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Let $M$ and n-dimensional Riemannian Manifold without boundary. Suppose we have an isometry $\tau_{x}: M \to M$ such that $\tau_{x}(x)=o$, for a fixed point $o$ in $M$.

My question is, how can understand $\tau_{x}$?

The first time I saw this, I thought $\tau_{x}$ was of the form $\tau_{x}(x)=x-x=o$, but this doesn't make sense, because $M$ is not a vector space. Then, I thought, if $M=\mathbb{S}^{1}\subset\mathbb{R}^{2}$, and define the polar angle for $x$ as $\theta_{x}$ with $\theta\in[0,2\pi)$. Now, if we take the fixed point $o=(1,0)\in\mathbb{S}^{1}$, we have something like $\tau_{\theta_{x}}(\theta_{y})=e^{i(\theta_{y}-\theta_{x})}$. But this is more complicated for Riemannian manifold.

Now, if $\tau_{x}: M \to M$ is an isometry, then is a smooth map of smooth manifolds. Given some $p\in M$, the differential of $\tau_{x}$ at $p$ is a linear map,

$$d\tau_{x}\vert_{p}:T_{p}M\to T_{\tau_{x}(p)}M$$

from the tangent space of $M$ at $p$ to the tangent space of $M$ at $\tau_{x}(p)$. Then the differential is given by

$$d\tau_{x}(X)(f)\vert_{p}=X(f\circ \tau_{x})\vert_{p}$$

Here $X\in T_{x}M$, therefore $X$ is a derivation defined on $M$ and $f$ is a smooth real-valued function on M.

Any idea will be appreciated, Thanks!

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    $\begingroup$ I'd suggest you first try to understand the notion of the differential $d \tau_x$. Then being an isometry means that $d \tau_x$ maps tangent vectors in $T_p M$ to tangent vectors in $T_{\tau_x(p)}M$ that have the same length with respect to the Riemannian metric. $\endgroup$ – Nate Eldredge Jul 9 at 4:17
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For the first sentence, I think it means:

Given a fixed point $p\in M$ (I took the liberty to replace $o$ by $p$ in case you confuse it with the origin of a vector space), suppose for any $x$, we have an isometry $\tau_x:M\to M$ such that $\tau_x(x)=p$.

Basically it is making the assumption that for any point you can find an isometry that maps this point to a previously given fixed point.

As for the differential part, it doesn't seem like you are asking anything so I don't really what to answer.

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