AlbertoGarcía-Raboso: Are you assuming that $A$ and $C$ are again discrete for the last statement? –  Ulrich Pennig Oct 21 '13 at 16:13
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This is all well, but I was hoping to understand the right hand side, i.e. $[BG,BH]_0$ with the topology on $H$, but maybe the equivalence of mapping spaces is all you get in general. –  Ulrich Pennig Oct 21 '13 at 16:17
    
@Ulrich: in the last statement you do not need to assume that $A$ and $C$ are discrete. –  Alberto García-Raboso Oct 21 '13 at 18:11

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