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I'm new to tensor analysis, and came across the topic of vectors and duals, and faced a massive confusion. Are vectors and duals different representations of the same object ? I had another doubt regarding this, as follows.

Let us say, we have a set of basis vectors $u_1,u_2 = (2,1), (1,1)$

Suppose we have a vector $V$ in this space, given by $V = 2u_1 +3u_2$.

As I've read, the components here are the 'contravariant' components of the Vector. However, we could express the same vector in its dual space, using the 'covariant' components. In order to do this, we would need to the 'dual-basis' Using $u^j u_i = \delta^i_j $ in matrix notation, i've calculated the dual basis vectors to be $u^1,u^2 = (1,-1), (-1,2)$

My question is, how do I express the vector $V$ in this dual basis. Do I use the simple change of basis matrix ? Moreover, the coefficients that I get, in this case, are these the 'covariant' coefficients ? I've seen expressions like this, $x_i = V.u_i$

This doesn't make sense to me, as $V.u_i = x^iu_i.u_i = x^i$

Can someone point out where I've made a mistake in this, and how can I correctly express the same vector in the co-variant and contra-variant way?

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    $\begingroup$ Dual vectors (also called one forms) are not the same objects as vectors. Why do you assume that an object $V$ given to be a vector "can be represented" using a dual basis? While something similar might be possible, it doesn't have to be the case always. $\endgroup$ – newtothis Jun 16 at 7:08
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Confusion about this goes on and on and on. And I've seen it go on for more than forty years. The word "dual" and the words "contravariant"/"covariant" correspond to different things. The problem is that in finite dimension, and working in coordinates, dual spaces look so much as the original space $V$ that there is a risk of confusion.

Dual Space

Dual space of a given linear space $V$: $$ \begin{array}{cccc} \omega: & V & \rightarrow & \mathbb{R}\\ & x & \mapsto & \omega\left(x\right) \end{array} $$ The space of all these things is called the dual space of $V$: $$ V^{*}:=\left\{ \omega:V\rightarrow\mathbb{R}/\omega\textrm{ is linear}\right\} $$ Here you have two different spaces, so it's obvious you can choose any basis you want to expand your linear forms. Why not choose the one in which the action of the $\omega$'s on the vectors is simplest?: $$ \alpha_{i}\left(e_{j}\right)=\delta_{ij} $$ This is called the dual basis.

Scalar product

Scalar product on a given linear space $V$: $$ \begin{array}{cccc} \left\langle \:,\:\right\rangle : & V\times V & \rightarrow & \mathbb{R}\\ & \left(x,y\right) & \mapsto & \left\langle x,y\right\rangle \end{array} $$ Here you have one space $V$, and one space only. Perhaps you do not care to consider linear mappings on the scalars. This context is more typical when you're thinking geometrically, and you wish to talk about distances and angles.

And the problem when you have a scalar product arises because you would like to expand vectors in terms of their projections on different 1-dim subspaces: $$ x=\left\langle x,e_{i}\right\rangle e_{i}\:\textrm{wrong!!!} $$ This formula doesn't work when your basis is not orthonormal. If you just naively project your vector on the basis, you get, $$ \left\langle x,e_{j}\right\rangle =\left\langle x^{i}e_{i},e_{j}\right\rangle =x^{i}\left\langle e_{i},e_{j}\right\rangle =x^{i}g_{ij} $$ This is not your contravariant coordinate --the coordinate $x^j$ such that, $$ x=x^{j}e_{j} $$ And the reason why it's not is that your basis is not orthonormal: $$ \left\langle e_{i},e_{j}\right\rangle =g_{ij}\neq\delta_{ij} $$ What you need is, so to speak, a basis that is "orthonormal with respect to $\left\{ e_{i}\right\} $" Enough of the formalism has been developed that I don't need to repeat. Let's go to your example. The basis you propose is not orthonormal: $$ \left\langle u_{i},u_{j}\right\rangle =g_{ij}=\left(\begin{array}{cc} 5 & 3\\ 3 & 2 \end{array}\right) $$ Let's call that the "covariant basis". The "contravariant basis" is: $$ u^{1}:=\left(a_{1},a_{2}\right) $$ $$ u^{2}:=\left(b_{1},b_{2}\right) $$ Solving for, $$ \left(a_{1},a_{2}\right)\cdot\left(2,1\right)=1 $$ $$ \left(a_{1},a_{2}\right)\cdot\left(1,1\right)=0 $$ and, $$ \left(b_{1},b_{2}\right)\cdot\left(2,1\right)=0 $$ $$ \left(b_{1},b_{2}\right)\cdot\left(1,1\right)=1 $$ You can easily prove,

$$ u^{1}=\left(1,-1\right) $$ $$ u^{2}=\left(-1,2\right) $$ Let us see that this is the proper contravariant basis. We must check that the vector $v$ that you provide can actually be expanded in terms or projections on the basis vectors, but by introducing the contravariant basis. That is,

$$ v\overset{?}{=}\left\langle v,u_{1}\right\rangle u^{1}+\left\langle v,u_{2}\right\rangle u^{2} $$ What is $v$?: $$ v=2u_{1}+3u_{2}=2\left(2,1\right)+3\left(1,1\right)=\left(7,5\right) $$ What are the projections on the given basis?: $$ \left\langle v,u_{1}\right\rangle =\left(7,5\right)\cdot\left(2,1\right)=19 $$ $$ \left\langle v,u_{2}\right\rangle =\left(7,5\right)\cdot\left(1,1\right)=12 $$ Check: $$ \left\langle v,u_{1}\right\rangle u^{1}+\left\langle v,u_{2}\right\rangle u^{2}=19\left(1,-1\right)+12\left(-1,2\right)=\left(7,5\right) $$ Now let's do the check by using the contravariant basis/covariant coordinates as the reference: $$ \left\langle v,u^{1}\right\rangle =\left(7,5\right)\cdot\left(1,-1\right)=2 $$ $$ \left\langle v,u^{2}\right\rangle =\left(7,5\right)\cdot\left(-1,2\right)=3 $$ And, $$ v\overset{?}{=}\left\langle v,u^{1}\right\rangle u_{1}+\left\langle v,u^{2}\right\rangle u_{2} $$ $$ \left\langle v,u^{1}\right\rangle u_{1}+\left\langle v,u^{2}\right\rangle u_{2}=2\left(2,1\right)+3\left(1,1\right)=\left(7,5\right) $$ It checks. Same vector, two different expansions in terms of different bases --and projections.

I hope that clarified the question a little bit.

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  • $\begingroup$ Thank you so much, this is exactly what I was looking for. However, there is one small doubt, as to a similar question regarding the geometrical picture of this. Would you mind if I ask it to you ? $\endgroup$ – Nakshatra Gangopadhay Jun 17 at 16:27
  • $\begingroup$ what I understand from most of the comments is that the confusion arises because of something called the reciprocal basis, which mimics the dual space, but unlike the dual space which is a different object, this reciprocal space is similar to our original vector space. Since the components in reciprocal space and dual space match exactly, we tend to call this the dual space instead, not knowing they are separate things. Is that right ? $\endgroup$ – Nakshatra Gangopadhay Jun 17 at 16:32
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    $\begingroup$ Sometimes you have a dual space in its own right because the linear forms that you're considering are very different things from your vectors. Typical example: functional spaces in quantum mechanics. If they're infinitely generated, it's particularly important to distinguish $V$ from its dual. If you just have one $V$ and a scalar product, and finite dimension, and your main concern are distances/angles, then you can relax about the whole thing and say that every vector $u$ has a unique linear form $\omega$ such that $\omega_{u}\left(x\right)=\left\langle u,x\right\rangle$. $\endgroup$ – joigus Jun 17 at 19:04
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Are vectors and duals different representations of the same object ?

Not generally. Duals are linear functionals of vectors, i.e. they are objects that take vector and produce real number, so they are different beasts.

But they do form vector space in their own right with same dimensionality as the original and every two vector spaces with same dimensionality are isomorphic to each other, meaning we can map each element of original vector space to its dual space and vice versa.

Having basis in original vector space $e_b$, we can find dual basis $e^a$ in dual space by requirement, that it is set of dual vectors which takes one basis elements and produces 1 and 0 for others, i.e. $$e^a(e_b)=\delta^a_b.$$ This defines an isomorphism between the two spaces (by identifying $e^0$ with $e_0$, etc.). Its not canonical though, because starting with another basis $e'_a$ would result in different isomorphism.

However, we could express the same vector in its dual space, using the 'covariant' components

This is incorrect. If we were to express vector in its dual space, then we are not expressing the vector, but mapping to another vector in different vector space. This is moreover not unique, and different basis would result in different dual vector to which it is mapped.

What different expression of the same vector means is that we are decomposing it relative to two different basis in the same vector space.

To define covariant and contravariant components, we first need canonical identification of vector space and its dual. This is usually achieved by metric and isomorphism is given by $e^a=g(e_a,.),$ i.e. we identify basis vector $e_a$ with dual vector $e^a$, that produces from vector $v$ the same number as operation $g(e_a,v)$.

Having a basis $e_a$ in vector space, canonical isomorphism that maps this basis into basis $e^b$ in dual space (this is not necessarily dual basis as defined in the first section!), we can define new basis $e'_c$ in the original vector space, by requirement, that $$e^b(e'_c)=\delta^b_c.$$

The result is, that the components of vector $v$ in this dual basis $e'_b$ will be the same as components of the dual vector $\alpha$ identified with $v$ by our isomorphism when expressed in coordinates $e^a$ that are identified with our original basis vectors $e_a$, i.e.: $$\alpha=\alpha_a e^a$$ $$e^a=g(e_a,.)$$ $$v=v^a e_a = v^b e'_b $$ $$ v^b =\alpha_b $$ $$e^a(e'_b)=\delta^a_b$$ $$e^a(e_b)=g_{ab}$$ where $\alpha \leftrightarrow v$ by the isomorphism.

Yes, it is a little confusing.

EDIT

To answer the comment

...can you tell me how to prove that the components in this third basis are the same as the components in the covector basis ?

Its property of isomorphism given by dual basis. The dual basis to $e^b$ is $e_b$ and the isomorphism maps $e^0$ to $e_0$ and analogically for every other index. For general covector the isomorphism is a map $$\Phi: \alpha \leftrightarrow v,$$ and on the basis we know that $$e_b=\Phi(e^b).$$ The map $\Phi$ is linear, since it is a morphism and preserves structure, i.e $$v=\Phi(\alpha)=\Phi(\alpha_a e^a)=\alpha_a\Phi( e^a)=\sum_a \alpha_a e_a,$$ i.e. $v^a=\alpha_a$ in dual basis.

For comparison, the isomorphism given by metric is $$\Phi_g: \alpha \leftrightarrow v,$$ and on the basis we know that $$f_b=\Phi_g(e^b).$$ Now, the metric is expressed as $g=g^{ab}e_a \otimes e_b,$ where $e_a$ is dual basis to $e^a$. It needs to be dual basis, because only then the formula $g(\alpha,\beta)=g^{ab}\alpha_a\beta_b$ holds, since $$g(\alpha,\beta)=\alpha_a\beta_bg(e^a,e^b)=\alpha_a\beta_bg^{cd}e_c(e^a)e_d(e^b)=\alpha_a\beta_bg^{cd} \delta^a_c\delta^b_d.$$ So if it would not be true that $e_a(e^b)=\delta_a^b$, then additional terms would appear.

Next we need to determine how $f_b$ is expressed in the dual basis. Acting with $f_b$ on basis covectors $e^a$ we get $$f_b(e^a)\equiv g(e^b,e^a)=g^{ba},$$ i.e the components of $f_b$ with respect to basis $e^a$ are $g^{ba}$. The first equality is simply the definition of canonical isomorphism $e_a\leftrightarrow f^a=g(e_a,.)$.

Now, for the general vector: $$v=\Phi_g(\alpha)=\Phi_g(\alpha_a e^a)=\alpha_a\Phi_g( e^a)=\alpha_a f_a=\alpha_a g^{ab}e_b.$$ I.e. now the componets are $v^b=\alpha_a g^{ab},$ which is well known raising of indexes.

Does not your head twirl from all this index gymnastics?

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  • $\begingroup$ I'm fairly new to this subject, so I'm a little slow at grasping the notations. But based on this, what I understand is, vectors and duals are different, and a certain vector corresponds to a certain dual in the dual space. However, we can define a third basis in the original vector space, which would relate to the dual of the original vector, and representing the original vector in this new basis would be similar to finding the dual. This is what we mean by the covariant component. Since the new basis is related to the dual, the component of this basis is the co-variant component. $\endgroup$ – Nakshatra Gangopadhay Jun 16 at 7:56
  • $\begingroup$ Sorry about the sloppyness. $\endgroup$ – Nakshatra Gangopadhay Jun 16 at 7:56
  • $\begingroup$ @NakshatraGangopadhay sorry, I cannot make out if you understood it or not. This whole thing is quite nuanced and proper terminology needs to be used. For example "Since the new basis is related to the dual", no it is related to one particular basis in dual space, not dual that you first claimed to be covector identified with some vector. The third basis is dual basis of the basis in covector space that is identified with original basis in original space. $\endgroup$ – Umaxo Jun 16 at 8:05
  • $\begingroup$ @NakshatraGangopadhay Anyway, while I disagree with Vincent that there is no third basis or covariant/contravariant components, I do think that this business with covariant/contravariant components is needlessly confusing and it would be much easier if we just forgot it and used simple maps between space and dual space instead of this basis shenanigans. $\endgroup$ – Umaxo Jun 16 at 8:12
  • $\begingroup$ @Umaxo okay so, components of the vector in this third basis, which is dual of the covector basis, which in turn is the dual of the original basis, are the same components of the dual to the original vector. However, since this maps back to our original space, we can call this the co-variant component of the original vector. Is this correct ? $\endgroup$ – Nakshatra Gangopadhay Jun 16 at 8:30
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You have several confusions here.

1.

The terms "contravariant" and "covariant" are simply wrong. There is no such thing as contravariant and covariant components of the same vector. 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.

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 $v = v^i \mathbf{e}_i$. The coefficients $v^i$ are the components the vector. 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 $\omega(v) \in F$. As long as $V$ is finite-dimensional, 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 $$\omega(v) = \omega_i v^i$$

The metric tensor $g$ can be used 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 you were trying to do. You are not representing the same vector with different components. What you are really doing is finding the components of the corresponding covector in the dual space. Vectors have components ("contravariant"), and covectors have components ("covariant"). That's it.

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, vectors and covectors have the same components.

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Let us say, we have a set of basis vectors $u_1,u_2 = (2,1), (1,1)$

This is also wrong. A basis vector in its own coordinate system is always $(1,0)$ or $(0,1)$. If it is not, then it means that it is defined relative to some other "background" coordinate system. Therefore, you need to be careful when raising and lowering indices because the components of $g$ depend on the coordinate system. You need to make sure that everything is expressed in the same coordinate system.

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