What are differences between affine space and vector space?

I know smilar questions have been asked and I have looked at them but none of them seems to have satisfactory answer. I am reading the book a course in mathematics for student of physics vol. 1 by Paul Bamberg and Shlomo Sternberg. In Chapter 1 authors define affine space and writes:

The space $\Bbb{R}^2$ is an example of a vector space. The distinction between vector space $\Bbb{R}^2$ and affine space $A\Bbb{R}^2$ lies in the fact that in $\Bbb{R}^2$ the point (0,0) has a special significance ( it is the additive identity) and the addition of two vectors in $\Bbb{R}^2$ makes sense. These do not hold for $A\Bbb{R}^2$.

Please explain.

Edit:

How come $A\Bbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $A\Bbb{R}^2$ does not make sense? Please give concrete examples instead of abstract answers . I am a physics major and have done courses in Calculus, Linear Algebra and Complex Analysis.

Solution 1:

Consider the vector space $\mathbb{R}^3$. Inside $\mathbb{R}^3$ we can choose two planes, $P_1$ and $P_2$. The plane $P_1$ passes through the origin but the plane $P_2$ does not. It is a standard homework exercise in linear algebra to show that the $P_1$ is a sub-vector space of $\mathbb{R}^3$ but the plane $P_2$ is not. However, the plane $P_2$ resembles a $2$-dimensional vector space in many ways, primarily in that it exhibits a linear structure. In fact, $P_2$ is a classical example of an affine space.

$\,\,\,\,\,\,\,\,\,$

One defect of the plane $P_2$ is that it has no distinguished origin. One can artificially choose a point and redefine the algebraic operations in such a way to give it an origin, but that is not inherent to $P_2$. Another problem is that the sum of two vectors in $P_2$ is no longer in $P_2$. One can think of $AR^{2}$ as being modeled on this situation.

Solution 2:

Consider an infinite sheet (of idealised paper, if you like). If it is blank, then there is absolutely no way to distinguish between any two points on the sheet. Nonetheless, if you do have two points on the sheet, you can measure the distance between them. And if there is a uniform magnetic field parallel to the sheet, then you can even measure the bearing from one point to another. Thus, given any point $P$ on the sheet, you can uniquely describe every other point on the sheet by its distance and bearing from $P$; and conversely, given any distance and bearing, there is a point with that distance and bearing from $P$. This is the situation that the notion of a 2-dimensional affine space is an abstraction of.

Now suppose we have marked a point $O$ on the sheet. Then we can "add" points $P$ and $Q$ on the sheet by drawing the usual parallelogram diagram. The result $P + Q$ of the "addition" depends on the choice of $O$ (and, of course, $P$ and $Q$), but nothing else. This is what the notion of a 2-dimensional vector space is an abstraction of.

Solution 3:

The easiest way for me to tell the two structures apart is their axioms.

A vector space is an algebraic object with its characteristic operations, and an affine space is a group action on a set, specifically a vector space acting on a set faithfully and transitively.

Why do we say that the origin is no longer special in the affine space? The issue is that both $V$ and $X$ are usually written as $\Bbb R^n$, although we are thinking of each of the two copies of this in different ways. The deal is that the set $X=\Bbb R^n$ really doesn't distinguish any of its elements... they're all the same. But in the vector space $\Bbb R^n$, you can spot the origin right away, called out in the axioms.

Why do we say affine points can be subtracted but not added? That makes it seem like there are indeed operations within the affine space just like there are in the vector space, blurring the picture.

The reason is precisely because of transitivity: if $V$ acts on $X$ so that $X$ is an affine space (written additively), then for any $x,y\in X$, there is a $v$ such that $v + x = y$. I've written the group action additively here, but it is suggestive to rewrite this as $y-x = v$ and confuse the element $v$ of the vector space with an element of $X$.