# Almost geodesic on non complete manifolds

Let $$M$$ be a connected manifold equipped with a connection $$\nabla$$. By Hopf-Rinow theorem, we know that if $$M$$ is complete then for any $$x,y$$ there exist a curve $$\gamma:[0,1] \to M$$ such that $$\gamma(0) = x, \gamma(1) = y$$ and $$\nabla_{\gamma'(t)} \gamma'(t)=0$$ for all $$t$$. This is a way to say that $$\gamma$$ is a geodesic.

Suppose now that $$M$$ is possibly non complete. Given a threeshold $$\varepsilon$$, is it always possible to find a $$\gamma$$ between two fixed points $$x,y$$ such that $$\frac{\lVert \nabla_{\gamma'(t)} \gamma'(t) \rVert }{\lVert \gamma' \rVert^2 } < \varepsilon$$

I kind of solved the case in which $$M$$ is of the form $$\mathbb{R}^n \setminus \cup_{i=1}^k N_i$$, where $$N_i$$ are submanifolds of codimension at least 2. In this case you can take a segment from $$x,y$$ and perturb it to be transverse to each $$N_i$$ in the $$C_2$$ topology (see Hirsch, differential topology, transversality chapter), so that $$\gamma''$$ will be almost zero and $$\gamma'$$ almost costantly $$(y-x)$$. Since $$\dim \gamma + \dim N_i < \dim M = n$$, transversality means $$\textrm{Im} \gamma \subset M$$. In my case this is enough to conclude, so this is just a curiosity :) Maybe something in the spirit of calculus of variations?

Start with the plane $$\mathbb R^2$$ and remove a slab, but keep a line going through the slab:

$$Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\}$$ $$M = \mathbb R^2 - Slab$$

  y

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Note that $$M_1$$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.

Now cut away a line-with-a-hole:

$$Line_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$ $$N_\delta = \mathbb R^2 - Slab - Line_\delta$$

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x


At the top of the slab (the point $$(1, 0)$$) we will always have $$\gamma'/|\gamma| = (0, 1)$$. If $$\delta$$ is small enough compared to your $$\epsilon$$, you shouldn't be able to turn fast enough to avoid crashing into $$Line_\delta$$.

Edit: Another answer is to take the manifold

$$ThickenedCircle_{r, \delta} = \{ p \in \mathbb R^2 : r-\delta < |p| < r+\delta \}.$$ First chose a sufficiently small $$r$$ so that the circle of radius $$r$$ does not obey your condition on the curve for $$\epsilon/2$$. Then if you chose $$\delta$$ small enough you get a flat incomplete 2-manifold where geodesics still must accelerate too much.

• Thanks. The second example is conceptually cristalline: if we were on the circle (embedded in R^2), the acceleration along the curve would be calculated by the ordinary second derivative and then *projecting * to the tangent space to $S^1$. The latter makes the acceleration zero. If, instead, you inflate the circle, you don't project anymore, and you get a big acceleration, no matter what you do. – Andrea Marino Nov 21 '20 at 9:10
• That's a good way of phrasing the connection between the intuition and the concrete math. – Tim Carson Nov 22 '20 at 0:52