0
$\begingroup$

Definition 1 If under a coordinate transformation $x^i\to \bar{x}^i(x^i)$ certain objects $A^i$ transform as $$A^i\to \bar{A}^{i}=\sum_{j}\frac{\partial \bar{x}^{i}}{\partial x^j}A^j,$$ those objects are called contravariant (components of a_ vector.

Definition 2 If under a coordinate transformation $x^i\to \bar{x}^i(x^i)$ certain objects $A_i$ transform as $$A_i\to \bar{A}_{i}=\sum_{j}\frac{\partial x^{j}}{\partial \bar{x}^i}A_j,$$ those objects are called covariant (components of a) vector.

Definition 3 The components of covariant and contravariant vectors are related by the metric tensor $g_{ij}$ as $$A_i=g_{ij}A^j.$$

I am confused about definitions 2 and 3 of covariant vectors. Def. 1 and 2 suggest that contravariant and covariant objects are very different things and probably not related to each other whereas definition 3 tells that they are related. How are definitions 2 and 3 related? Which one is more general? What is the connection?

$\endgroup$
7
  • 1
    $\begingroup$ “Definition 3” doesn’t seem to be a definition. (Do you have a book in which it is?) It states a relationship. $\endgroup$ – G. Smith May 25 at 5:23
  • $\begingroup$ @G.Smith This is what we use in the special relativity context. $A_\mu=\eta_{\mu\nu}A^\nu$. $\endgroup$ – mithusengupta123 May 25 at 5:25
  • $\begingroup$ Do you know how the metric transforms under a coordinate transformation? $\endgroup$ – Qmechanic May 25 at 6:00
  • $\begingroup$ I suspect the third one is a basis expansion $A=g(A,e_j)e^j$, but am not sure. $\endgroup$ – Emil May 25 at 6:15
  • 2
    $\begingroup$ @mithusengupta123 To that end you might find it helpful to note that $$\frac{\partial x^\mu}{\partial x'^\nu}\cdot \frac{\partial x'^\nu}{\partial x^\rho}=\delta^\mu_\rho$$ $\endgroup$ – Charlie May 25 at 13:51
4
$\begingroup$

The term "covariant vector" is a hideous misnomer. Many texts do not give a proper treatment of this (which I think is important to know). What we are actually dealing with are objects known as covectors, which are not quite the same as vectors. I will give a quick explanation below.

Consider a vector space $V$ over a field $F$. The field arises because $V$ has the operation of scalar multiplication such that for any $v \in V$ and $k \in F$, we have $kv \in V$. In other words, $F$ is just the set of scalars used in scalar multiplication. In practice, $F$ is usually $\mathbb{R}$ or $\mathbb{C}$.

If we define a basis $\mathbf{e}_i$ on $V$, then for every $v \in V$ there exists unique coefficients $v^i$ such that $v = v^i \mathbf{e}_i$. Now we introduce the dual space $V^*$ which is the set of all linear maps from $V$ to $F$. If $\omega \in V^*$, then we have $\omega(v) \in F$. Elements of $V$ are also linear maps from $V^*$ to $F$, so we also have $v(\omega) \in F$.

The elements of $V^*$ are known as covectors.

Now we define the dual basis $\mathbf{e}^i $, which is also often (incorrectly) referred to as the contravariant basis by $$\mathbf{e}^i (\mathbf{e}_j) = \delta^i_j$$

Therefore we can also decompose a covector $\omega$ into its components as $\omega = \omega_i \mathbf{e}^i$. With that, we have $$\omega(v) = \omega_i v^i$$

From here, how do we derive the transformation properties of the objects? Suppose we define an alternative basis $\bar{\mathbf{e}}_i = A^j_i \mathbf{e}_j$ where $A$ is any invertible matrix. Scalars, vectors and covectors exist independently of any basis and must therefore remain invariant under any change of basis. So we see that $v^i$ and $\mathbf{e}^i$ transform with $A^{-1}$ while $\omega_i$ transforms with $A$.

Now comes the key part. The metric tensor $g$ is defined as the symmetric tensor such that for any two vectors $u$ and $v$, $g(u,v)$ is their dot product (or inner product) and is a scalar. We can use it to "convert" any vector to a corresponding covector. If we have a vector $v$, its corresponding covector is $$g(v) = g_{ij} \mathbf{e}^i \otimes \mathbf{e}^j (v^k \mathbf{e}_k) = g_{ij} v^k \mathbf{e}^i \delta^j_k = (g_{ij} v^j) \mathbf{e}^i$$

Therefore we see that the components of the corresponding covector is $g_{ij} v^j$. This is what is really meant by raising and lowering indices. What we are really doing is finding the components of the corresponding covector in the dual space. We can apply the same procedure to arbitrary tensors by contracting one slot of $g$ with one slot of the tensor.

The important observation here is that once a metric is defined, we can "pretend" that vectors and covectors are the same. If I originally have tensor components $T^{ijk}$, we can automatically use $T^{\; jk}_i$, $T_{ijk}$, $T^{i \; k}_{\; j}$ and so on in calculations without ambiguity. But the tensors themselves live in different spaces. After all, $T^{ijk}$ is not a tensor, $T^{ijk} \mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{e}_k$ is.

A few extra things that might be useful to note:

  1. It is not necessary to have a coordinate system in order to introduce a basis. A coordinate system $x^i$ is part of a structure of a manifold $M$. For any point $p \in M$, its tangent space, $T_pM$, is a vector space. This vector space is the set of derivatives at $p$ along all smooth curves passing through $p$. This is also why tangent vectors are partial derivatives. The basis are then the derivatives along the coordinate curves themselves, $\mathbf{e}_i = \partial /\partial x^i$.
  2. Only if $\mathbf{e}_i = \partial /\partial x^i$ then the matrix $A$ above becomes the inverse Jacobian matrix $\partial x /\partial \bar{x}$, and $A^{-1}$ becomes the Jacobian $\partial \bar{x} /\partial x$.
  3. The terms "contravariant" and "covariant" are outdated and come from the fact that $v^i$ transform with $A^{-1}$ ("contra") while $\omega_i$ transform with $A$ ("co"). It is bad practice to continue using them in the modern context where tensors are handled using abstract index notation (or other coordinate-free methods).
  4. In Euclidean space with Cartesian coordinates, there is no difference between vectors and covectors because the components of the metric tensor is just the identity matrix. Only in this special case, there is no need to distinguish between vectors and covectors and index placement does not matter.
$\endgroup$
2
$\begingroup$

To give a more wordy answer to complement Vincent Thacker's nice, more detailed answer:

The starting point is that we have a manifold, $M$. The tangent space at the point $p$ in $M$ is denoted $T_pM$ and the cotangent space (the dual space of the tangent space) at the same point is denoted $T^*_pM$. Defining a (co)vector field on this manifold involves picking a (co)vector in each (co)tangent space (the precise term is a section). On our manifold we can define coordinates in the usual way with charts, which effectively labels each point in the manifold with a unique set of numbers and allows us to do physical calculations more conveniently, rather than working entirely in the abstract.

Hopefully the above is understood to some extent.

Once we have chosen coordinates on our manifold, there is a natural sense in which we can use those coordinates to define a basis for our tangent and cotangent space (through partial and exterior derivatives of the coordinate functions respectively). Note that when we change coordinate on our manifold the basis on our tangent and cotangent spaces will change too. So it is in this sense that a coordinate change will change the components of the vectors in a vector field on $M$, since the change in coordinate induces a change in the basis of the tangent spaces (the same is true for cotangent vectors).

In physics we often take this to define the vectors and covectors by "how they transform", and the two definitions are as you have given in your question.

The third definition you've given just highlights the fact that in physics we often have a metric on our manifold (for instance in general relativity), and this metric defines a linear isomorphism between the tangent space at each point and the cotangent space at the same point. This is the precise sense in which the metric "raises and lowers" indices on vectors and covectors, each vector $\vec v\in T_pM$ is being assigned by this isomorphism a unique$^{[1]}$ covector $\tilde\omega\in T^*_pM$.

So all of your definitions are valid to some extent, your confusion just derives from the (frankly, horrible) way that differential geometry is necessarily introduced in physics so as to not labour too much on the mathematical details.

Hopefully this is helpful.


$[1]$ - Note that an isormorphism is by definition a bijection (which is in particular a 1-to-1 map) which preserves some mathematical structure (in this case the vector space operations on $T_pM$ and $T^*_pM$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.