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Killing vector field => global isomorphisms?

  1. May 2, 2014 #1
    Suppose we have a vector field ##V## defined everywhere on a manifold ##M##. Consider now point ##p \in M##. As a consequence of the existence and uniqueness theorem of differential equations. this implies that ##V## gives rise to a unique local flow
    $$\theta:(-\epsilon,\epsilon) \times U \to M$$
    for ##(-\epsilon,\epsilon) \in \mathbb{R}## and ##p \in U## where ##U## is an open subset of M.

    Now if ##\theta_t(p) = \theta(t,p)##, and if ##V## is a killing field, then ##\theta_t: U \to M## should be isometries. But do they belong to isometry group ##Isom(M)##? I.e. are they global as a consequence of ##V## being globally defined and a killing field?

    Is there not a bijective correspondence between global isometries and globally defined Killing vector fields?
     
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  3. May 2, 2014 #2

    quasar987

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    The issue is that for a given t, no matter how small, [itex]\theta_t(p)[/itex] may not be defined for all p in M. The typical example is M= open upper half plane and [itex]V = -\partial/\partial y[/itex]. Given any t, you can always choose For p=(x,y) with y small enough so that the flow at time t is not defined (because following the flow would "take us out of M").

    But if M is compact, you can cover it with finitely many open sets U_i on which the flow is defined on [itex](-\epsilon_i,\epsilon_i)[/itex]. Then for [itex]|t|<\epsilon:=\min_i\epsilon_i[/itex], the flow is defined globally.

    Also, if V is complete, then its flow is defined on all M for all t.
     
    Last edited: May 2, 2014
  4. May 2, 2014 #3
    I see the problem here when ##M## is not compact. Does this then imply that ##\text{Isom(M)}## does not contain any globally defined one-parameter subgroups? For then it seems possible to use these to define a global action who's induced vector field would be Killing fields.
     
    Last edited: May 2, 2014
  5. May 2, 2014 #4

    quasar987

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    Even if M is not compact there can exist complete vector fields.

    For instance, let f:R-->R be a smooth "bump function" which is 0 on [0,½] and 1 on [1,oo), and let [itex]V'(x,y):=-f(y)\partial/\partial y[/itex]. This is a Killing field on our open half plane whose flow is defined on all of M for all t.
     
  6. May 2, 2014 #5

    Ben Niehoff

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    I have to disagree there. ##V' \equiv - f(y) \, \partial_y## is not Killing on the upper-half-plane with standard Euclidean metric (which I think you have been discussing):

    [tex]\mathcal{L}_{V'} \big( dx \otimes dx + dy \otimes dy \big) = - 2 \frac{\partial f}{\partial y} \, dy \otimes dy \neq 0.[/tex]
    Your attempt to "compress" the flow of ##V'## causes it to fail to preserve the metric between ##\frac12 < y < 1##.

    The upper-half-plane with Euclidean metric is geodesically incomplete, which is why ##V \equiv \partial_y## fails to be a global isometry.
     
  7. May 2, 2014 #6

    quasar987

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    Oh yes of course, bad example! Thx Ben.
     
  8. May 5, 2014 #7

    quasar987

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    I should maybe say then that d/dx is Killing complete on the open upper half-plane.
     
  9. May 5, 2014 #8
    After reading up on flows in John Lee's "introduction to smooth manifolds" at page 212 he states the fundamental theorem of flows which asserts that given a smooth vector field ##V## on ##M## there exists a unique smooth maximal flow ##\theta: D \to M##, where ##D## is an open subset of ##\mathbb{R} \times M##, such that ##\theta^{(p)}: (-\epsilon_p,\epsilon_p) \to M## is the unique maximal integral curve starting at ##p##.

    I might have misunderstood your argument for ##-\partial_y## on the upper half plane, but does it not imply that such a maximal flow does not exist?
     
  10. May 5, 2014 #9
    Never mind. I got it straighten out.
     
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