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Show that $f:\mathbb{R}\to S^1$ given by $f(t)=e^{i t}$ is a local isometry between Riemanninan manifolds.

So, basically we need to show that for each $p\in\mathbb{R}$ there exists $U\subseteq\mathbb{R}$ open with $p\in U$ such that $f:U\to f(U)$ is an isometry, i.e., $\left<{u,v}\right>_q=\left<{df_q(u),df_q(v)}\right>_{f(q)}$ for every $q\in U$ and $u,v\in T_q(U)$.

I assume we can just take $T_q(U)=\mathbb{R}$ for every $q$. But I don't really know how to start. Would anyone give me a hint? How can we compute the differential $df_p$?

Thank you.

Talexius
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  • Just nitpicking, but this will only be an isometry without the factor of $2\pi$ (if you pick the metric on $S^1$ induced by its embedding in the plane.) – Lukas Geyer May 16 '16 at 22:00
  • It was $q$ not $p$, I edited it. But I don't understand why it is a subset of $S^1$. We have $T_q(\mathbb{R})=\mathbb{R}$ for all $q\in U$. And since $U$ is an open subset of $\mathbb{R}$ we have $T_q(U)=T_q(\mathbb{R})$ for every $q\in U$. $T_q(U)$ is the tangent space at the point $q$. @Jake – Talexius May 16 '16 at 22:02
  • Sorry, you are right @LukasGeyer i'll edit – Talexius May 16 '16 at 22:02
  • @Talexius Ah, my bad. – Arkady May 16 '16 at 22:04

1 Answers1

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The Riemannian metric on $\mathbb R$ is $g=dt^2$ and $\mathbb C$ induces a metric on $S^1$ by $h=\frac 1 2 (dz\otimes d\bar z+d\bar z\otimes dz)$.

For $f$ to be a local isometry, it has to be a local diffeomorphism and $f^*h =g$.

First we check that $f$ is a local diffeomorphism. Since you had trouble computing the differential, I will explain this in more detail. But first, please have a look on my answer here.

We see that, as written in the trivial chart, $f$ is obviously smooth and $df$ can be computed using the vector space derivative: $$ df_t = (i e^{it}) \neq 0 \quad \forall t \in \mathbb R $$ Hence the rank of $df$ is 1 and so $df_t$ is an isomorphism for each $t\in\mathbb R$.

Secondly, we compute the pullback of $h$ under $f$. There are two ways of doing this. First, let us do it the "formal" way:

$T\mathbb R$ is spanned by $\partial_t = (1)$ and for that vector we find $$ df(\partial_t) = (i e^{it}) \cdot (1) = i e^{it}, $$ hence $$ \langle \partial_t, \partial_t \rangle = 1 = \frac 1 2 \left( (i e^{it})(-i e^{-it}) + (-i e^{-it})(i e^{it}) \right) = \langle df(\partial_t), df(\partial_t) \rangle. $$ The (in my opinion) easier and more instructive way is the note that $$ f^*dz := d(z\circ f) = d(e^{it}) = i e^{it} dt, $$ and $$ f^*d\bar z := d(\bar z\circ f) = d(e^{-it}) = -i e^{-it} dt, $$ see also my answer here. Thus we have $$ f^*h = \frac 1 2 \left(e^{it} e^{-it} dt^2 + e^{-it} e^{it} dt^2 \right) = dt^2 = g. $$ Hence, $f$ is a local isometry.

Markus Heinrich
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