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On $\{\vec{x}\in \mathbb{R^n}:x_1^2+x_2^2+\cdots+x_n^2<1\}$ in $\mathbb {R^n},$ what $\alpha$ can make the metric $$g=(1-x_1^2-x_2^2-\cdots-x_n^2)^{-\alpha}(dx_1\otimes dx_1+dx_2\otimes dx_2+\cdots+dx_n\otimes dx_n)$$complete?

I guess this is related to the Hopf-Rinow theorem, but I fail to solve it. Looking forward to your help. Thanks.

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up vote 4 down vote accepted

A natural first step would be to prove that the geodesics through $0$ are exactly the straight lines through $0$. Then Hopf-Rinow says that $B$ is complete if and only if any straight line segment from $0$ to $S^{n-1} = \{x\in R^n : \|x\| = 1\}$ has infinite length. Thus if $\|x\| = 1$, you must see whether or not the path $\gamma\colon [0,1]\to R^n$ given by $\gamma(t) = tx$ has infinite length. The length of $\gamma$ is $$\int_0^1 \|\gamma'(t)\|\,dt = \int_0^1 \frac{dt}{(1 - t^2)^{\alpha/2}}.$$ This integral converges if and only if $\alpha<2$. Thus $B$ should be complete if and only if $\alpha\geq 2$.

Edit: In response to the question of how we know the geodesics through $0$ are the straight lines. One should suspect this is the case upon observing that the metric under consideration is radially symmetric. In particular, reflections through hyperplanes through the origin are isometries, and hence preserve geodesics. We will use this fact.

Suppose that $\vec{v}$ is a tangent vector at $0$, and let $\gamma$ be the geodesic in the direction of $\vec{v}$. Let $H$ be any hyperplane through $0$ containing the line through $\vec{v}$. If $r_H\colon B\to B$ is the reflection through $H$, it is an isometry, and hence $\gamma = \gamma\circ r_H$. It follows that $\gamma$ is contained in $H$. But this is true for every hyperplane $H$ through $0$ containing the line through $\vec{v}$, and hence $\gamma$ lies in the intersection of all such $H$. The intersection of all such $H$ is of course the line through $\vec{v}$, as desired.

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Thanks. What does "$0$" mean in your answer? The origin? And, why does "Hopf-Rinow say that $B$ is complete if and only if any straight line segment from $0$ to $S^{n-1}$ has infinite length"? Where does "infinite" come from? – Vladimir May 6 '12 at 1:14
"0" does mean the origin, sorry. You can use the Hopf-Rinow theorem to see the geodesic has to have infinite length in two different ways. 1) If the $\gamma$ has infinite length, then the exponential map at $0$ is defined on the whole tangent space, since geodesics starting at $0$ can never "leave" the space $B$. 2) If the $\gamma$ has infinite length, this shows that the balls $\overline{B}(0,R) = \{x : d(x,0)\leq R\}$ is bounded away from $\partial B$, and thus is compact. It follows that all closed and bounded sets are compact. – froggie May 6 '12 at 12:51
Thanks! But I still don't know why "the geodesics through $0$ are exactly the straight lines through $0$". Could you explain it? – Vladimir May 8 '12 at 16:01
@Jack Witt: I edited my answer to give a reason for this. Hope it helps! – froggie May 8 '12 at 17:41

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