I have some conceptual confusion when thinking about the Lie algebra of a set of vector fields. Below, there are two questions which refer to the same problem, but from different viewpoints. Any comments/suggestions are greatly appreciated. Sorry if my questions appear silly.
Let $V=\{V_i\}$, $i=1,\dots,m$ be a set of vector fields on a smooth manifold $M$, $dim(M)=n\ge m$. Let $L(V)$ be a non-involutive Lie algebra of vector fields $V$.
- For any $p\in M$, $L(V)$ is an infinite-dimensional group acting locally on some neighborhood $U(p)\subset M$ of $p$. Any group action can be seen as $\exp(t\cdot v):U(p)\rightarrow U(p)$, where $v\in L(V)$. Here comes my first problem.
When we look at elements of $L(V)$, these are all vector fields. If we consider them is being elements of a vector space, these form an infinite dimensional set. However, vector fields belong to a more general object: a module over $C^\infty(M)$. As such, they form a finite-dimensional set. Thant is, we can always find a set of basis vector fields and express the remaining ones as linear combinations (over $C^\infty$) of the basis v.f.'s. I wonder, how infinite-dimensionality enters the picture? Is there something that cannot be expressed using a finite-dim. basis? - When we think about the Lie algebra of a Lie group $G$, it is basically the same at each point $g\in G$. This is not true for the Lie algebra of vector fields. Would it make sense to speak about rank of the Lie algebra $L(V)$ at some point $p\in M$, to say that $p$ is a critical point of the Lie algebra etc? If so, how should we consider the rank of $L(V)$ at $p$? As the rank of vector fields $v\in L(V)$ at $p$? Obviously, it cannot be higher than $n$. So, again, we have an inf-dim set whose rank does not exceed $n$ locally.
Best Answer
As for 1: it is infinite dimensional over the extremely small field $\mathbb{C}$ (which if you want can be seen as the tiny subset of constant functions within $C^\infty(M)$). Anything which is finite- but positive dimensional over $C^\infty(M)$ is infinite dimensional over $\mathbb{C}$ since $C^\infty(M)$ is itself infinite dimensional over $\mathbb{C}$.
For 2: The reason $Lie(G)$, viewed as a Lie algebra of vectorfields on $G$ is the same at every point is that we do not take ALL vectorfields, but only the $G$-invariant ones. This has no analogon in on the general manifold $M$, unless $M$ itself comes with a $G$-action.
What you describe here is looking at the Lie algebra structure at the tangent space $T_pM$to $M$ at $p$. An element of the tangent space is called a tangent vector. It is defined as an equivalence class of vector fields on $M$ under the equivalence relation 'having the same value at point $p$'. You can see that the Lie algebra structure on vector fields respects the equivalence relation so indeed it defines a Lie algebra structure on $T_pM$, turning it into an $n$-dimensional Lie algebra over $\mathbb{C}$.
Only looking at equivalence classes of vector fields in your Lie algebra $L(V)$ gives a smaller (at most $m$-dimensional but perhaps even smaller) Lie subalgebra. It could be any Lie subalgebra of $T_pM$ including the one that only consists of the single element $0$.