Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 2

A Hadamard space is a complete CAT(0) space. A useful equivalent definition, which makes the remaining assertions consequences rather than assumptions, is a complete metric space in which every pair $x,y$ has a midpoint $m$, with $d(x,m)=d(y,m)=d(x,y)/2$, satisfyingThis is the nonpositive-curvature midpoint inequality. Euclidean Hilbert spaces satisfy it with equality in the parallelogram formula. The comparison argument in (c) shows that this definition gives the usual CAT(0) triangle condition.

Write $L=d(x,y)$. If $m^{\prime }$ is any other metric midpoint, apply the inequality from (a) to the chosen midpoint $m$ with $z=m^{\prime }$. The right side is $\frac{1}{2}(L/2{)}^2+\frac{1}{2}(L/2{)}^2-L^2/4=0$, so $m=m^{\prime }$. Every pair has a unique midpoint.

Choose successive midpoints to define $\gamma (k/2^n)$ on all dyadic parameters. Consecutive points at level $n$ have distance $L/2^n$. The triangle inequality gives $d(\gamma (s),\gamma (t))\le (t-s)L$ for dyadic $s<t$. Conversely,which gives the reverse bound. Thus equality holds. Completeness extends this map uniquely and continuously to $[0,1]$, retaining $d(\gamma (s),\gamma (t))=|s-t|L$. It is a constant-speed geodesic.

Every other geodesic has the same midpoint, and then the same successive midpoints, so agrees on all dyadic parameters. Continuity gives agreement everywhere. Hence $\boxed{\text{there is exactly one geodesic segment between any two points}.}$

The comparison triangle must have third side $d(x,y)$: the printed $d(y,z)$ contains an undefined $z$. Use $A=d(a,x)$, $B=d(a,y)$ and $C=d(x,y)$.

First extend the midpoint inequality along any constant-speed geodesic $\gamma$ from $p$ to $q$. If $L=d(p,q)$, the function $d(z,\gamma (t){)}^2-L^2{t}^2$ is midpoint-convex by (a). Iteration proves its convexity bound for dyadic $t$, and continuity then gives the Hadamard squared-distance inequalityApply this first to $[a,x]$ with $z=y_t$, and then to $[a,y]$ with $z=x$:Simplification givesIn a Euclidean comparison triangle, the cosine rule gives the scalar product of the two vectors from the common vertex as $(A^2+B^2-C^2)/2$. The squared distance between their multiples $s$ and $t$ is exactly the right side displayed above. Taking nonnegative square roots proves the required triangle comparison, including degenerate Euclidean triangles.

Let $p,q$ lie in the closed ball of center $z$ and radius $R$. On their unique geodesic, the Hadamard squared-distance inequality givesThus every point of that geodesic lies in the ball. By uniqueness there are no other geodesics to consider, so every closed ball is geodesically convex.