Covariant vectors and Dual spaces

There are contravariant vectors and their duals called covariant vectors. But the duals are defined only once the contravectors are set up because they are the maps from the space of contravariant vectors to R and thus, it won't make sense of to talk of covariant vectors alone and according to this http://en.wikipedia.org/wiki/Covariant_transformation, we first have a basis for contravariant vectors, and then a corresponding basis for the covariant vector, and the fact that the covariant and contravariant vectors transform differently is based on the fact that covariant vector space is the dual space for contravariant vector space.

What if it was the other way around in our convention ? So, now their way of transforming will get changed. Then what sense does it make to distinguish vectors like these? A vector is simply an element of the vector space. How it transforms depends on the basis you chose for that space, and not on the nature of the vector.

Then what does it mean that the gradient is a covariantvector ? Now saying because it transforms in a certain way makes no sense.

Also see here arxiv.org/abs/1002.3217

• I don't think that the transformation laws follow from the abstract ideas of what tangent and cotangent vectors are, but more from the geometric interpretation of tangent vectors as those tangent to some curve and cotangent vectors as those orthogonal to some surface. Then, the transformation laws follow from the chain rule for a general remapping of positions. – Muphrid Mar 28 '14 at 14:42

Okay, there are two facts to be taken into account here:

Vectors are elements of a vector space. (Let's say a real, d-dimensional vector space $V$ for concreteness). If you use a basis $\lbrace e_i \rbrace \subseteq V$ you can express those vectors as a linear combination of elements. ie: for any $v \in V$: $$v = v^{i} e_{i}$$ where $v^{i}$ are real numbers, called the components of $v$ with respect to $\lbrace e_{i} \rbrace$.

Covectors are elements of the dual space $V^{*}$ of $V$. You can also choose a basis $\lbrace \epsilon^{i} \rbrace \subseteq V^{*}$ to express these objects as linear combinations in a similar fashion. ie: for any $\omega \in V^{*}$: $$\omega = \omega_{i}\epsilon^{i}$$ where $\omega_{i}$ are also real numbers, called the components of $\omega$ with respect to $\lbrace \epsilon^{i} \rbrace$.

The first choice.

In principle, the bases $\lbrace e_{i} \rbrace$ and $\lbrace \epsilon^{i} \rbrace$ are not related in any way. However, in order to simplify calculations, we often choose a very special basis for the dual space: The dual basis is the unique basis in the dual space such that: $$\epsilon^{i}(e_{j}) = \delta^{i}_{j}$$

(Note the two different workings of the word "dual": while dual space it means the space $V^{*} := Hom(V,\mathbb{R})$, dual basis refers to the uniquely defined basis on $V^{*}$ such that $\epsilon^{i}(e_{j}) = \delta^{i}_{j}$).

Now suppose we want to make a change of basis (AKA linear transformation) in $V$. Let's say our vector $v = v^{i}e_{i}$ can be written in terms of the new basis $\lbrace a_{i} \rbrace$ as: $$v = v^{i} e_{i} = w^{j}a_{j}$$ While the basis transforms with a certain matrix: $e_{i} = \Lambda^{j}_{i} a_{j}$, the components with respect to that basis transform with the inverse of that matrix: $v^{i} = w^{j} (\Lambda^{-1})^{i}_{j}$

Now, when we pair up the elements of the basis $\lbrace e_{i}\rbrace$ and the elements of the basis $\lbrace \epsilon^{i}\rbrace$ we would like our dual basis convention to remain true, so any change in basis $\Lambda$ on $V$ will induce a change of basis on $V^{*}$. $$\delta^{i}_{j} = \epsilon^{i}(e_{j}) = \epsilon^{i}(\Lambda^{k}_{j}a_{k}) = \Lambda^{k}_{j}\epsilon^{i}(a_{k}) = ...$$

The matrix that relates $\lbrace \epsilon^{i} \rbrace$ with the new basis (let's call it $\lbrace \alpha^{i} \rbrace$) on the dual space needs to be the inverse $\Lambda^{-1}$ of $\Lambda$ in order to satisfy the relation $\alpha^{i}(a_{j}) = \delta^{i}_{j}$. $$...=\Lambda^{k}_{j}\epsilon^{i}(a_{k}) = \Lambda^{k}_{j}(\Lambda^{-1})^{i}_{l}\alpha^{l}(a_{k}) = \Lambda^{k}_{j}(\Lambda^{-1})^{i}_{l} \delta^{l}_{k}$$ $$=\Lambda^{k}_{j}(\Lambda^{-1})^{i}_{k} = \delta^{i}_{j}$$

The second choice:

We use the word "covariant" to describe the way the basis of $V$ transforms.

From this, we start calling "contravariant" the way the basis of $V^{*}$ transforms, because it needs to use the inverse transformation in order to keep the duality convention.

We call the components of the elements of $V$ "contravariant" because, as we saw before, they need to transform inversely to the basis of $V$ in order to keep invariance.

Finally, we call "covariant" the components of the elements of $V^{*}$, because they need to transform with inversely the basis of $V^{*}$ and since their basis transforms contravariantly, they end up transforming with the same matrix as the basis of $V$.

So, in summary:

What if it was the other way around in our convention ? So, now their way of transforming will get changed.

No. Only the way we call the transformation behaviours would change. What matters is:

1. The components of the elements of a vector space need to transform in the opposite way as the basis of that vector space does.
2. The basis of $V$ transforms in the opposite way as the basis if $V^{*}$ does, in order to keep the duality convention.

Then what sense does it make to distinguish vectors like these? A vector is simply an element of the vector space. How it transforms depends on the basis you chose for that space, and not on the nature of the vector.

No. The components of a vector will always transform with the opposite transformation to the one that transformed the basis, regardless of what specific basis is that.

Then what does it mean that the gradient is a covariantvector ? Now saying because it transforms in a certain way makes no sense.

The gradient of a function has covariant components because it naturally is a map $TM \rightarrow \mathbb{R}$. It takes a vector and gives you the directional derivative of the function in that direction. So it is an element of $T^{*}M$ (the dual space to $TM$), whose basis has a transformation behaviour (contravariant) opposite to that of the basis of $TM$ (covariant).