Why the globe cannot be projected on the plane
A reality that is easy to see, but hard to prove
Anyone familiar with GIS will know. You cannot project the globe on the plane without giving anything up. Whatever projection you use there is always a draw back. UTM preserves distance but cannot map the full globe, Mercator preserves local angles but not distance or straight lines globally, the gnomic projections can preserve straight lines, but not angles.
If one would ever think of a way to project the globe to the plane without deforming core geometric features, many, very many, problems of GIS would be solved in an instant. No more complected re-projections, just one projection to rule them all. Unfortunately we can be sure that no one will ever find such a projection as it can be mathematically proven that such a projection cannot exist.
Intuitively this makes perfect sense, just try to flatten the globe and you will notice you cannot do it without deforming it’s surface in one way or another. However simple it is to visualize the impossibility of this, it is actually related to a deeper mathematical truth, and actually less straightforward to understand than you might initially suspect.
The mathematical field dealing with surfaces (like the globe) and their projections is called differential geometry. A field of which I noticed that actually very few people working on GIS have a deeper understanding of. This is actually a shame as in GIS the core thing we are working on is basically an instance of a Riemannian Manifold. And with some basic knowledge about this mathematical construct gives a really nice frame on GIS in general.
There is many general theorems from Riemannian geomtery that have a pure GIS interpretation, but in this article I will focus on the fact that one cannot project the globe (or sphere) on the plane while presevering it’s geometric features.
A fact for which we will actually require the full Riemannian geometric toolbox! I admit that the article is a pretty tough read, but once you worked your way through it you will know more then just why the sphere cannot be projected on the plane, you will actually have a more rounded understanding of the fundamentals of differential geometry. Fundamentals that can be found everywhere in GIS and beyond. In fact, much of general relativity is build on the very same concepts…
So… Why can the sphere not be projected on the plane without loss of geometric properties?
Reading tip: In case you are just interested in the general idea and not the mathematical details you can choose to just read the executive summaries below each section. These summaries should give you a good idea of what the reasoning looks like.
The inner product
The first thing we need is a notion of ‘geometric properties’. The genesis of geometric properties. We start out with defining these geometric properties on a 2 dimensional vector space (the plane). (In the whole article I will assume we are dealing with 2 dimensions, everything discussed actually works in any dimension, but for simplicity we will just assume 2 for now).
The genesis of all geometric properties is the inner product, which is a linear function that maps 2 vectors to a value.
If we define the dot product of two vectors as follows:
that is to say we simply multiply the entries of two vectors of each dimension and sum over the dimensions. We end up with our regular notion of distance in the pane.
So that’s great. The dot product gives rise to the geometric notions like distance and angles that we are after in case of a 2 dimensional pane. But what about none flat planes? Obviously the mercator projected world does not have this dot product as it’s distance measure. Otherwise Greenland would be 10 times the size of Congo, where in reality it is about the same size…
Executive summary: The inner product is the genesis of geometric notions on a vector space.
The general inner product
To facilitate not trivial spaces (like the mercator projected globe) we will define an innerproduct more generally as any linear function
It does not matter what you do here, as long as it is linear we allow it!
Each function that is linear in it’s entries can be written as a tensor (matrix). So a dot product is defined by a certain tensor η that is called the metric tensor.
We can now denote the dot product as:
Once we are given a metric tensor on a vector space we have a notion of length and angles (and therefore geometry).
The metric tensor of a ‘regular’ vector space is simply the identity matrix
But as we allow any matrix η we allow ourselves much more freedom (which we will need to deal with non trivial surfaces)
Having a metric tensor is a big step, as now we have a simple mathematical notion that equips our vector space with the geometric notions. But we will need quite a bit more in our toolkit.
Executive summary: The metric tensor gives us an inner product, which fixes the geometric notion on a space.
The metric tensor on a general surface
Ok great so we can equip a vector space with a metric tensor, giving us a real geometric space. This is nice and well but we are not quite there yet. Since we are looking at general surfaces, the sphere in particular, our surface in not really a vector space. As a matter of fact the only vector space in 2 dimensions is the plane, and the sphere is obviously not the plane. So how do we equip a general surface M with a geometric notion?
To understand how to do this let’s, for the sake of simplicity assume that our surface is is embedded in the 3 dimensional pane.
This surface could for example be the sphere or any other surface, the torus or any other surface you can think of.
Now how can we equip such a surface with a metric? Well since our definition of a metric needs a vector space to work we cannot directly apply this to the surface M itself. However. The surface M has at each point x a tangent space, which in fact is a real vector space! So maybe we cannot apply our metric on the surface itself, but we can apply it to the tangent space of the surface.
We can denote the tangent space at point x of surface M as:
In general a n-dimensional surface has a n-dimensional vector space as it’s tangent space at each point. (provided that the surface is differentiable (smooth enough) in the n+1 dimensional space in which it is embedded).
Now let’s have a look at a path over our surface M. We can regard a path as a function mapping the interval 0 to 1 to a point on the surface.
We can take the derivative of this path. The derivative maps each point x along the path to a tangent vector v
This tangent line v at each point x along is a vector within the tangent space of M at point x.
Now if we want to get the length of the line you can take the integral over the infinitesimal lengths of the tangent vector in each of the tangent spaces.
You can at this point probably feel where this is going. We can define not 1 metric tensor, however we can define a whole collection of metric tensors at each point on the surface M.
We will need to require that the metric tensor is only allowed to vary smoothly across the surface M, that is to say when going over the surface the metric is only allowed to change smoothly. But this is a rather technical requirement, that I won’t explore further here.
If we want to know the length of our path we can simply calculate
which looks complicated, but is simply the integral over the infinitesimal tangent line.
Executive summary: We can equip a general surface with a metric by defining a metric tensor in each of the tangent spaces of the surface.
Definition of a Riemannian Manifold
We found a way to equip a vector space with a notion of geometry (distances and angles) by using a metric tensor. We then generalized this to general smooth surfaces by equipping the tangent space along these surfaces with such a (varying) metric tensor.
Remember that all the drawing of surfaces and paths were just inspiration to come up with the correct definition. Now we know how to define it in the ‘visual’ case we can just generally state that a metric space consists of a surface M that can be parametrized by some coordinates x and a function mapping each point of the manifold to a metric tensor.
At last we have a real mathematical definition of what we mean by a metric space.
The mercator projected world can now simply be viewed as a Manifold. We have our coordinates x and y, that parametrize our surface. The metric tensor of our surface is not trivial, this is the reason that a straight line (line of shortest distance between two points) does not ‘look’ straight.
Remember that we can think of a 2 dimensional surface as a surface embedded in 3 dimensional flat space. This gave us the inspiration of our definition, but in a more abstract sense it does not need to be visualised this way. If we parametrize a surface with 2 coordinates and attach a metric tensor at each point it counts as a perfectly fine Manifold regardless of whether we can easily visualize it.
Executive summary: A surface together with a metric tensor on the tangent space at each location of the surface is called a (Riemannian) Manifold.
Equivalence of manifolds
Now that we have a notion of manifolds the logical next question to answer is.
When are two manifolds essentially the same?
We could parametrize the sphere as all points such that
Or we could use spherical coordinates and say
These are obviously both the same surfaces, but we just used different coordinates to describe them. (Or in GIS terminology projections)
A fair definition of equivalence would be that two manifolds are equal whenever their metrics are essentially the same.
A metric tensor can be said to be essentially the same whenever
for some matrix u. This is a good definition as in this case both metrics would be the same, but just expressed using different basis vectors.
The question of why there is no coordinate transformation mapping the globe to the plane now becomes:
Why are the sphere and the plane not equivalent manifolds?
In GIS we use many projections to display the globe, but all these projections are just equivalent representations of the same manifold. The metrics on each of the projections look different (and therefore the projections look very different) but the metrics are equivalent in an abstract sense. They can be turned into one another by a coordinate transformation (and with that a linear transformation in each of the tangent spaces).
Executive summary: Two manifolds are equivalent whenever their metrics are equivalent.
Our strategy for proving the claim
Now we have both a formal definition of what a manifold is and when we regard two manifolds as the same we can at last continue to answer our initial question. The answer will boil down the the following:
1 Two equivalent manifolds have the same curvature.
2 The curvature of the sphere does not equal the curvature of the pane.
3 This implies that the sphere and the plane are not equivalent. In other words there exists no coordinate transform between them that preserves the metric.
The notion of curvature
As you can see from our strategy we need one additional notion in order to make our point. We need to define what we mean by curvature. Unfortunately for us differential geometry is not exactly easy and the notion of curvature requires some work.
The notion of curvature is based upon the notion of parallel vector transportation. Which in turn depends on a notion of vector fields and derivatives. We will first set out to define this things and then revisit curvature when we have these tools at hand.
Executive summary: We need to define what we mean by the curvature of a surface.
Vector fields on manifolds
It is rather easy to construct a vector field on a surface. We just pick a vector in the tangent space on each point on our surface. Or in other words a vector field V is a smooth function mapping each point x on the surface M to a vector in it’s tangent space.
Of special interest are basis vector fields. If we can parametrize our surface M with parameters α and β the derivatives along α and β at each point x give us 2 basis vectors of the tangent space at point x.
That is to say a vector v can be described as a sum of basis vectors times some scalar.
In other words a parametrization of our surface M gives us a vector field of basis vectors for our tangent spaces. This trivially looking fact is actually a rather deep understanding allowing us to express vectors in the tangent space using the same coordinate parametrization of the surface M.
If we now think of a surface M in an abstract sense (not embedded in some flat space) we can still talk about abstract tangent spaces. Simply by using our coordinate parametrize as ‘directions’. A coordinate direction is interpreted as a basis vector and we can express any vector as a combination of it’s basis vectors.
Execute summary: We can define a vector field on a manifold by assigning to each point on the manifold a vector in it’s tangent space.
Derivatives of a vector field
Now we know what a vector field on our surface is. We can consider what it means for this vector field to ‘change’. If we follow a path across our surface how much does our vector change? Or in other words. Given a vector field V, what is
The answer is a bit more nuanced then you might initially think. We cannot simply take the derivative of the components of the vector. The answer is not
The reason for this is that actually the basis vectors themselves change as well! Think of what happens if we parametrize the sphere with spherical coordinates θ and φ. The basis vector in the φ direction on the pole and the vector in the φ direction on the equator will differ 90 degrees from one another!
To get the real derivative we will need to take the derivatives of the basis vectors themselves into account as well.
Instead of using a general path we can also take the derivate into μ coordinate direction
Now let’s define a connection symbol to make this formula look a bit better.
Giving us
So the derivative of a vector field consists of a derivative of it’s components and some connection symbol that corrects for the basis vectors. Of course we want our derivative to be independent of the particular coordinate paramterization that we chose. We therefore define the derviate of a vecor field to be
Which is basically it’s derivative corrected for changes in our basis vectors themselves.
By definition of the connection symbols the derivative of a basis vector field is zero.
Note that our derivative is independent of our coordinate parametrization, precisely because the connection symbols correct for the changing basis vectors. The derivative is therefore well defined (a function mapping a metric space to an equivalent metric space keeps the derivative intact)
In case we transport a vector over a path in such a way that
we say call it parallel transport. Parallel transport therefore means that we leave the vector unchanged with respect to the basis vector field.
Executive summary: Since basis vectors on a manifold change from location to location the derivative of a vector field needs to correct for this changing of the basis itself.
Connection symbols in terms of the metric
Since a metric space consists of a surface M and a metric η it makes sense to want to express the connection symbols in terms of the metric. This way, given a metric space we can calculate the needed connection symbols.
We start by realizing that the metric was defined as the inner product of the basis vectors
Since
We find
Cyclic permuting the parameters i,j and l gives us 3 formulas allowing to express the connection symbols in terms of the metric.
Executive summary: The connection symbols are a function of the metric tensor, meaning that the definition of a manifold fixes it’s connection symbols.
Curvature revisited
Now with the notion of parallel vector transportation at our disposal we are now able to define curvature.
In a flat space it does not matter via which path you transport a vector from point x to point y. No matter how the vector gets there, once there it will always look the same. However on a non-trivial curved surface it very much does matter via which path you move your vector. On the sphere you can end up with very different vectors.
For example we can move the vector from the pole to the equator in 2 ways:
We end up with a 90 degree difference in angle!
This ‘twisting’ of directions based on your path is exactly what we associate with our every day use of curvature.
We can define curvature locally by moving vectors only an infinitesimal step. One time in the μ direction and then the ν direction, and one time first in the ν direction and then in the μ direction. We can then have a look at the infinitesimal difference between these two situations. This gives us a definition of curvature R at a certain point of our surface:
Graphically this expression would be the difference of the vector V at point D:
On a non curved surface this expression would of course be zero, but in curved surfaces one can no longer simply swap the order of taking derivatives.
The curvature was defined using the Riemann tensor R.
The Riemann curvature tensor is well defined as we used the well defined notion of parallel transport and the well defined notion of a vector in the tangent space on a single point to define it.
The Riemann curvature tensor can be expressed in term of the connection symbols. We can first calculate the double derivative in terms of the connection symbols
Now if we subtract
We will lose the cross term
Giving us
Which is a long formula, but it does do the trick, given the connection symbols we can calculate the curvature. Since we can calculate the connection symbols from the metric this indirectly means that we can calculate the curvature from the metric as well!
This means that a metric space, which is a surface M with some metric η intrinsically defines the curvature.
Executive summary: the curvature around a point on a surface is the degree in which moving a vector around that point changes the direction of the vector.
A small lemma
For our proof we will need a small lemma about the trace of a tensor. A trace for a tensor is defined as the sum over it’s entries:
We claim that the trace of a tensor is invariant under any coordinate transform. In other words:
We can calculate that:
Now we can realize that
And therefore
Giving us the desired result.
Executive summary: Taking the trace of a tensor is invariant under transformation. That is to say if two manifolds have equivalent tensor fields on them, then the traces of these tensor fields will be equivalent as well.
Contracting the curvature tensor
In the above we saw that taking traces is invariant for coordinate transformation. We can apply this knowledge on the curvature tensor R.
By taking the trace twice we can contract the curvature tensor to a simple scalar which is called the Ricci scalar:
Since the Ricci scalar is just a number it does not, like the curvature tensor, depend on the basis vectors that we chose to describe it. I therefore follows that if two manifolds are equivalent they need to have the exact same Ricci scalar at every point. Likewise if there is a point in which the Ricci scalars are not equal the metric spaces cannot be equivalent.
Executive summary: We can contract the curvature tensor to a scalar by taking it’s traces. This scalar will be the same if two manifolds are equivalent.
Calculating the Ricci scalar for the sphere and plane
The Ricci scalar gives us an easy way to check if two metric spaces can be equivalent. If the sphere and plane turn out to not have the same Ricci scalar at every point we can conclude that they are not equivalent.
Calculating the Ricci scalar for both spaces is a rather elaborate procedure, but as we demonstrated we can calculate the connection symbols from the metric and consequtavely the curvature tensor from the connection symbols and lastly the Ricci scalar from the the curvature tensor.
The metric tensor of the sphere is given by
Meanwhile the metric tensor for the plane is simply given by
Using the formulas above we can find that the Ricci scalar of the sphere is 1 and the Ricci scalar of the plane is 0.
The fact that the Ricci scalars are different now implies that the metric spaces cannot be equivalent.
Executive summary: The Ricci scalar of the sphere 1 at every point, that of the plane is 0 at every point. The manifolds are therefore not equivalent.
Conclusion
Since the Ricci scalar of the sphere and the pane are distinct we can conclude that there is no metric preserving coordinate transformation between the two.
For if there were such a transformation their metrics would be equivalent. This would in turn imply their curvatures to be equivalent which in turn implies their Ricci scalar to be equivalent. Since the Ricci scalar is just a number this would imply the numbers to be equal. Which they are not.
Final thoughts
In GIS we are using many representations of the sphere. All these representations are members of the equivalence class of manifolds of that of the sphere.
Their metric tensors are described with respect to other coordinates causing the globe to have very different looks depending on your coordinate choice. However in an abstract sense the metrics of all of these projections are equivalent and can be transformed from one another.
Curvature is an intrinsic property of a manifold that can be calculated from the metric. Curvature is a rather complex object but can be contracted to a simple scalar. This scalar is EXACTLY the same for each and every GIS projection of the globe.
The fact of the matter is that the globe and therefore all of it’s equivalence class members are curved spaces and therefore you will never be able to transform it to the flat plane with trivial metric as this is a zero curvature manifold.
