# Showing that the “abstract” tangent space of a submanifold of the $\mathbb{R}^d$ is isomorphic to the tangent space that's a subset of $\mathbb{R}^n$

Let $M$ be an $n$-dimensional smooth submanifold of the $\mathbb{R}^d$, and $p \in M$. Let $T_p^{A}M$ denote the "abstract" tangent space of $M$ in a point $p$, given by $T_p^AM = \{\gamma: (-\epsilon, \epsilon) \to M: \gamma \text{ is a curve with } \gamma(0) = p\} / \sim$

where $\sim$ is the equivalence relation on the set of said curves, given by

$\gamma \sim \delta <=> \exists$ chart $(\phi, U)$ with $p \in U$, so that $(\phi \circ \gamma)^.(0) = (\phi \circ \delta)^.(0)$

(so $T_p^AM$ is the tangent space consisting of equivalence classes of curves in $M$ going through $p$, where two curves are identified with each other if there is a chart so that the differential (in $\mathbb{R}$) of the chart composed with either of those curves is the same.)

Next, let $T_p^{SMF}M$ denote the "submanifold" tangent space, given by

$T_p^{SMF}M = \{\dot{\gamma}(0): \gamma: (-\epsilon, \epsilon) \to M$ with $\gamma(0) = p\} \subseteq \mathbb{R}^d$

(Sorry for the slightly lenghty intro, but I wanted to give a precise description of what I'm talking about in the following. The actual question starts here:)

I first want to show that the mapping $F: T_p^{SMF}M \to T_p^AM, \dot{\gamma}(0) \mapsto [\gamma]$ is well-defined and linear.

Using that, I want to conclude that $F$ is an isomorphism. (In other words, the goal of the exercise is to show that both definitions describe exactly the same tangent space, except for isomorphism, when we have a submanifold of the $\mathbb{R}^n$.)

I'm afraid I haven't got too many ideas on how to get started. For showing that it's well-defined, I think I have to choose a suited chart for $M$, but would any chart work or does it need to be a specific one? I was given the hint that $F$ then can be expressed using suited differentials, but I'm not sure how.

Edit: Showing that these different tangent spaces are isomorphic sounds like something of which surely there must be proofs to be found, so if this has already been asked, I'd of course also appreciate being linked to it. Since I don't know if "abstract" and "submanifold" tangent space are the official/most commonly used names for these constructs, I couldn't find much regarding it so far.

Recall that subset $S$ of a manifold $X$ is a submanifold of dimension $k$ if and only if for every point $p$ of $S$ we can find a chart of $X$ as follows: an open neighbourhood is sent to $\mathbb{R}^n$ ($n$ the dimension of $X$) and the intersection of the open set with $S$ is sent to $\mathbb{R}^k$ (with $p$ sent to $0$, say).
Now, we must see that two curves $\gamma_1,\gamma_2:(-\epsilon,\epsilon)\to M$ with $0\mapsto p$ have the same tangent vector (the usual derivative) at $0$ if and only if they are equivalent. Equivalence means that for every chart $(U,\phi)$ of $p$ we have $(\phi\circ \gamma_1)'(0)=(\phi\circ\gamma_2)'(0)$ (to ask this for one chart is the same as asking it for every chart, since the property is invariant by diffeomorphisms in $\mathbb{R}^k$). So assume $\gamma_1'(0)=\gamma_2'(0)$. Pick an adapted chart $(U,\phi)$ as above centred ad $p$, then this is a true diffeomorphism between euclidean open sets, so $$(\phi\circ\gamma_1)'(0)=d\phi(\gamma_1'(0))=d\phi(\gamma_2'(0))=(\phi\circ\gamma_1)'(0).$$ The same argument can be used to do the other way, and gives the correspondence (if you want, we proved well definedness and injectivity, by definition the maps respect the vector space operations, the spaces have the same dimension $\Rightarrow$ isomorphism).
• You might want to add the dollar signs near the end of your post. Thanks, I must say, that very much makes sense; I had the most trouble seeing why the transformation is well-defined at first, but the rest seems logical. The linearity from the transformation follows because $T_p^AM$ is a vector space (which was already proven to me earlier), does it? (Because we then have $[\alpha \gamma_1 + \gamma_2] = \alpha [\gamma_1] + [\gamma_2]$?) – moran May 15 '16 at 15:04
• Yes, of course to talk about linearity of the map you already need to know that domain and codomain are vector spaces. So the proof of the linearity really depends on your definition of sum in $T^AM$ (of course, any 2 definitions will in the end be equivalent). One possible definition is this: $[\gamma_1+\gamma_2]$ is the class of the curve $\gamma$ that in some chart $(U,\phi)$ comes from the line $t\mapsto ((\phi\circ\gamma_1)'(0)+(\phi\circ\gamma_2)'(0))t$. With this definition, the linearity is a triviality. – Sonner May 17 '16 at 9:19