How to show the space of closed curve is Hilbert manifold?

Question

In the picture below ,$(M,g)$ is a Riemannian manifold.

Why $\mathcal L_M$ is a Hilbert submanifold of $L^{1,2}(S^1,R^r)$ ?

Besides, what is the inner and name of $L^{1,2}(S^1,R^r)$ ?

The picture below is from the 3 page of Kwangho Choi and Thomas H. Parker's Convergence of the heat flow for closed geodesics.

Answer 1

The fact that you get a Hilbert manifold is not trivial and involves some technical work but the basic intuition is that a chart around a smooth map $u \colon S^1 \rightarrow M$ should be modeled on an open neighborhood of the zero vector field inside the Hilbert space $\Gamma^{1,2}(u^{*}(TM))$ of vector fields of regularity $W^{1,2}$ along $u$. The chart map will then be given by

$$ X \mapsto \{ \theta \mapsto \exp_{u(\theta)}(X(\theta))\} $$

where $\exp$ is the exponential map induced by the Riemannian metric $g$. The size of the open set on which the chart is defined is determined by a lower bound on the injectivity radius on $u(S^1)$ (which is compact) in order to guarantee that the map is one-to-one.

There are various technical issues one must resolve one way or another:

1. One should check that the charts are indeed homeomorphisms onto their image. This involves understanding the continuity of solutions of an ODE (here, the one determining the geodesics) with respect to a small change of coefficients with respect to the $W^{1,2}$ metric.
2. One should check that the resulting transition maps (between open subsets of Hilbert spaces) are smooth.
3. One should understand what happens when $u$ is not necessarily smooth. This involves arguing that the charts above cover also all the non-smooth curves or generalizing your framework for the above to make sense even if $u$ is not necessary smooth.

For more details, you should consult any book or article in "global analysis" that discusses rigorously the construction of a smooth manifold structure on a space of maps between manifolds.