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I see the following lines in the lecture notes of my Riemannian geometry course: $x\in M, v\in T_x M, X\in T_v(T_x M), \langle X, v\rangle =0$. I get confused on the definition of “ the tangent space of a tangent space”. What’s going on here? Since it’s calculating the inner product of $X$ and $v$ with the given Riemannian metric, I think $X$ must also be a vector in the tangent space of $M$ at point $x$, but how can $X$ be, or in other words, what will be the form of $X$ if we set a local basis $\{u^i\}$?

Apocalypse
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2 Answers2

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Every (Euclidean) vector space $V$ is (trivially) a (Riemannian) manifold, the simplest chart given by the identity.

Then, if you stick to the definitions, a tangent vector to, say, a curve through some point $p\in V$ is an element of $T_{c(0)} V$, assuming $c(0)=p$. But in $V$ there is a natural way of identifying that space with $T_0 V$ through translation by $-c(0)$ (seen as the position vector of the point $c(0)$, and the latter, in turn can be naturally identified with $V$ itself.

Now the same reasoning can be applied if $V$ is the tangent space $T_pM$ of some manifold. So if $v\in T_p M$, an element $X$ of $T_v T_pM$ can be identified with an element of $T_pM$, which is typically also denoted by $X$.

Thomas
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  • If they can be identified, why do we bother introducing the “double tangent space” instead of simply using the original tangent space? – Apocalypse May 19 '20 at 10:28
  • @Apocalypse Because it arises automatically, if you apply the principles invented for dealing with manifolds on the tangent bundle (which is a manifold itself). – Thomas May 19 '20 at 13:30
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I may be wrong but it seems you are studying the exponential map of a riemannian manifold ? If so, then you can canonically indentify $T(T_xM)$ with $T_xM \times T_xM$ as it is a vector space. Thus, at a point $v \in T_xM$, you have $T_v(T_xM) = \{v\}\times T_xM$, which is just the vector space $T_xM$. You can then use the euclidean structure of $T_xM$.

In the context of the exponential map, choosing a tangent vector at $v \in T_xM$ is the same as choosing a tengent vector at $0 \in T_xM$ and moving it constantly throught $t \in [0,1] \to tv$, so you can study the differential of $\exp$ that way and show it is, whith the identification, the indentity map, at $0$.

Didier
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  • Indeed I meet this in the exponential map and Jacobi field section, but if $T_v(T_x M)={v}\times T_x M$ can be identified with $T_x M$, what’s the point of introducing this new notation? – Apocalypse May 19 '20 at 10:35
  • Which new notation ? As $\exp : T_xM \to M$, then, formally, $\mathrm{d}\exp : T(T_xM) \to TM$. It is the definition of the differential of a function. In this particular case, you have a trivial bundle on the left, and considering the differential at a point $v$ makes the identification caninical. – Didier May 19 '20 at 10:49
  • I mean the “double tangent space” notation. Why don’t use $T_x M$ directly but $T_v(T_x M)$ since they can be identified. – Apocalypse May 19 '20 at 11:02
  • The theorem "$\mathrm{d} \exp (0) = \mathrm{Id}$" isn't a priori a rigorous statement because it is a linear morphism between two different vector spaces, that is $T_0T_xM$ and $T_xM$. But in that case you can identidy canonically. This does not mean it is "trivial". You have to understand what you're doing.

    I guess that's why many books are doing a rigorous notation, then say you can identify and thus, you have an identity map.

     – Didier May 19 '20 at 11:17
  • Okay, I think I have a better understanding about it. Thank you for the detailed explanation! – Apocalypse May 19 '20 at 11:26