I am watching a lecture series on General Relativity and I am currently studying the mathematics required for rigorously understanding the subject. I am trying to self study the math and I am sorry if the following confusion is a silly one. I was going through the definition of tangent space. The lecturer motivates the audience by asking the question "what is the velocity of a curve $\gamma$ at point p?".

Then he gives the following definition.

Def: $(M,O,A)$ be a smooth manifold an the curve $\gamma:R \rightarrow M$ be at least $C^1$. Suppose $\gamma(\lambda_0)=p$.

THen, the velocity of $\gamma$ at p is the linear map $$v_{\gamma,p}: C^{\infty}(M) \rightarrow R$$ $$f \rightarrow (f o \gamma )^{'}(\lambda_0)$$

By difficulty is with understanding the introduction of the function f. If we are interested ion finding the velocity of the curve at p why is there an f in the definition?

I went on to read Wikipedia to get an alternative definition and I found the following definition under the heading "definition as velocities of curves":

$$(d\phi)_x:T_xM\rightarrow R^n$$ $$(d\phi)_x(\gamma^{'}(0))=(\phi o \gamma)^{'}(0)$$ where $\phi: M \rightarrow R^n$ is a chart. This definition makes sense to me because we just have the curve $\gamma$ and we have the chart $\phi$ to take it to $R^n$. But, again in the same Wikipedia article when the definition of velocity is given in terms of derivations they again make use of a smooth function $f$ on M. Why is it required?

I am not sure whether I have posed the question in the right manner and most probably it is a lame question but this has been bothering me for sometime.

  • 2
    $\begingroup$ $\nu_{\gamma, p}$ sends an arbitrary smooth function, $f$, to its "derivative in the direction of gamma at p". It is rather abstract to think of a vector as a map from the space of smooth functions to $\mathbb{R}$, but its convenient because it doesn't make use of any ambient euclidean space, i.e. its a definition that makes sense for an abstract manifold and clearly doesn't depend on how the manifold is imbedded in some $\mathbb{R}^k$ $\endgroup$ – Tim kinsella Oct 15 '16 at 7:43

The $f$ in the defenition you have used is an arbitrary function from $M$ to $\mathbb{R}$. Also known is that the function is a smooth one, as $f \in C^{\infty}(M) $.

So, you have chosen a smooth function on the manifold. You also have a curve $\gamma : \mathbb{R} \to M$. So, $f \circ \gamma$ is a map from $\mathbb{R}$ to itself, and has a derivative. By finding this derivative at the point $\lambda_0$, you are essentially computing the rate at which the values of $f$ change as we move along the curve $\gamma$. This is the very defenition of velocity of a curve.

As an analogy, assume $f$ gave you the temperature at every point on the manifold. Then I could describe my velocity at a point on a curve in terms of how quickly I felt the temperature changing. Note that this only works if the temperature function was a smooth function.

Also, note that $f$ is a fixed function; You wouldn't do very well to describe one velocity in terms of how the temperature was changing and the next in terms of how the air pressure was changing. This is the reason why when adding to velocities at $\lambda_0$ we define the addition as: $$v_{\gamma , p} + v_{\delta, p} = v_{\sigma, p}$$iff $$v_{\sigma, p}(f) = v_{\gamma , p}(f) + v_{\delta, p}(f).$$

It migh also help to notice that is if $v_{\sigma, p}(f) = v_{\gamma , p}(f) + v_{\delta, p}(f)$ then $v_{\sigma, p}(g) = v_{\gamma , p}(g) + v_{\delta, p}(g)$ (as long as $g\in C^{\infty}(M))$ . So, we can go on to provide the $v$'s vector space structure irrespective of what function you choose, as long as it is smooth.

So, to sum it all up: The velocity of a curve $\gamma$ at $\lambda_0$ is an object $v_{\gamma , p} $ that takes a smooth function as an argument and tells you how it is changing along the given curve $\gamma$. That is a pretty nice defenition, in my opinion.

On a similar note, consider the chart coordinate map $x: M \to \mathbb{R}^d $. In place of $f$, try using the components of $x$. ie. $x^i: M \to \mathbb{R}$

  • $\begingroup$ Thanks you. Now things are clear to me. I guess this whole confusion arose because I didn't have a proper idea of what is called as the velocity of a curve. Now, things are clear to me. Thanks once again. $\endgroup$ – Rajath Krishna R Oct 15 '16 at 9:17

Let $T_pM$ be the tangent space of $M$ in $p$. One of the definitions of the tangent space derives from the following.

Two manifold deifferentiable curves $γ,σ$ osculate on $p$ when (instead of zero you can put $λ_0$)


2)There is a chart $(U,φ)$ of $M$ so that $D(φογ)_0=D(φοσ)_0$

Osculation gives an equivalence relation and let's denote $v_{p,γ}=v=[(p,γ)]$ the equivalence classes.

Then these vectors/equivalence classes $v$ form the vector space we call $T_pM$.

One characteristic of these vectors is that they can be seen as derivations, meaning that they satisfy Leibniz' product law, and as such they can be seen as functions $v:C^{\infty}(M) \to \Bbb R$ like this


This definition of velocities is equivalent to the second one you give.

(Again wherever I put $0$ you can put $λ_0$)


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