Conceptual difference between Covariant and Contravariant tensors

Question

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 their basis vectors, and a common example is gradient of some scalar field, 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?

Answer 1

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., live in some affine space $A$, most frequently $3$-dimensional space or in $4$-dimensional space-time. Let us call $A$ the observation space. An entity $X$ is quantified by measurements which produce $n$-tuples $(x_1,\ldots,x_n)$ of real numbers. For example, quantifying mass produces $1$-tuples, quantifying velocity produces $3$-tuples. To perform a measurement we need a coordinate system on the observation space (a $3$-dimensional spatial coordinate system or a $4$-dimensional time-spatial coordinate system or something more complicated). For example, measuring velocity gives three components, one per each coordinate direction.

That is, we have an observation space $A$ 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 $\mathbb R^n$; it is an element of an "abstract state space" $S$ about nothing is known a priori. It may be a vector space or something more complicated. The dimension of $S$ depends on $X$, but the dimension of $A$ does not determine the dimension of $S$.

The measurement process can be regarded as a function assigning to $X \in S$ and a coordinate system $\mathfrak C$ of $V$ a value $\phi_{\mathfrak C}(X) \in \mathbb R^n$. The function $\phi_{\mathfrak C} : S \to \mathbb R^n$ is assumed to be injective.

Let us emphasize again that $X$ is not the same as $\phi_{\mathfrak C}(X)$; 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.

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 may regard the observation space $A$ as a vector space (e.g. three-dimensional space or four-dimensional space-time) in which our measurements are performed, and for each ordered basis $\mathfrak B$ of $A$ we get an associated injection $\phi_{\mathfrak B} : S \to \mathbb R^n$.

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.

A very nice situation occurs when $n = \dim A$ (i.e. the dimension of the measurement result space $\mathbb R^n$ agrees with the dimension of $A$) and all $\phi_{\mathfrak B}$ are bijections. This means that the state space has the same dimension as $A$ and that the elements $X \in S$ are "vectorial entities" in $A$.

What happens if we make a change of basis in the observation space $A$? Given ordered bases $\mathfrak B =(b_1,\ldots,b_n)$ and $\mathfrak B' = (b'_1,\ldots, b'_n)$, 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 $A = (a_{ij})$ is the matrix such that $b'_j = \sum_{i=1}^n a_{ij}b_i$, then $(\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1})(x) = A^t x$ for all $x$ (covariant) or $(\phi_{\mathfrak B'} \circ \phi_{\mathfrak B}^{-1})(x) = A^{-1} x$ for all $x$ (contravariant). In both cases there exists a unique vector space structure on $S$ such that all $\phi_{\mathfrak B}$ become isomorphisms of vector spaces. Physicists call the elements of $S$ covariant resp. contravariant vectors.

Covariance and contravariance are not intrinsic properties of the vectors of the state space $S$. The elements of $S$ are vectors and nothing else if we only look at $S$. Covariance and contravariance become visible only under the measurement process which gives isomorphisms between measurement result space and state space. I think it was a great achievement to recognize that measurement result space and state space of vectorial entities need not be identical although they have the same dimension.

In the covariant case the state space $S$ can be identified naturally with the dual of the observation space $A$, in the the contravariant case it can be identified naturally with $A$ 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.

In physics there are many important non-vectorial entities. Examples are tensors.

Answer 2

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.