Why can the "arrow" vectors be slid parallelly anywhere in space? So, I was studying for my first E&M course from Griffiths and the first chapter is mostly math preliminaries. Griffiths has a section starting page 10 on "How Vectors Transform?" where he gives a definition of vectors as things which transform nicely under some position transformations i.e. he is not using the general definitions of abstract vectors but also imposing a representation group on them. So, despite their simplicity, I am utterly confused with the so called "arrow" vectors.
One of his exercises is to show that translational transformations are vector-component invariant and that's where I began to seriously doubt things. What is the so-called "Arrow" Vector? Does it have a proper definition? And most importantly, why can arrow vectors like these be slid paralllelly anywhere in the space and they would still mean the same?
I don't know how any of these properties follow from the definitions of vector spaces. If anyone could give their opinion on this, I would be glad.
 A: I will try to answer this in terms of differential geometry. Other viewpoints might be better suited, as flat space/spacetime is also an affine space, but I cannot answer it from this point of view.
In the language of differential geometry, at each point of the manifold, there is tangent vector space in which vectors reside. These tells you, in which direction and how quickly you can move. As EM theory is theory of how charges move, this is the appropriate vector space to develop the theory in. The acceleration tells you velocity, force tells you acceleration and EM field tells you force.
In flat spacetime or flat space, there is also prescription how to parallel transport these vectors from one point to another, which does not depend on the curve by which you transport the vector.
Physically this means, that if you have an arrow attached to gyroscope which points in direction of some possible movement (and thus represent vector from tangent vector space) and you travel from A to B by whichever path, the resulting vector in B will be the same every time. This is not the case in general relativity, where spacetime is curved.
Anyway, in flat space/spacetime this parallel transport defines canonical isomorphism between all the tangent spaces and they can all be identified. So if you have some basis vectors and a vector in point A, the canonical isomorphism tells you, that coordinates of the vector in point A w.r.t to the basis are the same as identified vector in B w.r.t. to the identified basis. Thus for all intents and purposes you can just slid your vectors anyway you like.
The physics way of defining vectors in flat space/spacetime is through their components w.r.t to some orthonormal bases and how they transform between different orthonormal basis. But they will not tell you this, because the orthonormal basis can be identified with cartesian coordinate system and physicists "think" in terms of this coordinate system and not in terms of abstract vectors, parallel transports, canonical isomorphisms as these are usually just unnecesary complications.

One of his exercises is to show that translational transformations are vector-component invariant and that's where I began to seriously doubt things

I did not read the book, but I think this basically asks you to show wheter two cartesian coordinate systems, which are just translated w.r.t to each other are identified with the same basis vectors.
A: Vectors in mechanics are split into two categories based on their behavior. There are free vectors (what the book calls "arrow" vectors) and position-specific vectors.
Free vectors have the property that you don't need to specify a location to use them. An example is a force vector, where regardless of where it is applied, it will result with the same acceleration of the center of mass. The following is true regardless of where $\boldsymbol{F}$ is applied.
$$\boldsymbol{F} = m\, \boldsymbol{a}_{\rm COM}$$
Another way of saying this, is that if you measure a force vector from a different coordinate system that is axis aligned, but displaced by some amount, the components of the vector will remain unchanged.
By contrast, torque is position-specific as where it is measured is important. We always specify torque about the center of mass, or torque about the origin in order to use it
$$ \boldsymbol{\tau}_{\rm COM} = \mathbf{I}_{\rm COM} \boldsymbol{\alpha} + \ldots $$
and if you wanted to calculate torque about the origin for example (or any other point), then you need a transformation law
$$ \boldsymbol{\tau}_{\rm ORIGIN} = \boldsymbol{\tau}_{\rm COM} + \boldsymbol{r}_{\rm COM} \times \boldsymbol{F} $$
Another example is rotational velocity, which is shared by all parts of rotating frame, as well as rotational acceleration. Note that since a change of coordinate system origin does not affect the vector components, there is no need to specify the point where these quantities are measured.
Since
$$\begin{aligned} 
 \boldsymbol{\omega}_{\rm ORIGIN} & = \boldsymbol{\omega}_{\rm COM} \\
 \boldsymbol{\alpha}_{\rm ORIGIN} & = \boldsymbol{\alpha}_{\rm COM} \\
\end{aligned}$$
then when used in equations like in $\boldsymbol{\tau}_{\rm COM}$ above, there is no need to subscript the location for rotational motion.
Also note that linear velocity is position-specific, as to where it is measured is important. You will need similar transformation laws for velocities as you did for torques
$$ \boldsymbol{v}_{\rm ORIGIN} = \boldsymbol{v}_{\rm COM} + \boldsymbol{r}_{\rm COM} \times \boldsymbol{\omega} $$
Lastly, momentum is a free vector, and angular momentum is location-specific. If you examine their definitions, you will notice on the left-hand side which ones are location-specific and which ones are free.
$$ \begin{aligned}
\boldsymbol{p} & = m\, \boldsymbol{v}_{\rm COM} \\
\boldsymbol{L}_{\rm COM} & = \mathbf{I}_{\rm COM} \boldsymbol{\omega}
\end{aligned} $$
Also notice and interesting property where mass and mass moment of inertia seems to convert from one to the other. The convert location-specific velocity to free momentum, and free rotation velocity to location-specific angular momentum.
A: As used in electromagnetism to represent positions, velocities, accelerations, fields, and forces, "vectors" are an example of "a vector space" as defined in abstract algebra, but they are not the most general possible example. For example, electromagnetism is a theory with no more than three spatial dimensions.  You could construct a vector space with more than three dimensions --- there are perfectly reasonable infinite-dimensional vector spaces. But a theory in such a space wouldn't be electromagnetism.
We want to be able to use vectors to encode exactly two properties: a direction, and a magnitude.  But if you accidentally build a vector that "can't be slid," it has at least three properties: a direction, a magnitude, and a starting location.  We want to use vectors to represent ideas like "to the east."  A vector that can't be slid would represent, when I wrote it, "to the east from my house"; that is not useful to you because you do not know where I live.
