MathOverflow Asked by Andrea Marino on December 4, 2020

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
--------o-------------
--------o-------------
--------o-------------
x
```

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:

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

```
y
-----------------------------o-- -
| δ
--------o----------------------- -
--------o-----------------------
--------o-----------------------
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.

Correct answer by Tim Carson on December 4, 2020

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