First let $G$ be a topological group and $BG$ its classifying space. Let $\mathcal{L}BG=\text{Map}(S^1, BG)$ be the free loop space of $BG$.

We can see that $\mathcal{L}BG$ has the homotopy type of $G/^{\text{ad}}G$, the quotient of $G$ by the adjoint action of $G$ itself. One way to see it is to identify the homotopy classes $[S^1,BG]$ with the pullback principal $G$-bundle on $S^1$, while the later is 1-1 correspondence to the transition function at the origin.

Now let's try to achieve the isomorphism using the simplicial model of $S^1$ and $BG$. The simplicial model of $S^1$ is given by (see Loday's Free loop space and homology) $$ S^1_0=\{*\}\\ S^1_1=\{s_0(*),\tau\} $$ where $d_0\tau=*$ and $d_1\tau=*$, and for $n>1$ $$ S^1_n=\{s_0^n(*),s_{n-1}\ldots\widehat{s_{i-1}}\ldots s_0\tau,i=1,\ldots,n\}. $$

On the other hand, there is a standard simplicial model for $BG$ (see Weibel's An introduction to homologicial algebra, page 257 Example 8.1.7): $$ BG_n=\underbrace{G\times \ldots\times G}_n $$ and $$ s_i(g_1,\ldots,g_n)=(g_1,\ldots,e,\ldots,g_n) $$ $$ d_i(g_1,\ldots,g_n) = \left\{ \begin{array}{ll} (g_2,\ldots,g_n) & \textrm{if $i=0$}\\ (g_1,\ldots,g_ig_{i+1},\ldots,g_n) & \textrm{if $0< i<n$}\\ (g_1,\ldots,g_{n-1})& \textrm{if $i=n$} \end{array} \right. $$ In these models, both $S^1$ and $BG$ have a base point.

Given this model, it is clear that the simplicial maps (i.e. the maps which commutes with the face and degeneracy maps) from $S^1_{\bullet}$ to $BG_{\bullet}$ is uniquely determined by the image of $\tau\in S^1_1$ in $BG_1=G$. Hence we get the famous isomorphism $$ \pi_1(BG)\cong G, $$ given that $ \pi_1(BG)=[(S^1,*),(BG,*)]$.

$\textbf{My question}$ is: how can we get rid of the fixed base point and get the isomorphism for free loop space $\mathcal{L}BG \simeq G/^{\text{ad}}G$ in the frame work of simplicial set?

  • 2
    $\begingroup$ I think you've only identified the zero-cells of the mapping simplicial set. $\endgroup$ – S. Carnahan May 30 '14 at 0:50
  • $\begingroup$ You should be getting $\pi_1(BG) \cong \pi_0(G)$, not $G$. $\endgroup$ – Qiaochu Yuan May 30 '14 at 1:04
  • $\begingroup$ @S.Carnahan Yes you are right! After considering the mapping simplicial set I think I solved this problem. Thank you very much! $\endgroup$ – Zhaoting Wei May 30 '14 at 10:28
  • $\begingroup$ @QiaochuYuan Sure, and, as pointed out by S. Carnahan, the map I wrote in the question is just the $0$-cells of the mapping simplicial set. After considering the whole mapping simplicial set we can get the nerve of the action groupoid of the adjoint action of $G$ on itself. $\endgroup$ – Zhaoting Wei May 30 '14 at 10:31

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