2
$\begingroup$

Fix a closed connected manifold $Q$ and let $LQ$ denote its free loop space. We can get second homology classes on $Q$ by "doing things" to loops in $Q$.

For instance, if we have a loop of loops, it gives rise to a torus in $Q$, whose fundamental class lies in $H_2(Q)$.

Another way of getting a torus is to start with a constant loop, deform it into another contractible loop, then deform it some more until two of its points intersect, at which point we can cut the loop into two loops, then deform them separately, make them meet later at a point, concatenate them, get a contractible loop, and then contract it to a point. We again get a torus in $Q$, which again produces a class in $H_2(Q)$.

Generalizing this, if we have a map $\Sigma \to Q$, where $\Sigma$ is say, a closed orientable surface, we can cut it up into pieces looking like disks and pairs of pants. The disk parts come from contractible loops, while pairs of pants give a homotopy between one boundary component and the concatenation of the other two components, if we deform them so that they meet at one point. The class $[\Sigma]$ again gives rise to a class in $H_2(Q)$.

In this way, performing deformation and concatenation, we can get arbitrary classes in $H_2(Q)$ (say with integer coefficients).

Note that in general loops from all the connected components of the loop space of $Q$ are needed in order to produce all possible homology classes.

My question is the following: suppose you're only allowed to use a subset $S \subset \pi_0(LQ)$ which contains the contractible loops and is closed under concatenation of loops intersecting at a point, for example, a subset which is the image by the natural map $\pi_1(Q) \to \pi_0(LQ)$ of a normal subgroup $G \triangleleft \pi_1(Q)$. What part of $H_2(Q)$ will you be able to get using the above procedure?

One example: if you're only allowed to use contractible loops, it's easy to see that you will only get spherical classes in $H_2(Q)$.

$\endgroup$
  • 1
    $\begingroup$ Is should be equivalent to ask: What elements in $H_2(Q)$ can be represented by maps $f:M\to Q$ from a surface such that $Im(\pi_0(Lf))\subset S$. A simpler version might replace $\pi_0(LQ)$ by $H_1(Q)$. As degree $1$-maps are surjective in homology, you can only represent $[Q]\in H_2(Q)$ by a map $f: M\to Q$ if $f$ is surjective on $H_1$. It should be possible to solve the $H_1$-question for all surfaces completely. $\endgroup$ – Lennart Meier Nov 28 '14 at 18:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.