# Affine Spaces are?

1. Jan 21, 2013

### Tenshou

What exactly is an affine space I am not sure I mean I recently got a book on fractals and they are working with them, They look very similar to linear (vector) spaces! I am not sure what the difference is... If some one could explain in great depth and detail what exactly these spaces are, and what they are good for. It would be of great thanks :)

P.S.
I think I know, yet what I know about this is incomplete.

2. Jan 22, 2013

### HallsofIvy

Staff Emeritus
An "affine space" is essentially a "flat" geometric space- you have points, you can calculate the distance between them, you can draw and measure angles and the angles in a triangle sum to 180 degrees (pi radians). You cannot add points or multiply points by a number as you can vectors.

If you define a coordinate system on an affine space you can then define the 0 vector as given by the origin, other vectors as given by the coordinates of a point. So, basically, a "linear space" is an "affine space" with a specific coordinate system.

3. Jan 22, 2013

### Tenshou

I do not understand, can you explain it more concretely? because this kind of picture it paints in my head looks more euclidean. I always thought they were like euclidean spaces but sheared by some factor? Like non-orthonormal basis could create an affine space, is this at least some what true?

4. Jan 23, 2013

### HallsofIvy

Staff Emeritus
As soon as you have any basis at all, not just orthogonal bases, you will have a vector. An "affine space" is a set of points without any given coordinate system or basis. As soon as you choose a basis or coordinate system, you have changed the affine space into a vector space.

5. Jan 24, 2013

### Tenshou

so an affine space is not a "special" type of vector space?

Edit: I mean what is the difference between an affine transform and a linear transform? besides they way they map points? is there any other difference in this?

Last edited: Jan 24, 2013
6. Jan 24, 2013

### HallsofIvy

Staff Emeritus
No, it is not. An affine space is not a vector space at all. Given an affine space, you can create a vector space from it by choosing any point in the affine space to be the "0" vector.

A linear transformation must map the 0 vector to the 0 vector. An affine transformation can map any specific point in the domain space to any point in the range space.

7. Jan 24, 2013

### mathwonk

think of three dimensional euclidean (x,y,z) space. the x,y plane is a vector subspace because it has an origin. but some other plane not passing through the origin is just an affine subspace because it has no particular choice of origin.

a linear transformation of three space will take a plane through the origin to another plane through the origin, but an affine transformation may take it to a plane not through the origin.

i.e. even though euclidean space has an origin, an affine map does not see the origin, and treats
it as if there isn't one. in this way we can use euclidean space to represent affine space, by forgetting where the origin is.

linear transformations are made up of coordinate functions like this

ax+by+cz,

but affine transformations are made up of coordinate functions like this:

ax+by+cz + d.

then we could say that the affine structure on euclidean space is that part of the euclidean structure that is preserved by affine transformations.

Last edited: Jan 24, 2013
8. Jan 25, 2013

### Tenshou

Thank the both of you for your wonderful explanations! So does this mean that the an affine space is Hausdorff to a parallel space that is linear (or perhaps another affine space?)? Like does it meet the T_1 axiom?

9. Jan 25, 2013

### Fredrik

Staff Emeritus
An affine space S isn't required to be a topological space, but you can always use the euclidean metric on the associated vector space V to define a topology on V, and then use it and a bijection from S into V to define a topology on S. (If f is such a bijection, there's a unique smallest topology on S such that $f^{-1}(E)$ is open for all open E). This way S will have all the topological properties of $\mathbb R^n$ for some n.

10. Jan 26, 2013

### Fredrik

Staff Emeritus
I'm going to elaborate a bit on affine spaces. (I needed to get these ideas straight in my own head anyway).

Suppose that we want a set S to have the property that the difference between two arbitrary members of S is a vector, but other linear combinations as+bs' are undefined, and no member of S has an algebraic property that distinguishes it from the rest (like the 0 in a vector space).

We can use a map
$$\phi:V\times S\to S.$$ If this map has the property that for each $(s,s')\in S\times S$, there's a unique $v\in V$ such that $\phi(v,s)=s'$, then we can think of v as a "difference" s'-s. This interpretation is suggested by the notation $\phi(v,s)=v+s$.

Suppose that we also want to be able to, for each s in S, define a vector space structure on S that makes s the additive identity (i.e. 0). Then we need to require some other things from $\phi$. Suppose that $\phi$ has the property that for each s in S, the map $\phi_s:V\to S$ defined by $\phi_s(v)=\phi(v,s)$ for all v in V, is a bijection. Then we can try to use the map $\phi_s$ to define a vector space structure on V that makes s the additive identity (i.e. 0). Let $a,b\in\mathbb R$ and $s',s''\in S$ be arbitrary. We define the linear combination $as'+bs''$ by
$$as'+bs''=\phi_s\big(a\phi_s^{-1}(s')+b\phi_s^{-1}(s'')\big).$$ Does this make s the additive identity? No, not automatically.
$$s'+s=\phi_s\big(\phi_s^{-1}(s')+\phi_s^{-1}(s)\big).$$ Because of this, we also require that $\phi(0,s)=0$ for all s in S. This ensures that $\phi_s^{-1}(s)=0$. And this simplifies the right-hand side above to s'.

This is all we need. Now we can call the triple $(V,S,\phi)$ an affine space. And then we can start abusing that terminology by calling S an affine space.

If you look up the definition of "affine space" in a math book, you will see that the conditions on $\phi$ are different, but the main reason why those conditions are used, is that they make the conditions I mentioned in this post true.

Last edited: Jan 26, 2013
11. Feb 2, 2013

### Tenshou

Thank you Sir Fredrik for your input. It helped a lot good Sailings to all of you guys for your inputs :3

12. Feb 2, 2013

### Fredrik

Staff Emeritus
You're welcome. I see a couple of mistakes in my post, so I need to make a comment about that. I mentioned a map $\phi:V\times S$, but I didn't say that V is a vector space. When I said "to define a vector space structure on V", I meant "to define a vector space structure on S". We're using the vector space V and the map $\phi$ to define many different vector space structure on S.

13. Feb 2, 2013

### Tenshou

Yeah I get what you mean, like phi is a group action on the space and set into the set, I get what you mean I assumed that V stood for Vector Space and S stood for Set, although I still don't understand the reason for the notation of the phi_s, could you please explain this? (this helped me, because I recently found an article online which goes into a little more detail about this topic, If anyone would like they can message me, and then I can get around to sending them a link to the paper, it is rather interesting.) Again, many thanks to all the people who did help and post.

14. Feb 3, 2013

### Fredrik

Staff Emeritus
You may need to be more specific. What about the notation bothers you?

We have a function $\phi:V\times S\to S$, and we want to use it to define a function from V into S. How would you do it? The obvious way is to pick any point s in S, and then use the map $v\mapsto\phi(s,v)$. It seems natural to call it $\phi_s$, since I used the function $\phi$ and an $s\in S$ to define it.

(The words "function" and "map" are interchangeable in my posts, and in most books. There are however some authors who prefer to use "function" only when the codomain is ℝ or ℂ).

Why don't you post it here? If it's a serious math article, I don't think you will get in trouble with the moderators.

15. Feb 4, 2013

### Tenshou

It is from the University of Penn, I believe...

Affine Space

Why wouldn't you call it $\phi_v$ isn't that what the functor is acting on >.< I apologize if I am being a little picky. I just don't understand if the set of vectors or points are being acted on?

16. Feb 5, 2013

### Fredrik

Staff Emeritus
Functor? $\phi_s$ is just a function from V into S. A functor (in the category of vector spaces) would take vector spaces to vector spaces. $\phi_s$ takes vectors in a specific vector space V to points in a specific set S.

For each s in S, there's a map $v\mapsto \phi(s,v)$ from V into S. I chose to denote it by $\phi_s$. You're suggesting $\phi_v$. First of all, how would you choose v? Second, these are infinitely many functions, one for each s, and you want to use the same notation $\phi_v$ for all of them? Then how would you interpret an expression like $\phi_v(u)$, where u is a member of V?

17. Feb 5, 2013

### Tenshou

Okay never mind you answered my question. So, the elements $s$, the point in the "affine set" are the one which are being "acted upon" by $\phi_s$, correct? this is what I am confused about. 【・_・?】 Isn't the function $\phi$ a bi-linear "map"? or is it only a bi-linear where there is a tensor product between the two and not a Cartesian product?

$\phi:V\times S\to S$
<----- This right here.

One moment, is $\phi(s,v)$ like a "scalar product" type thing? if so, then wouldn't that be like adding a vector with a point? how does this work?

18. Feb 5, 2013

### Fredrik

Staff Emeritus
No. I said that for each s in S, we define $\phi_s:V\to S$ by $\phi_s(v)=\phi(s,v)$ for all v in V. So each $\phi_s$ takes a member of V to a member of S.

Suppose that X,Y,Z are vector spaces. A map $f:X\times Y\to Z$ is said to be bilinear if for all $(x,y)\in X\times Y$ both of the maps $f_x:Y\to Z$ and $f_y:X\to Z$ defined by $f_x(v)=f(x,v)$ for all v in Y and $f_y(u)=f(u,y)$ for all u in X, are linear.

When we define $\phi$, we can't say that it's bilinear, since S isn't even a vector space. However, once we have used one of the $\phi_s$ maps to turn S into a vector space, then it makes sense to ask if $\phi$ is bilinear with respect to that vector space structure on S. It should be easy to check if it is, but I haven't done it myself, so I don't know for sure.

It doesn't work, unless we define those sums. We could do this using the function $\phi$ (just define $v+s=\phi(v,s)$), but this is kind of pointless. We're not interested in those sums. We're only interested in differences s'-s (where s and s' are both in S) and we define those by saying that s'-s is the unique v such that $\phi(v,s)=s'$.

19. Feb 5, 2013

### Tenshou

Oh okay, thank for cleaning this up for me.

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