If $(M,g)$ is a geodesically complete Riemannian manifold of negative sectional curvature bounded below by $K<0$ then is it true that for any $x,y,x_0\in M$ $$ d_H( Log(x_0,x),Log(y,x_0) ) \leq d_M^2(x,y) \leq \|Log(x_0,y)-Log(x_0,x)\|_2^2 , $$ where $d_H$ is the hyperbolic metric on $\mathbb{R}^d$, $d_M$ is the metric induced by the Riemannian metric tensor $g$ and $Log(x_0,\cdot)$ is the Riemannian Log map on $M$ about $x_0$?
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$\begingroup$ The inequality has no sense --- please red/correct it. $\endgroup$– Anton PetruninSep 9, 2017 at 4:08
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$\begingroup$ I thnk you want to square the first $d_H$, so that this might be true when $M$ is hyperbolic space. Also, you want ${}\le{}$ instead of ${}<{}$. $\endgroup$– Ben McKaySep 9, 2017 at 8:02
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1$\begingroup$ For the first inequality, you want to map the "comparison triangle" in $(TM,g_H)$ to $M$. This gives you an inequality similar *but not identical!) to your first. You get an inequality similar to the second by regarding cosine theorems for hyperbolic and Euclidean geometry - maybe you can figure this out for yourself? $\endgroup$– Sebastian GoetteSep 10, 2017 at 17:09
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1$\begingroup$ Sorry, my comment above is rubbish. Rather, note two things. First, if $M$ has $\pi_1(M)\ne\{0\}$, then it is always possible that $d_M(x,y)$ is very small, even though preimages of $x$ and $y$ in $T_{x_0}M$ can be far apart. Second, the inequality from the very left (with a square inserted) to the very right is wrong. And it should be clear by considering very small triangles that as stated (without the square), the first inequality cannot hold. $\endgroup$– Sebastian GoetteSep 11, 2017 at 9:44
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$\begingroup$ Oh you're right. I was thinking something similar today but how can I use comparison triangels to prove (the reverse) type of inequality? $\endgroup$– ABIMSep 12, 2017 at 19:17
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