Let $(X,d)$ be an $n (\geq2)$ dim Alexandrov space with curvature $\geq k$. $B(x,r)$ is an open ball in $X$. Let $M_{k,n}$ be the $n$ dim space form of constant curvature $k$. $B_k(r)$ is an open ball with radius $r$ in $M_{k,n}$.

Question: It is true that $H^{n-1}(\partial B(x,r)) \leq H^{n-1}(\partial B_k(r)) $? Here $H^{n-1}$ means the $(n-1)$ dim Hausdroff volume.


Yes, it is true.

In fact there is is a distance non-contracting map $\ell\colon\partial B(x,r)\to \partial B_k(r)$.

If $x$ is a regular point then $\ell$ is the $k$-logarithm --- it sends point $z$ to the point $\tilde z$ such that geodesics $[xz]$ and $[\tilde x\tilde z]$ go in the same direction for some identification of tangent spaces $\mathrm{T}_xX=\mathrm{T}_{\tilde x}\mathbb{M}^n[k]$.

If $x$ is a singular point, approximate it by a regular points apply the construction as above and pass to the limit.


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.