# Parametrizations and coordinates in differential geometry - what's the difference?

From what I've read one can introduce the notion of a tangent vector to a point on a manifold in terms of an equivalence class of curves passing through that point (the equivalence relation being that they have the same tangent at that point). Now my confusion arises from the fact that in the texts that I've read, the author introduces a curve $\gamma :(a,b)\subset\mathbb{R}\rightarrow M$ that is parametrized in terms of some real parameter $t\in (a,b)$, with $a<0<b$ and $\gamma (0)=p\in M$. From this a tangent vector at a point $p\in M$ can be defined in a coordinate independent manner, in terms of the directional derivative of function $f:M\rightarrow\mathbb{R}$, as $$\frac{df}{dt}\bigg\vert_{p}$$ I know that if one introduces a coordinate chart $(U,\phi)$ (where $p\in U$) then the curve can be represented in terms of the local coordinates, $$(\phi\circ\gamma)(t)=\gamma (x^{1}(t),\ldots ,x^{n}(t))$$ So what distinguishes a parameter $t$ (parametrizing the curve $\gamma$) from a coordinate? Is it that the parameter is defined in terms of an intrinsic property of the curve (such as arc-length) and thus is independent of any coordinate system, or am I completely misunderstanding things?

• At best, a parameter (in this setting) describes a point on a curve in $M$, not a general point in some neighborhood of $p$. Further, nothing prevents $\gamma$ from crossing itself, so "the parameter value of a point" needn't be well-defined. But perhaps these observations miss the point of your question. Does the question arise because, in some settings, a parametrization of a neighborhood $U$ of $p$ is the inverse map of a coordinate system in $U$...? – Andrew D. Hwang Jun 29 '15 at 11:29
• @AndrewD.Hwang Yes to your last part. As far as I understand, when one introduces a coordinate map $\phi:U\subset M\rightarrow V\subset\mathbb{R}^{n}$ for a patch on a manifold each point $p\in U$ in this given patch is labelled by a unique value $\phi (p)=(x^{1},\ldots ,x^{n})\in\mathbb{R}^{n}$ and we can use the inverse of this map $\phi^{-1}:V\rightarrow U$ to analyse the points in the patch directly by their parametrization in terms of their coordinate labels, i.e. $p=\phi^{-1}(x^{1},\ldots ,x^{n})=(u^{1},\ldots ,u^{n})\in U$, right? So what distinguishes the two?! – Will Jun 29 '15 at 11:39
• The term "parametrization" is getting used in two distinct senses: 1. As a description of a curve representing a tangent vector at $p$; 2. As the inverse of a coordinate chart. That's terminological coincidence. So, to clarify, your question is about the distinction between a "parametrization" in the second sense and a coordinate chart? If that's right: The two certainly differ formally. Just as you say, a chart $\phi$ maps a subset $U$ of $M$ to a subset $V$ of $\mathbf{R}^{n}$ while a parametrization $\phi^{-1}:V \to U$ maps in the other direction. I fear I'm still missing the point. – Andrew D. Hwang Jun 29 '15 at 12:05
• @AndrewD.Hwang My confusion really is that what distinguishes the parametrization of a curve from the "second sense". Why is the parameter $t$ not a coordinate? Is it because, in general there is no local inverse map $\gamma^{-1}:M\rightarrow (a,b)$? I got confused when trying to explain to a friend why the definition of a tangent vector in terms of curves on the manifold is coordinate independent - he argued that isn't one just introducing a new coordinate $t$ to parametrize the curve and so the argument is circular? – Will Jun 29 '15 at 12:24

Generally in geometry, the term "coordinate (function)" near a point $p$ of a manifold $M$ refers to one component of a coordinate chart $\phi$ defined in some open neighborhood $U$ of $p$.

In the preceding sense, even if $\gamma$ is a simple smooth curve through $p$ (i.e., $\gamma$ is one-to-one, infinitely differentiable, defined on some neighborhood of $0$, and satisfies $\gamma(0) = p$), the parameter $t$ of $\gamma$ is not a coordinate on $M$ because $\gamma$ is not (of itself) a parametrization of some open neighborhood of $p$ in $M$ (unless $M$ is a curve).

That said, there might well exist a parametrization of some neighborhood $U$ of $p$ that extends $\gamma$, in which case $t$ would become a coordinate at $p$ in the resulting chart.

Incidentally, when one says "the definition of a tangent vector at $p$ is coordinate-independent", it means in a loose philosophical sense that there is a definition, possibly referring to a coordinate chart near $p$ or a parametrization of a neighborhood $U$ of $p$, that defines precisely the same criterion no matter which chart is used. (Physicists often express the same idea as "the object transforms correctly" under change of coordinates.)

This is not the meaning one might read from the non-mathematical semantics of the phrase. :)

For example Wikipedia's definition of a tangent vector decrees that two smooth curves $\gamma_{1}$ and $\gamma_{2}$ through $p$ are equivalent if, with respect to some coordinate chart $\phi$ defined near $p$, $$\frac{d}{dt}\bigg|_{t = 0} (\phi \circ \gamma_{1})(t) = \frac{d}{dt}\bigg|_{t = 0} (\phi \circ \gamma_{2})(t).$$ In concrete terms, if the curves are expressed as $n$-tuples of smooth, real-valued functions, they have the same velocity vector at $0$ in the sense of elementary calculus. It's easy to see this definition doesn't depend on the chart, though it does refer to an arbitrary chart.

• So is the point that $\gamma$ simply defines a path along the manifold and is not coordinate chart for any points on the manifold and thus when one parametrizes it, such a parameter $t$ is not a coordinate of the curve (in general the map $t\mapsto\gamma (t)$ is not one-to-one)... – Will Jun 29 '15 at 16:20
• ...Is this how one could justify the coordinate independence of this definition - as coordinate maps label a unique point on the manifold, whereas the map $\gamma$ does not label points along the curve uniquely (in general), i.e. $\gamma (t_{1})=\gamma (t_{2})$ does not imply that $t_{1}=t_{2}$, whereas for a coordinate map $\phi$, $\phi (p)=\phi (q)\Rightarrow p=q$? Is it also correct to say that the definition of a tangent vector (as in terms of curves) is independent of the parametrization of the curve? – Will Jun 29 '15 at 16:20
• You might say there are two issues. 1. It's true $\gamma$ might not be simple; in an extreme case $\gamma$ could be constant (representing the zero vector). 2. Even if $\gamma$ is simple, the image of $\gamma$ is not a neighborhood of $p$ unless $M$ is a curve. [...] – Andrew D. Hwang Jun 29 '15 at 16:47
• Regarding your second comment, it might be more accurate to say that a reparametrization of $\gamma$ is not generally equivalent to $\gamma$. That is, the tangent vector represented by $\gamma$ is not invariant under reparametrization. For example, traversing a curve twice as quickly multiplies the represented tangent vector by $2$, even though the same set of points is traced in each case. – Andrew D. Hwang Jun 29 '15 at 16:50
• Ah, so is the term coordinates of a point on a manifold exclusively referring to the case in which a map maps a neighbourhood of a point on the manifold to a subset of $\mathbb{R}^{n}$ in a one-to-one fashion? – Will Jun 29 '15 at 16:56