# Isometry on Riemannian Manifold

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!

• 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. – Nate Eldredge Jul 9 at 4:17

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 $$\tau_x(x)=p$$.