Question about ratio of Riemannian densities

35 Views Asked by At

I am studying the paper "On the MCP property of metric measure spaces" by S. Ohta (https://ems.press/content/serial-article-files/1508) and I have a question regarding a passage at page 813 of the paper, proof of Theorem 3.2 therein.

Let $(M,g)$ be a $n$-dimensional, complete Riemannian manifold without boundary with $n\geq2$ and let $K\in\mathbb{R}$. Fix $p\in M$, $\xi\in T_pM$ with norm $1$ and an orthonormal basis $\{e_1,\dots,e_n\}$ in $T_pM$ with $e_1=\xi$. Denote by $k_i$ the sectional curvature of the plane spanned by $e_1$ and $e_i$ for each $i=2,\dots,n$. For a small $r>0$, it follows from$$\frac{s_K(r)}{s_K(2r)}=\frac{1}{2}\bigg(1+\frac{K}{2}r^2+O(r^4)\bigg)$$that$$\frac{A_p(r,\xi)}{A_p(2r,\xi)}=\frac{1}{2^{n-1}}\prod_{i=2}^n\bigg(1+\frac{k_i+O(r)}{2}r^2+O(r^4)\bigg),$$where $A_p(r,\xi)$ is the density of the Riemannian measure on the vectors in $T_pM$ with norm $1$ and it holds that $\mathrm{d}\mathrm{vol}_g(q)=(\exp_p)_\sharp[A_p(r,\xi)\mathrm{d}r\,\mathrm{d}\xi]$ and $q=\exp_p(r\xi)$ and $s_K(r):=\frac{1}{K}\sin(\sqrt{K}r)$ for $K>0$, with $\sinh$ for $K<0$ and as $r$ for $K=0$.

Where does the formula for the ratio of the densities come from, especially the part where the sectional curvatures appear? Does someone have a reference for that?