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By examining numerous examples I have become quite convinced that the following statement is true.

Let $x$ and $y$ be distinct points in the hyperbolic plane $\mathbb{H}$. Let $\gamma$ be the geodesic passing through $x$ and $y$ and let $v_x$ and $v_y$ be the unit tangent vectors to $\gamma$ at $x$ and $y$ respectively. We assume that $v_x$ points from $x$ toward $y$ and $v_y$ points from $y$ toward $x$.

Let $\mathcal{F} = (f_t)_{t \in \mathbb{R}}$ be any one-parameter family of isometries of $\mathbb{H}$. Let $d_x$ be the derivative at zero of $\mathcal{F}$ evaluated at $x$, so that $f_t(x) = x + td_x+o(t)$ at $t \to 0$. Make a similar definition of $d_y$. Then we have $\langle v_x,d_x \rangle = - \langle v_y, d_y \rangle$.

Is this true in general, and if so is there a geometric interpretation for it?

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  • $\begingroup$ This is also true in the Euclidean plane, where I have verified it with explicit computation. $\endgroup$
    – burtonpeterj
    Commented Aug 26, 2019 at 2:23
  • $\begingroup$ In the Euclidean plane this is visually obvious for translations but not so obvious for rotations when $x$ and $y$ are at distinct distances from the center of the rotation. $\endgroup$
    – burtonpeterj
    Commented Aug 26, 2019 at 2:24
  • $\begingroup$ In the Euclidean case when $\mathcal{F}$ is a rotation about a point $c$ with $|c-x| > |c-y|$ and such that $\angle cyx$ is right then $|d_x| > |d_y|$ but $d_y$ is parallel to $v_y$ while $d_x$ is at a nonzero angle to $v_x$. However, the formula I asked about still holds. $\endgroup$
    – burtonpeterj
    Commented Aug 26, 2019 at 2:37
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    $\begingroup$ This is a special case of the first variation of arc length. But in can be proved in a more elementary way by differentiating $\langle x, y \rangle = \mathrm{const}$, (Minkowski scalar product) in the hyperbolic case and $\|x - y\|^2 = \mathrm{const}$ in the Euclidean case. $\endgroup$
    – Ivan Izmestiev
    Commented Aug 26, 2019 at 4:48
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    $\begingroup$ This is true in any Riemannian manifold. Experts known that the projection of a Killing vector field on a totally geodesic submanifold gives a Killing vector field of the totally geodesic submanifold. In your case the totally geodesic submanifold is the geodesic and the projection is a traslation (the minus sign in your equation is due to the way you took the tangent to the geodesic at x and y). $\endgroup$
    – Holonomia
    Commented Aug 26, 2019 at 15:27

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