I Cordinates on a manifold

kiuhnm

Let $M$ be an $n$-dimensional (smooth) manifold and $(U,\phi)$ a chart for it. Then $\phi$ is a function from an open of $M$ to an open of $\mathbb{R}^n$. The book I'm reading claims that coordinates, say, $x^1,\ldots,x^n$ are not really functions from $U$ to $\mathbb{R}$, but from $\mathbb{R}^n$ to $\mathbb{R}$. In other words, we should write $$\phi(p) = (x^1(\phi(p)), \ldots, x^n(\phi(p)))$$ instead of $$\phi(p) = (x^1(p), \ldots, x^n(p)).$$ That doesn't sound right to me. $\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}$, which means that an element of $\mathbb{R}^n$ is already an $n$-tuple, so why should we define an extra coordinate system on it? If we want a different coordinate system we just pick a different chart and not the same chart with a different coordinate system for $\mathbb{R}^n$.
Do you agree or am I missing something?

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fresh_42

Mentor
2018 Award
Let $M$ be an $n$-dimensional (smooth) manifold and $(U,\phi)$ a chart for it. Then $\phi$ is a function from an open of $M$ to an open of $\mathbb{R}^n$. The book I'm reading claims that coordinates, say, $x^1,\ldots,x^n$ are not really functions from $U$ to $\mathbb{R}$, but from $\mathbb{R}^n$ to $\mathbb{R}$. In other words, we should write $$\phi(p) = (x^1(\phi(p)), \ldots, x^n(\phi(p)))$$ instead of $$\phi(p) = (x^1(p), \ldots, x^n(p)).$$ That doesn't sound right to me. $\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}$, which means that an element of $\mathbb{R}^n$ is already an $n$-tuple, so why should we define an extra coordinate system on it? If we want a different coordinate system we just pick a different chart and not the same chart with a different coordinate system for $\mathbb{R}^n$.
Do you agree or am I missing something?
We do not have coordinates on $M$, which is why we use those of the chart, which are coordinates at $\phi(p)$. A coordinate $x^i(p)$ is simply not defined, resp. will be defined as $y^i(p) := \phi^{-1}(x^i(\phi(p))$. So the question is: Where are we, on the manifold or in the Euclidean space of the chart?

kiuhnm

We do not have coordinates on $M$, which is why we use those of the chart, which are coordinates at $\phi(p)$. A coordinate $x^i(p)$ is simply not defined, resp. will be defined as $y^i(p) := \phi^{-1}(x^i(\phi(p))$. So the question is: Where are we, on the manifold or in the Euclidean space of the chart?
An element of $\mathbb{R}^n$ has the form $(x_1,\ldots,x_n)$ by definition, so we must have $\phi(p) = (\phi_1(p),\ldots,\phi_n(p))$. Why should we define a coordinate system for $\mathbb{R}^n$ when $\mathbb{R}^n$ is already a cartesian product? I would agree with you if we were talking about the vector space $\mathbf{R}^n$, but we're not.

fresh_42

Mentor
2018 Award
I don't know which context we have here, but e.g. if the $\mathbb{R}^n$ is the tangent space, then the coordinates are functions themselves. This is also generally true: if we vary $\phi(p)$ we get different coordinates. And they do depend on the image point on the chart, and not on the one of the manifold, if we write it correctly.

kiuhnm

I don't know which context we have here, but e.g. if the $\mathbb{R}^n$ is the tangent space, then the coordinates are functions themselves. This is also generally true: if we vary $\phi(p)$ we get different coordinates. And they do depend on the image point on the chart, and not on the one of the manifold, if we write it correctly.
We're not talking about tangent spaces here. The manifold could be non-smooth. We're talking about the charts $(U,\phi)$ where $\phi:U\to\mathbb{R}^n$ is an homeomorphism. The point is whether we need to introduce coordinates $x_1,\ldots,x_n$ explicitly or if $x_i$ is just $\phi_i$. I'm saying it's the latter, whereas the author of the book and you (if I understand you correctly) say it's the former. My argument is that $\mathbb{R}^n$ is a cartesian product so we don't need to introduce a basis and coordinates to indicate points in $\mathbb{R}^n$. The points can already be indicated through tuples thanks to $\mathbb{R}^n$ being a cartesian product, so I don't think that identifying $\phi_i$ with $x_i$ is an abuse.

fresh_42

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2018 Award
We're not talking about tangent spaces here. The manifold could be non-smooth. We're talking about the charts $(U,\phi)$ where $\phi:U\to\mathbb{R}^n$ is an homeomorphism. The point is whether we need to introduce coordinates $x_1,\ldots,x_n$ explicitly or if $x_i$ is just $\phi_i$. I'm saying it's the latter, whereas the author of the book and you (if I understand you correctly) say it's the former. My argument is that $\mathbb{R}^n$ is a cartesian product so we don't need to introduce a basis and coordinates to indicate points in $\mathbb{R}^n$. The points can already be indicated through tuples thanks to $\mathbb{R}^n$ being a cartesian product, so I don't think that identifying $\phi_i$ with $x_i$ is an abuse.
$\phi = (\phi_i)\, : \, M \supseteq U \longrightarrow \mathbb{R}^n$ and $(x_i) \, : \, \mathbb{R}^n \longrightarrow \mathbb{R}^n\,.$ Since $p \notin \mathbb{R}^n\,,$ $x_i(p)$ is a sloppy notation.

Infrared

Gold Member
I agree with @kiuhnm here. In my experience, $x^i$ means the $i$th component of $\phi$, so is a function $U\to\mathbb{R}$. Can you show exactly what your textbook says?

stevendaryl

Staff Emeritus
I don't understand, either. If $\phi$ is a function from $U$ to $R^n$, and the coordinates are frunctions from $R^n$ to $R$, why not just say that coordinates are a function from $U$ to $R$?

What I do understand that a coordinate transformation is a function from $R^n$ to $R^n$. So the transformations no longer involve the manifold.

kiuhnm

$\phi = (\phi_i)\, : \, M \supseteq U \longrightarrow \mathbb{R}^n$ and $(x_i) \, : \, \mathbb{R}^n \longrightarrow \mathbb{R}^n\,.$ Since $p \notin \mathbb{R}^n\,,$ $x_i(p)$ is a sloppy notation.
Whether something is an abuse or not depends on the definitions you choose. You decided to define the $x_i$ the way you did, which makes my notation an abuse. But the point is this: what do you gain by using your definition? Or in other words, what do I lose by not using it?

pasmith

Homework Helper
I agree with @kiuhnm here. In my experience, $x^i$ means the $i$th component of $\phi$, so is a function $U\to\mathbb{R}$. Can you show exactly what your textbook says?
This would be my understanding also (as long as the chart is clear from the context).

fresh_42

Mentor
2018 Award
Or in other words, what do I lose by not using it?
It depends on whether you want to distinguish the Euclidean coordinates from their pullbacks to the manifold. As I mentioned earlier: it is all a matter of where you are. If the $x_i$ are Euclidean coordinates, then they do not apply to points of the manifold, but on their image under the chart instead. If they are meant to be the local coordinates on the manifold, then they apply to points of it, but aren't Cartesian anymore. If you prefer a mishmash then you can switch back and forth and risk the loss of rigor. You seem to identify base points with points anywhere in the fiber, a bit too arbitrary if you ask me, because you reduce notation to a single section.

kiuhnm

I agree with @kiuhnm here. In my experience, $x^i$ means the $i$th component of $\phi$, so is a function $U\to\mathbb{R}$. Can you show exactly what your textbook says?
It says:

--- starts ---
Let $(U,\phi)$ be a coordinate chart with $p \in U$ and suppose that $\phi(p)=q$. If $x^1,\ldots,x^n$ are the standard coordinate functions on $\mathbb{R}^n$ then $q$ has coordinates $(x^1(q),\ldots,x^n(q))$. Thus we can write $$\phi(p) = (x^1(q),\ldots,x^n(q)).$$ It is a convenient abuse of notation to view the $x^i$ as coordinate functions on $U$ instead of $\phi(U)$ and to write $$\phi(p) = (x^1(p),\ldots,x^n(p)).$$ As $U$ generally does not live in Euclidean space, the second expression above makes no sense unless it is interpreted to mean the first one. (Of course, if $U$ is a subset of Euclidean space then $\phi$ is just the identity map, so the second expression makes perfect sense; this is the justification for the convention.)
--- ends ---

While it's true that the $(x^i)$ can't be global coordinates on $U$ if $U$ is not Euclidean, they can certainly be local coordinates on $U$. Wasn't that the whole point of using the charts?

fresh_42

Mentor
2018 Award
While it's true that the (xi)(xi)(x^i) can't be global coordinates on $U$ if $U$ is not Euclidean, they can certainly be local coordinates on $U$. Wasn't that the whole point of using the charts?
Yes, as long as you are talking about a certain section such as tangent spaces, which is why I asked.
You seem to identify base points with points anywhere in the fiber, a bit too arbitrary if you ask me, because you reduce notation to a single section.
It works in what was the whole point of using the charts but terminology developed and was generalized.

kiuhnm

It depends on whether you want to distinguish the Euclidean coordinates from their pullbacks to the manifold. As I mentioned earlier: it is all a matter of where you are. If the $x_i$ are Euclidean coordinates, then they do not apply to points of the manifold, but on their image under the chart instead. If they are meant to be the local coordinates on the manifold, then they apply to points of it, but aren't Cartesian anymore. If you prefer a mishmash then you can switch back and forth and risk the loss of rigor. You seem to identify base points with points anywhere in the fiber, a bit too arbitrary if you ask me, because you reduce notation to a single section.
Maybe I'm starting to see the problem with my definition: the coordinates would just be a local parametrization of the curve but maybe I'd lack enough structure to do regular calculus over them. The cleanest way is to define $\phi$ from $M$ to $\mathbb{R}^n$ and then the coordinates on $\mathbb{R}^n$ itself. That way, it's clear we can do regular calculus with them. I don't know anything about fibers and such so please try to keep the discussion a little less formal for my sake :)

lavinia

Gold Member
Maybe I'm starting to see the problem with my definition: the coordinates would just be a local parametrization of the curve but maybe I'd lack enough structure to do regular calculus over them. The cleanest way is to define $\phi$ from $M$ to $\mathbb{R}^n$ and then the coordinates on $\mathbb{R}^n$ itself. That way, it's clear we can do regular calculus with them. I don't know anything about fibers and such so please try to keep the discussion a little less formal for my sake :)
Right, An n-tuple of functions might not be a coordinate chart.

kiuhnm

Right, An n-tuple of functions might not be a coordinate chart.
I'm still not completely sure. We assumed that $(U, \phi)$ is a coordinate chart from the start. Let's say $\phi$ goes from the manifold $M$ to $\mathbb{R}^n$ and we also have some coordinates $x^i:\mathbb{R}^n\to\mathbb{R}$. Then let's define $\phi^i = x^i\circ\phi$. Would $(U, (\phi^1,\ldots,\phi^n))$ still be a valid chart? I suspect we threw away some structure from a strictly formal and maybe overly pedantic point of view but in practice we know it's still there and so we don't care.

lavinia

Gold Member
I'm still not completely sure. We assumed that $(U, \phi)$ is a coordinate chart from the start. Let's say $\phi$ goes from the manifold $M$ to $\mathbb{R}^n$ and we also have some coordinates $x^i:\mathbb{R}^n\to\mathbb{R}$. Then let's define $\phi^i = x^i\circ\phi$. Would $(U, (\phi^1,\ldots,\phi^n))$ still be a valid chart? I suspect we threw away some structure from a strictly formal and maybe overly pedantic point of view but in practice we know it's still there and so we don't care.
Yes, it would be a chart. $(U, (\phi^1,\ldots,\phi^n))$ is the same as$(U, \phi)$. Otherwise put, $(\phi^1,\ldots,\phi^n))$ is the same mapping of $U$ into $R^{n}$ as $\phi$.

All the book is pointing out is that the coordinates functions are compositions of $\phi$ with the coordinate functions on $R^{n}$. If one had different coordinates on $R^{n}$ there would be different coordinate functions on $U$. If one had a different chart on $U$ the coordinate functions on $U$ would be different.

I thought you were saying that if you just had an n-tuple of functions on $U$ that is would not necessarily make a chart. That is true.

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kiuhnm

Yes, it would be a chart. $(U, (\phi^1,\ldots,\phi^n))$ is the same as$(U, \phi)$. Otherwise put, $(\phi^1,\ldots,\phi^n))$ is the same mapping of $U$ into $R^{n}$ as $\phi$.

All the book is pointing out is that the coordinates functions are compositions of $\phi$ with the coordinate functions on $R^{n}$. If one had different coordinates on $R^{n}$ there would be different coordinate functions on $U$. If one had a different chart on $U$ the coordinate functions on $U$ would be different.

I thought you were saying that if you just had an n-tuple of functions on $U$ that is would not necessarily make a chart. That is true.
In my head I was identifying $x^i$ with $\phi^i$ and so the $x^i$ were local coordinates directly on $U$.

In the meantime I'd like to thank you all for your patience!

"Cordinates on a manifold"

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