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Let $X$ be a based space. Then the Moore loop space $MX$ is defined to be the topological monoid whose points are based loops $[0,a] \to X$ where $a \ge 0$ is allowed to vary. Composition is gotten by concatenating loops.

Since $MX$ is a topological monoid, we can form the bar construction $BMX$. This is geometric realization of the simplicial space $$ [n] \mapsto (MX)^{\times n} $$ where the face and degeneracy maps are defined using: projection, multiplication, and insertion of the identity element.

If $X$ is a connected based CW complex, then there is a homotopy equivalence $$ BMX \simeq X . $$ A standard way to prove this is to construct a quasi-fibration $EMX \to BMX$ whose fiber at the basepoint is identified with $MX$ up to homotopy equivalence, in which $EMX$ is contractible. The quasi-fibration is functorially associated with $MX$. However, it seems to me that the equivalence $BMX \simeq X$ gotten in this fashion depends on a contractible space of choices. In particular it doesn't seem to be natural.

Question:

Is there a zig-zag of natural transformations $$ BMX = f_0(X) \leftarrow f_1(X) \to f_2(X) \leftarrow \cdots \to f_n(X) = X $$ which yields a chain of equivalences when $X$ is a connected CW complex?

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I wrote down two quick and simple solutions in Lemmas 14.3 and 15.4 on pages 84 and 90 of "Classifying spaces and fibrations ([15] on my web page). The first is the evident zigzag of natural weak equivalences $$ X \leftarrow B(PX,MX,\ast) \rightarrow B(\ast,MX,\ast) = BMX, $$ where $PX$ is the Moore path space. The left arrow is obtained from an obvious map of simplicial spaces to the constant simplicial space at $X$: Just compose loops and paths and evaluate at the end point. The second is induced from $PX\to \ast$.

The second starts with a very slight modification of what Charles wrote down. For a point $u=(t_0,\cdots, t_p)\in \Delta^p$ with $0\leq t_i\leq 1$ and $\sum t_i = 1$, let $u_{i} = t_0 + \cdots + t_{i-1}$. For $\gamma_i\in MX$ of length $a_i$, $1\leq i\leq p$, define $$ \xi |(\gamma_1,\cdots,\gamma_p),u| = (\gamma_1\cdots\gamma_p)(\sum_{1\leq i\leq p} u_i a_i). $$ This modification makes the compatibility check with the simplicial identities a bit more trivially trivial. To see that $\xi$ is a weak equivalence, take its ordinary loops. There is a natural inclusion $\Omega X \to MX$, which is a weak equivalence; as for any grouplike monoid, there is a natural weak equivalence $MX \to \Omega BMX$; and we have $\Omega \xi\colon \Omega BMX\to \Omega X$. The composite of these three maps is the identity map $\Omega X\to \Omega X$, hence $\Omega \xi$ is a weak equivalence. If $X$ is connected, it follows that $\xi$ is a weak equivalence.

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  • $\begingroup$ @Peter: I very much like both Charles' answer and your answer. But as you already documented the proof in [15], I am going to mark your answer as the accepted one. $\endgroup$ – John Klein Mar 20 '13 at 2:27
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Given $n$ Moore loops $(\gamma_1,a_1),\dots,(\gamma_n,a_n)$ on $X$, and a point $0\leq t_1\leq \cdots \leq t_n\leq 1$ in the standard $n$-simplex, I can obtain a point $$ (\gamma_1*\cdots *\gamma_n)(a_1t_1+\cdots +a_nt_n) $$
in $X$, by evaluating the composite Moore loop. Note that $0\leq a_1t_1+\cdots+a_nt_n\leq \sum a_i$. I'd better also poiint out that $\beta*\alpha$ means "first do $\alpha$, then do $\beta$".

I'd like to say that this gives a natural transformation $BM(X)\to X$, in which case I need to check that it is compatible with face and degeneracy maps.

Degeneracy maps correspond (in the simplicial space $MX^n$) to inserting $(\mathrm{const},0)$ into the sequence of loops, and (on the standard simplices) to omitting the corresponding $t_i$, so this looks good.

Face maps correspond (in the simplicial space) to composing adjacent $(\gamma_i,a_i)$ and $(\gamma_{i+1},a_{i+1})$, and (on the standard simplices) to inserting $t_{i+1}=t_i$ in the sequence of $t$s. Note that $$ (\gamma_i,a_i)*(\gamma_{i+1},a_{i+1})=(\gamma_i*\gamma_{i+1},a_i+a_{i+1}). $$

I feel unequal at the moment to writing down the identity the face map relation implies in general; when $n=2$, it's basically $$ (\gamma_1*\gamma_2)(a_1t+a_2t)=(\gamma_1*\gamma_2)((a_1+a_2)t)) $$ which is a certainly a tautology. So I'm guessing my formula does give a natural transformation, which superficially looks like it should give the expected weak equivalence.

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  • $\begingroup$ @Charles: I am very happy with answer, and I feel ambivalent about not marking your answer as the accepted one (see my comment under Peter's answer). $\endgroup$ – John Klein Mar 20 '13 at 2:27
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I think the following will do, but it is not pretty.

Write an element of $\Delta^n$ as a tuple $0 \leq x_1 \leq \cdots \leq x_n \leq 1$. Identifying $[0,1]$ with $[-\infty, \infty]$, we may as well consider an element of $\Delta^n$ as a tuple $-\infty \leq x_1 \leq \cdots \leq x_n \leq \infty$.

Write a point $x$ in $\Delta^n \times (MX)^n$ as a tuple $(x_1, \ldots, x_n; (a_1, \gamma_1), \ldots, (a_n, \gamma_n))$.

We produce a map $f_x : \mathbb{R} \to X$ as follows. We send the interval $(-\infty, x_1]$ to the basepoint. On the interval $[x_1, x_1+a_1]$ we apply $\gamma_1$. Then we send $[x_1+a_1, x_2+a_1]$ to the basepoint. On the interval $[x_2+a_1, x_2+a_1+a_2]$ we apply $\gamma_2$. ``And so on".

Adopts the convention that if $x_1=-\infty$ then the interval $[x_1, x_1+a_1]$ is the singleton $\{-\infty\}$ and does not contribute to $f_x$. Similarly at the other end.

Finally, one checks that the association $x \mapsto f_x$ respects the simplicial identities, so yields a continuous map $BMX \to \mathcal{C}(\mathbb{R}, X)$, which we can compose with the map which evaluates at the origin.

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