# Prove isometry preserving excision is Killing-like?

(If you think thia is e.g. not well expressed you already understand the request for help.)

Theorem: Given a manifold $M$ equipped with a metric $g$ and possessing at least one non-trivial isometry $\phi$ generated by a Killing field $K$, for $M' = M\setminus W$, where $W$ is an open subset in $M$ $\phi:M'\rightarrow M''$ is an isometry if and only if $\forall p \in \partial \overline{W}$, $T_{p}(p)M'$ is parallel to $K(p)$.

Corollary: The symmetries of $W$ determine the isometries of $M'$: if tangents to $\partial \overline{W}$ are in $K_i$ everywhere tangent to $\partial \overline{W}$, $K_i$ remains a Killing field of $M'$.

If not actually false, this may be known/trivial but I can't find a proof - and I am hampered by lack of knowledge & notation in constructing one (so the statement might not be very good either; conditions on the manifold missing, for example).

Having thought about this in the specific context of the timelike Killing fields of Minkowski space (Lorentzian metric) the general case (Riemmanian or Lorentzian) stated above seemed plausible, but a proof of the special case is what I really need.

Sketches. The theorem is that if a region is excised from a manifold, if the boundary of that region follows the integral curves of a Killing field of the original manifold that field is a Killing field of the resulting manifold.

Proof (by contradiction in the case of the timelike Killing fields of Minkowski space... Lorentzian metric). Assume that $\phi: M'\rightarrow M''$ is an isometry; choose a point $p \in \partial \overline{W}$; since $W$ is timelike there is always one Killing vector field $K$ parallel to $T_{p}(p)M'$; choose some other $q \in \partial \overline{W}$, then either the tangent at $q$ is parallel to $K$ or it is not: if it is not, $\partial \overline{W}$ must intersect the integral curves of K and the excision therefore breaks the bijection (by deleting image points) and there can be no isometry at all - a contradiction. Thus $\partial \overline{W}$ must be ruled by the integral curves of K. (Probably needs extending/theorem reformulating for the general case because there's no guarantee that he there is a Killing vector anywhere tangent to the excision boundary)

Pedagogical answers will be doubly welcome - it's one thing to have an answer, another to understand it!

(reposted with minor improvements from math.se)

## 1 Answer

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 $t_0 \in \mathbb R$. 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 and is finite (because the set is not empty, as it contains $0$, and $t_0<+\infty$ is an upper bound), strictly positive not greater than $t_0$, and again $\phi_s(x_0) \in \partial A$.

(Indeed, if $\phi_s(x_0)\in A$ there is an open neighborhood of $\phi_s(x_0)$ completely included in $A$ so that $\phi_\tau(x_0) \in A$ also for some $\tau>s$ which is impossible for the very definition of $\sup$, if $\phi_s(x_0)\in M\setminus \overline{A}$, since this set is open, there would be an open neightborhood of $\phi_s(x_0)$ completely included in $M\setminus \overline{A}$ so that $\phi_\tau(x_0) \not\in A$ in some $(s-\epsilon, s]$ which is again impossible for the very definition of $\sup$; the only remaining case id $\phi_s(x_0) \in \partial A$.)

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 $\partial A$ 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 $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$ (see the final ADDENDUM). 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 result concerns the associated one-parameter group of isometries.

ADDENDUM. I prove here that

Lemma. If $S$ is an embedded $n-1$-dimensional smooth submanifold of the $n$-dimensional smooth manifold $M$, and $X$ is a smooth vector field over $M$ which does not vanish at $x_0\in S$ and is not tangent (i.e., is transverse) to $S$ at $x_0$, then there is a coordinate patch $x^1,x^2,..., x^n$ around $x_0$ in $M$ such that $x_0 \equiv (0,0,\ldots, 0)$, $S$ is the portion of the plane $x^1=0$ contained in the image of the chart, and the integral curves of $X$ are (restrictions around $t=0$ of) the curves $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.

Proof. As $S$ is embedded, there is a coordinate patch $(U, \psi)$ in $M$ around $x_0\in S$ such that $\psi(S \cap U) = \{ (y^1,\ldots, y^n) \in \psi(U) \:|\: y^1=0\}$ and we can always assume $\psi(x_0)= (0,\ldots,0)$. Now $X = \sum_a Y^a\frac{\partial}{\partial y^a}$ is such that $Y^1(0,\ldots, 0) \neq 0$ just because $X$ is transverse to $S$ at $x_0$ (the coordinates $y^2,\ldots, y^n$ are coordinates on $S$). The integral lines of $X$ in coordinates satisfy $\frac{dy^a}{dt} = Y^a(y^1(t),\ldots, y^n(t))$. We are free to fix $t=0$ exactly on $S$ for all curves. Now introduce the coordinates $x^2 =y^2,\ldots, x^n=y^n$ on $S$ and write the said integral curves as smooth functions $y^k = y^k(t,x^2,\ldots, x^n)$, where $x^2,\ldots, x^n$ denotes the initial point on $S$ (at $t=0$) of the considered integral curve. The said map is smooth as well known from standard theorems on smooth dependence from initial data of Cauchy problems. Finally define $x^1=t$. Since the Jacobian matrix $J=[\frac{\partial y^a}{\partial x^b}]$ exactly at $x_0$ satisfies $$\det J(x_0) = \frac{\partial y^1}{\partial t}|_{x_0} = Y^1(0,\ldots, 0) \neq 0$$ Dini's theorem proves that $x^1=t$, $x^b= y^b$ define an admissible smooth coordinate system in $M$ around $x_0$. In local coordinates $x^1,\ldots, x^n$, the portion of $S$ entering the domain of the coordinates is still represented by $x^1=0$ (because $x^1=t$ and all integral curves intersect $S$ at $t=0$) and, locally, the integral curves of $X$ are trivially restrictions around $t=0$ of the curves $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.

• A quick initial acknowledgment & copious thanks! Will need to study that carefully but will probably accept later. – Julian Moore Jul 8 '17 at 4:55
• Nonetheless... as promised - and as a matter of principle & good faith - I have added my own bounty of 100 for your considerable effort (with request but not conditional upon extension to A closed... is it sufficient to say that it follows for A closed by symmetry since which set the boundary belongs to is not relied upon??) – Julian Moore Jul 15 '17 at 6:07
• It is clear that (a) -> (a') because, since $A$, $M\setminus \overline{A}$, $\partial A$ are disjoint sets whose union s $M$, per difference (a) implies $\phi_t(\partial A) \subset \partial A$ and hence $\phi_t(\partial A) =\partial A$ because every $\phi_t$ is bijective. Now suppose that (a') holds. If $X$ is not tangent at $x_0 \in \partial A$, exploiting the same argument as I used in (2) you can construct a integral line exiting from $A$ passing for $X(x_0)$. So (a') implies in particular that $X$ is tangent to $\partial A$ and so (b) holds, thus (a')->(b) <->(a). – Valter Moretti Jul 15 '17 at 8:51
• Referring to (a)->(a') I was a bit implicit. To be more explicit: as soon as you know that $\phi_t(\partial A) = \partial A$, from the hypothesis (a) you also have $\phi_t(M \setminus A) = \phi_t(M \setminus \overline{A} \cup \partial A) = \phi_t(M \setminus \overline{A}) \cup \phi_t(\partial A) = M \setminus \overline{A} \cup \partial A = M \setminus A$. This complete the proof that (a)-> (a'). – Valter Moretti Jul 15 '17 at 9:25
• Well, it is nothing but the theorem I have already used to define $\phi_t(x)$. The solution of a first order differential equation defined by a smooth vector field on a smooth manifold is a jointly smooth function of the parameter along solutions and initial condition. – Valter Moretti Jul 19 '17 at 20:22