2
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ – Andrea Marino Nov 21 '20 at 9:10
  • $\begingroup$ That's a good way of phrasing the connection between the intuition and the concrete math. $\endgroup$ – Tim Carson Nov 22 '20 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.