Edit: According to interesting comment of Thomas Rot to the previous version of the question, we revise the question as follows:

First note that if a manifold $M$ is a parallelizable manifold , then it gets a natural Riemannian metric which is independent of the base point $x\in M$.In fact $TM \simeq M \times \mathbb{R}^n$ enable us to carry the standard Euclidean inner product to each fiber of $TM$. In the following question we apply this obvious fact to $M=TS^n$ as follows:

It is well known that the tangent bundle $TS^n$ of $S^n$ is a parallelizable manifold, then once we fix a trivialization for its tangent bundle(The tangent bundle of $TS^n$), $TS^n$ gets s a natural Riemannian metric, the Euclidean one along each fiber of its tangent bundle. Now we can restrict this metric on $T S^n$ to the zero section $S^n$.

Is there a trivialization of $TS^n$ for which the corresponding restricted metric coincide to the standard metric of $S^n$?

  • 3
    $\begingroup$ "The euclidean one along each fiber" depends on the trivialisation you choose, hence there is no such natural one. $\endgroup$
    – Thomas Rot
    Sep 24 '17 at 21:29
  • $\begingroup$ @ThomasRot Yes you are right. We should fix a trivialization. I am not aware of a canonical one.Is there any? On the other hand it would be possible that all trivialization leads to isometric metrics on $S^n$, so the comparison of the restricted metric with the standard one (up to isometry) is a well defined question $\endgroup$ Sep 24 '17 at 21:44

You describe only a fiber metric on $TS^n$, an inner product on each fiber, but not a Riemannian metric on the total space. If you consider the metric induced from the embedding $Ti:TS^n \to T\mathbb R^{n+1} = \mathbb R^{n+1}\times \mathbb R^{n+1}$ (where $i:S^n\to \mathbb R^{n+1}$), then the answer is yes.

Further explanation:

$i:S^n\subset\mathbb R^{n+1}$ is the standard embedding. The image of the tangent mapping $Ti$ consists of all $(x,\xi)\in \mathbb R^{n+1}\times \mathbb R^{n+1}$ such that $|x|=1$ and $\langle x,\xi\rangle = 0$. The Riemannian metric induced on $TS^n$ from the Euclidean metric on $\mathbb R^{n+1}\times \mathbb R^{n+1}$ is Euclidean on each fiber $T_xS^n$ (fix $x$), and the metric on the zero section is the usual one (set $\xi=0$).

  • $\begingroup$ Dear Prof. Michor thank you for your answer. But I search for a Riemannian metric on $TS^n$. Every parallelizable manifold $M$ admit a Riemanian metric with constant coefficient, independent of base point $x\in M$. $\endgroup$ Sep 27 '17 at 13:03
  • $\begingroup$ @Ali Taghavi: This is a Riemannian metric on $TS^n$, it is the Euclidean one on each fiber as you specified, and it has the property you asked for: The embedding of the 0-section is an isometric immersion. $\endgroup$ Sep 29 '17 at 8:21
  • $\begingroup$ I do not understand the first 3 lines of your answer.May I ask you to explain about that? $\endgroup$ Sep 30 '17 at 23:00
  • $\begingroup$ I am not sure that $S^n$ is parallelizable except $n=1,3,7$; check e.g., $S^2$ (say, for non-zero section). Parallelization is just diffeomorphism $TM \to M\times R^n$, it does not provide any immersion into, say $R^N$; so you have to chose one. And of course taking on the product $M\times R^n$ a product metric you always have zero section $M\times 0$ isometric to $M^n$. $\endgroup$
    – valeri
    Oct 1 '17 at 8:45
  • $\begingroup$ Dear Prof. Michor thanks again for your attention to my question and further explanation. $\endgroup$ Oct 3 '17 at 16:19

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