Group of diffeomorphisms and its tangent space i.e. its Lie algebra

Question

So I feel like there are many questions and also many sources on what I am asking, but I still don't understand what I think is a very basic thing in my head:

It is known, that for a Lie group $G$ (for both, infinite and finite dimensional), its tangent bundle is isomorphic

$$TG \cong G \times T_\mathrm{Id}G. $$

Also, it is known that $T_\mathrm{Id}G $ is isomorphic to the Lie algebra, i.e. the set of left invariant vector spaces on the Lie group.

Now let's consider the diffeomorphism group. I am interested in diffeomorphims on $\mathbb{R}^2$ that differ from the identity only on a compact set, but let's consider the diffeomorphism group on a compact manifolds (I don't understand the explanations that I found for this space either) $\operatorname{Diff}(M)$.

Now, I have read in different sources, that the Lie algebra of $\operatorname{Diff}(M)$ is given by vector fields on $M$, but I don't really understand why. Is it so trivial that there is no explanation for this anywhere? How do we go from left invariant vector fields on the diffeomorphism group to vector field on $M$ itself?

EDIT: I'm sorry if I'm just dense now but I fail to understand your answers concerning my question. As far as I see, you mostly explain why tangent vectors at the identity correspond to left invariant vector fields on the diffeomorphism group. But what I fail to understand is: Why does the the following hold:

$\mathfrak{X}(M)= $ left invariant vector fields on diff. group $\subset \mathfrak{X}(\operatorname{Diff}(M))$

Answer 1

I will try to explain a bit more. An element of the Lie algebra of a Lie group $G$ is, by definition, an element $v$ of $T_e(G)$, where $e$ is the identity element of $G$. Construct some smooth curve $\gamma(t)$ in $G$, for $t \in (-\epsilon, \epsilon)$ such that $\gamma(0) = e$ and $\gamma'(0) = v$.

Now let us specialize to the case where the Lie group is $\operatorname{Diff}(M)$, where $M$ is a compact manifold. Let us not worry about the topology on $\operatorname{Diff}(M)$ for the time being, because it doesn't seem that this is what is troubling the OP's author. So, now, an element of the Lie algebra of $\operatorname{Diff}(M)$ is represented by $\gamma(t)$, a $1$-parameter family of diffeomorphisms of $M$. Map $v$, an element in the Lie algebra of $\operatorname{Diff}(M)$, to the following vector field $X_v$ on $M$, defined as follows.

Given $p \in M$, note that $\gamma(t)(p)$ defines a curve in $M$, passing through $p$ at $t = 0$. So its derivative with respect to $t$ at $t = 0$ is an element $X \in T_p(M)$. Moreover, $X$ depends smoothly on $p$, so it defines a vector field, say $X_v$ on $M$. One can check that the map $v \mapsto X_v$ is well defined, independent of our choice of $\gamma(t)$, as above.

Conversely, given a vector field $X$ on $M$, its flow $\Phi^t_X$ is well defined for $t \in (-\epsilon, \epsilon)$ (here, we have used compactness of $M$ and the theory of ODEs). But $\Phi^t_X$ is a $1$-parameter family of diffeomorphisms of $M$ (which is the identity at $t = 0$), so it defines a smooth curve $\gamma(t)$ in $\operatorname{Diff}(M)$. The derivative $\gamma'(0)$ is an element of the Lie algebra of $\operatorname{Diff}(M)$. So we have defined a map which maps a smooth vector field on $M$ to an element of the Lie algebra of $\operatorname{Diff}(M)$.

There remain some details to check, namely that the two maps I have defined are inverses of each other. I hope I explained things better than I did previously.

Answer 2

For compact manifolds the identification is a consequence of the exponential law for smooth functions. It states that $$f\colon L\rightarrow C^\infty (K,N)$$ is smooth ($K$ a compact smooth manifold) if and only if the associated map $$f^\wedge\colon L\times K\rightarrow N, f^\wedge (\ell,k) = f(\ell) (k)$$ is smooth.

The idea really is that $\operatorname{Diff}(K)$ is an open subset of $C^\infty (K,K)$ whence the exponential law is applicable. This allows you to identify curves running through the identity morphism with vector fields. The key is to observe that the derivative of a curve $f$ taking values in $C^\infty (K,K)$ gets identified by the exponential law with the partial derivative (in the $L$ component) of the associate map $f^\wedge$.

Note that the exponential law is a non-trivial theorem. Its proof and a derivation of the Lie algebra of $\operatorname{Diff}(K)$ can be found in the compact case in my book "An Introduction to Infinite-Dimensional Differential Geometry" (it is hardly the only source, there are many other older sources) [1] (link leads to the free open access version, Exponential law and function spaces are treated in chapter 2.2 and 2.3 and the diffeomorphism group is one of the main examples in chapter 3).

Addendum: For the non-compact case the exponential law is no longer valid in the form stated (continuity of the mappings becomes a problem here) and one has to work around this. In particular, one can prove that the vector fields are NOT the Lie algebra of $\operatorname{Diff}(N)$ if $N$ is a non-compact manifold. Instead one has to use the compactly supported vector fields as Lie algebra. See e.g. [2] for the details. In the special situation of $N=\mathbb{R}^n$ you might want to look at H. Glöckner's article [3] for an account (it is easier than the general case).

For the general case of a non-compact manifold, another way of doing it is by changing the calculus to convenient calculus (sacrificing the continuity of smooth maps but preserving a stronger exponential law), see [4] for an account and the proof details.

Addendum 2: As per the OPs edit, one question is still how the vector fields on $K$ can be identified with left invariant vector fields on $\operatorname{Diff}(K)$. The first step for this is to note that due to the exponential law one has $TC^\infty (K,K)\cong C^\infty (K,TK)$. In particular, this identification takes the tangent space of $\operatorname{Diff}(K)$ at the identity to the set of vector fields on $K$. Now the tangent space at the identity of a Lie group can be identified as usual with the left invariant vector fields on the Lie group (the usual books have a proof how one constructs the isomorphism taking the tangent space at the identity to the left invariant vector fields or check the book I mentioned for details).