It is well-known that one can detect based loopspaces using the machinery of operads. Namely, given a group-like space $X$ with an action of $\mathbb{E}_n$-operad, then it is homotopy equivalent as an $\mathbb{E}_n$-space to $\Omega^n Y$ for some space $Y$.

Is anyone aware of a similar recognition principle for free loopspaces? Namely, is there some operad-like structure that if a space $X$ admits an action of this structure, then it is an $n$-fold free loopspace $\mathcal{L}^n Y$ for a space $Y$, which up to a weak equivalence is recovered from this action?

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    $\begingroup$ You should check this arxiv.org/pdf/1303.6989v2.pdf $\endgroup$
    – Ilias A.
    Feb 22, 2015 at 19:49
  • $\begingroup$ @Amrani Thanks for the reference. I was hoping for a more ``manageable'' model. Is there a less abstract model $\Theta_{S^1_+}$-space? @Sanath The use of operads is not strictly necessary, however, I am looking for something in the same spirit. $\endgroup$ Feb 22, 2015 at 21:01
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    $\begingroup$ Great question. Some thoughts: (1) the free loop space of $X$ carries a $\mathrm{Diff}(S^1) \simeq S^1 \rtimes \mathbb Z/2$ action by rotating (and reflecting) the circle, and (2) the target, I think, can be reconstructed as the homotopy fixed points of this action. If I were to drop the orientation reversal, you would be asking: given a homotopy $S^1$-space $X$, when is $X = \mathrm{maps}(S^1,\mathrm{maps}_{S^1}(*,X))$? $\endgroup$ Feb 23, 2015 at 0:45
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    $\begingroup$ Being a bit cavalier with hom-tensor adjunctions, I guess this is asking: give $G = S^1$ the trivial left $G$ action; then when is $\mathrm{maps}_G(G,X) \simeq X$ as $G$-spaces, where the $G$-action on $\mathrm{maps}_G(G,X)$ is via the usual (right) $G$-action on $G$. I am reminded of the following characterization: a monoid $M$ is a group iff the diagonal $M$ action on $M \times M$ is isomorphic (via some probably-non-identity map on $M \times M$) to the action which acts just on the left factor and trivially on the right. $\endgroup$ Feb 23, 2015 at 0:49
  • $\begingroup$ @Nerses: what motivates you to recognise free loop spaces? Why not something in a different direction, like suspension recognition? $\endgroup$ Feb 24, 2015 at 19:34

1 Answer 1


Edit, 2/25/15: Well, after discussing it with Nerses in the comments, it seems I was wrong; taking homotopy fixed points does work! (If you're already guaranteed that the space is a free loop space, that is.) The argument is at the end. Most of my original answer is below.

First, there is a general machine you can try to run in situations like this to see what structure is available: in any category with finite coproducts, the functor $\text{Hom}(c, -)$ naturally acquires the structure of a model of the Lawvere theory whose $n$-ary operations are given by maps $c \to nc$, where

$$nc = \bigsqcup_{i=1}^n c$$

denotes the coproduct of $n$ copies of $c$. This is because $\text{Hom}(nc, -) \cong \text{Hom}(c, -)^n$, and by the Yoneda lemma, natural transformations $\text{Hom}(c, -)^n \to \text{Hom}(c, -)$ are naturally in bijection with homomorphisms $c \to nc$.

If you run this machine in the homotopy category of pointed spaces with $S^1$ the circle, you get that $\text{Hom}(S^1, -)$ naturally acquires the structure of a model of a Lawvere theory which turns out to be the Lawvere theory of groups, as expected. Similarly, for $S^n, n \ge 2$ you get the Lawvere theory of abelian groups. (Details here; other examples here and here.)

Unfortunately, any version of this machine in unpointed spaces suffers from the same problem: in unpointed spaces, the coproduct is the disjoint union, so e.g. morphisms from a circle $S^1$ into a disjoint union $n S^1$ of $n$ circles factor through the inclusion of one of the connected components. This means that the resulting Lawvere theory has no interesting $n$-ary operations, and that everything is determined by the unary operations, or by morphisms $S^1 \to S^1$.

And this is just not a lot of data. Consider the case that $Y = BG$ is the classifying space of a discrete group $G$. Then $LY = LBG$ is the classifying space of the groupoid $G/G$, the adjoint quotient of $G$, which has objects the elements of the group and morphisms given by conjugation. This is equivalent to the disjoint union of the classifying spaces $B C_G(g)$ of the centralizers of a representative of each conjugacy class of $G$. A map $S^1 \to S^1$ of degree $k$ sends the conjugacy class of $g \in G$ to the conjugacy class of $g^k$, but otherwise the different conjugacy classes don't interact, so I don't see any hope for gluing them back together via a procedure like taking homotopy fixed points.

Edit, 2/25/15: Now for the argument. I was being silly. It's very straightforward:

Theorem: Let $S$ be a topological group and let $Y$ be a space. Then the space of $S$-homotopy fixed points for the natural action of $S$ on the mapping space $[S, Y]$ is weakly equivalent to $Y$.

Sketch. By the universal property, $S$-homotopy fixed points for the natural action of $S$ on $[S, Y]$ is weakly equivalent to the mapping space $[S/S, Y]$, where $S/S$ now denotes the homotopy quotient of the natural action of $S$ on itself by translation (not conjugation). But $S/S$ is a point. $\Box$

But because I was so confused about the case $S = S^1, Y = BG$ I want to go through it in detail. First let's describe explicitly what the action of $S^1$ on

$$LBG = B(G/G) = \bigsqcup_{[g]} B C_G(g)$$

is, where the union runs over all conjugacy classes $[g]$ of $G$. One of my confusions is that I forgot that taking $g$ to be the identity gave a copy of $BG$ sitting inside $LBG$ to aim for; what happens when we take homotopy fixed points is that all of the other connected components just disappear, rather than becoming spliced back together.

Giving an action of $S^1$ on each component amounts to writing down a map $S^1 \to \text{Aut}(B C_G(g))$ for each $[g]$, hence (after taking $\pi_1$) writing down a map $\mathbb{Z} \to Z(C_G(g))$. Of course there is only one natural candidate for such a map, which is the map sending a generator $1 \in \mathbb{Z}$ to $g \in Z(C_G(g))$. The action of $S^1$ on $B C_G(g)$ obtained in this way is encoded in a fiber sequence

$$B C_G(g) \to E \to B S^1$$

where $E$ is the homotopy quotient of $B C_G(g)$ by this action, and the homotopy fixed points of the action are homotopy sections of the map $E \to BS^1$. The long exact sequence in homotopy applied to this fiber sequence is

$$1 \to \pi_2(E) \to \pi_2(BS^1) \cong \mathbb{Z} \xrightarrow{g} C_G(g) \to \pi_1(E) \to 1$$

and we conclude that $\pi_2(E) \cong |g| \mathbb{Z}$ if $g$ has finite order $|g|$ in $C_G(g)$ and that $\pi_2(E) \cong 0$ if $g$ has infinite order. It follows that the map $\pi_2(E) \to \pi_2(BS^1)$ has no homotopy sections unless $g$ is the identity, so in fact only the conjugacy class of the identity contributes to homotopy fixed points. In this case the action of $S^1$ is trivial, $E = BG \times BS^1$, and hence the space of homotopy sections is the space of maps $BS^1 \to BG$. But since $\pi_1(BS^1)$ vanishes this is just $BG$ itself.

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    $\begingroup$ The paper linked to by Amrani Ilias works in pointed spaces and allows a bit more structure than the approach above, using not just coproducts of copies of $c$ but also suspensions (in a setting where that makes sense, e.g. an $\infty$-category). But the fact that it's working in pointed spaces is crucial; it's using the based suspension. $\endgroup$ Feb 23, 2015 at 5:14
  • $\begingroup$ I haven't looked at the paper very carefully, but does the space have to be connected? Can't we just take the maps from $S^1_+$? $\endgroup$ Feb 23, 2015 at 5:31
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    $\begingroup$ @Nerses: that doesn't fix anything. The copy of $S^1$ is disconnected from the basepoint so the wedge sum is just the disjoint union with a basepoint added. You still don't get any interesting $n$-ary operations. $\endgroup$ Feb 23, 2015 at 5:55
  • $\begingroup$ Regarding the comment by Theo. Isn't it true that if we take $Y$ an $S^1$-space, then $Y^{hS^1}$ is the homotopy limit of the cosimplicial space $\mathcal{L}^\bullet Y$, where we use the $S^1$-action to twist the coface maps (in the discrete case this is the complex that computes the group cohomology)? If it is true then in the case, when $Y=\mathcal{L}X$, we get that $(\mathcal{L}X)^{hS^1}$ is equivalent to $\mathrm{Map}(\mathrm{hocolim}({(S^1)^{(\times \bullet+1)}}),X)$, where $\mathrm{hocolim}({(S^1)^{(\times \bullet+1)}})$ is just a point. $\endgroup$ Feb 25, 2015 at 6:54
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    $\begingroup$ @Nerses: it seems I spoke too soon. I think you're right and I'm wrong! My apologies. I'll edit in details. I failed to recognize that homotopy fixed points on some connected component could be empty. $\endgroup$ Feb 25, 2015 at 8:09

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