Conceptual difference between Covariant and Contravariant tensors I am having some confusion over the concept of covariant and contravariant vectors. Most text books on tensors define contravariant vectors/tensors as objects whose components vary inversely to the changes in its basis vectors. Most common example given is that of velocity.
Covariant vectors (or one-forms), on the other hand, are defined as entities that transform in the same way as the changes in its basis vectors, and a common example is gradient of some scalar filed, like a temperature gradient.
But my understanding is that a vector is a vector. Intrinsically, it is neither covariant nor contravariant. It is the vector components that are either co or contra variants. If the vector is written in covariant bases, than its components are contravariant. If the same vector is written in contravariant bases, then its component becomes covariant.
As we can see, a vector can be written as
$$ \vec V \equiv v^ie_i \equiv v_ie^i $$
So, the same entity is both covariant and contravariant, depending on the basis vectors.
Intuitively also it makes sense. The velocity of an object can be expressed as n meters per second, (meaning the object travels n meters in a second), or as 1/n seconds per meter (meaning the object takes 1/n seconds to travel a meter). In the first way it will be a contravariant vector, and in the second way it will be covariant.
So, is my understanding correct, or are physical entities intrinsically covariant or contravariant?
 A: A vector in contexts where co- and contravariant are used is not just any vector in $\mathbb R^n$. No, in this context, we're working on a manifold $M$, that is, roughly speaking, a smooth (hyper)surface. At each point $p$ of the surface we can construct a tangent (hyper)plane. This tangent plane is spanned by a set of tangent vectors, and these vectors together form the tangent space $T_pM$ to $M$ at $p$.
A vector is an element of such a tangent space. A covector, on the other hand, is an element of its dual space. Meaning that a covector is a linear map which sends vectors to real numbers. That's an entirely different object from a vector! Vectors aren't maps! At least not in this context.
Of course, using some linear algebra, we can construct covectors from vectors and vice versa. Given a vector $v$ and a scalar product $\langle\cdot,\cdot\rangle$, define the linear map $v^\ast: w\mapsto\langle v,w\rangle$. $v^\ast$ is a covector which in some sense corresponds to $v$, if we distinguish this chosen scalar product. In a similar way we can construct vectors from covectors. If you've come that far in your studies of physics: this scalar product is the metric $\eta$, which you use to lower or raise indices. This lowering and raising is just shorthand for the construction described above. Crucially, lowering and raising indices is not a change of basis, so it's not just a change of the vector components. It's the construction of a completely different object. If $v$ is a velocity, then $v^\ast$ is not! What's more, these two objects behave differently if we do actually change the coordinates. The vector varies contravariantly, meaning that if we change the coordinate system via applying the linear map $\Lambda$ to the basis vectors, then we need to apply the map $\Lambda^{-1}$ to the coordinate vector of $v$ in order to obtain the correct coordinates. A covector, however, requires that we apply $\Lambda$ to its coordinate vector in order to obtain the correct coordinates. That's what makes the difference between co- and contravariant vectors. They're different objects, which consequently behave differently under coordinate transformations. They are not just different representations of the same object. Physical quantities are intrinsically co- or contravariant, even though covariant quantities can be constructed from contravariant ones and vice versa. Velocities are always covariant. The object with lowered indices is not a velocity anymore.
A: You are right, an element $v$ of a vector space $V$ is a vector and nothing else. It does not make sense to call it covariant or contravariant if we only look at $V$.
However, the origin of these concepts is in physics. Physical entities $X$ like velocity, electric field, etc., are quantified by measurements which produce an $n$-tuple $(x_1,\ldots,x_n)$ of real numbers. For example, quantifying a velocity produces $3$-tuples. Each measurement is based on a measurement space in which we observe the entity and on a coordinate system on the measurement space (a $3$-dimensional spatial coordinate system or a $4$-dimensional time-spatial coordinate system or something more complicated) with respect to which we measure the components $x_i$ of the entity $X$. Usually one makes one measurement per coordinate direction.
That is, we have a measurement space $\mathbb R^n$ and describe $X$ by a measurement result which is an $n$-tuple $(x_1,\ldots,x_n) \in \mathbb R^n$. But we must be aware that the entity $X$ is not an element of the measurement space $\mathbb R^n$; it is an element of a state space $V$ which has the same dimension as the measurement space. Anyway, the state space $V$ is isomorphic to the measurement space $\mathbb R^n$, an isomorphism $\phi : V \to \mathbb R^n$ being determined by our measurements.
Let us emphasize again that $X$ is not the same as $\phi(X)$; the latter is just a measurement result. Obviously we may get different values $(x_1,\ldots,x_n)$ of our measurements if we use different coordinate systems, and this is the subtlety: The isomorphism $\phi$ depends on the chosen coordinate system $\mathfrak C$. Thus we obtain a whole family of isomorphisms $\phi_{\mathfrak C} : V \to \mathbb R^n$.
In the simplest case we consider coordinate systems having the same origin; we do not take into account translations or moving coordinate systems (as in special relativity). In other words, we have the measurement space $\mathbb R^n$ (e.g. three-dimensional space or four-dimensional space-time) in which our measurements are performed, and for each ordered base $\mathfrak B$ of $\mathbb R^n$ we get an associated isomorphism $\phi_{\mathfrak B} : V \to \mathbb R^n$.
Of course there must be a definite relationship between the quantifying tuples associated to different coordinate systems (resp. ordered bases), but it is a priori not clear what this relationship looks like and what its mathematical description is.
Although the elements of the state space $V$ are vectors and nothing else if we only look at $V$, they have additionally a certain relationship with the measurement space $\mathbb R^n$, described by the isomorphisms $\phi_{\mathfrak B}$. We must not forget this relationship.
What happens if we make a change of basis of the measurement space $\mathbb R^n$? Given ordered bases $\mathfrak B$ and $\mathfrak B'$, we get an automorphism $\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1}$ on $\mathbb R^n$. As already mentioned, it is not a priori clear what this automorphism looks like, and the answer depends on the given physical entity. The most elementary two cases are that $\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1}$ behave covariant or contravariant. This means that if $A$ is the matrix such that $b'_i = A b_i$, then $(\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1})(x) = A x$ for all $x$ (covariant) or $(\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1})(x) = (A^{-1})^T x$ for all $x$ (contravariant).
So you see that covariance and contravariance are not intrinsic properties of the vectors of the state space, but properties which become visible only if we consider the relationship between measurement space and state space. I think it was a great achievement to recognize that measurement space and state space need not be identical although they have the same dimension.
In the covariant case the state space $V$ can be identified naturally with the dual of the measurement space, in the the contravariant case it can be identified naturally with the measurement space itself (see Why is tensor from a vector space covariant, not contravariant?). But this is a purely mathematical point of view, not all physicists will perhaps be interested  in these identifications.
