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$M$ is a Riemannian manifold, $f$ a (smooth) function from $M$ to itself, $v$ only vectors for which the exponential map is defined.

Then, Is it true that

if $f(p)=p$ and $\mathrm{d}f_p=\operatorname{id}$ then $f(\exp_p(v))=\exp_p(v)\ \forall v\in T_pM$?

Michael Hardy
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Has
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  • What are your thoughts on the problem? What have you tried? – msteve Sep 07 '15 at 13:44
  • Tbh, I cannot see why it should be true. I don't see how the differential of $f$ has anything to do with $f(exp_p(v))$. – Has Sep 07 '15 at 13:48

1 Answers1

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The answer is no. For example take the function: $f(x)=x^2+x$. $f(0)=0,f'(x)=2x+1 \Rightarrow f'(0)=1$, however $\text{exp}_0(v)=v$, so $f(\text{exp}_0(v))=f(v)=v^2+v \neq v = \text{exp}_0(v)$.

Note:

Any Riemannian isometry whose domain is a connected manifold, is determined by its value and its differential at a single point.

So, if you assume $f$ is an isometry, then it follows that $f=Id$, so the claim is true.

This is proved for instance in Do-carmo's book "Riemannian Geometry". (see pg 163 lemma 4.2).

Asaf Shachar
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