It seems to me that your question has not so much to do with Killing fields. It is a more general question. Consider a smooth vector field $X$ over a smooth (Hausdorff) manifold $M$ and suppose that the **one-parameter group of local diffeomorphisms** $\phi$ associated to $X$ is **global** (which is equivalent to saying that $X$ is **complete**). In other words, if $x\in M$ the differential equation $$\dot{\gamma}_x(t) = X(\gamma_x(t))$$ with initial condition $$\gamma_x(0)=x$$ admits a (unique) maximal solution $\gamma_x= \gamma_x(t)$ defined for *all* $t \in \mathbb R$.

There are sufficient conditions assuring that $\phi$ is global (for instance it happens provided $M$ is compact).

This way, $\phi : \mathbb R \times M \ni (t,x) \mapsto \phi_t(x):= \gamma_x(t) \in M$ is smooth and well-defined. Moreover

(1) $\phi_0 = id$

and

(2) $\phi_t \circ \phi_\tau = \phi_{t+\tau}$ for every $t,\tau \in \mathbb R$.

The case you are considering also requires that $M$ is equipped with a nondegenerate metric $g$ and $X$ is a *complete* $g$-Killing vector field.

In this case every $\phi_t : M \to M$ is an isometry.

Well, coming back to the general case, the following proposition is valid.

**PROPOSITION**. *Let $A \subset M$ be an open set whose boundary $\partial A$ is a smooth codimension-$1$ embedded submanifold of the smooth manifold $M$ and $X$ a smooth complete vector field on $M$.
Then the following two facts are equivalent.*

*(a) $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for every $t \in \mathbb R$.*

*(b) $X$ is tangent to $\partial A$.*

**Proof**.

(1) We prove that not (a) implies not (b).

If it is false that $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for all $t$, then there must exist a point $x_0 \in A$ such that $\phi_{t_0}(x_0) \not \in A$ or a point $x_0 \in M\setminus \overline{A}$ such that $\phi_{t_0}(x_0) \not\in M \setminus \overline{A}$ for some some $t_0 \in \mathbb R$. Assume the former is valid (the latter can be treated similarly). Assume $t_0>0$ the other case is analogous.
There are now two possibilities for $\phi_{t_0}(x_0) \not \in A$. One is $\phi_{t_0}(x_0)\in \partial A$ and in this case define $s:= t_0$. The other possibility is $\phi_{t_0}(x_0)\in M \setminus\overline{A}$. In this case, define
$$s := \sup\{t \in [0,+\infty) \:|\: \phi_\tau(x_0) \in A\:, \quad \tau < t\}\:.$$ This number exists is finite (the proof is trivial), strictly positive not greater than $t_0$, and again $\phi_s(x_0) \in \partial A$ (the proof is easy). Let us prove that such $s$ (in both possibilities) cannot exist if (b) is valid. Indeed, $X|_{\partial A}$ is a well-defined smooth complete vector field on the smooth manifold $\partial M$ and thus the associated Cauchy problem *over $\partial A$* with initial condition $\dot{\gamma}(s)= \phi_s(x_0) \in \partial A$ at $t=s$ admits a complete solution *completely contained in* $\partial A$ *also for $t<s$*, but this curve now viewed as a integral line of $X$ in $M$ is uniquely determined and we know by hypothesis that *it starts at $x_0 \not \in \partial A$* finding a contradiction.

(2) We prove that not (b) implies not (a).

Let us assume that (b) is false finding that (a) is false as well.
Assume now that there is $x_0 \in \partial A$ such that $X(x_0)$ is transverse to $\partial A$. As $\partial A$ is an embedded smooth manifold, $X$ is smooth and does not vanish at $x_0$, it is not to difficult to prove that there is a coordinate patch $x^1,x^2,..., x^n$ around $x_0$ in $M$ ($n = dim(M)$) such that $x_0 \equiv (0,0,\ldots, 0)$,
$\partial A$ is the portion of the plane $x^1=0$ contained in the image of the chart, and
the integral
curves of $X$ are the curves $\mathbb R \ni t \mapsto (t,0,\ldots,0)$. Since the plane separates $A$ from $M \setminus \overline{A}$, it is evident that there are points in $A$ which are moved into $M \setminus \overline{A}$ by $\phi$ and viceversa. Therefore $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for every $t \in \mathbb R$ is false.

QED

Evidently, if $X$ is a complete Killing field, the results concerns the associated group of isometries.

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Valter Moretti