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In mathematics the concept of a vector can be made quite general and abstract with the idea of vector space. We define what is a vector space $V$ and we say that a vector is an element of a vector space $V$.

This concept is algebraic, and the important thing is the behavior of the operations performed on elements of $V$ inasmuch as it happens with groups, rings and fields. In that case, a vector space might include concepts quite far from the geometric idea of an arrow, including spaces of functions and much more.

On the other hand there is a quite particular case of this concept that is indeed related to geometry. That would be the tangent spaces to a smooth manifold $M$.

Given a smooth manifold $M$, for each $x\in M$, the tangent space $T_x M$ is a vector space comprising all directions tangent to $M$ at $x$. This implies that $T_x M$ is actually defined precisely so that we can picture its elements as arrows at $x$.

Physicists usually define vectors in a different way, with transformation laws. Quoting Arfken:

The set of $N$ quantities $V_j$ is said to be the components of an $N$-dimensional vector $\mathbf{V}$ if and only if their values relative to the rotated coordinate axes are given by

$$V_i'= \sum_{i=1}^N a_{ij}V_j, \quad i=1,2,\dots,N$$

As before, $a_{ij}$ is the cosine of the angle between $x_i'$ and $x_j$.

Now my question here is: when this standard definition is made, nothing is mentioned about what are the assumptions made over the space of all vectors.

Certainly it should be a vector space, however I ask: when physicists perform this standard definition of a vector, what is being defined is an element of a general vector space, that could be simply anything, or one has in mind exactly the tangent spaces to a smooth manifold?

Is this definition, defining a vector in a general vector space, or it is defining a vector that belongs to the tangent space to a manifold?

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There are in fact several additional interpretations of tangent vectors. An equivalent definition of a tangent vector involves equivalence classes of curves.

The appropriate interpretation largely depends on the context of a problem. As QMechanic stated, we may take a linear algebra viewpoint, or view them in terms of differential geometry.

In the latter case, one has the bundle $TM \to^\pi M$ and sections are maps $s : M \to TM$ such that for the projection, $\pi \circ s = \mathrm{id}_M$. Thus a tangent vector is a section of $TM$. It can be viewed as,

$$TM = \bigcup_{p \in M}T_pM$$

that is, the union of all tangent spaces on the manifold. A tangent vector at a point $p$ lies in $T_pM$ which is a fibre of the bundle $TM \to^\pi M$.

This interpretation is useful in general relativity, wherein physics is cast in the language of differential geometry. On the other hand, if say, I were solving a problem in classical mechanics involving firing a cannonball, I would think of its velocity vector as simply some $v\in \mathbb R^3$.

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  • $\begingroup$ Thanks @JamalS. I do know these definitions. Also here I'm not considering fields, so when I talk about smooth manifolds, I'm thinking about a vector at a point (an element of $TM$ rather than a section). What I want to know is: when the definition via transformation law that I stated is performed, Physicists are assuming that there is a background manifold? In other words, they are assuming that the vector being defined is an element of a tangent bundle of a manifold? Or they are not assuming anything like that, and thinking of a general vector space? Thanks again for the aid! $\endgroup$ – user1620696 Feb 25 '17 at 14:54
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OP is essentially asking if tensors$^1$ in physics should be understood

  1. within the category of vector spaces and multilinear maps, i.e. linear algebra;

  2. or within the category of vector bundles and bundle maps, i.e. differential geometry?

The answer is: Both, depending on context. In the latter case, tensors are more properly called tensor fields.

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$^1$ A vector is a (1,0) tensor.

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