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When we're introduced to $\mathbb{R}^3$ in multivariable calculus, we first think of it as a collection of points. Then we're taught that you can have these things called vectors, which are (equivalence classes of) arrows that start at one point and end up at another.

At this point $\mathbb{R}^3$ is an affine space, not a vector space: for two points $x, y \in \mathbb{R}^3$, the operation $x + y$ is meaningless (my professor likes to say: "You can't add Chicago and New York!") but the operation $x - y$ gives a vector (the vector which points from New York to Chicago). You can also add a point and a vector, which gives you a translated point.

The distinction between the point $(0, 1, 2)$ and the vector $\langle 0, 1, 2 \rangle$ is sometimes made.

But then we quickly move on to treating $\mathbb{R}^3$ as a vector space, where instead of a point $A$, you have vectors starting at the origin with their tip at $A$. For example, parameterized curves such as

$$r(t) = (t, t^2, 3t)$$

are called "vector-valued functions" and not "point-valued functions". So, my question is, what is the reason that we historically don't define two spaces -- $\mathbb{R}^3$ and $\mathbb{R}^3_{\text{affine}}$? (I'm sure there's better notation).

For example, my "point-valued function" $r(t)$ would be a function $\mathbb{R} \rightarrow \mathbb{R}^3_{\text{affine}}$, but its derivative $r'(t)$ (the velocity vector) would be a function $\mathbb{R} \rightarrow \mathbb{R}^3$. What would this make more difficult?

In particular, I know that $\mathbb{R}^3 \iff \mathbb{R}^3_{\text{affine}}$ is a bijection, and that we use this sometimes, but how often in multivariable calculus? If we are using it all the time, then it wouldn't make sense to emphasize the distinction.

Eli Rose
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2 Answers2

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Note that $\mathbb{R}^3$ explicitly introduces coordinates for three dimensional space, and coordinates are not necessary to have three dimensional affine space. You are right in observing that once we identify points by coordinates, there is an ambiguity in whether we mean a point or a vector when coordinates are given. But this scarcely means we don't consider $\mathbb{R}^3$ to be an affine space.

In general classic geometry deals with Euclidean space in affine terms and without coordinates. The magnitude of lengths and angles does not require a special designation of an origin, such as we have in synthetic geometry (introducing coordinates and making many arguments easier).

Abstractly we can consider any vector space to be an affine space by "forgetting" the designation of a distinguished origin. Many of the operations in multivariable calculus can be viewed as shifting the origin to a convenient point, e.g. a point on a surface, and carrying out analysis "locally" in a neighborhood of that point. In this sense we are making frequent use of the affine structure of $n$-dimensional Euclidean space.

hardmath
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  • I'd noticed how we shift the origin around all the time -- thanks for making that point! I suppose my question is then a pedagogical one. Why in undergraduate multivariable calculus do we mostly not distinguish between points and vectors? I imagine the answer will be "it's easier not to", which I'm fine with: I'm just looking for some examples of what it makes easier. – Eli Rose Oct 13 '15 at 20:15
  • As you described, your professor makes an effort with "We can't add New York and Chicago." Of course we can add vectors, and it is of some importance that we can "add" a vector to a point (translating it). It might be hard to justify making more of an issue of this in an undergraduate analysis course. After all it may be a termination of sorts for math classes taught to engineers and physics majors by math faculty. – hardmath Oct 13 '15 at 20:25
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In differential geometry, we deal with tangent spaces to manifolds. Let's make a precise distinction in our case. There are more than one definition of tangent space, amd one can show certain equivalences between them.

Take $p\in \Bbb R^3$ and define: $$T_p(\Bbb R^3) = \{(p,v) \mid v \in \Bbb R^3\}.$$ So far, $p$ and $v$ are just triples of real numbers, but we think of $p$ as the point and $v_p \equiv (p,v)$ as a tangent vector to $\Bbb R^3$ in $p$. The set $T_p(\Bbb R^3)$ has a natural vector space structure given by $(p,v_1)+\lambda(p,v_2) = (p, v_1+\lambda v_2)$, and is then isomorphic to $\Bbb R^3$ via: $$T_p(\Bbb R^3) \ni (p,v)\mapsto v \in \Bbb R^3.$$ The issue is that we have $\Bbb R^3 \cong T_p(\Bbb R^3)$, but it is not true that given a manifold $M$, we have $M \cong T_p(M)$, since $M$ is may not even be a vector space. And thus, confusion arises. Also, note that $T_p(\Bbb R^3) \neq T_q(\Bbb R^3)$ if $p\neq q$, so we can't "add Chicago to New York", but we'll have these tangent spaces isomorphic all right. For example, if you have a curve $r: I \to M$, now in general, we have that $r'(t) \in T_{r(t)}(M)$. When you start working with ambients other than $\Bbb R^3$ the distinction point/vector will be more clear and we'll be unable to hide all these isomorphisms under the carpet (when they exist).

Ivo Terek
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