# Killing vector field => global isomorphisms?

1. May 2, 2014

### center o bass

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?

2. May 2, 2014

### quasar987

The issue is that for a given t, no matter how small, $\theta_t(p)$ may not be defined for all p in M. The typical example is M= open upper half plane and $V = -\partial/\partial y$. 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 $(-\epsilon_i,\epsilon_i)$. Then for $|t|<\epsilon:=\min_i\epsilon_i$, 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
3. May 2, 2014

### center o bass

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
4. May 2, 2014

### quasar987

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 $V'(x,y):=-f(y)\partial/\partial y$. This is a Killing field on our open half plane whose flow is defined on all of M for all t.

5. May 2, 2014

### Ben Niehoff

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):

$$\mathcal{L}_{V'} \big( dx \otimes dx + dy \otimes dy \big) = - 2 \frac{\partial f}{\partial y} \, dy \otimes dy \neq 0.$$
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.

6. May 2, 2014

### quasar987

Oh yes of course, bad example! Thx Ben.

7. May 5, 2014

### quasar987

I should maybe say then that d/dx is Killing complete on the open upper half-plane.

8. May 5, 2014

### center o bass

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?

9. May 5, 2014

### center o bass

Never mind. I got it straighten out.