# how to define complete pseudo-riemannian manifold

In riemannin geometry, we define distance function by minimizing the length of curves. However we have nondefinite metric on psedu-riemannian manifold, so we cannot define a length of a curve as riemannian manifold $$L(\gamma)=\int_a^b<\dot\gamma,\dot\gamma>^{\frac{1}{2}}dt$$ Then we also cannot define the distance function, and complete concept on pseudo-riemannian manifold.

So how we define a complete pseudo-riemannian manifold? Can we just substitute the length with $$L(\gamma)=\int_a^b|<\dot\gamma,\dot\gamma>|^{\frac{1}{2}}dt$$

• @AloizioMacedo So geodesically complete equals that $\exp_p$ can be evaluated for all $v\in T_pM$ in the pseudo-riemannian situation? – gaoxinge Jun 23 '16 at 7:36
This can presumably be done without the need for defining geodesic lengths. Observe that the notions of connections and parallel transport are perfectly valid for pseudo-Riemannian manifolds, and therefore we have the notion of the exponential map $\exp:TM \to M$ given by $\exp(p,v) = \exp_p(v) = \gamma(p,1,v)$ where $\gamma$ is the unique geodesic starting at point $p$ with initial tangent vector $v$, and the geodesic evaluated at "time" $1$. We can therefore define completeness of a pseudo-Riemannian manifold to mean that $\exp_p$ can be evaluated for all $v \in T_pM$. Note that this is equivalent to geodesics being extended to all of $\mathbb{R}$ as the real parameter.
• So geodesically complete equals that $\exp_p$ can be evaluated for all $v\in T_pM$ in the pseudo-riemannian situation? – gaoxinge Jun 23 '16 at 7:37
• @gaoxinge Correct. This is equivalent to the concept of geodesics being extended to $\mathbb{R}$ as their domain because of the homogeneity property of geodesics. You can check Barrett O'Neill's book on semi-Riemannian geometry if you're interested. – Mnifldz Jun 23 '16 at 7:56