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For any space $X$ there is a fibration $$ \Omega X\to LX\stackrel{ev}{\to} X $$ where $LX=Map(S^1,X)$ is the free loop space, $\Omega X = Map_*(S^1,X)$ is the based loop space, and $ev:LX\to X$ is the evaluation map sending $\omega$ to $\omega(1)$. There is a canonical section $\sigma_0:X\to LX$ sending $x\in X$ to the constant loop at $x$.

I'm interested in the classification of sections of this fibration up to fiberwise homotopy, which seems to be a difficult problem in general. In particular, the standard obstruction theory methods (such as Corollary VI.6.16 on p.302 of Whitehead's "Elements of homotopy theory") don't seem to apply.

So I'm asking about what I believe to be the simplest example where this fibration does not split, namely $X=S^2$.

Have the sections of $ev: LS^2\to S^2$ been classified up to fiberwise homotopy?

The only reference I could find on this problem was

Samson Saneblidze, MR 1255935 On the homotopy classification of sections in the free loop fibration, J. Pure Appl. Algebra 91 (1994), no. 1-3, 317--327.

from which I can probably deduce an answer for the rationalization $S^2_\mathbb{Q}$.

In case the answer turns out to be "no", let me ask an easier question. There is an action $S^1\times S^2\to S^2$ given by rotation around an axis. The adjoint of this action gives a section $\sigma: S^2\to LS^2$.

Is the section $\sigma$ fiberwise homotopic to the canonical section $\sigma_0$? If not, how do I see this?

Added later: I think I see a way to distinguish between $\sigma$ and $\sigma_0$. Denote the adjoints of these sections by $\sigma',\sigma_0':S^1\times S^2\to S^2$. Let $S^1\subseteq S^2$ be a diameter which passes through the two fixed points of the rotation. Then $\sigma'$ and $\sigma_0'$ when restricted to $S^1\times S^1\subset S^1\times S^2$ have different degrees (1 and 0), hence they are not homotopic.

I wonder if this can be formalized somehow.

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  • $\begingroup$ Why can't one use obstruction theory? And in the $S^2$-example, can one use the adjoint map and ask whether it extends to $D^2\times S^2$ (and use obstruction theory here)? $\endgroup$
    – user95545
    Aug 25 '16 at 11:21
  • $\begingroup$ @user95545: Obstruction theory can surely be used, but I don't see how to use it directly. If the fiber $\Omega S^2$ were 1-connected, then sections would be classified by $H^2(S^2;\pi_2(\Omega S^2))$. Alas, it's only 0-connected. I'll have to think about your idea for the specific example. $\endgroup$
    – Mark Grant
    Aug 25 '16 at 12:51
  • $\begingroup$ @TomGoodwillie: Thanks for your comment. I have a few questions. For the conclusion of your second sentence, how do you know the connecting morphism $\pi_1(\Omega S^2)\to \pi_0(F)$ is trivial? And in your third sentence, which bundle is it you're pulling back to $D^2$? $\endgroup$
    – Mark Grant
    Aug 25 '16 at 20:14
  • $\begingroup$ Sorry. That comment was largely nonsense. I'll try again. $\endgroup$ Aug 26 '16 at 0:02
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The space of sections of $\Lambda S^2\to S^2$ is a subspace of the space of all maps $$S^2\to \Lambda S^2=Map(S^1,S^2).$$ When viewed as a subspace of the space of all maps $$S^1\to Map(S^2,S^2),$$ it becomes $\Omega_1 Map(S^2,S^2)$, the based loopspace of the space of self-maps of $S^2$, with the identity map as basepoint. Thus the problem is to compute $\pi_1(Map(S^2,S^2),1)$. I claim that it has order two.

The space of all degree one maps $S^2\to S^2$ is sometimes denoted by $SG(3)$; it is the "structure group" for oriented spherical fibrations of rank $3$. It is part of a fibration sequence $$ SF(2)\to SG(3)\to S^2 $$ with $SF(2)$ the space of degree one based maps $S^2\to S^2$, a component of $\Omega^2S^2$. Mapping into this is another fibration sequence $$ SO(2)\to SO(3)\to S^2. $$ Thus $SG(3)/SO(3)$ is homotopy equivalent to $SF(2)/SO(2)$.

The latter is $1$-connected and has finite $\pi_2$. This follows as soon as one knows that $\pi_1SO(2)\to \pi_1SF(2)=\pi_3S^2$ is an isomorphism, which I think is not hard to see.

Thus $\pi_1SG(3)$ is the cokernel of the map $\pi_2(SG(3)/SO(3))\to \pi_1SO(3)=\mathbb Z/2$.

This map is zero because it factors through a map $\pi_2(SG(3)/SO(3))\to \pi_1SO(2)$ from a finite group to $\mathbb Z$.

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Two sections are fiber wise homotopic if the adjoint maps $S^1\times S^2\to S^2$ are homotopic rel base point$\times S^2$. Cut $S^1$ open at the base point, then you obtain maps $I\times S^2\to S^2$ which are the identity of $S^2$ at each endpoint of the interval $I$. For a homotopy between two such maps $f_1,f_2$ rel endpoints $\times S^2$, we have to extend a map $\partial (I^2)\times S^2\to S^2$ to $I^2\times S^2$.

So we need to extend over a 2-cell $I^2\times pt$, where there is no obstruction, but a choice in $\pi_2S^2=\mathbb Z$, and over the top-dimensional cell where the obstruction is an element in $\pi_3 S^2=\mathbb Z$.

I guess one would now need to understand how this element $o(f_1,f_2,g_{12})$ depends on the choice of extension $g_{12}$ over the 2-cell.

If one has overcome this difficulty, one has an addition formula $$o(f_1,f_2,g_{12})+o(f_2,f_3,g_{23})=o(f_1,f_3,g_{12}\cup g_{13}),$$ see for example Baues, LNM 628, Thm 4.2.7. This could lead to a classification.

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Some comments.

1) The space of sections of $LX \to X$ is a grouplike topological monoid, by pointwise multiplication. So, its path components do form a group. In fact it is a loop space $\Omega_1F(X,X)$ of the space of functions $F(X,X)$, where loops are based at the identity map $1\!\!\! : X \to X$.

2) For $A\to X$ a map, let $\Gamma(LX|A)$ denote the space of sections of $LX \to X$ along $A$. This is the same as the space of sections of the pullback $A \times_X LX \to A$. This is also a loop space, this time of $F(A,X)$.

When $A = X$ we set $\Gamma(LX) = \Gamma(LX|A)$.

3) The restriction map $$ F(X,X) \to F(A,X) $$ is a fibration and it loops to the restriction map $\Gamma(LX) \to \Gamma(LX|A)$ . Its fiber at the constant map $A\to X$ is the based function space $F_*(X/A,X)$.

4) Let's apply the above to the inclusion $\ast \subset S^2$ basepoint. This gives a homotopy fiber sequence $$ \Omega^2 S^2 \to F(S^2,S^2) \to S^2\, , $$ where the second map is given by evaluation.

5) Take $\pi_*$ (with respect to the correct basepoint "1") to get an exact sequence of groups $$ \cdots \to \pi_2(S^2) \to \pi_1(\Omega^2S^2;1) \to \pi_1(F(S^2,S^2);1) \to 1 $$ where $\pi_1(F(S^2,S^2);1) = \pi_0(\Gamma(LS^2))$, and $\pi_2(S^2) = \Bbb Z = \pi_1(\Omega^2S^2;1)$. It follows that there's a short exact sequence of groups $$ \Bbb Z \to \Bbb Z \to \pi_0(\Gamma(LS^2)) \to 1\, . $$ It therefore suffices to compute the homomorphism $\Bbb Z \to \Bbb Z$. To compute this, you'll need to understand how the boundary operator works. This amounts in the end to the computation discussed by "user95545" above.

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