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Consider a $2l$-dimensional Riemannian manifold $(M,g)$ without a boundary and let $V=V^{\mu}\frac{\partial}{\partial^{\mu}}$ be a Killing vector field, i.e. $$ \mathcal{L}_Vg_{\mu \nu} = 0 \Leftrightarrow \nabla_{\mu}V_{\nu} + \nabla_{\nu} V_{\mu}=0 $$ and assume it generates a $U(1)$ isometry.

How exactly can I see this isometry from the equations above? Usually one finds the symmetries from solving the differential equations above. What must these solutions (or equations to be solved) need to look like in order to see this $U(1)$ isometry?

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In general, any vector field $V$ on a manifold determines a flow $\phi_t$, which is the solution to the differential equation $$ \frac{d}{dt}\vert_{t=t_0} \phi_t(p) = V_{\phi_{t_0}(p)}, ~~ \phi_0(p) = p. $$
$\phi_t$ is just the collection of integral curves and $V$ and for every $t \in\mathbb R$, $\phi_t$ is a diffeomorphism of $M$ and $\phi_{t+s} = \phi_t \circ \phi_s$. This means that $\phi$ determines an $\mathbb R$ action on $M$. But now it is possible that this $\mathbb R$ actions descends to a $U(1) = S^1$ action. This happens if $\phi_t = \phi_{t+c}$ for some constant $c$, for then we get an action of $\mathbb R / c\mathbb Z \simeq S^1$. Intuitively, this happens when all of the integral curves of $V$ close up after time $c$.

Also note that $V$ is a Killing field if and only if $\phi_t$ is an isometry for all $t$.

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  • $\begingroup$ So should I interpret $V$ as the tangent vector in any point on the flow $\phi_t$? Because if I parameterize the flow by $t$ then $V$ looks like a tangent vector i.e. the velocity of the flow. $\endgroup$
    – Marion
    Mar 16, 2015 at 18:32
  • $\begingroup$ That's correct, $V$ is tangent to the flow at any point. This is what the equation defining $\phi_t$ says. $\endgroup$ Mar 16, 2015 at 23:55

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