Affine Spaces and Vector Spaces

[vanhees71]

I will bring a definition from Analyse mathématique by Laurent Schwartz

We shall say that the nonvoid set $E$ is an affine space (over the field $\mathbb{R}$ or $\mathbb{C}$) if
there is a vector space $V$ and a mapping $h: E\times E\to V$ such that
1) $h(A,B)+h(B,C)+h(C,A)=0$
2) for any fixed element $A\in E$ the mapping $B\mapsto h(A,B)$ is a bijection of $E$ onto $V$.

$\vec{AB}:=h(A,B)$

That's in a nutshell precisely the general definition of an affine space that I always have in mind, when I talk about affine spaces, and that's how it's used in Physics to describe both physical space in Newtonian mechanics (with the addition that the vector space in this case has also a scalar product and thus inducing the usual notions of Euclidean lengths and angles in addition to the most general affine space described above) as well as special-relativistic spacetime (Minkowski space, with the additional structure of a indefinite fundamental form of the vector space with signature (1,3) or (3,1), making it a specific pseudo-Euclidean affine space).

In this formulation, I'd refer to the point $A$ as the origin of the chosen reference frame, and given this arbitrary choice of $A$ you can identify the vectors uniquely with each point (by definition), but I'd still not identify simply the vectors with the points (though you can of course do so, and there seems Dale's and my mutual misunderstanding come from; it's nothing wrong with either view in my opinion).

 

[cianfa72]

Sorry, I've a doubt about the properties of the set of displacements in an affine space.

From the definition of affine space the set of displacements commute since they form a vector space. Is the reverse true ? In other words if we have a set of abstract points $A$ and a set of commutating displacements on it can we always endow the set $A$ with an affine structure turning it into an affine space ? Thank you.

 

[vanhees71]

By definition an affine space consists of a set of points and a vector space, and for any two points $AB$ there's a unique vector $\overrightarrow{AB}$ and for any vector $\vec{V}$ and any point $A$ there's always a point $B$ such that $\overrightarrow{AB}=\vec{V}$. Vector addition in terms of vectors described with two points is defined by $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. Vector addition is by definition commutative.

 

[cianfa72]

Take for example a 2D paraboloid with the exponential map as mapping at each point. On the paraboloid it is 1-to-1 from the tangent space at each point and the points on the 2D surface.

Now in order to endow it with an affine structure what is needed is a path-indipendent affine connection such that displacements commute.

Is the above requirement actually sufficient to endow the paraboloid with an affine structure ?

 

[jbergman]

Take for example a 2D paraboloid with the exponential map as mapping at each point. On the paraboloid it is 1-to-1 from the tangent space at each point and the points on the 2D surface.

Now in order to endow it with an affine structure what is needed is a path-indipendent affine connection such that displacements commute.

Is the above requirement actually sufficient to endow the paraboloid with an affine structure ?

The question is if you can fine an affine connection with the properties you listed. My guess is no if the space isn't flat.

 

[pervect]

The question is if you can fine an affine connection with the properties you listed. My guess is no if the space isn't flat.

Well, to determine whether a space is flat or not, we need, as far as I know, either a connection or a metric. If we have both, one of the questions that comes to mind is if they both agree on whether the "space" is flat or not. One reason I'm concerned is that I know we can make a flat slab (space) non-flat, by changing the metric on it.

The specifics are Einstein's "heated slab" argument in "Relativity, the special and general theory", chapter 24, https://www.bartleby.com/173/24.html. Basically, a "flat" heated slab becomes "curved" via the expansion and contraction of rulers used to measure distances on the slab via a "temperature field" which affects the rulers and hence the distance between points and the metric.

Some clarity on what makes a space "the same space" seems to be needed here, but I'm not sure of what the answer to this question is.

Personally, I tend to have intuitions about curvature based on the idea that a metric exists and represents something physical. Then there is an equivalence class of metrics that represents the physics of the space, and metrics not in this equivalence class don't.

I'm assuming the situation with connections is similar, but I suppose I don't have a formal proof handy that my ideas are correct even for the case of metrics. Probably some details need to be filled in, a metric seems more like it determines a chart to me than determining the whole space.

 

[jbergman]

Well, to determine whether a space is flat or not, we need, as far as I know, either a connection or a metric. If we have both, one of the questions that comes to mind is if they both agree on whether the "space" is flat or not. One reason I'm concerned is that I know we can make a flat slab (space) non-flat, by changing the metric on it.

The specifics are Einstein's "heated slab" argument in "Relativity, the special and general theory", chapter 24, https://www.bartleby.com/173/24.html. Basically, a "flat" heated slab becomes "curved" via the expansion and contraction of rulers used to measure distances on the slab via a "temperature field" which affects the rulers and hence the distance between points and the metric.

Some clarity on what makes a space "the same space" seems to be needed here, but I'm not sure of what the answer to this question is.

Personally, I tend to have intuitions about curvature based on the idea that a metric exists and represents something physical. Then there is an equivalence class of metrics that represents the physics of the space, and metrics not in this equivalence class don't.

I'm assuming the situation with connections is similar, but I suppose I don't have a formal proof handy that my ideas are correct even for the case of metrics. Probably some details need to be filled in, a metric seems more like it determines a chart to me than determining the whole space.

https://en.m.wikipedia.org/wiki/Affine_connection has a section on affine spaces with some good information. I haven't digested it all but it says the following, "an affine space is a manifold equiped with a flat cartan connection.

 

[jbergman]

Take for example a 2D paraboloid with the exponential map as mapping at each point. On the paraboloid it is 1-to-1 from the tangent space at each point and the points on the 2D surface.

I doubt this claim about the 1-1 mapping. Imagine a geodesic that starts at a point p on the side of the parabola and goes around the bowl of the parabola and returns to the same point, similar to a great circle on a sphere or a geodesic around a cylinder.

 

[Orodruin]

Well, to determine whether a space is flat or not, we need, as far as I know, either a connection or a metric.

No, you need a connection and that’s it. However, if you have a metric, the Levi-Civita connection is the unique metric compatible and torsion free metric and is often implied as the connection.

The question is if you can fine an affine connection with the properties you listed. My guess is no if the space isn't flat.

Flatness is a property of the connection as per above.

Regarding the paraboloid, it has the same topology as the affine 2D space so it is indeed possible to endow it with an affine structure simply by adopting the affine structure of the projection on the plane. However, it will not be compatible with the metric induced by the embedding in $\mathbb R^3$.

 

[cianfa72]

Regarding the paraboloid, it has the same topology as the affine 2D space so it is indeed possible to endow it with an affine structure simply by adopting the affine structure of the projection on the plane. However, it will not be compatible with the metric induced by the embedding in $\mathbb R^3$.

As far as I can tell, the projection $\pi$ of the paraboloid on the plane is actually an homeomorphism. Then you are saying that we can use $\pi^{-1}$ to "inherit" the affine structure of the plane on the paraboloid. It turns out that the inverse map of straight lines on the plane are defined as geodesics on the paraboloid. Since the mapping is 1-1, starting from an arbitrary point they do not intersect as the shealf of straight lines from the corresponding point on the plane.

So I think the implied connection on the paraboloid may look a bit "weird" however it works as required.

 

[cianfa72]

So if the above #80 is correct then we can always endow any manifold (globally) homeomorphic to the $\mathbb R^2$ plane with an affine structure turning it in an affine space.

 

[Orodruin]

Well, yes. That does not mean that that affine structure has any sort of meaning or that it is metric compatible.

 

[jbergman]

So if the above #80 is correct then we can always endow any manifold (globally) homeomorphic to the $\mathbb R^2$ plane with an affine structure turning it in an affine space.

I think we would need them to be diffeomorphic not just homeomorphic since connections require a smooth structure. For examples where these don't coincide see https://en.wikipedia.org/wiki/Exotic_R4.

In addition, since the paraboloid is diffeomorphic to the plane they are essentially the same thing in the category of smooth manifolds. In other words a paraboloid could be considered just a different coordinate system for the plane.

The same as a sphere and an ellipsoid are the same objects in the category of smooth manifolds.

They become different objects when we add additional structures to differentiate them like differing connections or metrics.