# Local flow of a vector field, local one-parameter subgroup of diffeomorphism group?

I arrived at the following question about vector-field-induced flows on a smooth manifold.

Let $M$ be a smooth manifold, $X \in \Gamma(TM)$ a smooth vector field and $\Phi_X: \{(p,I_p)\ |\, p \in M\} \to M$ its maximal local flow, i.e. such that an integral curve $\gamma$ of $X$ with $\gamma(0)=p$ can be maximally extended to the open interval $0 \in I_p$.

I read a claim in a book I'm currently studying, that for any such flow $\Phi_X$ the map $\hat \phi : (\mathbb{R},+) \to (\mathrm{Diff}(M),\circ),\, t \mapsto \Phi_X(\cdot,t)$ is a local group homomorphism. (In the book it is also mentioned that any such flow is a diffeomorphism for some open neighbourhood of $t=0$, that I also don't understand in first place and intuitively would not agree with.)

Clearly, if $X$ (and thus also $\Phi_X$) is complete, i.e. $\forall p \in M: I_p = \mathbb{R}$, then $\hat \phi$ is even a group homomorphism, the one-parameter group of diffeomorphisms generated by the flow of $X$.

I, however, don't see how the local group homomorphism property is satisfied for a non-complete vector field.

(May someone provide me with a pretty definition of a local group homomorphism between topological groups, btw?)

As a counterexample I had $M = (-1,1)$ with the identity chart $(M,id)$ and the vector field $X = \frac{\partial}{\partial id^1}$ in mind. since in that case there clearly doesn't even exist any $t \neq 0$ such that $\Phi_X(\cdot,t)$ is a diffeomorphism, as it is not even defined on the entire $M$?

Am I disregarding something here?

• After having posted the question, stackexchange now recommended me another useful thread whose question already pretty much answers my question to quite some extent already... Seems like the book I'm reading was not very careful there... math.stackexchange.com/questions/1371287/… This however would mean that there is simply no such "local group homomorphism" for non-complete vectorfields? – Lars D. Aug 22 '17 at 14:51
• Yes, you're correct I think. NB in this setting one still retains some features of a local group homomorphism, i.e., $\Phi_X(\,\cdot\,,0) = \operatorname{id}_M$ and $\Phi_X(\Phi_X(x, t),s) = \Phi_X(x, s + t)$ wherever both sides are defined. – Travis Nov 8 '18 at 14:15