0
$\begingroup$

Let $(M,g)$ be a Riemannian manifold. A smooth vector field $X$ is called a Killing vector field if the flow of $X$ acts by isometries, or, equivalently, if $L_X g = 0$. Now why is the sum of Killing vector fields a Killing vector field?

$\endgroup$
  • 1
    $\begingroup$ What is the rule for sums in Lie derivative definition? $\endgroup$ – DiegoMath Jan 21 '15 at 21:39
  • $\begingroup$ If you want, you can use the following definition instead: $$g(\nabla_YX,Z)+g(Y,\nabla_ZX)=0,\ \forall Y,Z\in\Gamma(TM).$$ $\endgroup$ – DiegoMath Jan 21 '15 at 21:47
  • $\begingroup$ @DiegoMath The problem is I don't really know what the flow of a sum of two vector fields is. Is $L_{X+Y}g = L_x g + L_Y g$ true? I also don't see why. $\endgroup$ – Balerion_the_black Jan 21 '15 at 22:09
  • 1
    $\begingroup$ Use this computation for Lie derivatives:$$(L_Xg)(Y,Z)=X(g(Y,Z))-g(L_XY,Z)-g(Y,L_XZ).$$ Can you conlude the statement? Remember that $L_XY=[X,Y]$. $\endgroup$ – DiegoMath Jan 21 '15 at 22:18
1
$\begingroup$

In the comments there are more than one good answers to the question. However, there is some other way to think of it.

Recall that $\mathcal{X}(M)$, the space of vector fields on $M$, is the Lie algebra of the Lie group $\mathrm{diff}(M)$, the group of automorphisms of $M$. The space $\mathcal{K}(M)\subset\mathcal{X}(M)$, which consists of the Killing vector fields, is the Lie algebra of the group of isometries on $M$. Hence, $\mathcal{K}(M)$ is a vector space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.