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Let $M$ be a Riemannian manifold and $p$ be a point of $M$.

Let $v$, $v'$ be tangent vectors to $M$ at $p$. Of course we have $\langle v,v'\rangle_p$ defined.

Let $u$, $w$ be tangent vectors to $T_p(M)$ at $v$.

How is $\langle u,w\rangle_v$ defined? How is it related to the metric of $M$?

Michael Albanese
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user120713
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  • what do you mean for $u,w$ tangent to $T_pM$?? – janmarqz Jan 12 '14 at 01:27
  • $T_p(M)$ is a vector space and a manifold . v is a point of this new manifold so we can have tangent vectors to this manifold at point v, just like v is a tangent vector to M at point p. – user120713 Jan 12 '14 at 01:32
  • ok.. if we name it $TM=\bigcup_p(T_pM)$, is this the manifold that you are talking?? – janmarqz Jan 12 '14 at 01:36
  • i ask this, because $T_pM$ only is a vector space – janmarqz Jan 12 '14 at 01:37
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    I think this may related to the exponential map. We can get Guass Lemma. This is from the book "Riemannian Geometry" of Do Carmo. – gaoxinge Jan 12 '14 at 01:41
  • isn't this [https://en.wikipedia.org/wiki/Double_tangent_bundle]?? – janmarqz Jan 12 '14 at 01:45
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    thanks for all your feedback. – user120713 Jan 12 '14 at 01:55
  • $u,w\in T_{(v,p)}TM=(v,T_pM)$, right?? – janmarqz Jan 12 '14 at 02:02
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    Each tangent space is a manifold. $T_p(M)$ is therefore a manifold. I am not talking about tangent bundle here. Just a single tangent space at p. i guess you are right, if i understand your notation correctly – user120713 Jan 12 '14 at 02:05
  • when you say $u\in T_pM$ then $(u,p)\in TM$. Then as Igor said the natural choice to do is $\langle u,w\rangle _v=\langle u,w\rangle_p$ when $u,v$ are tangent at $p$... but if they are $u\in T_pM$ and $w\in T_qM$ for two different position $p$ and $q$??? :D greets!! – janmarqz Jan 12 '14 at 02:35
  • $u \in T_v(T_p M)$. u is tangent at v, not at p. w is also tangent at v, not at p or q. – user120713 Jan 12 '14 at 02:47
  • let me draw your attention to [http://math.stackexchange.com/questions/52509/riemannian-metric-of-the-tangent-bundle/635357#635357 ] I mistook posting here. – janmarqz Jan 12 '14 at 03:44

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As you say, $T_p(M)$ is a vector space, and the tangent vectors to $v$ are just $u+v$ and $w+v.$ The natural scalar product on the tangent space is therefore one inherited from $T_p(M)$ (if you think of $T_p(M)$ as "a vector space and a Riemannian manifold") which is thus exactly the one given by the Riemannian metric on $M$ at $p.$

Igor Rivin
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