# Relation between component and algebraic definition of covariant vectors

I studied contravariance and covariance concepts in following way: For any vector if we get its components by parallelogram way we achieve contravariant components, and if we want to get its components by orthogonal projection we achieve covariant components. For latter case we need reciprocal basis in order we can expand the vector by its covariant components. But I read somewhere, in fact covariant vectors are elements of dual space and they expanded by dual basis. My question is: how is the latter approach to covectors equivalent to previous approach? Furthermore, I need some good explanation for latter approach to covariant vectors.

"Covariant vectors as expressed in a dual basis" is exactly the same thing that "orthogonal projections to achieve covariant components"

Choose one generating system $\vec e_1, \vec e_2$, non necessary orthogonal, a vector $\vec v$ may be expressed by $\vec v = v^1 \vec e_1 + v^2 \vec e_2$. The coordinates $v^1, v^2$ are called contravariant coordinates of the vector $\vec v$

Now, let a new generating system $\vec f^j$ be a dual of the generating system $\vec e_i$. This is expressed by $\vec f^j.\vec e_i = \delta^j_i$ (this is the definition of "dual")

This means that $\vec f^2.\vec e_1 = \vec f^1.\vec e_2 = 0$

So, this means that $f^2$ is orthogonal to $\vec e_1$, and that $\vec f^1$ is orthogonal to $\vec e_2$

The coordinates $v_1, v_2$ of the vector $\vec v$, relatively to the generating system $\vec f^j$ are the covariant coordinates of $\vec v$ : $\vec v = v_1 \vec f^1 + v_2 \vec f^2$

Now, to find $v_1$ for instance, you have to take, from the arrow of the vector, the parrallel to $\vec f^2$, but we have seen above that $\vec f^2$ is orthogonal to $\vec e_1$, so " take the parrallel to $\vec f^2$" is the same thing as "take an orthogonal to $\vec e_1$"

That's how you get your orthogonal projections to achieve covariants components.

UPDATE

Considering covariant dual vectors as linear functions from vector space to $\mathbb R$, $f^i$ are promoted to functions : $f^i( \vec e_j) = \delta ^i_j$

So, you have :

$$w(\vec v) = ( w_1 f^1 + w_2 f^2)(v^1 \vec e_1 + v^2 \vec e_2) = (w_1.v^1 + w_2 v^2)$$

UPDATE 2

Now, in a finite dimensional space, there is an isomorphism between the vectors space and the dual space of the vector space . Let $\vec w$ be an element of the vector space and $w$ be an element of the dual space of the vector space. This isomorphism is :

$$w(\vec v) = \vec w.\vec v$$

Applying this isomorphism to the generating system, gives an isomorphism between $\vec f^i$ ("reciprocal basis") and $f^i$ (basis of the dual space):

$$f^i(\vec e_j) = \vec f^i.\vec e_j= \delta^i_j$$

• Thank you for your attention. You explain the first approach as I mentioned in my question. But the latter approach as I mentioned tells us covariant vectors are elements of dual vector space that is linear functions from vector space to scalar field (called functional). In fact my qestion is how two approaches are equivalent. – user27058 Jul 18 '14 at 11:15
• I update my answer – Trimok Jul 19 '14 at 9:11
• Thank you for updating. But one question remains: To obtain covariant component of a vector we shouldconstruct base which are orthogonal to original bases. We can do that by constructing reciprocal bases in our vector space and I think it is non necessary to extract such bases from the dual of the vector space, and decompose the vector on those bases. Can you help me, more ? – user27058 Jul 19 '14 at 11:32
• "Reciprocal basis" and "Basis from the dual of the vector" space are the "same thing". More precisely, there is an isomorphism. See my UPDATE 2 in the answer. – Trimok Jul 21 '14 at 8:02
• Stricly speaking, the isomorphism is valid only for finite dimensional (vector) spaces. The dual vector space is a more general formalism, and it is ofen used in differential geometry, too. So try to use it and become familiar with it, because it will be useful to you in the future. – Trimok Jul 23 '14 at 8:16

In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space.

It's a typical school phrase... "Do not add apples and pears", and it's true!

If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the usual axioms of a vector. However, you can not add a column and a row vector.

With this you now understand that all the time you've been working with two types of vectors.

The thing is that, they are not related with each other until you impose a relation between them, e.g., you identify their bases through the introduction of a scalar product $$\left<e^i\mid e_j\right>= \delta^i_j.$$ Since this identification can be seem as a map from vectors to scalars, one says that covectors are objects living in the dual space of vectors.

It seems to me that you want to know the "difference" between (and relation among) the coordinates draw in blue and green in the picture below The green coordinates are the contravariant components of the vector, defined by the "parallel" projection wrt the other axis. The blue coordinates are the covariant components of the vector, defined by "orthogonal" projection along each axis.

NOTE: in a rectangular frame these two coincide, and there is no ambiguity.

## Why are the blue one called "covariant"?

The usual answer is... because they transform as the basis! (Yes, sure... but I don't still get it!)

A base vector defines a direction, and a flow in that direction is defined via the (directed) gradient. The gradient is represented by the set of perpendicular (hyper)planes orthogonal to the reference direction. Thus, perpendicular to that direction... sounds familiar, THE BLUE COORDINATES!!!

Since the derivative can be written as $\partial_i$ let's denote this type of components by $V_i$

## What about the green ones?

I could try to explain heuristically why these are denoted with an upper index, but in this stage let's mathematics do its job.

I'd do the following (I did it once and convince myself... so try to do it!):

• By defining a vector basis in the non-rectangular frame (in term of the rectangular one), say $\{\vec{e}_1,\vec{e}_2\}$ in terms of $\hat{i},\hat{j}$ and $\theta$ the angle between the non-rectangular axis, Find the metric of the non-rectangular frame. $$g_{ij} = \vec{e}_i\cdot\vec{e}_j.$$
• Then invert the metric,
• Raise the index of the covariant vector using the inverse metric.

You will get exactly the component represented by the green projection.

## Why are they called dual?

Because you can always define the action of one type over the other... resulting into a field ($\mathbb{K}$). Mathematically, these define the elements of the dual space.

In other words, if you say that $\{V^i\}\in \mathbf{V}$, then $\{V_j\}\in \mathbf{V}^*$

It is not my idea, but you can read about that in (for example) the books by B. Schutz ("A first course in Genral Relativity" or "Geometrical Methods of mathematical physics", if I remember well).

A geometrical interpretation is that contravariant vectors are arrows, while covariant vectors are perpendicular planes to a given direction, the action of one over the other is the number of hyper-planes you intersect with your arrow. That is the connection.

In other words, you have related two unrelated vector spaces (say $\mathbf{V}$ and $\mathbf{U}$), by introducing a scalar product ($\cdot:\mathbf{V}\times\mathbf{U}\to \mathbb{K}$). This is equivalent to defining an action of $\mathbf{U}$ over $\mathbf{V}$ by $$\mathbf{U}:\mathbf{V}\to\mathbb{K}.$$

With the last map, you conclude that the introduction of a scalar product is equivalent to the identification of $\mathbf{U}$ with $\mathbf{V}^*$

• Thank you for your attention. I agree with you, but I think my question is still unanswered. I presented two ways for introducing covariance concept. I need demonstrating the equivalency of the two approaches. – user27058 Jul 17 '14 at 15:39
• @aminliverpool I just understood your edited question! I'll update my answer tomorrow ;-P – Dox Jul 17 '14 at 19:39
• Thank you for your answer. Ok. I agree with your description of covectors by projecetion, but in last section of your answer, I cant understand how you settled relation between projection description of covectors and dual space. I cant understand your statement: "Because you can always define the action of one type over the other... resulting into a field (K). Mathematically, these define the elements of the dual space.". Is it possible for you to explain more. – user27058 Jul 18 '14 at 5:26
• @aminliverpool What about now? I included an extra paragraph to explain the duality. – Dox Jul 18 '14 at 12:48
• Thank you. I feel I near to solution of my problem. But one question remains: To obtain covariant component of a vector we should construct bases which are orthogonal to original bases. We can do that by construct reciprocal bases in our vector space and I think it is not necessary to extract such bases from the dual of the vector space, and decompose the vector on those bases. Can you help me, more ? – user27058 Jul 18 '14 at 14:40

You may find this question and especially Emilio Pisanty's answer to be enlighting with regards to what co- and contravariant vectors really are.

Now, if I may rephrase your original question a bit: *How do parallelograms and lines in $\mathbb{R}^3$ relate to co- and contravariant vectors in $T_p \mathbb{R}^3$ and $T^*_p\mathbb{R}^3$?

The first one is easy. Take a line in $\mathbb{R}^3$. It has a direction $v = v^i e_i\in\mathbb{R}^3$. Identify $v\in T_p\mathbb{R}^3$ in the natural way (since $T_p\mathbb{R}^3$ is just $\mathbb{R}^3$, just give it the same basis $\{e_i\}$). You've got yourself a (contravariant) vector.

The second one is a bit trickier: Let the parallelogramm be spanned by $v,w \in \mathbb{R}^3$. Then, there is the natural notion of associating this area with the exterior product $v \wedge w$, which actually represents as $v \wedge w = v^i w^j e_i \wedge e_j$ an element of $T_p\mathbb{R}^3 \wedge T_p\mathbb{R}^3$, and could be seen as a constant 2-form. Now, identify the basis $\{e_i \wedge e_j\}$ of $T_p\mathbb{R}^3 \wedge T_p\mathbb{R}^3$ with the Hodge dual of the cross product as $e_i \wedge e_j = \star (e_i \times e_j) = e^k$, which is the dual basis of $T^*_p\mathbb{R}^3$. You've got yourself a (covariant) covector.

The above approach generalises nicely to the $\mathbb{R}^n$ equipped with some (Euclidean) metric (i.e. a scalar product, essentially):

Let $\{e_i\}$ be a basis of the $\mathbb{R}^n$ as a vector space. Let $v \in \mathbb{R}^n$ be a vector.

The contravariant components of $v$ are obtained by $v^i := (v,e_i)$, which is your orthogonal projection.

A basis for the hyperplanes $\Lambda^{n-1} \mathbb{R}^n$ (or "hyperparallelograms", if you prefer) is given by $\{e_{i_1} \wedge \dots \wedge e_{i_{n-1}}\}$. To be able to "project" $v$ onto it, you must either turn $v$ into an element of $\Lambda^{n-1}\mathbb{R}^n$ or turn the hyperplane into an element of $\mathbb{R}^n$, since you cannot take the scalar product of a plane with a vector. So, we let Hodge duality do its job, and say that the covariant components of $v$ are obtained by $v_i := (v,\star(e_{i_1} \wedge \dots \wedge e_{i_{n-1}}))$, where $i$ is precisely the index not ocurring in $\{i_1, \dots ,i_{n-1}\}$. Why is this covariant, you ask?

Because the defining property of the Hodge dual is $v \wedge \star w = (v,w)(e_1 \wedge \dots \wedge e_n)$. From plugging in $e_i$ for $v$ and $\star(e_{i_1} \wedge \dots \wedge e_{i_{n-1}})$ for $w$, you can see that $\star(e_{i_1} \wedge \dots \wedge e_{i_{n-1}})$ precisely fulfill the relation for the dual basis to $\{e_i\}$ (you have to use that $\star^2 = -\mathrm{id}$ for $(n-1)$-vectors)

Therefore, we have shown by translating the geometric idea of projecting onto hyperplanes into precise mathematical language that it is exactly the same as expanding in the dual basis.

• Thank you for your attention and your useful link. But I think my question is still unanswered. I presented two ways for introducing covariance concept. I need demonstrating the equivalency of the two approaches. – user27058 Jul 17 '14 at 15:40
• @aminliverpool: I have expanded my answer. If it is still not what you seek, I advise that you give a precise definition of what you understand under "projecting onto hyperplanes" in the OP. – ACuriousMind Jul 17 '14 at 16:16
• Thank you for your answer. I am not familiar with "exterior product" and "Hodge dual", and so I cant understand your answer. My understanding of projection is the same as Dox explained in his answer. Is it possible for you to reanswer my question in a simple way? – user27058 Jul 18 '14 at 5:39
• @aminliverpool: It seems Dox (and others) do a far better job of thinking along geometric lines and easing down on the abstraction. All I could do is to explicitly construct why the exterior product and areas are intrinsically related, but I'm afraid that that would be more math that probably wouldn't help you. – ACuriousMind Jul 18 '14 at 12:01