Prove isometry preserving excision is Killing-like? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2022-01-21T06:06:32Z https://physics.stackexchange.com/feeds/question/341968 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/341968 4 Prove isometry preserving excision is Killing-like? Julian Moore https://physics.stackexchange.com/users/1280 2017-06-28T19:13:45Z 2017-07-16T08:10:37Z <p>(If you think thia is e.g. not well expressed you already understand the request for help.)</p> <p><strong>Theorem</strong>: 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 $K(p)$.</p> <p><strong>Corollary</strong>: 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 $M'$.</p> <hr> <p>If not actually false, this may be known/trivial but I can't find a proof - and I am hampered by lack of knowledge &amp; notation in constructing one (so the statement might not be very good either; conditions on the manifold missing, for example).</p> <p>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.</p> <p><strong>Sketches</strong>. 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.</p> <p>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)</p> <p>Pedagogical answers will be doubly welcome - it's one thing to have an answer, another to understand it!</p> <p>(reposted with minor improvements from math.se)</p> https://physics.stackexchange.com/questions/341968/-/343540#343540 7 Answer by Valter Moretti for Prove isometry preserving excision is Killing-like? Valter Moretti https://physics.stackexchange.com/users/35354 2017-07-07T19:32:09Z 2017-07-16T08:10:37Z <p>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 <strong>one-parameter group of local diffeomorphisms</strong> $\phi$ associated to $X$ is <strong>global</strong> (which is equivalent to saying that $X$ is <strong>complete</strong>). 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 <em>all</em> $t \in \mathbb R$.</p> <p>There are sufficient conditions assuring that $\phi$ is global (for instance it happens provided $M$ is compact).</p> <p>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 </p> <p>(1) $\phi_0 = id$ </p> <p>and </p> <p>(2) $\phi_t \circ \phi_\tau = \phi_{t+\tau}$ for every $t,\tau \in \mathbb R$.</p> <p>The case you are considering also requires that $M$ is equipped with a nondegenerate metric $g$ and $X$ is a <em>complete</em> $g$-Killing vector field. </p> <p>In this case every $\phi_t : M \to M$ is an isometry.</p> <p>Well, coming back to the general case, the following proposition is valid. </p> <p><strong>PROPOSITION</strong>. <em>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.</em> </p> <p><em>(a) $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for every $t \in \mathbb R$.</em></p> <p><em>(b) $X$ is tangent to $\partial A$.</em></p> <p><strong>Proof</strong>.</p> <p>(1) We prove that not (a) implies not (b).</p> <p>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 $t_0&gt;0$ 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 $$s := \sup\{t \in [0,+\infty) \:|\: \phi_\tau(x_0) \in A\:, \quad \tau &lt; t\}\:.$$ This number exists and is finite (because the set is not empty, as it contains $0$, and $t_0&lt;+\infty$ is an upper bound), strictly positive not greater than $t_0$, and again $\phi_s(x_0) \in \partial A$.</p> <p>(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 $\tau&gt;s$ 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 $\phi_s(x_0) \in \partial A$.)</p> <p>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 <em>over $\partial A$</em> with initial condition $\dot{\gamma}(s)= \phi_s(x_0) \in \partial A$ at $t=s$ admits a complete solution <em>completely contained in</em> $\partial A$ <em>also for $t&lt;s$</em>, but this curve now viewed as a integral line of $X$ in $M$ is uniquely determined and we know by hypothesis that <em>it starts at $x_0 \not \in \partial A$</em> finding a contradiction. </p> <p>(2) We prove that not (b) implies not (a).</p> <p>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 <strong>ADDENDUM</strong>). 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.</p> <p>QED</p> <p>Evidently, if $X$ is a complete Killing field, the result concerns the associated one-parameter group of isometries. </p> <hr> <p><strong>ADDENDUM</strong>. I prove here that</p> <p><strong>Lemma</strong>. <em>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 $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.</em></p> <p><strong>Proof</strong>. 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 <strong>on</strong> $S$). 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 $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.</p>