
#1
Jan2113, 08:12 PM

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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
Jan2213, 10:38 AM

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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
Jan2213, 09:59 PM

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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 nonorthonormal basis could create an affine space, is this at least some what true?




#4
Jan2313, 07:18 AM

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Affine Spaces are..?
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
Jan2413, 04:26 AM

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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? 



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Jan2413, 07:01 AM

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#7
Jan2413, 04:42 PM

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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. 



#8
Jan2513, 01:42 AM

P: 150

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
Jan2513, 05:58 AM

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#10
Jan2613, 03:08 AM

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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 righthand 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. 



#11
Feb213, 02:54 PM

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




#12
Feb213, 04:17 PM

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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
Feb213, 09:38 PM

P: 150

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
Feb313, 03:19 AM

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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). 



#15
Feb413, 11:56 PM

P: 150

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
Feb513, 06:35 AM

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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
Feb513, 09:46 AM

P: 150

##\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
Feb513, 11:28 AM

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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. 


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