1

The problem is trivial yet I got stuck and have a problem with justification of the fact $\text{Isom}(\mathbb{R}^n,g=\sum_{i=1}^n (dx_i)^2)=\mathbb{R}^n \rtimes O(n)$. Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a diffeomorphis, in order to be isometry it must satysfie $f^*g=g$ which can be translated to: $$[\frac{\partial f_i}{\partial x_j}]_{i,j}^{T}[\frac{\partial f_i}{\partial x_j}]_{i,j}=I_n.$$ Hence derivative must be orthogonal in each point. Yet I don't see why that derivative must be constant?

EDIT: I know it can be done by considering Riemannian manifold as metric space, yet for that purpose one have to develop concept of metric and integration is needed. I would like to be able to do it directly so the above conditions fallow easily from definition of isometry yet I cant finish that. Is that argument with metric really needed here?

J.E.M.S
  • 2,608
  • 1
  • 19
  • 32
  • 2
    You need only to assume that $f$ is an isometry in the metric sense (assumptions on differentiability not needed). See here –  Aug 03 '16 at 21:21
  • And as group, it should be a semi direct product instead of direct product. –  Aug 03 '16 at 21:23
  • Hmmm actually I would like to show it directly that why I am asking – J.E.M.S Aug 03 '16 at 21:30
  • The word isometry comes from the idea that there is a metric to preserve. Maybe there’s a definition independent of a metric, but I don’t know one. Can you please include your metric-free definition of isometry in the question? – David K Oct 27 '18 at 18:20

1 Answers1

1

The result follows easily if you think in terms of geodesics. Recall that any isometry $\phi$ of a complete Riemannian manifold $M$ is uniquely determined by $\phi(o)$ and $d\phi(o)$, at some point $o \in M$. This follows because isometries map geodesics to geodesics, according to $$ \phi(\exp_o(v)) = \exp_{\phi(o)}\left(d\phi(o)\cdot v\right) $$ In particular, if two isometries $\phi$ and $\psi$ of $M$ verify $\phi(o) = \psi(o)$ and $d\phi(o) = d\psi(o)$, then $\phi \equiv \psi$.

Now, let $M$ be $\mathbb{R}^n$ (with the Euclidean metric, which is complete) and $\phi$ an isometry. Let $t = \phi(0)$ and $R = d\phi(0)$ where $0$ is the origin. As you pointed out, $R$ is an orthogonal linear mapping of $\mathbb{R}^n$ (identified with its tangent spaces). Then, $\phi$ must coincide with the affine transformation $\psi(x) = Rx + t$, since this $\psi$ is clearly an isometry.

-- Salem