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i have this question :

Let $(M,g)$ be a lorentzian manifold, and $\gamma:[0,1]\rightarrow M$ be a spacelike curve in $M$, between two different point $A$ and $B$, so :

can: $\underset{\gamma}{inf}\int_0^1\sqrt{g(\gamma'(t),\gamma'(t))}dt$

be zero ??? (the "$inf$" is on the set of spacelike curves)

with $\gamma(0)=A, \gamma(1)=B$ and $A\neq B $

thanks for every answer

kamerove
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  • Do you have any assumptions on the causal structure of $(M,g)$? And which curves do you take into account, when you take the $\inf$? Any curve? – Thomas Jul 02 '18 at 15:57
  • Thomas; edited , is it clear the question like that ? yes the inf is on any spacelike curve on the surface$ – kamerove Jul 02 '18 at 16:01
  • Not really. First $M$ was the Lorentz manifold, now it is a spacelike hypersurfae. The assumption that $\gamma$ is spacelike is then redundant, $M$ is Riemannian and the, clearly, the distance of any two points is $>0$. I was, in the beginning, under the impression that you have two points which admit one spacelike curve $\gamma$ joining them. Then it makes sense to ask whether the $\inf$ about some family of curves joining the points can be $0$. The answer to this will depend on the class of curves you will allow and, possibly, on the causal structure of the Lorentz manifold. – Thomas Jul 02 '18 at 16:07

1 Answers1

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Without assumptions on causality the answer is yes. Take the strip $$\{(x,y): 0\le y\le1\}$$ and identify $(x,0)$ with $(x, 1)$. Then let the direction of $(0,1) = e_2$ be timelike, i.e.

$$(g_{ij} ) =\left(\array{-1 & 0 \\ 0 & 1} \right)$$

Then, first of all, $c(t) = (t,t ) $ is a null geodesic from $(0,0)$ to $(1,1)$.

Let $\bar c(t) =(1+t, 1-t),\, 0 \le t \le 1$. This is again a null geodesic, this time joining $ (1,1) $ and $(2,0) =:B$

Clearly there is a spacelike curve ($\gamma(t)=(t,0),\, 0\le t \le 2$) from $A$ to $B$.

It's now not difficult to see that there is a sequence of spacelike curves joining $A$ and $B$ such that the lengths tend to zero (choose a sequence approaching the joint curve $c+ \bar c$ in the strip $0\le y \le 1$ from below)

I'm quite sure the answer will be different if you assume that the Lorentzian manifold obeys some reasonable causality assumptions. I'd assume that the $\inf$ in question will be positive in globally hyperbolic manifolds, but don't have the time to follow up on this right now. Also I'm quite sure the same is true with assumptions not as strict as globally hyperbolic (maybe the Wiki page even has some information on this).

Thomas
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  • Thomas: "I'd assume that the inf in question will be positive in globally hyperbolic manifolds" -- Consider a spacelike curve consisting of several straight line segments, each of positive length, connecting $A$ and $B$, in infinite plain flat spacetime. Each segment can be replaced by two segments of much smaller length (both being "closer to lightlike segments"). And so on; therefore the inf of curve lenghts, over the infinite sequence of curves, must be zero. (There are however other intuitions about spacelike curves etc..) – user12262 Nov 07 '20 at 07:05