In the description of a manifold, we often start with the mathematical definition that $M=\cup M_i$ and if $m\in M_i \subset M$, where m is a point on the manifold, then it is mapped by a one-to-one map to $\mathbb{R}^n$. Then another case where point $m \in M_i \cap M_j$, then it is defined by an extra more map. Where I want to go is towards the compound map $\phi_j \circ \phi_i^{-1}$ which is from $\mathbb{R}^n$ to $\mathbb{R}^n$. Books and all known references (any reference on manifolds) say that the compound map is then specified by the set of functions $x'^{\mu}(x^{\nu})$. I want to know why are we able to make this link between a compound map and those functions. What does this mean?


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  • $\begingroup$ 1. This seems to be a pure math question 2. The $x'^\mu(x^\nu)$ are the "compound map" from one (open subset of) $\mathrm{R}^n$ to another, there is no "link" to make $\endgroup$ – ACuriousMind May 26 '15 at 14:46
  • $\begingroup$ 1. No I asked because this $x'^{\mu}(x^{\nu})$ appears in general relativity. 2. What is the significance of $x'^{\mu}(x^{\nu})$, I maybe don't get the meaning of the mathematical structure of $x'^{\mu}(x^{\nu})$. @ACuriousMind $\endgroup$ – PhilosophicalPhysics May 26 '15 at 14:54
  • $\begingroup$ Another way to say @ACuriousMind's point 2 is that the $x'^{\mu}(x^{\nu})$ are the components of the map. You said yourself that the map is $\phi_j \circ \phi_i^{-1}:\mathcal{U}=M_i\cap M_j\to\mathbb{R}^n$ to $\mathbb{R}^n$. So it spits out an array of numbers, which are functions of the co-ordinates $x_\nu$ of the in the open set $\mathcal{U}\subseteq \mathbb{R}^n$. In physics, I like to think of the meaning of the transition map (which is another name for your $\phi_j \circ \phi_i^{-1}$) as the following: the boundaries of our patches $M_i$ are not "real" in the sense that ... $\endgroup$ – WetSavannaAnimal May 27 '15 at 2:38

A lot of your question really boils down to how to switch coordinates, which already happens in $\mathbb{R}^n$, but usually purely for convenience. In General Relativity, you might be forced to changed coordinates. And there is also the physical implications of general coordinate transformations. This answer will try to address all these issues.

Firstly, let's address coordinates, how to switch and why in General Relativity you sometimes are forced to switch coordinates. We will start with a single coordinate system. For an event $m$ (a point in your manifold) in a region of spacetime $M_i \subset M$, where $M$ is the total spacetime, then there can be a coordinate map $\phi_i$ which is a one-to-one mapping from all of $M_i$ to $\mathbb{R}^4$. So specifically there are $4$ numbers called coordinates for each event $m$, and every distinct event in $M_i$ gets a different set of $4$ numbers and vice versa. So if you look at different coordinates you get different points in $M_i\subset M$, but no matter how you merely change your coordinates you are always in $M_i$ since $\phi_i^{-1}$ always maps to $M_i$.

But in General Relativity, sometimes you are forced to have multiple coordinate systems because no coordinate system works for the whole spacetime. This sometimes happens if your spacetime has a nontrivial topology. This is technically possible even without curvature, and even without matter of any kind. An example is an empty Pac-Man type universe. Or a universe that is flat and infinite (and empty) spatially but time is like an angle, after $2\pi$ units of time, you are back to the past. I'll go into some detail about that last one.

So let's look at that example spacetime. Your spacetime can be $\mathbb{R}^3\times\mathbb{S}^1$, it is like a cylinder with time wrapping around, but you have three dimensions of flat space extending out, not one. If you want an even more explicit set you can consider: $$M=\left\{(v,w,x,y,z)\in \mathbb{R}^5: v^2+w^2=1\right\}.$$

However, you should not take literally the mere appearance that $M$ is a subset of $\mathbb{R}^5$. Spacetime just is what it is so as to produce the results we see experimentally and observationally. There can be multiple mathematical objects that predict the same experimental and observational outcomes, it would be unscientific to read too much into an accidental feature of just one of the many such objects. An object that makes the same predictions as $M$ above is:

$$\left\{\left\{(w+n2\pi,x,y,z):n\in\mathbb{Z}\right\}\subset \mathbb{R}^4: w,x,y,z \in \mathbb{R}\right\}.$$

So let's look at $$M=\left\{(v,w,x,y,z)\in \mathbb{R}^5: v^2+w^2=1\right\},$$ and define $$M_1=\left\{(v,w,x,y,z)\in \mathbb{R}^5: v^2+w^2=1, v \leq \sqrt{2}/2\right\}, \text{ and}$$

$$M_2=\left\{(v,w,x,y,z)\in \mathbb{R}^5: v^2+w^2=1, v \geq -\sqrt{2}/2\right\}.$$

By construction, $M_i\subset M$, and so now we can define our coordinate maps by $\phi_1^{-1}(t,x,y,z)$ = $(-\cos(3\arctan(t)/2),\sin(3\arctan(t)/2),x,y,z)$ and by $\phi_2^{-1}(t,x,y,z)$ = $(\cos(3\arctan(t)/2),\sin(3\arctan(t)/2),x,y,z).$ Now we see that you can look at any value of $t$ whichever one you like, but as long as you use just one of those coordinate maps, $\phi_i^{-1}$, you will always be in $M_i$.

To be a manifold, technically we only need continuity of $\phi_j\circ \phi_i^{-1}$ and $\phi_i\circ \phi_j^{-1}$. But that's not good enough to do physics, we want to be able to do calculus too. But those compound maps are actually differentiable, so we can actually do calculus in a coordinate patch and transfer over our derivatives and such as well. So now if you want to describe a curve in $M$ from $(-1,0,0,0,0)\in M_1$ to $(1,0,0,0,0)\in M_2$ that moves in a steady way and leaves $x=y=z=0$ and the whole way has $w\geq 0$, then you can imagine it as a curve in $\mathbb{R}^4$ that starts at the origin and goes in the $-t$ direction. Use $\phi_1{-1}$ to turn that curve in $\mathbb{R}^4$ into a curve in $M_1$. Eventually the curves enters $M_1\cap M_2$ but no matter how what value of $t$ we put into $\phi_1^{-1}(t,0,0,0)$, we are never going to be outside $M_1$. But there is a value of $t$ ($t=\tan (\pi/3)$) for which $\phi_1^{-1}(t,0,0,0)$=$(0,1,0,0,0)$, and at this point we are in $M_1\cap M_2$ so we could start using the other coordinate map $\phi_2$ and have it turn the curve in $M_1\cap M_2$ into a curve in $\mathbb{R}^4$. At this point it might help to imagine two copies of $\mathbb{R}^4$, one for each coordinate map. And note that when $t$ is close enough to zero ($|t|\leq\tan (\pi/6)$), you are not in $M_1\cap M_2$, but as soon as $t$ is too far away from zero ($|t|>\tan (\pi/6)$) in one copy of $\mathbb{R}^4$ any point in one copy of $\mathbb{R}^4$ is effectively in $M_1\cap M_2$ and hence also in the other copy of $\mathbb{R}^4$. The topology is requiring that you switch between the two coordinates.

This is no different than changing coordinates in $\mathbb{R}^n$, you get different numbers but it is the same point. So your question comes down to how to switch from coordinates in one map to coordinates in a different map.

If you have coordinates in $M_i$ you have $ n$ numbers $ (x_1,x_2, \dots x_n)$ and if you use the coordinate map $ \phi_i^{-1}$ you can take those coordinates and get the point $m=\phi_i^{-1}(x_1,x_2, \dots x_n)$ they correspond to in that coordinate system. Then you can use the coordinate map $\phi_j$ to get the $n$ numbers that same point corresponds to in the $ M_j$ coordinate system, the coordinates $\phi_j(m)=\phi_j(\phi_i^{-1}(x_1,x_2, \dots x_n))$.

Now looking at $\phi_j \circ \phi_i^{-1}$ we see it takes the coordinates in the $M_i$ coordinate system and gives us the coordinates in the $M_j$ coordinate system. Since it gives us the $n$ coordinates in the $M_j$ coordinate system we could instead have $n$ functions. Each function giving us one of the $n$ numbers we need.

And each of those functions needs to know which point we are, so they need to know all the coordinates in the other system. So $x'^{\mu}(x^{\nu})$ is a function that gives us the $\mu$-th coordinate in the $M_j$ coordinate system. And it uses all $n$ coordinates from the $M_i$ coordinate system, and the term $x^{\nu}$ is a placeholder for all the coordinates $x^0, x^1, \dots x^n$ in the $M_i$ coordinate system.

As a simple example you can switch from Cartesian coordinates to spherical coordinates in $\mathbb{R}^3$. For instance $x'^{\mu}(x^{\nu})=x'^{\mu}(x^0,x^1,x^2)=x'^{\mu}(x,y,z)$ and if we select $\mu$ to be the radial coordinate we get $x'^{\mu}(x,y,z)=r(x,y,z)=\sqrt{x^2+y^2+z^2}.$

Again, in the overlap region all you are doing is effectively changing coordinate systems in that same overlap region. It's just that you need to change coordinate systems to get to a different region of your manifold. Whereas in $\mathbb{R}^n$ when you change coordinate systems it's usually convenience, not because you have to.

Since I mentioned that my example can come up in General Relativity, I'll spell it out. Basically, in any coordinate patch $M_i$, you can pretend you are in $\mathbb{R}^4$ and treat $3\arctan (t)/2$ as if it were time so use the metric $$ds^2=\frac{9}{4}\left(\frac{1}{1+t^2}\right)^2dt^2-dx^2-dy^2-dz^2.$$

Computing the Einstein Tensor in either coordinate patch we get that it is zero, so this corresponds to no matter (no energy, no momentum, no stress). So this is a vacuum solution of Einstein's Field Equation in either coordinate patch, the patches overlap and the transition is differentiable. So it is a solution. It's basically a boring empty space with nothing in it, and time just repeats itself sometimes (which was the simplest way to force you to use two coordinate patches).

So now let's look at the physical implications of general coordinate transformations. Surely it shouldn't matter which coordinate map you use, and it shouldn't matter if you use spherical coordinates, cylindrical coordinates or cartesian coordinates, the physics should be the same. This is literally the physical principle that allowed us to switch between coordinate systems. But once we have other things (such as that metric tensor), then we also need to make sure that they produce the same physics in the two coordinate systems.

This is not always obvious. For instance, I happened to choose a metric and a coordinate system where the metric looked the same in both coordinate systems. In general you need to write a metric for each coordinate system. For instance the metric

$$ds^2=dw^2-dx^2-dy^2-dz^2$$ and the metric $$ds^2=dw^2-dr^2-r^2\left(d\theta^2+\sin^2(\theta)d\phi^2\right)$$ are both actually the Minkowski metric with a transition $z=r\cos (\theta)$, $y=r\sin (\theta) \sin (\phi )$ and $x=r\sin (\theta) \cos (\phi )$, so they are also vacuum solutions, but time does not repeat itself, and the cartesian coordinate system actually covers the whole spacetime, so we don't need to use coordinate transformations if we don't want to.

There is a whole method to write your equations in terms of tensors defined on your coordinates so that it doesn't matter which coordinate system you use. And this method has the added feature that the equations themselves (as tensor equations) have the exact same form in every coordinate system, provided the compound map that transitions from one coordinate system to the other is differentiable as a map from $\mathbb{R}^4$ to $\mathbb{R}^4$.

  • $\begingroup$ Thank you for your reply. "Change you coordinates a little bit" in your first paragraph, what exactly do you mean by that? And in the last paragraph A gain, in the overlap region all you are doing is effectively changing coordinate systems in that same overlap region. It's just that you need to change coordinate systems to get to a different region of your manifold." You mean here tby a different region, a different region but in the overlap area, no? $\endgroup$ – PhilosophicalPhysics May 26 '15 at 18:06
  • $\begingroup$ Also one more thing to add, in books this $$x'^{\mu}(x^{\nu})$$ is referred to "General Coordinate Transformation" that well-known transformation in GR. Does this ring a bell on why is it the case? May be this question to me is more important than the two that preceded. Hope you can help me with that! $\endgroup$ – PhilosophicalPhysics May 26 '15 at 19:06
  • $\begingroup$ @PhilosophicalPhysics Edited. $\endgroup$ – Timaeus May 27 '15 at 2:31
  • $\begingroup$ Excellent! Thank you. $\endgroup$ – Beyond-formulas May 27 '15 at 13:29

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