Affine Spaces and Vector Spaces

[PeterDonis]

There had been some strange statements about the word origin. I think it should be clear that the zero vector is meant.

It was clear that that's what @Dale meant by "origin". But @vanhees71 was using a different definition of origin, the origin of a reference frame. You can certainly construct a correspondence between the two, as I described in post #27. But you do have to construct it.

 

[fresh_42]

This doesn't make sense. The difference of two vectors is another vector, not a point. If you have a set of points, the difference between two points is a vector, but that's going from the affine space to the vector space, not the other way around.

That's why it makes sense, the difference is a vector, which makes the endpoints of the difference building vectors a part of the affine set.

 

[fresh_42]

It was clear that that's what @Dale meant by "origin". But @vanhees71 was using a different definition of origin, the origin of a reference frame. You can certainly construct a correspondence between the two, as I described in post #27. But you do have to construct it.

Yes, that's what I meant with my example of tangent spaces if you avoid the vector field terminology.

 

[PeterDonis]

I think the formal definition is

The UPenn reference @Dale gave defines 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.

This seems to be basically the definition @vanhees71 was using.

 

[PeterDonis]

the difference is a vector, which makes the endpoints of the difference vectors a part of the affine set.

I'm not sure this makes sense either. If all you have is a vector space, the vectors don't have "endpoints". You have to already have a set of points to attach endpoints to the vectors at all.

Note that the UPenn definition I described in my previous post includes the set of points independently of the set of vectors; it doesn't derive the points from the vectors.

 

[fresh_42]

This seems to be basically the definition @vanhees71 was using.

The point is that you have to find a phrasing which doesn't use $\vec{v}_0$ in the definition, because this would bond the entire construction to an outer vector space where everything takes place. With the definition as difference, you avoid the kick-off and only need a vector space as reference.

It is simple: $y=mx$ is linear, $y=mx+b$ affine linear.

 

[PeterDonis]

The point is that you have to find a phrasing which doesn't use $\vec{v}_0$ in the definition

The definition I referenced from the UPenn article does not have $\vec{v}_0$ anywhere. I don't understand what issue you are raising here.

Do you agree with the UPenn definition or not? If you do, then for the sake of having a common definition to work from in this discussion, I would suggest that we use that one, since I understand what it's saying and I evidently don't understand what yours is saying.

If you don't agree with the UPenn definition, can you give a reference that you think has a better one?

 

[fresh_42]

O.k., let's go formal then.

An affine space is a pair $(\mathbb{A},\mathbb{V})$ of a set of points $\mathbb{A}$ and a vector space $\mathbb{V}$, where $a\in \mathbb{A}$ if there are vectors $\vec{u},\vec{v} \in \mathbb{V}$ such that $a=\vec{u}-\vec{v}$.

The point is, that you formally leave $\mathbb{V}$ and define a set of points instead. The difference does no longer belong to $\mathbb{V}$. It is an attempt to save the arrow model, which is lost in the definition of vector spaces, since $\vec{u}-\vec{v}$ doesn't start at the origin if we use arrows.

 

[fresh_42]

If you don't agree with the UPenn definition

I do. It is exactly what I have said. Just with other letters.

 

[PeterDonis]

let's go formal then

So, then, you do not agree with the UPenn definition, since you are now giving a different one?

Can you give a reference that uses your definition? And an explanation of what is wrong with the UPenn one?

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

I still don't understand how this defines $a$ as a "point" instead of as a vector. If all you have is a vector space, the difference of two vectors is a vector, not a point. Your definition here, as far as I can tell, just defines the affine space $\mathbb{A}$ as the vector space $\mathbb{V}$, since the difference of any two vectors in $\mathbb{V}$ is just another vector in $\mathbb{V}$.

 

[PeterDonis]

The point is, that you formally leave $\mathbb{V}$ and define a set of points instead. The difference does no longer belong to $\mathbb{V}$.

I don't see how this follows at all. As I said in my previous post just now, the difference of any two vectors in $\mathbb{V}$ is just another vector in $\mathbb{V}$. I don't see how you can then wave your hands and say the differences somehow no longer belong to $\mathbb{V}$.

 

[Dale]

This just shows that this claim is fairly common. It doesn't show why it is true (if it is true)

Nonetheless, it does indicate that many scientists agree that a vector space has an origin.

Do I have to point out that Wikipedia is not a valid source?

It is when it agrees with valid sources, which it does in this case.

In other words, this source is arguing against interpreting a vector space as a space of points, not for it.

No, it is arguing against representing a space of points as a vector space. I.e. it is arguing that the set of points is an affine space not a vector space precisely because a set of points do not have an origin and a vector space does. The source supports the claim that a vector space has an origin.

This source does not argue against interpreting a vector space as a set of points and, even if it did, it still clearly considers a vector space to have an origin.

I'm sorry, but I don't accept any of these references as supporting your position.

Then it is time to close this thread. I have given valid references supporting my position and you have given none supporting the opposite, namely that a vector space does not have an origin. If we cannot even agree on what the literature clearly says then there is no point.

You have to make the qulification that you interpret the vector space as an affine space in the above explained way. Otherwise it's not understandable, at least not for one trained in the German mathematical tradition.

No qualification is needed. A vector space satisfies all of the axioms of an affine space, so it is both. I don’t think that German affine spaces have different axioms.

 

[fresh_42]

I still don't understand how this defines aa as a "point" instead of as a vector. If all you have is a vector space, the difference of two vectors is a vector, not a point.

UPenn is exactly the same, although terribly incomplete as quoted. I assume we are talking about the same vector of →EE→ which defines all the points. Then have ei+→v0∈Eei+v→0∈E and e1−e2∈→Ee1−e2∈E→. If you have varying vectors, then you get again the entire space including the origin.

Your definition here, as far as I can tell, just defines the affine space AA as the vector space VV, since the difference of any two vectors in VV is just another vector in VV.

No, I define the affine space as →v0+Vv→0+V. This isn't a vector space anymore. But the difference of two elements of this set is a vector.

Edit: I forgot to explicitly mention that $\vec{v}_0 \notin_{i.g.} \mathbb{V}$. But I didn't say it belongs to $\mathbb{V}$ anyway.

 

[PeterDonis]

No, it is arguing against representing a physical space of points as a vector space. I.e. it is arguing that the set of physical points is an affine space not a vector space precisely because physical points do not have an origin and a vector space does.

Hm, ok, I see the distinction you are making. The issue is not with "a vector space is a set of points", it is with "a set of points is a vector space".

you have given none supporting the opposite, namely that a vector space does not have an origin.

Just to be clear: my argument has not been that a vector space does not have an origin. My argument has been that the definition of "origin" you are using when you say that a vector space has an origin is different from the definition of "origin" that @vanhees71 was using. You can construct a correspondence between them, as I've described (post #27), so, as I pointed out in response to @vanhees71, it is not "absurd" to make use of such a correspondence, but they're not the same definition.

A vector space satisfies all of the axioms of an affine space, so it is both.

If we use the UPenn definition of an affine space (in the article you referenced), I agree with this. On this definition, all vector spaces are affine spaces (with additional structure), but not all affine spaces are vector spaces. (Or, more precisely, not all sets of points $E$ in the UPenn definition are also vector spaces, but all vector spaces can be treated as sets of points $E$.)

 

[fresh_42]

Just to be clear: my argument has not been that a vector space does not have an origin. My argument has been that the definition of "origin" you are using when you say that a vector space has an origin is different from the definition of "origin" that @vanhees71 was using. You can construct a correspondence between them, as I've described (post #27), so, as I pointed out in response to @vanhees71, it is not "absurd" to make use of such a correspondence, but they're not the same definition.

Yes. That is why $T_pM$ is strictly speaking not a vector space, as its "origin" is $p$. But we identify $p$ with the origin and call it a vector space. But this vector space $\mathbb{V}$ is actually $T_pM=p+\mathbb{V}$, a vector space kicked off $0$ into $p$. That's where the word origin gets confusing, and what you correctly mentioned: you have to construct $p=\vec{0}$. In case of tangent spaces we have such a natural point, in GR not so much.

 

[PeterDonis]

UPenn is exactly the same, although terribly incomplete as quoted

Then I am missing something, because I don't see it as the same as yours.

I assume we are talking about the same vector of $\vec{E}$ which defines all the points.

In the UPenn definition, there is no such vector. There is a set of points $E$ and a vector space $\vec{E}$. Both are given. Neither is defined from the other.

I define the affine space as →v0+Vv→0+V. This isn't a vector space anymore.

Isn't $\vec{v}_0$ just the zero vector of the vector space? And isn't the space of $0 + V$ for all $V$ in the vector space, just the same vector space again? (Since adding two vectors gives another vector, and adding the zero vector to any vector just gives the same vector.)

 

[PeterDonis]

That is why $T_pM$ is strictly speaking not a vector space, as its "origin" is $p$.

Huh? If @Dale is correct that the origin of a vector space is the zero vector, then the origin of $T_p M$ is the zero vector of $T_p M$. This vector maps $p$ to $p$ (if we treat the manifold $M$ as flat, so we can map vectors to points), but that doesn't make it the same as $p$.

this vector space $\mathbb{V}$ is actually $T_pM=p+\mathbb{V}$, a vector space kicked off $0$ into $p$.

I'm not following you. What you are describing is not $T_p M$, it is the mapping I described above, of vectors in $T_p M$ to points in $M$ that we can construct if we treat $M$ as flat.

 

[fresh_42]

In the UPenn definition, there is no such vector. There is a set of points $E$ and a vector space $\vec{E}$. Both are given. Neither is defined from the other.

The UPenn reference @Dale gave defines 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.

This seems to be basically the definition @vanhees71 was using.

This is exactly when it becomes incomplete. If we have a set $E$ and a vector space $\vec{E}$ such that $E=E+\vec{E}$, then how do you get rid of the self reference?
I have $E=v_0+\vec{E}$ with a specified point $v_0$ which I considered as a vector in $\mathbb{V}\oplus \mathbb{F}\cdot \vec{v}_0$. I get $\vec{v}_0 \in E$ by the requirement $\vec{v}_0+\vec{0}\in E.$ so whether you look at this in one way or another doesn't make a difference.

 

[fresh_42]

I was referring to the discussion about origin as of a vector space in comparison to a reference frame.

 

[PeterDonis]

If we have a set $E$ and a vector space $\vec{E}$ such that $E=E+\vec{E}$,

That's not what the definition says. The definition says that we have a set of points $E$, a vector space $\vec{E}$, and an operation $+$ that combines a point from $E$ and a vector from $\vec{E}$ to get another point from $E$. The operation does not define $E$.

 

[fresh_42]

That's not what the definition says. The definition says that we have a set of points $E$, a vector space $\vec{E}$, and an operation $+$ that combines a point from $E$ and a vector from $\vec{E}$ to get another point from $E$. The operation does not define $E$.

Then what defines $E$? When does a point belong to $E$?

I use location vectors, meaning their endpoints as point set, and UPenn abstractly says set. In the end it is a plane hanging from the hook meant for the candelabrum. I considered the room and tried to get rid of the rope by using the difference. We obviously don't know what UPenn does, as you just said it did not define $E$. So why do you want to use it all the time, if it doesn't define the points of $E$?

I still think UPenn means the end of the rope as the one point the entire plane is attached to.

 

[PeterDonis]

Then what defines $E$?

Why does anything have to define $E$ over and above saying that it's a set of points?

If you can ask this question, I can equally well ask what defines $\vec{E}$? If one requires additional justification, so does the other.

Or, to put it another way, if you think $\vec{E}$ can just be defined by saying "here's a vector space", but $E$ can't just be defined by saying "here's a set of points", how is it that I'm allowed to conjure a vector space out of nowhere, but I'm not allowed to conjure a set of points out of nowhere?

I use location vectors, meaning their endpoints as point set

To me this means that, instead of just considering an affine space as an abstract mathematical structure, you are considering a particular semantic model of it, and incorrectly insisting on additional properties that are present in your particular semantic model, but not in the abstract mathematical structure itself.

So why do you want to use it all the time

I suggested using the UPenn definition for purposes of this discussion because @Dale linked to it and it seemed to me to capture the abstract mathematical structure of an affine space reasonably well.

if it doesn't define the points of $E$?

It doesn't define what the points of $E$ refer to precisely because it is defining an abstract mathematical structure, not a particular semantic model of it.

I still think UPenn means the end of the rope as the one point the entire plane is attached to.

I don't think the UPenn definition is committing itself to any particular semantic model. It's just trying to define the abstract mathematical structure of an affine space.

 

[fresh_42]

I don't think the UPenn definition is committing itself to any particular semantic model. It's just trying to define the abstract mathematical structure of an affine space.

Sure, that's what I'm saying: we have to find a definition which doesn't require the set-off $v_0$. But we also need a criterion to decide whether a given point belongs to $E$ or not. We cannot list them. If we cannot decide membership to $E$, then it is a useless definition.

If we define $E=e_0 + \vec{E}$ or as $E=\vec{e_0}+\vec{E}$ is indeed semantics.
If we allow $e_0$ to be any point, then we get $E=E+\vec{E}$ which doesn't help us.

The criterion is: $e,f\in E \Longleftrightarrow e-f \in \vec{E}$. If we want to define it for a single point, then we have to specify a certain element $e_0$ and define $E=\{\,f=e_0+\vec{e}\,\}$. We can use any point of $E$ as $e_0$, but we have to fix it. Otherwise we have only the difference.

 

[Dale]

In my experience the UPenn type of definition basically says that anything, for convenience labeled $E$, that has the subsequent properties is an affine space.

 

[PeterDonis]

we have to find a definition which doesn't require the set-off $v_0$.

There is no "set-off" in the UPenn definition that I can see.

we also need a criterion to decide whether a given point belongs to $E$ or not.

A particular semantic model might have points that are in the set $E$ and points that aren't, so you would need such a criterion; but that's a property of that specific semantic model, not of the abstract structure itself. There can be other semantic models in which all the points in the model are in the set $E$, so no such criterion is needed.

In other words, the criterion you are talking about is a property of particular semantic models, not of the abstract mathematical structure itself.

 

[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).