Let $X$ be a based connected space. The space of based continuous morphisms $Top_{\ast}(S^1,X)$ is the space of loops $\Omega X$. Since $S^1$ is homotopy equivalent to the Eilenberg-Mac Lane Space $K(\mathbf{Z},1)$ I was wandering if there is any "geometric interpretation" of the space of based continuous morphisms $Top_{\ast}(\mathbf{C}P^{\infty},X)$. $\mathbf{C}P^{\infty}$ can be identified with the Eilenberg-Mac Lane Space $K(\mathbf{Z},2)$ More generally, what would be the "geometric interpretation" of the space of based continuous morphisms $Top_{\ast}(K(\mathbf{Z},n),X)$.
4
-
3$\begingroup$ $S^1$ has two interesting kinds of extra structure as an object of the homotopy category of based spaces: first, it's a cogroup object, but second, it's a group object. The first kind of extra structure is the one related to loop spaces, while the second kind of extra structure is the one related to its description as $B \mathbb{Z}$. The Eilenberg-MacLane spaces $B^n \mathbb{Z}, n \ge 2$, on the other hand, are only group objects, and have no cogroup structure (since their cohomology has interesting cup products). $\endgroup$– Qiaochu YuanCommented Jul 24, 2015 at 17:59
-
1$\begingroup$ There is an homotopy theoretic interpretation (I don't know if I can call it geometric) for maps from $CP^{\infty} \to X$ when $X = BY$ for an $A_{\infty}$ H-space $Y$. They are precisely the maps from $S^1 \to Y$ which are $A_{\infty}$ maps. (I apologies for recklessly putting something wrong as answer and convey my thanks to Dylan Wilson for pointing it out.) $\endgroup$– PrasitCommented Jul 26, 2015 at 19:06
-
$\begingroup$ @Prasit Thanks, I think your argument is nice. I think you need to add that Y is group-like. I think I see how to think about the interpretation of $Top_{\ast}(K(\mathbf{Z},n),X) $ using $E_{n}$ operads. $\endgroup$– MaxCommented Jul 27, 2015 at 17:01
-
$\begingroup$ I know Dylan pointed out group-like, but I am not sure if we actually need the condition 'group-like'. $\endgroup$– PrasitCommented Jul 27, 2015 at 22:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
[Edited after the comment below] Well, I don't know exactly what you mean by "geometric interpretation", but for topological group $G$ the unpointed mapping spaces $Map(BG,X)$ are examples of homotopy fixed points. and the pointed mapping spaces $Map_*(BG,X)$ fits in the fibration $$Map_*(BG,X) \rightarrow Map(BG,X) \rightarrow X$$ where the last map is the evaluation map.
Now, $$Map(BG,X)\cong Map^G(EG,X)$$ where we provide $X$ with trivial action, and this latter is nothing but the homotopy fixed points of $X$ with trivial $G$-action.
If $G$-action on $X$ is trivial, $X$ is just the fixed point set, so one can consider $Map_*(BG,X)$ as the space that measures the difference between the genuine fixed point set and the homotopy fixed point set.
As one can construct a model for $K(A,n-1)$ that has a structure of a topological group, you can take $G=K(A,n-1)$ in the above.
-
$\begingroup$ Are you sure about your definition of homotopy fixed points. Your are using the pointed maps. $\endgroup$– MaxCommented Jul 27, 2015 at 16:59
-
$\begingroup$ @Max you are right, I corrected the answer. $\endgroup$ Commented Jul 27, 2015 at 17:23