# How to precisely distinguish vectors and points? [duplicate]

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Distinction between vectors and points

I have a doubt about the distinction between points and vectors. I know there's already a topic about that here in the web site, but i thought the correct was to create a new one. Well, the question is: in euclidean space we identify both points and vectors with elements of $\mathbb{R}^n$, but I know they're different things.

And I know that when dealing with general manifolds the situation gets worse and it's needed to define precisely the notion of a tangent space at each point of the manifold. So my question is: how is it possible to define precisely the distinction between points and vectors first in euclidean space and then in general manifolds ?

I've seem a book on differential geometry where the author introduces the operation of addition of points and multiplication of point by scalar, but i did think that these operations are meaningless geometrically speaking.

I've heard about the notion of an affine space, is that the correct way to make a rigorous distinction between vectors and points?

Thanks in advance for the help.

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## marked as duplicate by Rahul, sdcvvc, Michael Greinecker♦, William, Guess who it is.Aug 31 '12 at 10:03

I'm not sure I understand what you are looking for when you say "define precisely the distinction". It might help to explain how the answers to the previous question on the distinction between vectors and points are not what you are looking for. –  Rahul Aug 24 '12 at 19:11

I don't know that point/vector is best way to make this distinction. Think about position and velocity. We can think of both of these things as vectors. Position only has meaning when it is measured relative to something, we can call position the displacement from that location (usually the origin). If we measure relative to ourselves, the position changes when we move. However the velocity of an object is the same no matter where we are relative to it. We would say that velocity is translation invariant. (Please do not bring up physics)

When we say point we usually meaning something like position. When we say "vector" we usually mean something that isn't dependent on a distinguished point like the origin. Or at least, we make clear what the distinguished point is.

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At a very high level, mathematicians tend not to bother distinguishing between things when they're basically the same. It's quite hard to define precisely what I mean by 'basically the same', but it's a bit like having two isomorphic groups. Most mathematicians would - even if they didn't say so explicitly - have in the back of their mind the notion that the group of symmetries of a triangle is the same as the group of permutations of three elements. All you're doing is giving a different label to the elements of the group and possibly rearranging them. Similarly, it's easy to show that equivalence classes on a set X form a partition of X. But we can go in the other direction, and take any partition of X, and call two elements of X equivalent if they are in the same part of the partition. So it makes sense not to distinguish between equivalence relations on X and partitions of X - there's a clear bijection between them, and it's useful to use tools from one area to help with the other.

You mention that elements of $\mathbb{R}^n$ can be thought of either as vectors or as ordered $n$-tuples (or points). Since these are clearly basically the same thing, there is no point differentiating between them - if we treat them at the same object then we can use all the linear-algebra properties of vectors and the finite-sequence properties of ordered $n$-tuples at the same time. There's nothing to be gained from treating them as distinct objects.

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I'm guessing you can tell a point and an arrow apart if I draw tham on a paper. If that really is the case, then you understand the geometric difference between the two.

My guess is that you notice we use ordered tuples to write both, and you're confused because they look the same. Well they do look exactly the same sometimes!

In fact, you can use points to represent vectors! If you plot a point on an x-y chart, and draw an arrow from the origin to the point, you call that the position vector of the point. In this case, the components for the vector and the coordinates for the point are identical, and its easy to confuse. But the point itself is really just that dimensionless point, and the vector itself is the 1 dimensional oriented line segment (=arrow) from the point to the origin.

The bottom line is that you'll have to desensitize yourself to this, because things that look alike but aren't the same abound in mathematics. You can tell ordered pairs and open intervals in the real line apart, right?

Just hang on to the geometric interpretation, and you should be OK. A point has no dimensions, but we keep track of its location with those coordinates.

A vector is extant in a certain direction, with a certain length. The way we represent vectors is to slide their sources to the origin, and then write down what point their arrowhead lands on. The numbers we record vectors with record their length and direction, but not location. For vectors in ordinary Euclidean space, location is not important: just direction and length.

So you see, even though the tuple of numbers looks the same, it's actually keeping track of different types of information.

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rschwieb, thanks for your answer. I understand the affine structure, where we can add point to vector and subtract points. But for example, if I have a function $F : U \subset \mathbb{R}^n \to \mathbb{R}^m$, should I treat the objects of the image as points or vectors? Because if I treat as points I won't be able to add two of these functions for example, because addition of points is not defined. Thanks once more, and sorry if I said something silly. –  Leonardo Aug 24 '12 at 19:40
@Leonardo No no, I'm sure this question and sentiment is very common. Actually I won't be able to help you as much with affine space: I don't have a good feel for it. My feeling is that affine space is another layer of intuition to develop, separate from points and vectors (or from your intuition of vectors). I only know affine space is like "a vector space where you can't remember where you put the origin". –  rschwieb Aug 24 '12 at 19:44

The words "point" and "vector" are quite mathematically overloaded.

If your mean "point" and "vector" to be terms with geometric content, then "vector" most likely means to you "an element of a finite dimensional real vector space." Although one can attach geometric significance to elements of infinite dimensional spaces and elements of vector spaces defined over other fields. But let's stick with the finite dimensional real case. "Point" is a bit easier. It almost certainly means to you "an element of a manifold" (smooth or otherwise).

In this context, every vector is a point. But not all points are vectors. A vector is a point plus more structure.

Specifically, to each point in a smooth manifold, $p \in M$, one can attach a tangent space, $T_pM$. Then putting these tangent spaces together (say disjoint union) one gets $TM = \cup_{p\in M} T_pM$ (the tangent bundle). It turns out that the tangent bundle itself has the structure of a smooth manifold. Thus the elements of $TM$ are simultaneously points (of the tangent bundle manifold) and vectors (each element lives in a tangent (vector) space $T_pM$ for some $p\in M$). So the "points" in $TM$ have more structure than the points of $M$. If we choose an element $p \in M$, it may or may not make sense to perform "scalar multiplication". However, this always makes sense for the points in $TM$ (since each "point" belongs to some vector space). Next, given two points in $M$ it may or may not make sense to "add" them. This is also generally true in $TM$. However, if we carefully select $v,w \in TM$ so that $v \in T_pM$ and $w \in T_qM$ where $p=q$, then $v+w$ is defined (since both $v$ and $w$ belong to the same vector space).

Now when we start dealing with $\mathbb{R}^n$ everything collapses down. If you let ${\bf x} \in \mathbb{R}^n$. Then $T_{\bf x}\mathbb{R}^n$ is canonically isomorphic to $\mathbb{R}^n$ itself. Since everything is "flat" one can "parallel transport" all of the vectors tangent at ${\bf x}$ back to the origin. So $T_{\bf x}\mathbb{R}^n$ is essentially the same as $T_{\bf 0}\mathbb{R}^n$ (the tangent space at the origin). This in turn can be identified with $\mathbb{R}^n$ (identifying terminal points of vectors based at the origin with the terminal point itself). So in the end we essentially have $T_{\bf x}\mathbb{R}^n=T_{\bf 0}\mathbb{R}^n=\mathbb{R}^n$. So for most purposes we can totally ignore the distinction between "point" and "vector" when working with $\mathbb{R}^n$.

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Bill Cook, thanks for you answer. It was this kind of doubt I had. There's still one thing I'd like to ask: if I have a map $F : M \to N$ where $M$ and $N$ are manifolds, as I see elements of manifolds as points then this map receives a point and produces a new point. But what about $F : \mathbb{R}^n \to \mathbb{R}^m$ ? Should I treat the elements of the image as points or vectors? Because once treated as vectors I can add and multiply by scalars while if I treat as points those operations are geometrically meaningless if I understood well. Thanks again for your support. –  Leonardo Aug 25 '12 at 16:38
@Leonardo there's really no way to answer your question except, "It depends." Most of the time the difference between treating elements as either points or vectors is just a matter of taste. For your map $F$, there are cases where it makes most sense to think of it as a map from points to vectors and other times from vectors to points or points to points or vectors to vectors...it depends on the map! –  Bill Cook Aug 27 '12 at 0:55

Yes, the affine space is the correct concept to look at for the distinction of points and vectors.

However, there's a simple way you can get an uniform definition of vectors and points: Define your space as $S = \mathbb R^n\times\{0,1\}\subset\mathbb R^{n+1}$. Now you define the ordinary vector space operations on $R^{n+1}$, but with a restriction: The operations are only defined ibn $S$, and only if the result again ends up in the set $S$. You'll find that for the elements $v\in S$ with $v_{n+1}=0$ you basically face no restrictions; indeed, they are just forming the vector space $\mathbb R^n$. However, if $v_{n+1}=1$, there are severe restrictions: For example, you cannot multiply them with an arbitrary number $\lambda$, because $\lambda v_{n+1}=\lambda\notin\{0,1\}$. You also cannot add them (because then $v_{n+1}+w_{n+1}=2$. However, you can add to it a vector with $v_{n+1}=0$. You can also do affine combinations of them, that is, for two vectors $v,w$ you can create $\lambda v+(1-\lambda)w$ (because, again, the last component becomes $1$), and similarly for more than two vectors (note that this is now an "atomic" operation because the intermediate steps are not defined; however you can rewrite it so that all intermediate steps are defioned, too, as $v + \lambda(w-v)$). You also can subtract two of those vectors with last component $1$, which gives you a vector with last component $0$.

Now you probably can already guess what those vectors with $v_{n+1}=1$ represent: They represent the points in $\mathbb R^n$! You can add an $\mathbb R^n$-vector (i.e. an $S$-vector with last component $0$ to a point, and get another point. You can subtract two points, and get a vector. And the affine combinations of two points form the straight line through those points (and similarly, if you have three points not on a straight line, you get a plane from their affine combinations, and so on).

Also note that there's an $1:1$ relation between points and vectors: For each vector $(v,0)$ there's a point $(v,1)$ which you get by adding the vector $v_0$ to the origin $(0,1)$.

OK, now it looks as if the origin was somehow special: After all, we've got an unique association from the (special) null vector. However this is not really the case: We could do the same restriction using any other point $(o,1)$ as origin; then we get the $1:1$ association $(v,0)\leftrightarrow(v+o,1)$. Note that all above constructions are completely unaffected by this choice, because all the vector space operations which leave the last component fixed are actually operations which leave the constant vector fixed; also, where the last component cancels out, so does the fixed vector. Note that by changing the basis, using $(o,1)$ instead of $(0,1)$ as $n+1$st basis vector, you even recover the original form $(0,1)$ for the new origin expressed in the new basis.

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