Is the group of diffeomorphisms a Lie Group?

Solution 1:

This is shamelessly ripped from "Milnor, Remarks on infinite dimensional Lie Groups. In: Relativity, Groups and Topology II."

Endow $M$ with an arbitrary Riemannian metric (always possible via the wizardy of partitions of unity) and let $V = \operatorname{Vect}(M)$ be the space of smooth vector fields on $M$. This space $V$ is the locally convex topological vector space upon which we are going to base the smooth structure of $\operatorname{Diff}(M)$. Let $\epsilon >0$ be sufficiently small so that all geodesics are uniquely defined by their endpoints (i.e. defines a geodesically convex neighbourhood around each point). Define $$V_0 = \{ (x \mapsto v(x)) \in V: \| v(x) \| < \epsilon, \forall x \in M \}$$ and for each $v \in V_0$ define the map $\phi_v: M \to M$ which takes $p \in M$ to the endpoint of the geodesic segment with initial point $p$, initial velocity $v(x)$, and length $\| v(x) \|$. Our condition on $\epsilon$ guarantees that this is well defined.

This defines a homeomorphism $\phi: V_0 \to U_0\subseteq C^\infty(M,M)$ where the elements of $U_0$ are the $f:M \to M$ sufficiently close to $\operatorname{id}_M$ so that $f(p) \neq -p$. Let $U_1 \subseteq U_0$ be the diffeomorphisms of $U_0$ and $V_1 = \phi^{-1}(U_1)$, so that $\phi^{-1}: U_1 \to V_1$ is a chart of $\operatorname{Diff}(M)$. As $\operatorname{Diff}(M)$ is a group, one can construct an atlas by taking all left-$\operatorname{Diff}(M)$ translates of $U_1$, and this gives the smooth structure.