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In subspaces of $U \subseteq \mathbb{R}^n$ the goal is to define the tangent space in a geometric sense as a space that is orthogonal to the space $U$ at a point $p \in U$. This can be visualized in cases such as $S^2 \subseteq \mathbb{R}^3$, where the tangent space at a point $p \in S^2$ is just the space of points that are orthogonal to $p$, that is, $$O_p=\{x \in \mathbb{R}^3 \ | \ \langle x,p \rangle =0\}.$$ I am not sure if this is the way it's done, but I guess a standard way to generalize this could be to define the tangent space of a subspace $U \subseteq \mathbb{R}^n$ at $p \in U$ to be the set $\{x \in \mathbb{R}^n \ \vert \ \langle x,p \rangle=0\}.$ When trying to generalize the spaces we are working with, one doesn't have the nice structure of a subspace of $\mathbb{R}^3$, which is why one needs to adapt definitions. One way that can be done is by working with smooth functions as the following standard way does. Let $M$ be a smooth manifold. Then the tangent space at $p \in M$ is $$T_pM=\{ v:C^{\infty}(M) \to \mathbb{R} \ \vert \ v \ \text{linear and satisfies the product rule}\}.$$

I wonder, in what sense the intuitive geometric view and the abstract manifold definition are compatible. What one probably wants is for these two definitions to be isomorphic, such that they share the same properties as vector spaces and we have a way to identify them by an isomorphism. In the case of the sphere this is quickly done: the sphere is a $2$-dimensional manifold and it is well known that the tangent space $T_pS^2$ is then also $2$ dimensional. Since $O_p$ is $2$-dimensional as well, we get an isomorphism $O_p \cong T_pS^2$. Can this be done for all subspaces of $\mathbb{R}^n$ and is a linear isomorphism suffienct to capture the properties that we want to capture?

user3118
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  • I guess, think about tangent spheres. Any $C^1$ curve on a sphere has derivative orthogonal to itself, everywhere, so the tangent space is literally the set of tangent vectors – FShrike Jan 25 '23 at 20:10
  • I don't think I can make sense of the first sentence in your question. Also the description of the tangent space of $S^2$, while correct, makes use of special properties of the unit sphere. In general, the position vector of a point in a set (you probably want to talk about manifolds) is not directly related to the tangent space of that set in that point. – Thomas Jan 25 '23 at 20:25
  • @Thomas Could you specify what is unclear? I will gladly edit the question. As for your second remark. Do you mean that I precisely use that the sphere is centered around the origin? That could be the case, I need to think a bit more about this. I can't remember seeing this definition using the inner product, its just what I came up with and which, to my mind, is the most intuitive definition in this setting. – user3118 Jan 25 '23 at 20:31
  • Here are some answers I've written where I talk about how to identify the abstract tangent space with actual subspaces of $\Bbb{R}^n$: here, here, and here. Hopefully that sheds some light. Regarding your last paragrah, an isomorphism itself is not sufficient: all vector spaces of the same dimension are isomorphic. So, we want the isomorphism to be "special" relative to whatever additional choices we make (e.g a specific choice of chart). – peek-a-boo Jan 25 '23 at 20:34
  • @peek-a-boo Reading your first post I had the following thought. For simplicity consider $S^2$. Then we can identify $T_pS^2 \cong \mathbb{R}^2$ which identifies $\gamma$ with $\gamma '(0)$. However, in this definition of $T_pM$ one has that $\gamma(0)=p$ iirc. Also I think it is a standard proposition that $\langle \gamma(t),\gamma'(t) \rangle=0$. So in particular, using $\gamma(0)=p$, one gets that the identification gives a vector to $p$. I need to work out some details, but this may give an identification to the space I suggested using the inner product. – user3118 Jan 25 '23 at 20:47
  • @peek-a-boo Do you know whether the inner product definition is a valid one? I am not sure if I read it anywhere, but it is the most intuitive way to define tangency, which is why I wanted an identification to this space. It has already been pointed out that it might be incorrcet, though, which is why I am looking for reference here. – user3118 Jan 25 '23 at 20:48
  • To identify the tangent space of the sphere with $\Bbb{R}^2$, you can't say identify $\gamma$ with $\gamma'(0)$. The latter is an element of $\Bbb{R}^3$. Sure, it lives inside a 2-dimensional subspace of $\Bbb{R}^3$, but there are infinitely many such subspaces and you can't identify them all simultaneously. The correct subspace is the orthogonal complement of the point in question. There is no canonical isomorphism between this 2-dimensional subspace and $\Bbb{R}^2.$ – peek-a-boo Jan 25 '23 at 20:52
  • @peek-a-boo I am sorry I fear that I don't get what you mean. The identification I suggest would identify $T_pM$ with the orthogonal complement as long as its an isomorphism, which I would need to check the details for. Could you please be a bit more precise? – user3118 Jan 25 '23 at 20:59
  • I suggest you read the links I gave you again, and spend more time with it. – peek-a-boo Jan 25 '23 at 21:08
  • @peek-a-boo Let $M$ be an $n$-dimensional submanifold of $\mathbb{R}^N$ and $p \in M$. Then there is a function $T_pM \to \mathbb{R}^N$ that sends $[\gamma]$ to $\gamma'(0)$. This gives an isomorphism of $T_pM$ and an $n$-dimensional subspace of $\mathbb{R}^N$. This subspace should not only be isomorphic to the orthogonal complement of $p$ I mentioned above, but actually equal to it. Hence this gives the desired isomorphism and the two definitions agree in that they are isomorphic. What I meant with identify was to say that there is an isomorphism to avoid having to write compositions. – user3118 Jan 26 '23 at 09:57
  • That's certainly correct. I was just emphasizing that in the case of the sphere, $n=2\neq 3 =N$, because I thought you had mixed up what lives where. – peek-a-boo Jan 26 '23 at 14:42
  • @peek-a-boo Thanks a lot, this was more straight forward than I thought it would be, probably because I forgot the orthogonality lemma of the $\gamma'(0)$ before reading your posts. And I may have mixed up where they live to be honest, so your reminder to write this out helped a lot. – user3118 Jan 26 '23 at 14:51
  • See https://math.stackexchange.com/questions/4421786/tangent-vector-of-manifold/4421904#4421904 – Paul Frost Jan 26 '23 at 23:46

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