13

I am reading this introduction to Mechanics and the definition it gives (just after Proposition 1.1.2) for an affine subspace puzzles me.

I cite:

A subset $B$ of a $\mathbb{R}$-affine space $A$ modelled on $V$ is an affine subspace if there is a subspace $U$ of $V$ with the property that $y−x \in U$ for every $x,y \in B$

It later says that this definition is equivalent to to the usual one, namely that of closeness under sum with elements of a $U$, but it seems to me that there is a problem with the first definition. Just imagine the usual $\mathbb{R}^2$ plane as an affine space modeled on $\mathbb{R}^2$. According to this definition the subset $\{(0,0);(0,1)\}$ is an affine subspace, while this is not so according to the usual definition of an affine subspace. Is there an error in the book?

Thom
  • 133
  • 1
  • 1
  • 4

2 Answers2

31

If you take a subspace and shift it away from the origin, you get an affine subspace.

In other words, an affine subspace is a set $a+U=\{a+u \;|\; u \in U \}$ for some subspace $U$.

Notice if you take two elements in $a+U$ say $a+u$ and $a+v$, then their difference lies in $U$: $(a+u)-(a+v)=u-v \in U$. [Your author's definition is almost equivalent to the one I've given above. The author mistakenly says "for all $x,y$ when it should be for any fixed $x$ all $y$ lie in there iff $x-y$ lie in the subspace.]

If you are familiar with a bit of modern algebra, affine subspaces are just elements of quotient vector spaces. So for example, given $U$ a subspace of $V$, the set $V/U = \{ a+U \;|\; a \in V\}$ is the quotient of $V$ by $U$. It is a vector space itself (briefly, its operations are $(a+U)+(b+U)=(a+b)+U$ and $s(a+U)=(sa)+U$).

More concretely, the affine subspaces associated with $U=\{0\}$ are $a+U=\{a+0\}=\{a\}$ so $V/\{0\}$ is essentially just the points of $V$ itself.

A one dimensional subspace of $\mathbb{R}^n$ is a line through the origin. The corresponding affine subspaces are all lines (not just those through the origin). Specifically, if $U$ is a line through the origin, then $a+U$ is a line parallel to $U$ which passes through $a$.

Likewise, two dimensional subspaces of $\mathbb{R}^n$ are planes through the origin whereas the two dimensional affine subspaces are arbitrary planes.

Bill Cook
  • 30,374
  • -1; this doesn't answer the question. – Qiaochu Yuan Feb 24 '12 at 01:23
  • OK, so the "definition" above is not really a good definition, right? It is just a necessary (and not sufficient) condition for a set $B$ to be an affine subspace. – Thom Feb 24 '12 at 01:24
  • @Thom: it's actually a trivial condition as written. – Qiaochu Yuan Feb 24 '12 at 01:25
  • Bill, thanks you for your interesting explanation. I choose Qiaochu's answer because it addresses my question more directly, but I think I will look at these concept more deeply, so your answer will be useful to me – Thom Feb 24 '12 at 01:30
  • 8
    While this may not answer the question, it is a very neat and nicely put description of the difference between linear and affine regarding vector spaces – Marcos Oct 19 '15 at 02:10
  • Complementary to this very neat description: From the perspective of universal algebra, it is safe to work with the elementary definition of a substructure: An affine subspace $(B, W, \oplus)$ of an affine space $(A, V, +)$ is an affine space (!) such that $B \subseteq A$, such that $W$ is a vector subspace of $V$ and such that $\oplus$ is the restriction of $+$ to $B \times W$. – Max Flow Aug 17 '23 at 09:39
6

The definition you cite is incorrect (so yes, there is an error). Indeed, letting $U = V$ every subset is an affine subspace according to this definition.

Qiaochu Yuan
  • 468,795
  • By the way, for future reference this "modeled after" terminology is not standard in pure mathematics. One would say that an affine space is a torsor or principal homogeneous space for a vector space $V$. – Qiaochu Yuan Feb 24 '12 at 18:53
  • 1
    Although this question is old, let me add an example certifying falseness of the cited definition: $(\mathbb{R}_0^+, \mathbb{R}, +)$ is not an affine subspace of $(\mathbb{R}, \mathbb{R}, +)$ because it is not an affine space because $\mathbb{R}_0^+ + \mathbb{R} \not\subseteq \mathbb{R}_0^+$. Yet, it meets the condition of the cited definition as $\mathbb{R}_0^+ - \mathbb{R}_0^+ = \mathbb{R}$ which is identical to and thus a vector subspace of $\mathbb{R}$. – Max Flow Aug 17 '23 at 09:24