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I would like to solve next problem

A Killing vector field $X$ on a Riemannian manifold $(M, g)$ ($g$ is metric) has constant length if and only if every integral curve of the field $X$ is a geodesic in $(M, g)$.

I found here http://arxiv.org/pdf/math/0605371.pdf (Proposition 1) solution, but I don't understand what is $L$, how we define and use that $L$, and how from that (equation in proof) follows statement.

Or there is some alternative solutions?

My definiton of Killing vector field: Let $X$ be vector field on a Riemannian manifold $(M,g)$ and $U$ neighbourhood of a point $p \in M$. Let $\varphi: (-\varepsilon, \varepsilon) \times U \to M$ is flow of vector field $X$. Then $X$ is Killing vector field if for every $t_0 \in (-\varepsilon, \varepsilon)$ mapping $\varphi_{t_0}:U \to M$ is isometry.

user110822
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  • Where did you get that definition? Ordinarily $X$ is a Killing field if $\mathscr L_Xg=0$. – Ted Shifrin Jan 20 '14 at 18:19
  • I was wondering the same thing, since I am familiar with $\mathscr L_X g = 0$ as well. Perhaps the OP means isometric immersion? Wouldn't we then have $\mathscr L_X g = 0$? – Robert Lewis Jan 20 '14 at 18:49
  • @TedShifrin That is definition from my class. But, I see now, there is problem in my book: Prove that $X$ is Killing vector field (my definition) iff $g(\nabla_Y X, Z)+g(\nabla_Z X,Y)=0$ for all $Y,Z \in \chi(M)$. Maybe to first show that? (I don't know how) One note: we didn't introduce Lie derivate $\mathscr L$ on class. – user110822 Jan 20 '14 at 19:06
  • @RobertLewis I answered TedShifrin above. And YESSS, $\varphi_{t_0}$ is isometric, not immersion. My bad. – user110822 Jan 20 '14 at 19:06
  • Sure. The flow by an infinitesimal isometry is an isometry :) – Ted Shifrin Jan 20 '14 at 19:52
  • @user110822, since $\mathscr L_X g = \frac d{dt}\big|_{t=0} \phi_t^*g$, the Lie derivative condition is immediately equivalent to yours. That should be enough for you to figure out the proof in that paper. – Ted Shifrin Jan 20 '14 at 20:00

1 Answers1

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Your definition of a Killing vector field $X$ implies that
\begin{equation} g_t(Y,Z)=g_0(Y,Z)+O(t^2)\\\lim_{t \to 0}\frac{g_t(Y,Z)-g_0(Y,Z)}{t}=0 \end{equation}
the LHS is the definition of the Lie derivative i.e.
\begin{equation} (L_Xg)(Y,Z)=0 \end{equation} for any vector fields $Y$ and $Z$. So
\begin{equation} (L_Xg)(X,Y)=0\\ Xg(X,Y)-g(L_XX,Y)-g(X,L_XY)=0\\ g(\nabla _XX,Y)+g(X,\nabla _XY)-g([X,X],Y)-g(X,[X,Y])=0\\ g(\nabla _XX,Y)+g(X,\nabla _YX)=0 \end{equation} If the integral curve of $X$ is a geodesic, then $\nabla _XX=0$ and so
\begin{equation} g(X,\nabla _YX)=0\\ 0.5Yg(X,X)=0 \end{equation} Then $X$ has constant length. Conversely, if $X$ has constant length then $Yg(X,X)=0$ and so $g(\nabla _XX,Y)=0$ for any vector field $Y$ and so $\nabla _XX=0$. the result.

Semsem
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  • Nice answer. Also, you might want to know that @-tagging only works in comments, not answers, unfortunately. – Potato Jan 21 '14 at 08:32