# Why the surface of the sphere is not a Euclidean space?

In wiki, I see a description in manifold:

"Although a manifold resembles Euclidean space near each point, globally it may not. For example, the surface of the sphere is not a Euclidean space, but in a region it can be charted by means of map projections of the region into the Euclidean plane."

My questions:

1. Why is surface of the sphere not a Euclidean space? (I think points on surface can be represented by $x$,$y$,$z$ for example)

2. What does it mean after 'but'?
in a "region", is this region different from surface of sphere?

• An Euclidean space is $\Bbb R^n$ for some $n$. $S^n$ is not this. – Pedro Tamaroff Sep 16 '14 at 2:28
• I will begin with number 2. Consider the $2$-sphere $S^2$. Then, by stereographic projection, we may form a bijection between the plane and the region $S^2-\{0,0,1\}$. From this, can you see a reason why there exists no Euclidean space such that there is a bijection between the $S^2$ and $\mathbb{R}^n$? – Eoin Sep 16 '14 at 2:29
• One of the properties of Euclidean space is that one has proper translations, isometries of the space that leave no point fixed. This cannot be done on a sphere. (It could be done on a circle though, calling rotations "translations", but another property is that translations form a vector space; here the circle fails too.) – Marc van Leeuwen Sep 16 '14 at 6:52

Possibly what's confusing you is that the statement refers to the surface of the sphere, not the space the sphere is sitting in. If we ignore relativistic effects (and the fact that the Earth isn't quite a sphere), yes the sphere is in a Euclidean 3-dimensional space. In that space all the convenient Euclidean things apply. But the statement didn't refer to the sphere, or the space it's sitting in: it referred to the surface of the sphere. The surface of the sphere is a 2-dimensional space, not a 3-dimensional one, and points on it only need 2 coordinates (latitude and longitude, for example).

To help see the difference, consider a straight line between London and New York. In the 3-dimensional Euclidean space in which the Earth is embedded, that straight line goes through the Earth. But if we're only considering the surface of the Earth, that line doesn't exist. The straight line (shortest distance between the two points) on the surface lies along the great circle. Now consider drawing the lines from both New York and London to, say, Capetown, to make a triangle. Yes, if you draw the lines through the Earth you will get a nice Euclidean triangle with angles that add up to 180 degrees. But those lines don't exist in the space you are considering: you can only draw lines on the surface of the Earth. The angles of the triangle drawn on the surface of the Earth add up to more than 180 degrees, so the space must be non-Euclidean.

Edit: The bit after the "but" just seems to be saying that for most purposes you can treat a smallish bit of the Earth as if it were flat. You probably don't need to worry about the curvature of the Earth when looking at a street-map of your town.

• Nice and simple answer; even people majoring in other field can understand the key point. But I want to ask a small question: what does the "great circle" here mean? – sleeve chen Sep 16 '14 at 14:41
• A "great circle" is any circle on the surface of the Earth that has its centre at the centre of the Earth. The equator is a great circle, but other lines of latitude are not (their centres are north or south of the centre of the Earth). Lines of longitude are halves of great circles. There's more information at en.wikipedia.org/wiki/Great_circle – digitig Sep 16 '14 at 14:52
• just wondering; the surface of any $n$-dimensional set always has dimension $n-1$ ?? (I reason from the earth the map logic) – ElleryL Jun 13 '18 at 4:28
• @ElleryL I think so, if it's smooth (in the intuitive sense, not any strict mathematical sense). But if it's crinkly you can run into fractal dimensions where, for example, the coastline of a (2D) country can have a dimension between 1 and 2. en.wikipedia.org/wiki/Coastline_paradox – digitig Jun 15 '18 at 13:51

Intuitively, Euclidean geometry is flat. The sphere, globally, is clearly not flat. But locally, if you tear out a tiny piece of it it's pretty much flat.

More accurately, in a Euclidean space Euclid's fifth postulate holds. Thus, given a line and a point not on that line there exists a unique parallel to the line through that point. This fails on the sphere since there are no parallel lines at all. Every two straight lines (where straight should be interpreted as geodesics, that is straight relative to the curved geometry of the sphere, thus straight lines are in correspondence with great circles) intersect. However, the geometry of a small piece of the sphere is locally Euclidean, that is it looks and behaves just like a little piece of $\mathbb R^2$.

• In Euclidean space, the sum of angles of a triangle is $\pi$, while a spherical triangle always has angle-sum greater than $\pi$. It’s curious that “geometry” means earth measurement, but the surface of the earth is roughly spherical, and the axioms of Euclidean Geometry don’t fully apply. – Lubin Sep 16 '14 at 2:37
• @Lubin In spherical geometry the sum of a triangle's angles is related to its area. So you can't have triangles that are similar but have different areas. – Rosie F Mar 4 '20 at 12:28

(1) Topologically, Euclidean space is retractable to a point. That means there is a continuous map: $f:\mathbb R^n\times I\to \mathbb R^n$ such that $f(\textbf v,0)=\textbf v$ and $f(\textbf v,1)=\textbf 0$.

A sphere is not retractible to a point. There is no similar map $f:S^{n}\times I\to S^{n}$ (not a trivial result.) So $S^{n}$ is not topologically the same as any $\mathbb R^{m}$.

(2) Any point on $S^{n}$ has a neighborhood that is homeomorphic to some open subset of $\mathbb R^n$. Such a homeomorphism from a subset of $\mathbb R^n$ to an open subset of a space $M$ is often called a "chart" because it gives a coordinate system in that open subset.

• Short question: What does it mean intuitively in 2nd line? i.e. f: R^n*I -> R^n such that f(v,0) = v and f(v,1) = 0. – sleeve chen Sep 16 '14 at 14:44
• If you consider at "time" $t\in[0,1]$, the function $\phi_t:\mathbb R^n\to\mathbb R^n$, then it means that there is a continuous path from the identity function $\mathbb R^n\to\mathbb R^n$ and the $0$ map $\mathbb R^n\to\mathbb R^n$. This is an example from topology of a general idea called "homotopy." – Thomas Andrews Sep 16 '14 at 15:01
• So, first, v is a vector in R^n. Second, the reason for "A sphere is not retractible to a point" is because, the so called "deformation" depends heavily on the properties of surface; it is possible that there is no definition on the points or region through which deformation will pass. Correct? – sleeve chen Sep 18 '14 at 3:33
• Or roughly speaking, Euclidean space is concrete, all points inside the space have been defined; however, surface of a sphere is not. – sleeve chen Sep 18 '14 at 3:36
• Proving that the sphere is not retractable takes quite a bit of work. Your intuition about "concrete" doesn't make any sense to me. – Thomas Andrews Sep 18 '14 at 3:39

One way to think about the statement is this:

If you literally cut out a little disk out of the sphere, you can stretch it out (albeit infinitely far) to make it look like a plane (Euclidean 2-space).

However, no matter how you deform/stretch the entire sphere, there's no way to make it into a plane.

Often the INTERIOR of a sphere is referred to as a "ball" and the surface as a "sphere". The surface of a sphere is non-euclidean because the nearest analogs to straight lines, create circles, cross TWICE and no great circles can be parallel. Also, the circumference of any circle on the surface of a sphere is NOT 2 *pi * the radius, but is always SMALLER.

We can also think of a sphere as a compact set while the Euclidean is not compact. Hence the sphere is not even like the Euclidean to be Euclidean.