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
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).