# Vector fields, diffeomorphism subgroups and lie group actions

Let $M$ be a compact smooth manifold. Since any vector field is complete we get a $1$-parameter subgroup for each vector field. Consider the following generalization:

Let $\{X_j\} \in Vect(M)$ be a finite "basis" of some integrable subbundle of $TM$ (meaning that their locally linearly independent and closed under lie bracket). We have $Span\{X_j\} \cong \mathfrak{g}$ for some finite dimensional lie algebra. The exponential map $\varphi : \mathfrak{g} \to G$ gives a group equipped with a natural action $\rho: G \to Diff(M)$. Denoting the flow of $X$ by $\varphi_t^X$ the action looks like:

$$\rho : g=e^{X_1}e^{X_2}\dots e^{X_n} \mapsto \varphi_1^{X_1}\circ e_1^{X_2}\circ\dots \circ\varphi_1^{X_n}(-)$$

Question 1: Is this map well defined?

In any case if $\mathfrak{g}$ is abelian the action is well define and we get a product of circles and lines inside $Diff(M)$. For every point $x \in M$ the action of the torus part will carve an embedded submanifold of $M$ and the action of the euclidean part will carve an immersed submanifold.

Question 2: Is there anything more substantial to say here? When will the action of the torus part yield an actual torus?

Question 3: Does this construction give a $G/\ker\rho$-fiber bundle? (does it help if $G$ is compact?).

• For question 1, you have to be more specific. For example, what do you mean by a finite basis being closed under Lie brackets? If you mean that $[X_i,X_j]=\sum_kc_{ijk}X_k$ with constants $c_{ijk}$ everywhere on $M$, then I guess the answer to Q1 is "yes" if $G$ is connected and simply connected. And for Q3, you first of all need $G$ to be compact. Also note that $\ker\rho$ may change over $M$, so some fibres might be quotients of others, as in a Seifert fibration. – Sebastian Goette Apr 11 '16 at 15:01

• The map is well defined. The group $G$ is defined as a group of diffeomorphisms of $M$, so its action as diffeomorphisms of $M$ is defined. – Ben McKay Apr 8 '16 at 14:04
• So basically (if $G$ is compact) for every such collection of vector fields I get a free action of some quotient group of $G$ on $M$ and thus a fiber bundle for this group? – Saal Hardali Apr 8 '16 at 14:15
• No. The fibers are all tori, but (even if all of your vector fields $X_i$ are linearly independent) some fibers might not be as long'' as others. Consider the Klein bottle as a quotient of the flat 2-torus by an involution. In a picture, we draw a square, and identify top and bottom by arrows pointing the same direction, left and right in opposite directions. The middle of the square, going from left to right, gets quotiented to itself by diffeomorphism. The circle action of horizontal translation is different on that middle fiber than on the other fibers. It is not a fiber bundle. – Ben McKay Apr 8 '16 at 14:51