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I don't understand why vectors can't be curved. For example when specifying the angle vector for an object in cylindrical/spherical coordinates, I think a curved vector would make more sense.

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    $\begingroup$ Vectors aren't straight, either $\endgroup$ – Hagen von Eitzen Apr 7 '17 at 6:14
  • $\begingroup$ What does curved mean? $\endgroup$ – Alekos Robotis Apr 7 '17 at 6:24
  • $\begingroup$ By "curved vectors", do you mean wrapping an "ordinary, straight" vector on the curved side of a cylinder, or on a sphere, for example? $\endgroup$ – Frenzy Li Apr 7 '17 at 6:25
  • $\begingroup$ Maybe this is what the OP hoped for: en.wikipedia.org/wiki/Curvilinear_coordinates $\endgroup$ – J.G. Apr 7 '17 at 6:45
  • $\begingroup$ @HagenvonEitzen I thought they were straight?! $\endgroup$ – samjoe Apr 7 '17 at 6:48
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Well, a short answer is that we use vectors to do linear algebra and geometry, and we use limits and calculus to apply those concepts to curved objects. So, a vector is a basic building block, and shouldn't be too broadly defined.

Here's a non-answer that might help, or might be confusing: if vectors are curved, how will you take a dot product: that is, how will you measure the angle between two vectors numerically?

However, I will try to answer what I think you're getting at as I think it's a very interesting question. If you think of a "vector" as "a magnitude and direction" and would like to use a vector to say how far one has traveled in a given direction, then if you're talking about directions on a curved object like the surface of the Earth, for example, it makes sense to talk about them being curved. The best way of encapsulating this in my opinion is the notion of geodesic, which is like a line segment but it is curved to fit the space. (Essentially, a geodesic segment between points $A$ and $B$ is the shortest path from $A$ to $B$; in a flat space that's a line segment.) Unfortunately, geodesic segments with a direction don't behave like vectors in very important ways, one of which is: you can't add them.

Yes, you can fit two geodesic segments head to tail. Well, maybe--in a non-flat world, it's hard to agree on a consistent idea of "direction", so moving vectors around is suddenly complicated. But even if you pick some coordinates that line up nicely, under the "fit head of $\vec w$ to tail of $\vec v$" definition of addition, then in curved spaces, for most vectors, $\vec v + \vec w \ne \vec w + \vec v$. In fact, differential geometers use these discrepancies to measure curvature! (As well as to decide how coordinate systems should work for curved spaces.) Although I think to start out learning about geodesics and curvature it is easier to read about how Gauss thought of curvature for surfaces, specifically about the angles of triangles.

What we do instead for curved spaces is use tangent vectors, which I think of as the idea of a direction "if you could go 'straight' in that direction". We use derivatives to relate tangent vectors from one place to another, and integrals (line integrals) to measure the distance traveled along a curve. I would argue we use vectors to define not only coordinates but what the idea of "straight" vs "curved" means in the first place.

Finally, while I personally think of linear algebra as geometric, and vectors as geometric objects, even if they're in many dimensions, and I encourage others to do so: for many many applications, vectors and matrices are simply useful ways of organizing numbers that belong to some data. And matrix multiplication, dot products, and other concepts of linear algebra, which are geometric, have other applications. So, you can think of vectors as just a bunch of coordinates, or as objects with a magnitude and direction, but to make those definitions agree we need something else for curves.

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Since you tagged this question with linear algebra tag, I'll answer this question based on Vector spaces.
If you want vectors to represent curves, you'll lose the linearity in Vector spaces. For example: in $\mathbb R^2$, we can't define a sub-space as $V = (t^2,t^3): t \in \mathbb R$, because $ (1^2,2^3)\in V $, but $2*(1^2,2^3)=(2,2^4)\notin V $. So, non-trivial subspaces consist of the form $S=\{(x,mx):x \in \mathbb R \}$. So, curves cannot be subspaces of $\mathbb R^2$.

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I think the best answer so far is the one in the comments remarking that vectors aren't straight, so I'll elaborate on that a bit.

It's commonly taught to high school students that a vector is a line with a length and a direction. This is not true.

A vector is an element of a vector space (no kidding, that's the formal definition). It is almost always denoted as a coordinate $n$-tuple, such as $(3,-2,1)\in\mathbb{R}^3$. This is not a line segment of length $\sqrt{3^2+(-2)^2+1^2}=\sqrt{14}$ that is pointed in a certain direction. We will sometimes imagine or draw a line segment of that length in a certain direction as a visual representation when the geometric aspect of vectors are important, but the vector is the point at the end of that line and not the line itself. This is why (from a formal POV) your question makes no sense. How can a point be curved (or straight for that matter)?

You say

For example when specifying the angle vector for an object in cylindrical/spherical coordinates, I think a curved vector would make more sense.

If by that what you mean is that it makes more sense to you to represent vectors with arcs along the surface of a cylender or sphere instead of straight lines, then by all means draw them like that. If you don't change the endpoint of the line, you haven't changed the vector because the visual picture is just a representation to help you grasp what is going on.

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    $\begingroup$ So you're saying a coordinate is a vector? I'm not sure how Force (which is a vector) can be considered a coordinate $\endgroup$ – Goldname Apr 7 '17 at 16:57
  • $\begingroup$ @Goldname There's a very important difference between a physical object or phenomenon and the mathematics we use to model it. You're right that force isn't a coordinate, but it's also not a line! Force is modeled as a vector quantity, where we understand that the direction of the vector is the direction of the force, and the magnitude of the vector is the amount of force applied. That information is all encoded in a single coordinate. $\endgroup$ – Stella Biderman Apr 7 '17 at 17:00
  • $\begingroup$ I disagree that a vector is "not" a line itself, and I strongly disagree that one should ever draw "curved vectors". It is true that the magnitude of a vector can convey other things besides "length". However, the geometry of vectors treated as physical objects with lengths and directions is not something to casually discard. $\endgroup$ – Elizabeth S. Q. Goodman Apr 7 '17 at 17:54
  • $\begingroup$ @ElizabethS.Q.Goodman Vectors are elements of s vector space, which are commonly represented as coordinates: they are definitely not lines. If drawing them curved improves the OP's intuition, why shouldn't they draw them curved if it gets them the right answer? $\endgroup$ – Stella Biderman Apr 7 '17 at 21:20
  • $\begingroup$ Not all vector spaces even have a well defined notion of a norm, so idk how you even want to define the "length" of that line in general. $\endgroup$ – Stella Biderman Apr 7 '17 at 21:26
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As long as we don't have precise definition of "curve" your question does not make any sense!

If you have geometric intuition of curve then I can say vector can have curvature.

for example the set of all real valued functions defined on a nonempty set (like sphere) makes a vector space. In particular $f(x) = x^2$ is a vector in the space of all real valued function defined on $R.$

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I hope it's worth something 2 years later:

A vector is defined as a property at a point of a curve, and only at that point. Avoiding the fundamental mathematical definition of vectors, we can look at the intuitive definition. It is the direction in which something is headed at that point at that moment; where will it be an infinitesimal time after that moment? We arrive at the definition of the derivative, which takes the difference of position between two moments (if your parameter is time) that are infinitesimally apart in time (or any other parameter you work in). $\frac{df}{dt}=\frac{f(t)-f(t+dt)}{dt}$ might clear up what I'm talking about. So you look at two points (at $t$ and at $t+dt$), and if you want to represent any property of two points in space (Euclidean or whatever) it's gonna be a straight line. To introduce curvature, you need a third point, that's why the second derivative involves also a point at $t-dt$.

You could introduce an entity that takes the second derivative and makes a vector out of it, but it would be a whole different thing (and mathematically it would be difficult, if not impossible, to define it on a manifold other than simply a vector in the vector space at that point, because vectors are basically found to be no different from derivatives, so the phrase "make a vector out of a derivative" is non-sensical actually).

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