(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 ϕ generated by a Killing field K, for M′=M∖W, where W is an open subset in M ϕ:M′→M″ is an isometry if and only if ∀p∈∂¯W, Tp(p)M′ is parallel to K(p).
Corollary: The symmetries of W determine the isometries of M′: if tangents to ∂¯W are in Ki everywhere tangent to ∂¯W, Ki 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 ϕ:M′→M″ is an isometry; choose a point p∈∂¯W; since W is timelike there is always one Killing vector field K parallel to Tp(p)M′; choose some other q∈∂¯W, then either the tangent at q is parallel to K or it is not: if it is not, ∂¯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 ∂¯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)
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 ϕ associated to X is global (which is equivalent to saying that X is complete). In other words, if x∈M the differential equation ˙γx(t)=X(γx(t)) with initial condition γx(0)=x admits a (unique) maximal solution γx=γx(t) defined for all t∈R.
There are sufficient conditions assuring that ϕ is global (for instance it happens provided M is compact).
This way, ϕ:R×M∋(t,x)↦ϕt(x):=γx(t)∈M is smooth and well-defined. Moreover
(1) ϕ0=id
and
(2) ϕt∘ϕτ=ϕt+τ for every t,τ∈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 ϕt:M→M is an isometry.
Well, coming back to the general case, the following proposition is valid.
PROPOSITION. Let A⊂M be an open set whose boundary ∂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) ϕt(A)=A and ϕt(M∖¯A)=M∖¯A for every t∈R.
(b) X is tangent to ∂A.
Proof.
(1) We prove that not (a) implies not (b).
If it is false that ϕt(A)=A and ϕt(M∖¯A)=M∖¯A for all t, then there must exist a point x0∈A such that ϕt0(x0)∉A or a point x0∈M∖¯A such that ϕt0(x0)∉M∖¯A for some t0∈R. Assume the former is valid (the latter can be treated similarly). Assume t0>0 the other case is analogous. There are now two possibilities for ϕt0(x0)∉A. One is ϕt0(x0)∈∂A and in this case define s:=t0. The other possibility is ϕt0(x0)∈M∖¯A. In this case, define s:=sup 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, 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).
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