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Let $M$ be an $m$-dimensional simply connected Riemannian manifold that is not geodesically complete. Suppose $M$ has constant sectional curvature.

Because the curvature is constant, locally $M$ looks like a homogeneous space. In particular, locally we can define a full set of $n(n+1)/2$ Killing vectors. However, because $M$ is not geodesically complete, it cannot be globally homogeneous. Thus there is no way to construct a full set of killing vectors on $M$ by stitching together local Killing vectors.

My question: In these circumstances, is it sometimes possible to construct a partial set (fewer than $n(n+1)/2$) of Killing vectors by stitching together local Killing vectors?

Note that if $M$ were geodesically complete, then the Killing-Hopf theorem implies $M$ would be a homogeneous space.

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  • $\begingroup$ Simply connectedness should imply that your manifold is isometric to a subset of hyperbolic (or euclidean or sphere) space. So the Killing vectors should be defined everywhere, just their integral curves do not exist for all times. $\endgroup$
    – ThiKu
    Commented Feb 26, 2015 at 5:46
  • $\begingroup$ Even in the case where $M$ is simply-connected $(M,g)$ need not be an open subset of a complete constant curvature manifold: consider for instance the universal cover of an open subset of the plane with several holes. $\endgroup$
    – Thomas Richard
    Commented Feb 26, 2015 at 9:18
  • $\begingroup$ @ThiKu your point is well-taken. However I wonder if there is a terminology issue here. There are apparently two kinds of "Killing vectors", those with complete flows and those without. The ones with complete flows are special because they are elements of the Lie algebra of the isometry group. Are both types of vectors commonly referred to as "Killing vectors"? $\endgroup$
    – Josh Burby
    Commented Feb 26, 2015 at 15:04
  • $\begingroup$ Apparently we have an issue of terminology here. For me a Killing vector field $X$ is one that satisfies $L_Xg=0$, where $g$ is the Riemannian metric and $L_X$ the Lie derivative. This is true if the vector comes from a locally defined isometry, one doesn't need a globally defined isometry for this equality to hold. In other words, it's a local definition and doesn't relate to completeness of the flow. $\endgroup$
    – ThiKu
    Commented Feb 27, 2015 at 5:41

1 Answer 1

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Let $k$ denote the value of the constant sectional curvature of $M^m$ and let $X_k$ denote the unique simply-connected complete $m$-dimensional manifold of the constant curvature $k$. Then, since $M$ is simply-connected, there exists a map $dev: M^m\to X_k$, called a developing map of the metric on $M$, which is a local isometry (it is not, in general 1-1). Now, pull-back Killing fields from $X_k$ to $M^m$ via $dev$.

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  • $\begingroup$ Thank you. This answer together with the comments above have greatly cleared up a lot of my confusion. But how can we tell which, if any, of the pulled-back killing vectors are genuine Killing vectors on $M^m$? For instance, suppose $m=2$, $k=1$, and $dev(M^m)$ is the domain of the stereographic projection $S^2\rightarrow\mathbb{R}^2$. The killing vector that fixes the north pole will pull back to a genuine killing vector, but the others would not. $\endgroup$
    – Josh Burby
    Commented Feb 26, 2015 at 14:51

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