Affine Spaces and Vector Spaces

[fresh_42]

The abstract structure is a linear space with a shift. Normally it is all part of a vector space $\mathbb{F}^{n+1}$ if the affine space is $n$ dimensional. And again, if UPenn doesn't define $E$ in a way I can decide $e\in E$, then give me a definition which does. I'm absolutely certain, that the full definition of UPenn and mine are equivalent. The proof is trivial, as long as you allow an embedding.

The term affine variety is by far more interesting than affine space. It is as I said: $(x,mx)$ are linear, $(x,mx+b)$ are affine, and each definition is just a more general way to say this.

 

[PeterDonis]

The abstract structure is a linear space with a shift.

Which just pushes the question back to "what is a linear space with a shift?"

if UPenn doesn't define $E$ in a way I can decide $e\in E$, then give me a definition which does.

I don't see why you need such a thing to define the abstract structure of an affine space. You only need it if you have a particular semantic model and you want to know which set of points in the model is to be considered as the set $E$. And the place to look for that is not in the definition of an affine space, but in the definition of the semantic model. For example, if I tell you that I am considering the points in a plane embedded in Euclidean 3-space as an affine space, then I am defining the set $E$ to be the set of points in that plane. But that's not part of the definition of an affine space. It's part of the definition of my particular model.

 

[PeterDonis]

I'm absolutely certain, that the full definition of UPenn and mine are equivalent.

I don't see how to check this without asking the authors of the UPenn article. Failing that, I would rather take their definition as it is given, than speculate on what they actually meant but didn't say.

 

[fresh_42]

I don't see why you need such a thing to define the abstract structure of an affine space.

It is as abstract as a vector space is. I don't see a reason for semantic play games which the definition without the shift is. Affine spaces are intuitively easy. I can't see a reason to make it more complicated than $v_0+\mathbb{V}$ what they simply are.

I don't see how to check this without asking the authors of the UPenn article. Failing that, I would rather take their definition as it is given, than speculate on what they actually meant but didn't say.

$A:$

... an affine space (Definition 2.1.1) as (omitting the degenerate case of the empty set) a triple $< E, \vec{E}, +>$, where $E$ is a nonempty set, $\vec{E}$ is a vector space, and $+$ is a binary operation that adds a vector (member of $\vec{E}$) to a point (member of $E$) to get another point. The addition operation then has to have some simple properties.

$B:$ $E=e_0+\vec{E}$.

$A\Longrightarrow B$:
This is trivially true, as $E=e_0+\vec{E}$ for any point $e_0\in E$:
$e_0+\vec{E}\subseteq E$ and $e\in \{\,e\in E\, : \,e=e+\vec{0}\,\}$ hence $E\subseteq e_0+\vec{E}$.

$B\Longrightarrow A$:
Let $E :=\vec{e}_0 +\vec{E}$ for a given vector space $\vec{E}$ and a given vector $\vec{e}_0 \in \vec{E}\oplus \mathbb{F}.$ Then $E$ defines a set of points by varying the vectors of $\vec{E}.$ We also have a binary operation $E \times \vec{E}\longrightarrow E$ given by
$$
(e,\vec{v}) \stackrel{\beta}{\longmapsto} \vec{e}_0+ (\vec{v}+\vec{v}_0) \in E\text{ given }e=\vec{e}_0+\vec{v}_0
$$
by definition $B$. Thus all properties of an affine space hold.

The definitions are equivalent. Debating semantics is mathematically irrelevant. Or do you want me to apply the forget functor on vectors in order to make "a vector defines a point" formal?

 

[PeterDonis]

Thus all properties of an affine space hold.

This shows that any vector space $\vec{E}$ satisfies all the properties of an affine space. Nobody has disputed that that is the case; @Dale pointed it out many posts ago.

But you appear to be claiming something more than that. You are saying that, in the UPenn definition, any point $e$ in the set of points $E$ must be the difference of two vectors $\vec{u}$ and $\vec{v}$ in the vector space $\vec{E}$. But the difference of two vectors is a vector. So you are claiming that, for any affine space, the set of points $E$ must be a set of vectors--in fact, it must be identical to the vector space $\vec{E}$.

If that were true, it would mean that any affine space is a vector space. But the whole point of the mathematical concept of an affine space is that it is not the same as a vector space. The UPenn definition leaves open the possibility that there are affine spaces in which the set $E$ of points is not a set of vectors. Yours does not. So I don't see how the two can be mathematically equivalent.

 

[fresh_42]

But you appear to be claiming something more than that. You are saying that, in the UPenn definition, any point $e$ in the set of points $E$ must be the difference of two vectors $\vec{u}$ and $\vec{v}$ in the vector space $\vec{E}$.

I claim $e-f=\vec{v}_e-\vec{v}_f \in \vec{E}.$

 

[fresh_42]

If that were true, it would mean that any affine space is a vector space. But the whole point of the mathematical concept of an affine space is that it is not the same as a vector space.

Exactly. The difference is exactly the shifting point: $\vec{0}\in \vec{E}$ is translated into some point $e_0\in E$. And all elements of $E$ share this shift, which is why it is no additional group (in general, as any vector space is an affine space trivially). Affine spaces lack the zero. But the difference of any two points is in $\vec{E}.$

 

[PeterDonis]

I claim $e - f = \vec{v}_e - \vec{v}_f \in \vec{E}$.

Here you are saying that the difference of two points is a vector. Nobody is disputing that; I believe it has already been pointed out earlier in this thread.

What you said earlier was that a point is the difference of two vectors, which is a different statement:

$a\in \mathbb{A}$ if there are vectors $\vec{u},\vec{v} \in \mathbb{V}$ such that $a=\vec{u}-\vec{v}$.

This statement is what I have been questioning, since the difference of two vectors is a vector, not a point.

 

[wrobel]

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 $V$.

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

 

[vanhees71]

I think the formal definition is: An affine space $\mathbb{A}$ with an underlying vector space $\mathbb{V}$ is the set of all points, which can be written as a difference of vectors from $\mathbb{V}$.

This is only a special case, making a vector space to an affine space by interpreting the vectors of a given vector space as the set of points of an affine space with the vector space also as the vector space of this affine space, introducing the difference of vectors as the vectors connecting the corresponding "points". That's of course a valid description, but it's not what affine spaces are good for nor what they are used for in geometry and physics.

You an define standard high-school Euclidean geometry formally simply as an affine space with a Euclidean vector space (2D for planar geometry or 3D for spatial; it's of course generalizable to any dimension). The points are not the same as the vectors, and that's how it's also used in physics.

Minkowski space is also an affine space but this time with a pseudo-Euclidean vector space (usually in 4D with a fundamental form of signature (1,3) or (3,1), but generalizable to any dimension too). Also here the points are not the same as the vectors.

So the physical application use the general definition of an affine space as a pair $(M,V)$ with a set of points, $M$ and a vector space $V$ with the specific algebraic structure described in Wikipedia (which may not be a "valid source", but in this case both the English and ther German entries are in my opinion just fine).

 

[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.