# A A question about coordinate distance & geometrical distance

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1. Jul 19, 2016

### Frank Castle

As I understand it, the notion of a distance between points on a manifold $M$ requires that the manifold be endowed with a metric $g$. In the case of ordinary Euclidean space this is simply the trivial identity matrix, i.e. $g_{\mu\nu}=\delta_{\mu\nu}$. In Euclidean space we also have that the parallel postulate holds as well as the Pythagorean theorem. Furthermore, it is found that the extremal path between to points is a straight line and so the Pythagorean theorem holds for finite distances between points also.
Since the space is Euclidean we can use the Cartesian coordinate map, which is a global coordinate map (since Euclidean space is globally isomorphic to $\mathbb{R}^{n}$), and as such, considering $\mathbb{R}^{2}$ for simplicity, for two points $p=(x_{1},y_{1})$ and $q=(x_{2},y_{2})$, the distance between them is given by their so-called "coordinate distance" $$\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}$$, where $\Delta x = x_{2}-x_{1}$ and $\Delta y = y_{2}-y_{1}$.

My question is, why is this not the case in general? I mean, even in Euclidean space this is not the case in other coordinate systems (for example, in polar coordinates, $p=(r_{1},\theta_{1})$ and $q=(r_{2},\theta_{2})$, but the distance between them is not given by $\sqrt{\left(\Delta r\right)^{2}+\left(\Delta\theta\right)^{2}}$). Is it simply to do with the coordinate transformation between Cartesian and polar and the coordinate scaling that it introduces, or the fact that the basis vectors are not constant in the polar coordinate system?

I get that on a more general manifold, where the geometry is non-Euclidean (and as such the metric $g_{\mu\nu}(x)$ is non-trivial), such that the parallel postulate no longer holds and the Pythagorean theorem holds only locally, one must integrate the differential line element, $ds^{2}$ (of the form $ds^{2}=g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$ along a geodesic connecting the two points $p$ and $q$ to determine the distance between them, but is there an argument for why the line element is not always of the form $$ds^{2}=(dx^{1})^{2}+(dx^{2})^{2}+\cdots +(dx^{n})^{2}$$

Finally, why can one simply not take the difference between the coordinates of two points $p$ and $q$ to gain the coordinates of another point $p'$? Is this only true in Euclidean space because it is an affine space and so coordinate differences correspond to vectors (and hence satisfy the vector space axioms), and furthermore the Cartesian coordinate system is global, whereas, in a more general manifold, such a calculation is very much coordinate dependent and thus not well-defined, and indeed the difference between the coordinates of two points may not correspond to the coordinates of any other point on the manifold?!

2. Jul 19, 2016

### RUber

The distance formula, i.e. the 2-norm, is based on the idea that your basis is orthonormal. When you do not have an orthonormal basis your 2-norm begins to include some cross terms and your norm squared is not equivalent to the sum of squared differences. Luckily, you should be able to transform your coordinate system into one that is orthonormal (as you mentioned for polar-to-Cartesian)...if that is easier than working out the distance in your current basis.

3. Jul 20, 2016

### Frank Castle

Is there a reason though for why coordinate distance doesn't correspond to actual geometrical distance between points on a manifold in general? Also, why the simple difference between coordinates (of points on a manifold) does not, in general, correspond to the coordinates of another point on the manifold?

4. Jul 20, 2016

### Orodruin

Staff Emeritus
Why would they? Coordinates are just numbers describing where you are on a manifold. You can pick any numbers, including numbers that do not have units of length. Why would you expect coordinate distance to be equal to geometrical distance?

5. Jul 20, 2016

### Frank Castle

I agree with this intuitively, but what "throws" me on the matter, is that the coordinate difference in Cartesian coordinates (in Euclidean geometry) is equal to to the geometrical distance, in the sense that $\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$ (in the 2D case, for example).

Furthermore, the coordinates of another point can be obtained from coordinates of other points, in the sense that $p'=(x +\Delta x, y +\Delta y)$ (again in the 2D case, for example). But I think the reason for this is the "specialness" of Euclidean space as a manifold, in that it is an affine space and can be covered by a single global coordinate chart, i.e. the Cartesian coordinate chart (which is the trivial identity map)?!

6. Jul 20, 2016

### Orodruin

Staff Emeritus
Well, there is nothing stopping you from using the distances in orthogonal directions as coordinates in a Euclidean space.

7. Jul 20, 2016

### Stephen Tashi

Since coordinates employ numbers, you might be interested in attempts to define numbers in terms of "geometry" (in contrast to the usual approach of defining geometry in terms of numbers). This suggestion just popped into my mind because you said "throws" and, by coincidence, the starting point for such attempts appears to be the "Theory of Throws" of Karl von Staudt, https://en.wikipedia.org/wiki/Karl_Georg_Christian_von_Staudt

I, myself, haven't studied how one might carry out such a project. My guess is that one defines certain geometric operations without mentioning the concepts of "length" or "coordinates" and then defines "number" in terms of these geometric operations.

8. Jul 22, 2016

### Frank Castle

In terms of distances between points on a manifold, is the point that one needs a metric in order for such a notion to be meaningful. One then integrates this metric along the minimal path connecting two given finitely separated points to find the geometric distance between them. In Euclidean space, this is particularly straightforward since the metric is trivial, $g_{\mu\nu}=\delta_{\mu\nu}$ and is not position dependent (on the manifold). Furthermore, due to the parallel postulate holding in Euclidean space (and hence Pythagora's theorem holding for finite distances), the minimal path between two given points in a straight line. Thus, in Cartesian coordinates (where the basis is constant and orthogonal), the result is particularly simple $\Delta s=\sqrt{(\Delta x^{1})^{2} +(\Delta x^{2})^{2}+\cdots +(\Delta x^{n})^{2}}$. Out of interest, in other coordinate systems in Euclidean space, such as polar coordinates, is it true that $\Delta s=\sqrt{(\Delta r)^{2}+r^{2}(\Delta \theta)^{2}}$? (I have a feeling it's not, but I'm not too sure?!)

I've read that, in the case of the 2-sphere, there is no "natural" way to add two points on the sphere and end up with another point on the sphere. I assume this extends to a general statement about manifolds, but what is meant by the statement "no natural way"?!

Also, I know that in Euclidean space one can show that the minimal path between to points is a straight line, as follows (in 2D for simplicity):
Let $\sigma$ parametrise the path between two points $A$ and $B$ such that for a curve $\gamma(\sigma)$ between the two points, $\gamma(\sigma_{1})=A$ and $\gamma(\sigma_{2})=B$. We have then, that the distance between them, in Cartesian coordinates, is given by $$S=\int_{\sigma_{1}}^{\sigma_{2}}\frac{ds}{d\sigma}d\sigma=\int_{\sigma_{1}}^{\sigma_{2}}\sqrt{\left(\frac{dx}{d\sigma}\right)^{2}+\left(\frac{dx}{d\sigma}\right)^{2}}d\sigma$$ Varying the path $\gamma$ around the extremal path between these two points and requiring that such variations vanish at the boundary points leads us to the following Euler-Lagrange equations $$\frac{\partial L}{\partial x}-\frac{d}{d\sigma}\left(\frac{\partial L}{\partial x'}\right)=0 \\ \frac{\partial L}{\partial x}-\frac{d}{d\sigma}\left(\frac{\partial L}{\partial x'}\right)=0$$ where $L=L(x,x',y,y')=\sqrt{\left(\frac{dx}{d\sigma}\right)^{2}+\left(\frac{dx}{d\sigma}\right)^{2}}$ (with $x'=\frac{dx}{d\sigma}$ and likewise for y'). Thus, we see that $$\frac{\partial L}{\partial x}=0\\ \frac{\partial L}{\partial x'}=\frac{x'}{L}$$ with similar expressions for the $y$ coordinate. As such, we find that $$\frac{d}{d\sigma}\left(\frac{\partial L}{\partial x'}\right)=0 \\ \frac{d}{d\sigma}\left(\frac{\partial L}{\partial y'}\right)=0$$ which are equivalent to the equations $$\frac{\partial L}{\partial x'}=\text{constant}\\ \frac{\partial L}{\partial x'}=\text{constant}$$ Upon taking the ratio of these two equations we find that $$\frac{\frac{\partial L}{\partial y'}}{\frac{\partial L}{\partial x'}}=\frac{y'}{x'}=\frac{dy}{dx}=\text{constant}$$ Which is the equation of a straight line (apologies for the slap-dashness of this derivation, I've haven't got the free time at the moment to give a detailed run-through, but hopefully it comes across ok).

My question in relation to this is, what assumptions are made, i.e. about how one integrates on the manifold? I'm not particularly familiar with partitions of unity, but are they trivial for Euclidean space?

Last edited: Jul 22, 2016
9. Jul 23, 2016

### Stephen Tashi

If we turn that into a mathematical question, I probably won't know the answer, but at least I'll have contributed that service.

One interpretation of your words seems to set the scenario : "Given a manifold M and a distance function D(x,y) defined on M..." (After all, if we were "given" a distance function D(x,y) then we could say the concept of "the distance between x and y" was "meaningful".)

Then we are asking about needing a "metric". So, by that interpretation, we'd have to distinguish between "a metric" and a "distance function". I don't know how to do that.

From you examples, perhaps you are asking about a relationship between a distance function and the concept of integration along a path.

My relevant areas of ignorance on that question include:

1) If we are given a distance function on a manifold M, does that imply that M is a connected manifold ? -i.e. would there necessarily be a path "on the manifold" between two arbitrary points on the manifold ? Sorting out that question probably depends on how the concept of continuity is involved in the definition of "distance function" and the definition of "path".

2) If we are given a distance function D(x,y) on a manifold M and the existence of at least one path on M between x and y, does that imply the existence of a function phi(x) defined on the manifold such that D(x,y) is equal to the integral of phi(x) along some path in the manifold ? - and can we also add the requirement to phi(x) that that D(x,y) is equal to the integral along a path P between x and y that minimizes the integral ?

Perhaps 2) is what you are asking.

10. Jul 23, 2016

### lavinia

A few unrelated points

- There is not necessarily a length minimizing path between any two points on a manifold even if it is path connected. For instance, the plane minus the origin with the Euclidean metric will not have a length minimizing path between a point on the positive x-axis and a point on the negative x-axis. For any path connecting them, there will be a path that is shorter, this because any path connecting the two points will have to veer around the origin.

- A distance function determines a topology on a space. A path is a continuous function from an interval into the space with that topology.

- A manifold is a topological space and is defined without a distance function. It is just a space that looks like an open set in Euclidean space near any point. A distance function on the manifold means the topology that the distance function determines is the same as the topology of the manifold. One needs to prove that a Riemannian metric determines the same topology.

- One can put a metric on a manifold that does not come from a Riemannian metric. For instance the metric on a tetrahedron - viewed as a topological 2 sphere - is not smooth at the edges of the four faces.

Here are some questions:

- If one turns each face of the tetrahedron into a little pyramid by subdividing it into three triangles and raising the common central vertex a small distance, one gets a new space made of smaller triangles and with many more edges. Again the metric will not be smooth at any of the edges. Repeat this process for each new triangle and take the limit. Can one do this in such a way that the resulting metric is not smooth anywhere?

- Suppose one has a positive definite inner product on the tangent bundle of a smooth manifold that is nowhere differentiable. Does it determine a metric on the manifold?

- The Cantor function on the unit interval can not be expressed as the integral of a Lebesque measurable function. But it is not 1-1. Perhaps one could modify its construction to get a 1-1 function that is still not the integral of a Lebesque measurable function. This would give a distance function on the interval which is not an integral.

Last edited: Jul 24, 2016
11. Jul 24, 2016

### Frank Castle

I think this is probably more along the lines of what I'm thinking of.

I'm also unsure about defining distance between points in the first place. Does defining a particular form to a distance function define a particular metric space?
In Euclidean space, is it true that for any given coordinate system, the distance between two given points is given by $\sqrt{(\Delta x^{1})^{2}+\cdots(\Delta x^{n})^{2}}$?

Thanks for the information. I'm fairly new to differential geometry, so I'm not sure whether I'll be able to answer the questions you've posed quite yet, but it's good to have the information.

12. Jul 25, 2016

### lavinia

You are welcome.

One of the things to keep in mind is that a manifold can be defined without a distance function. It is a special kind of topological space. Around each point, it has a neighborhood that can be coordinatatized as a region of Euclidean space. In Differential Geometry, one gives a metric to the manifold via a smooth inner product on tangent vectors. This is an additional structure.There are many possible different inner products.

It can be shown that an inner product defines a distance function on the manifold and that this distance function determines that same topology as the topology of the manifold. Different inner products determine different distance functions but always the same topology. For instance, Euclidean space is a manifold. and it can have many different inner products. The inner product that yields Euclidean geometry is only one possibility.

Around each point on a manifold there is always a neighborhood - perhaps quite small - where the distance function is the length of the shortest path contained in the neighborhood. But globally such a path may not exist as the case of the plane minus the origin illustrates.

- Not all possible distance functions on a manifold come from a smooth inner products on tangent spaces.

Last edited: Jul 25, 2016
13. Jul 25, 2016

### Frank Castle

Ah ok, so it's possible to have a manifold in which there is literally no meaning of distance between points within it then?!

So does defining a particular distance function on a given manifold in some sense determine the geometry on that manifold? Also, how is the metric, $ds^{2}=g_{\mu\nu}(p)dx^{\mu}\big\vert_{p}dx^{\nu}\big\vert_{p}$,related to the distance function $d(p,q)$?

By the way, is the proof I gave (in post #8) for the shortest distance between two points being a straight line in Euclidean space correct at all?

14. Jul 26, 2016

### lavinia

Yes. By itself the manifold has no metrical relations.

A distance function must be added. There are many ways to do this, many different possible metrics.

Here is an example:

Start with a region,D, in the plane with coordinates $(u,v)$. So far it is just a coordinate domain. Now map it into 3 space by a smooth vector valued function $X(u,v)$. $X(u,v)$ is a surface is three space.

Now define a Riemannian metric on $D$ by $<∂/∂u,∂/∂u> = ∂X/∂u⋅∂X/∂u$, $<∂/∂u,∂/∂v> = ∂X/∂u⋅∂X/∂v$, $<∂/∂v,∂/∂v> = ∂X/∂v⋅∂X/∂v$

or in index notation $g_{ij} = ∂X/∂u_{i}⋅∂X/∂u_{j}$ where $u_{1} = u$ and $u_{2} = v$.

For each choice of the vector valued function $X(u,v)$ one gets a different Riemannian metric on $D$.

In the case where $X(u,v) = (u,v,0)$, $g_{ij} = δ_{ij}$ and one has the standard Euclidean metric. Here $D$ is not warped by $X$ but is just included in 3 space as part of the $(u,v)$- plane

In the case where $X(u,v) = (sin(v)cos(u), sin(v)sin(u),cos(v)$, $D$ is a region on the sphere of radius 1.

$∂X/∂u = (-sin(v)sin(u),sin(v)cos(u),0)$ and $∂X/∂v = (cos(v)cos(u),cos(v)sin(u),-sin(v))$

Taking the dot products give the metric

$g_{11} = sin^2(v)$, $g_{12} = 0$, $g_{22} = 1$

This is a second metric on $D$ and it determines the intrinsic geometric of the sphere not of Euclidean plane.

Notice that this metric forgets the function $X$ and makes no reference to 3 space. It is just a positive definite inner product on the coordinate tangent vectors $∂/∂u$ and $∂/∂v$. Yet it still determines the intrinsic geometry of the unit sphere.

Locally the distance between two points is just the length of the shortest path connecting them. This is also true globally if there exists a shortest path but sometimes there may not be a shortest path. In that case one takes the greatest lower bound of the lengths of paths between the two points.

The length of a path is obtained by integrating the length of its velocity vector, which is in turn, determined by the metric.

Last edited: Jul 28, 2016