Tangent space of a tangent space

Question

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 $\{u^i\}$?

Answer 1

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 $0$.

Answer 2

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 $X$.