The Sphere is not an Affine Space

Question

In Eric Gourgoulhon's "Special Relativity in General Frames", it is claimed that the two dimensional sphere is not an affine space. Where an affine space of dimension n on $\mathbb R$ is defined to be a non-empty set E such that there exists a vector space V of dimension n on $\mathbb R$ and a mapping

$\phi:E \times E \rightarrow V,\space\space\space (A,B) \mapsto \phi(A,B)=:\vec {AB}$

that obeys the following properties:

(i) For any point O $\in E$, the function

$\phi_O: E \rightarrow V,\space\space\space M \mapsto \vec {OM}$

is bijective.

(ii) For any triplet (A,B,C) of elements of E, the following relation holds:

$\vec {AB} + \vec {BC} = \vec {AC}.$

I would like to show that the sphere is not an affine space using this definition. My approach has been to assume that such a $\phi$ exists and then seek a contradiction. I can construct specific $\phi_O$'s that are bijective and I can show that a contradiction arises if I use the same construction centered at a new point A, wtih $\phi_A$, but this only invalidates the specific construction I made. I am having trouble generalizing this to any $\phi$.

Answer 1

The thing is, you might want to get some topology in the picture. In fact, if you do not, you can choose any bijection between the sphere and a $\mathbb R$-vector space, and you end up with a structure of vector space on your "sphere" (by transporting the structure). My point is, there exist such $\varphi$, but what you really want is not for $ \varphi_O$ to be only bijective : if your space already has a shape, you want it to be a homeomorphism.

And there is no homeomorphism between the sphere and a $\mathbb R$-vector space (for example because a vector space is contractible - you can shrink it continuously into a point - whereas the sphere is not ; you can look that up in any basic course of algebraic topology)

Answer 2

We could start from an easier problem---show that the $1$-dimensional sphere, i.e. the circle $S^1$, is not an affine space.

This one seems pretty easy---we can pick a point $A$ in $S^1$, a nonzero vector $v$ and a sequence of $n+1$ points of $S^1$, $(A_i)_{0\leq i \leq n}$ say, such that $$A_0 = A, \quad \overrightarrow{A_i A_{i+1}} = v \text{ for }0 \leq i \leq n, \quad A_n = A.$$ Then (ii) implies $$\sum_{i=0}^{n-1} \overrightarrow{A_i A_{i+1}} = \overrightarrow{AA} = 0.$$ From (ii) follows$$\overrightarrow{AA} = 0$$ as well, since$$\text{(ii)} \implies \overrightarrow{AA}+\overrightarrow{AA} = \overrightarrow{AA} \implies \overrightarrow{AA}=0.$$ And to conclude, we have thus $$n v = 0$$ which contradicts $v$ being nonzero.

Update: I think the issue is simple in principle. An affine space is just a space where we may subtract $2$ vectors---effectively a weakened linear space without an origin. So some coordinates---not unique but coordinates---must go to infinity. Compact spaces such as spheres cannot be affine spaces.

In my proof for the $1$-sphere, i.e. circle, I placed the points regularly, and then saw that the coordinate should have returned to the original one but it did not.

We can just do the same for the $2$-sphere and others, too, no? There are $1$-spheres inside $2$-sphere etc. If the difference were defined and smooth, then we could search for the $N$ points along the $2$-spheres such as the difference between the $N$th and $(N-1)$st point would be the same for all of them, just for the $1$-sphere. This must be possible if we pick a sufficiently small difference---many points on the circle around the sphere.

I would organize the thing by proving a stronger assertion, at least for all $D$-spheres but maybe for all compact manifolds. I am positive that at least for the complex affine spaces, the statement about non-compactness of affine spaces is true.

On the $2$-sphere, if we pick $2$ nearby points $A_1$ and $A_2$, they have some difference which is "small". So there must exist a point near $A_2$ such that$$A_3 - A_2 = A_2 - A_1.$$It must be somewhere on a topological circle surrounding by directions, and one of them must have the same direction, too. By taking a short-distance limit, we may be able to reconstruct a whole circle inside the sphere on which all the differences have the same direction, and then we adapt the length to get back after $N$ steps, and then it reduces to my circle proof above.

But there may be more elegant and much stronger proofs for all compact manifolds. There just are not any regions inside compact spaces where the coordinates could go to infinity, so it cannot be an affine space, I think.

Answer 3

Your definition is different from the definition here. (Though they are probably equivalent.)

Using that definition it is straight forward that $E$ is "flat". That is, given $A,B\in E$, the line joining them, $A+t\cdot (B-A)$ is contained in $E$.

To see this, note that $A=x+u$ and $B=x+v$, for $x\in E,u,v\in\vec{E}$. Then a typical point on the line connecting them is $x+u+t\cdot (v-u)\in E$.

Of course, $S^2$ doesn't have this property. (One may have to note that $S^2$ would have to be an affine subspace of $\mathbb R^3$, and conclude it would have to be a $2$-plane.)