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Prove isometry preserving excision is Killing-like?

+ 3 like - 0 dislike
34 views

(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)

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Julian Moore
asked Jun 28 in Theoretical Physics by Julian Moore (40 points) [ no revision ]
There is a point I do not understand: What is $M''$?

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Valter Moretti
Crossposted from math.stackexchange.com/q/2335558/11127

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

1 Answer

+ 2 like - 0 dislike

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
answered Jul 7 by Valter Moretti (2,025 points) [ no revision ]
Most voted comments show all comments
A quick initial acknowledgment & copious thanks! Will need to study that carefully but will probably accept later.

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Julian Moore
Just 1 initial question: Proposition (a) - you're removing the closure of (open) A, yes? Doesn't that leave M without a boundary? I intended A to be removed not its boundary as well. NB if you are available in chat anytime over the weekend please let me know.

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Julian Moore
I remove also the boundary. The picture is this way symmetric under swap of A and M. Sorry I am too busy and I cannot chat.

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

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Julian Moore
I added some further details...

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Valter Moretti
@Julian Moore What about the 50 points of the bounty??

This post imported from StackExchange Physics at 2017-07-11 20:34 (UTC), posted by SE-user Valter Moretti
Though I still have a couple of Q's I've accepted the answer - with many thanks. After some study I find the answer is based in fundamentals of Lie theory... so I have an entree, but I'm still mathematically ignorant so (if you are inclined to answer) Q1 : how does the proof carry over to M\A rather than M\clos(A)? (This is key for me and I don't see how the symmetry of openness between M & A helps.) And the 3 intermediates: "not difficult to prove", "is easy", "trivial" - where might I find (in descending order of importance!) further guidance on filling in the details? Again, my sincere thx!

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

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