# What is the free loop space $\mathcal{L}M$ of a manifold of a manifold $M$ for which the energy functional has no critical points?

What is the free loop space $\mathcal{L}M$ of a manifold $M$ for which $E:LM\to\mathbb{R}$ for $E:\gamma\mapsto\int_{S^1}\|\dot\gamma(t)\|^2dt$ has no non-degenerate critical points? Is it simply the empty set (is this possible)?

Thanks in advance!

• I think $E$ is always critical for trival loops $\gamma(t) = p \in M, \; \forall t$. I think $\gamma$ is always critical if $E(\gamma) = 0$. What do you think about this? Cheers! – Robert Lewis Sep 1 '18 at 19:03
• Yes, the existence critical points at level $0$, i.e. constant geodesics $\gamma(t)=p$, for $p\in M$ for all $t\in S^1$, means that $M=\mathcal{L}M$. @RobertLewis – Sergio Charles Sep 1 '18 at 19:14
• This is a good question, I think, endorsed (+1), but it's pretty broad. How do you want to characterize $\mathcal LM$? In raw terms, the free loop space $\mathcal LM$ is, after all, $\mathcal LM$. What kind of specific characterization are you looking for? – Robert Lewis Sep 1 '18 at 19:15
• The gist of my comment(s) is that $E(\gamma)$ always has some critical points. One for each $p \in M$, at least, taking $\gamma(t) = p$. We might need to look at something like homotopy classes of free loops to get more details. – Robert Lewis Sep 1 '18 at 19:18
• Of course such constant geodesics are degenerate. I define $\mathcal{L}M$ by the space of smooth loops in $M$, $C^{\infty}(S^1,M)$. @RobertLewis – Sergio Charles Sep 1 '18 at 19:19

## 1 Answer

The energy functional on the free loop space $LM$ only has degenerate critical points if $M$ is not zero dimensional. Of course all the constant loops are degenerate critical points (the critical set is a copy of $M$)

Here is the reason why all non-constant closed geodesic are degenerate:

There is a non-trivial action of the circle $S^1$ on the free loop space that leaves the energy functional invariant. If $\gamma:S^1\rightarrow M$ is a loop we define a new loop $s\cdot \gamma:S^1\rightarrow M$ by $c\cdot \gamma(t)=\gamma(t+s)$. If $\gamma$ is a non-constant loop then $c\cdot \gamma$ is also a non-constant loop (and different for most values of $c$!).

The functional $E$ is invariant under this action. Thus for any closed geodesic $\gamma$ for almost all $c$ the loop $c\cdot \gamma$ is also a closed geodesic (this is just a reparametrization). This shows that there cannot be isolated critical points. The best one can hope for is that the critical points come in $S^1$ families. I think it is true that this is true for a generic set of metrics, although I cannot give a reference from the top of my head.

You might wonder if there are manifolds without non-trivial closed geodesics. There is a Theorem of Lyusternik and Fet that this can only occur for non-compact manifolds (I take my manifolds without boundary here). Euclidean space is a simple example of a manifold without closed geodesics.