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I've recently started studying differential geometry and I'm a bit unsure on the notion of a tangent vector on a manifold. Is the point that we can no longer thing of a vector as an arrow (a straight line) extending between two points (in general we cannot compare two points on a manifold), as there is no well-defined concept of an origin, or indeed, what a straight-line is. Hence we need a new definition of a vector?! Given this, we can define a curve $\gamma :[0,1]\subset\mathbb{R}\rightarrow M$ (Question: doesn't this implicitly define a 1-D coordinate system, or is the point that a curve on the manifold exists independently of any coordinate system and hence we can choose a real parameter $t$ such that each value of t maps to a point on the manifold, therefore tracing out a curve on the manifold?). We then define a tangent vector in terms of a differentiable function $f$ by stating that tangent vector at a point $p\in M$ is the directional derivative of $f$ along the curve $\gamma$ at this point, i.e. $$\mathbf{v}_{p}[f]=\frac{d}{dt}(f\circ\gamma)\bigg\vert_{t=0}$$ where $\gamma (0)=p$. Clearly this definition is independent of coordinates, and it can also be shown to be independent of the curve passing through $p\in M$ (also, it is independent of the differentiable function $f$ as this was chosen arbitrarily). A tangent vector at $p$ is then defined to be the equivalence class of curves that have the same derivative at $p$, i.e. $$\mathbf{v}_{p}=[\gamma]=\lbrace \tilde{\gamma}\vert\;(\phi\circ\tilde{\gamma})'(0)=(\phi\circ\gamma)'(0)\rbrace$$ I'm really trying to understand the motivations and intuition behind this definition.

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  • $\begingroup$ It is not completely clear what you ask for. Yes, this is one possible definition of a tangent vector. The intuition is very simple: it is the set of all directions in which you can move, if (a) you live in this manifold and the manifold is all of your universe, and (b) you just happen to be in point $p$. $\endgroup$ – Peter Franek Mar 2 '15 at 14:34
  • $\begingroup$ I guess my real questions are: do we have to define a direction at a point (via the curve $\gamma$) in order to construct the notion of a vector at that point (such that the directional derivative of $f$ along $\gamma$ defines a the rate of change of $f$ along the vector tangent to $\gamma$ at point $p$). Also, a possible stupid question, but by paramterising $\gamma$ in terms of $t$ are we not implicitly defining a (kind of) 1-dimensional coordinate system? $\endgroup$ – Perpetual learner Mar 2 '15 at 14:50
  • $\begingroup$ (1) This is one way how to define tangent vectors. There are many other ways, equivalent. For example, you can define it just as vectors in $\mathbb{R}^n$ (with a fixed "origin" that corresponds to $p$) using a coordinate frame, just you need to identify the same vector when using a different coordinate frame. (2) No, there is no such thing as $1$-dimensional coordinate system, if the manifold is higher dimensional. A curve is.. just a curve :) $\endgroup$ – Peter Franek Mar 2 '15 at 14:55
  • $\begingroup$ (2) What if the manifold is itself 1-dimensional though? $\endgroup$ – Perpetual learner Mar 2 '15 at 14:59
  • $\begingroup$ Then the manifold consists just of disjoint copies of intervals and circles.. Not very interesting. Then an (imbedded) curve is a coordinate frame, yes. $\endgroup$ – Peter Franek Mar 2 '15 at 15:01
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I'm not sure how you understand a vector, but to me its simply an element of some vector space. The vector space is more important.

Here is a very conceptual answer to what I personally consider a motivation behind the definition of tangent vectors.

If you think of a manifold embedded in $\mathbb{R}^n$, intuitively the tangent space at a point is the "tangent hyperplane" "tangent" to that point. Any vector in this plane gives you a directional derivative.(you differentiate "stuff" along that "direction").

However without being embedded in $\mathbb{R}^n$, it is not clear how to differentiate functions on a manifold. What you can do is by utilizing a curve $\gamma : [-1,1]\to M$ and get a function $f\circ \gamma :\mathbb{R}\to \mathbb{R}$, which you know how to differentiate. One then fiddles around and realizes that the good condition to distinguish "different"(in the sense that they give "different" directional derivatives) curves is what you gave in your question.

But if one thinks abstractly what really is a directional derivative, it's something that takes a function $f\in C^\infty(M)$ as an input and gives you a number. Furthermore it should be a linear operator, and should satisfy the Leibniz rule. It then turns out these are exactly the "nice conditions". These are called derivations and is an equivalent way of defining the tangent space.

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Whitney’s Embedding Theorem says that we can smoothly embed any smooth manifold into $ \mathbb{R}^{n} $ for $ n $ large enough. In other words, any smooth manifold can be viewed as a smooth sub-manifold of a Euclidean space of sufficiently high dimension. Once we have this picture of a smooth manifold, we can draw a tangent plane to any point $ p $ on the smooth manifold and view tangent vectors as vectors in this hyperplane, where the origin is taken to be $ p $.

If we have a smooth curve $ \gamma: (-1,1) \to M $ in the original abstract manifold $ M $, we can transfer it to a smooth curve $ \gamma': (-1,1) \to \mathbb{R}^{n} $ whose image lies in the corresponding sub-manifold of $ \mathbb{R}^{n} $. As you know, the time derivative $ \dfrac{d}{dt} [\gamma'(t)] \Bigg|_{t = 0} $ of this curve in $ \mathbb{R}^{n} $ is a vector that is tangent to $ \gamma' $ at $ t = 0 $.

The reason why we need an abstract definition of a tangent vector is that we must be able to discuss differential geometry in a coordinate-free language. Manifolds have a reality that is independent of any choice of coordinates.

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