7
$\begingroup$

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 $[BG, B(A \to C)]_0$.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 10:16
  • $\begingroup$ @RicardoAndrade: I have to check what I really meant in the topological abelian case :-) and deleted the statement for now. Sorry for causing confusion. $\endgroup$ Oct 21, 2013 at 11:50
  • 2
    $\begingroup$ With regard to the first question, the answer is much more complicated, even for compact Lie groups; see the answers to this question. $\endgroup$ Oct 21, 2013 at 14:08
  • 2
    $\begingroup$ 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. $\endgroup$
    – Dan Ramras
    Oct 21, 2013 at 18:16
  • $\begingroup$ @DanRamras: Thanks! I will look into it. $\endgroup$ Oct 22, 2013 at 5:23

2 Answers 2

5
$\begingroup$

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).

$\endgroup$
5
  • 3
    $\begingroup$ 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). $\endgroup$ Oct 21, 2013 at 14:48
  • 1
    $\begingroup$ @Ricardo: I was just using the discrete case for illustration purposes. I'm sure there are proofs that don't require that much technology. $\endgroup$ Oct 21, 2013 at 14:59
  • $\begingroup$ @AlbertoGarcía-Raboso: Are you assuming that $A$ and $C$ are again discrete for the last statement? $\endgroup$ Oct 21, 2013 at 16:13
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 16:17
  • $\begingroup$ @Ulrich: in the last statement you do not need to assume that $A$ and $C$ are discrete. $\endgroup$ Oct 21, 2013 at 18:11
0
$\begingroup$

It may help to look at some of the papers of Ronnie Brown for the last part, and I would also mention two papers by Ronnie Brown, Marek, Golasinski, myself, and Andy Tonks,Spaces of maps into classifying spaces for equivariant crossed complexes, that interprets a similar problem in the case of $G$-equivariant maps where $G$ is either discrete or in the second paper is a general topological group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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