Given a smooth (differentiable) manifold $M$ and intervals $I_1,I_2\subset\mathbb{R}$, two curves \begin{equation} \gamma_1:I_1\to M\quad\text{and}\quad\gamma_2:I_2\to M~~, \end{equation}

are tangent at $p\in M$ if there exists $\lambda_0\in I_1\cap I_2$ such that \begin{equation*} \gamma_1(\lambda_0)=p\quad\text{and}\quad\gamma_2(\lambda_0)=p~~, \end{equation*}

and the derivatives of the images of $\gamma$ under some coordinate map $\varphi:M\to\mathbb{R}^n$ around $p$ are equal at $\lambda_0$: \begin{equation*} \dfrac{d}{d\lambda}(\varphi\circ\gamma_1)\bigg|_{\gamma_1(\lambda_0)}=\dfrac{d}{d\lambda}(\varphi\circ\gamma_2)\bigg|_{\gamma_2(\lambda_0)}~~. \end{equation*}

At this point, we say that the equivalence classes of all curves tangent at $p$ are the tangent vectors at $p$. However, if the objects in the equivalence class are the curves themselves, wouldn't it be better to define equivalence classes of the derivatives of the curves, and then say those classes are the tangent vectors (since we already know the tangent vectors are the velocities)? As it is now, if tangency is an equivalence relation among curves (as it is in the usual treatment), then we must introduce a non-standard equivalence relation, call it $\stackrel{T_p}{=}$, such that \begin{equation*} \gamma_1\stackrel{T_p}{=} \gamma_2\qquad\longleftrightarrow\qquad \dfrac{d}{d\lambda}(\varphi\circ\gamma_1)\bigg|_p=\dfrac{d}{d\lambda}(\varphi\circ\gamma_2)\bigg|_p~~, \end{equation*}

and then we say there exists an equivalence class, call it $[\gamma_k(\lambda_0)]^T$ consisting of all curves related by $\stackrel{T_p}{=}$.

THIS IS MY QUESTION: Why not use the ordinary equivalence relation $=$ so that the objects in $[\gamma_k(\lambda_0)]^T$ are the derivatives of the curves? This would make much more sense since we're going to end up saying that the tangent vectors are the derivatives later.

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    – ACuriousMind
    Jan 12 at 22:13

2 Answers 2


Your question sounds very similar to the first part here. So I suggest reading that first. I hope that addresses most of the issues, but if not, keep reading.

Why not use the ordinary equivalence relation $=$ so that the objects in $[\gamma_k(\lambda_0)]^T$ are the derivatives of the curves?

By ‘derivatives of the curves’, I hope you mean that of the chart-composed curve, because that is the thing which is $\Bbb{R}^n$-valued and hence already has an existing definition of derivative (a-priori, there is no definition for $\gamma’(0)$ or $\dot{\gamma}(0)$ or $\frac{d\gamma}{d\lambda}(0)$ when $\gamma:I\to M$ is a curve with values in $M$).

Ok, so taking this interpretation, the question now is why not think of tangent vectors to a curve $\gamma$ as the derivative $(\phi\circ\gamma)’(0)\in\Bbb{R}^n$? Well, the answer is that this alone is not a well-defined quantity, because this element of the vector space $\Bbb{R}^n$ very clearly depends on the choice of chart $\phi$. But, of course, this vector is not completely arbitrary, because if we had a different chart map $\psi$, then we would have \begin{align} (\psi\circ\gamma)’(0)&=((\psi\circ\phi^{-1})\circ (\phi\circ\gamma))’(0)=D(\psi\circ\phi^{-1})_{\phi(p)}[(\phi\circ\gamma)’(0)], \end{align} i.e two two vectors $(\psi\circ\gamma)’(0)\in\Bbb{R}^n$ and $(\phi\circ\gamma)’(0)\in\Bbb{R}^n$ are related by the linear map $D(\psi\circ\phi^{-1})_{\phi(p)}:\Bbb{R}^n\to\Bbb{R}^n$ which is the (first) derivative (at the appropriate point) of the chart-transition map. This means there is a very specific relationship between the tangent vectors of the two chart-induced curves, and this is something we can use to create a chart-independent equivalence relation.

So you see that if you were to pursue the idea that a tangent vector out to be a derivative of a curve, in the sense of the usual multivariable calculus derivative of the chart-induced curve, then naturally, the things we should be considering are equivalence classes of triples $(p,(U,\phi),v)$ where $p\in M$ is a given point, $(U,\phi)$ is a given chart for $M$ with $p\in U$, and $v\in\Bbb{R}^n$ is a given vector. The equivalence relation is then that \begin{align} (p,(U,\phi),v)&\sim (q,(V,\psi),w)\quad\iff\quad\text{$p=q$ and $w=D(\psi\circ\phi^{-1})_{\phi(p)}(v)$.} \end{align}

So, to reiterate, for a smooth curve $\gamma:I\to M$ with $\gamma(0)=p$, we don’t yet have a definition for $\gamma’(0)$ via difference quotients. So, we concede that and tell ourselves that “all the information of tangency is contained in the curve itself”, so we decide to consider a certain equivalence class of curves themselves. This is the most geometric way of discussing things.

A further thought might be that equivalence classes of curves sounds like a very weird object (whatever that means) while familiar vectors $(\phi\circ\gamma)’(0)$ are well… familiar. This leads us to the above equivalence relation, so if you want to formulate things in this manner, then you need to find the correct set and correct equivalence relation on that set. In the above case, it was an equivalence relation on $M\times\mathcal{A}_{\text{max}}\times\Bbb{R}^n$ (where $\mathcal{A}_{\text{max}}$ is the maximal atlas defining the smooth structure on $M$). This is of course nothing but the old school idea of transformation laws of vectors expressed formally. This is not as geometric if you just look at it plainly at face value.

All these definitions are equivalent (and in some sense, it doesn’t really matter ‘what’ an object exactly is, but more so how it behaves).


In this lecture tangent vectors at point $p$ are taken to be linear maps $X_{p, \gamma}: C^{\infty}(M) \to \mathbb{R}$ which are induced by curves $\gamma: \mathbb{R}\to M$. That is, consider a curve $\gamma$ which passes through $p\in M$ such that $\gamma(0)=p$. We can then define a linear map $X_{p, \gamma}:C^{\infty}(M)\to \mathbb{R}$ by defining its action an arbitrary $f\in C{^\infty}(M): M\to \mathbb{R}$.

$$ X_{p,\gamma}(f) = (f\circ\gamma)'(0) $$ The tangent space at point $p$, $T_pM$ is then

$$ T_pM = \{X_{p, \gamma}: \gamma \text{ is a smooth curve passing through } p \} $$

So in other words, to each $\gamma$ passing through $p$, we assign a linear map $X_{p, \gamma}:C^{\infty}(M)\to \mathbb{R}$ as described above. We then collect all such linear maps induced by all curves $\gamma$ passing through $p$ and call this the tangent space. Now, there will be many curves passing through $p$ which correspond to the same $X_{p, \gamma}$. That is $X_{p, \gamma} = X_{p, \gamma'}$ does NOT imply $\gamma = \gamma'$ (the mapping between curves and linear maps is not injective).

In this way tangent vectors are defined not using equivalence classes, but rather using a non-injective function between curves and linear maps. The usual results are proven later in the lecture.

  • $\begingroup$ ...and both definitions are equivalent. See e.g. Isham's differential geometry book or Wikipedia. +1 for Schuller (jk) $\endgroup$ Jan 12 at 21:43
  • $\begingroup$ Yes, this approach to constructing the tangent space appears in many places. Most authors call refer to these linear maps as "derivations," (Isham, Lee, Wald, etc.), but this is an independent construction of the tangent vectors, totally divorced from the present approach to tangent vectors as equivalence classes of curves. So, thank you for your input, but this is unrelated to my question about equivalence classes of curves. $\endgroup$ Jan 12 at 21:48
  • $\begingroup$ @hodopsmith no, in that construction tangent vectors are defined to be the abstractly set of derivations $C^{\infty}(M)\to\mathbb{R}$. That construction need make no reference whatsoever to curves. In the definition I have given the tangent vectors are defined to be those maps induced by curves. Clearly the definition I have given can easily be shown to be equivalent to the derivation definition, but the course of the definition is slightly different, and, importantly for your question, critical involves curves. $\endgroup$
    – Jagerber48
    Jan 12 at 21:57
  • $\begingroup$ @jagerber48: It's not clear that the two definitions are equivalent. That they are is a miracle of finite dimensional manifolds, they yield differing notions of tangency in the infinite dimensional case. $\endgroup$ Jan 12 at 22:01
  • $\begingroup$ @Jagerber48 My question was about equivalence classes. Thank you. $\endgroup$ Jan 12 at 22:01

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