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I asked the same question in Math.SE, but I was suggested to ask it here as well.

I am studying relativity, and as you know the theory extensively uses the notion of covariant and contravariant component of vectors.

My question is the following. Let say $\vec{x}$ is a vector which belongs to vector space $V$ with basis vectors ${e}_{i}$. We know that $\vec{x}$ can be written as: ${x}^{i} {e}_{i}$. The same vector $\vec{x}$ can be written in terms of the basis vectors ${e}^{i}$ of the dual space $V^{*}$ as: ${x}_{i} {e}^{i}$, why is this true? Isn't ${e}^{i}$ a basis set for the dual space $V^{*}$ and not the vector space $V$? How can the same vector belong to both the vector space $V$ and it's dual $V^{*}$? How can $\vec{x}$ which belongs to $V$ be written in terms of the basis vectors of another vector space $V^{*}$?

The Wikipedia article on Covariant transformation says that:

...so the dual space has the same dimension as the linear space itself. It is "almost the same space", except that the elements of the dual space (called dual vectors) transform covariantly and the elements of the tangent vector space transform contravariantly.

What does "almost the same" mean?

Again, along the same question, you see that in the Wikipedia article on Curvilinear coordinates, both bases (vector basis $e_{i}$ and covector basis $e^{i}$) are plotted in the same graph, although we know that they are bases for different vector spaces.

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closed as off-topic by John Rennie, Kyle Kanos, JamalS, user10851, ACuriousMind Mar 3 '15 at 17:08

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    $\begingroup$ The relevant cross-post: math.stackexchange.com/q/1172022 I know a user there told you you should post here, but cross-posting is generally discouraged, especially if you're well on your way to an answer on the other site. $\endgroup$ – user10851 Mar 2 '15 at 17:16
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    $\begingroup$ I'm voting to close this question as off-topic because it has been answered, and the answer accepted, on the Math SE. $\endgroup$ – John Rennie Mar 2 '15 at 17:18
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The same vector x⃗ can be written in terms of the basis vectors ei of the dual space V* as: xiei, why is this true?

It's not true.

The elements of the dual space are not vectors as we ordinarily conceive of them geometrically, e.g., directed line segments.

Rather, they are (geometrically) a set of oriented surfaces with a density proportional to their magnitude.

For example, consider a vector in the $x$ direction of length $L$. The associated element of the dual space, a one-form, is the set of surfaces in the $yz$ plane with density $L$.

The contraction of a vector and its dual gives the real number $L^2$ and, geometrically, is the number of surfaces of the one-form pierced by the vector.

For unit basis one-forms, the density is 1.

Now, picture graph paper with the coordinate axes drawn:

enter image description here

The vertical surfaces represent the one-form dual to the $x$ basis vector while the horizontal surfaces represent the one-form dual to the $y$ basis vector.

Draw a vector from the origin to the point, e.g., (3,4). The vector is then given by

$$3\hat{\mathbf x} + 4\hat{\mathbf y}$$

Note that the $x$ and $y$ components are precisely the numbers given by the contraction of the vector with the respective basis one-forms.

In other words, the vector pierces 3 vertical surfaces and 4 horizontal surfaces.

For cartesian coordinates in a flat space, a vector and its dual one-form have the same components. Thus, one might be led to falsely conclude that they represent the same object - they don't. Generally, the vector and its dual one-form do not have the same components.

Recalling matrix representation, a one-form is represented by a row vector while a vector is represented by a column vector. The row vectors 'live' in a different space than the column vectors.

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  • $\begingroup$ Thanks for the comment. I understand your point. What I don't understand is this line in one of the wiki pages: A vecto $\vec{v}$ can be given in terms of either basis, i.e.: $\vec{v}$= $v_{i} b^{i}$=$v^{i} b_{i}$. Where $b_{i}$ is the regular basis (tangent vectors to coordinate lines) and $b^{i}$ is the covector bais. Here is the link to the page: en.wikipedia.org/wiki/Curvilinear_coordinates $\endgroup$ – Ben Mar 2 '15 at 18:10
  • $\begingroup$ @Ben, I see that the factual accuracy of the article is disputed. As I understand it, it should be $\vec v = v^ib_i$ where $v^i = (b^i, \vec v)$ and $\tilde{v} = v_ib^i$ where $v_i = (\tilde{v}, b_i)$. $\endgroup$ – Alfred Centauri Mar 2 '15 at 18:19
  • $\begingroup$ Thanks Alfred. What you said makes complete sense to me. However, this is being mentioned in various pages, such as : en.wikipedia.org/wiki/Covariant_transformation, my only guess is that the two vector spaces V and V* might be the same in some specific case, for example in : en.wikipedia.org/wiki/Dual_basis. it says that: in the case of cartesian coordinate system $e_{i}$=$e^{i}$. In fact it gives you a formula to find the covector basis in the 3d Euclidean space. So the two space should be the same in this case!!! $\endgroup$ – Ben Mar 2 '15 at 18:28
  • $\begingroup$ @Ben, I suspect that what is meant is that the components are the same but the notation certainly is confusing. $\mathbf e_i$ is the $i^{th}$ basis vector while $\mathbf e^i$ is the $i^{th}$ basis one-form; they're not the same geometric object. $\endgroup$ – Alfred Centauri Mar 2 '15 at 18:54
  • $\begingroup$ This was exactly my confusion and that's why I raised the question. However, it is kind of making sense to me now that this might be tight in case of a vector space with an inner product. The action of the elements in V* (dual space) on an element in the vector space, is defined by dot product in the original space V. So, there is a link between them. $\endgroup$ – Ben Mar 2 '15 at 19:12
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If $V$ is a (finite dimensional) vector space with no additional structure, then $V^*$ is a different vector space of the same dimension, hence isomorphic to $V$ --- but it would be a great mistake to think of $V$ and $V^*$ as "the same'' or even ``almost the same'', because there is no preferred isomorphism between them. In other words, you can pick a basis for $V$, pick a basis for $V^*$, map the elements of the one basis to the other, and get an isomorphism, but there a gazillion different ways to do this, and no reason to prefer one to another. So there is no single natural way to identify elements of $V$ with elements of $V^*$.

On the other hand, if $V$ is not just a vector space, but a vector space equipped with a metric, then there is one very natural way of identifying $V$ with $V^*$ ---- first pick an orthogonal basis for $V$ (note that the very notion of ``orthogonal'' makes no sense without the extra structure), then map the elements of that basis to the corresponding elements of the dual basis, and then check that you'll get exactly the same map no matter which orthogonal basis you started with. So in this case, $V$ and $V^*$ are "almost the same" in the sense that there is one preferred way to identify an element of $V$ with an element of $V^*$.

Of course in the usual applications to relativity, your vector space comes equipped with a metric, so the second paragraph above applies instead of the first.

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  • $\begingroup$ Thank you so much, I understand that in the case where the vector space is equipped with a metric there is a preferred way to define the elements of V* from V. What I don't understand is, if these two are different vector spaces (Although related, but different). How can a vector in V (with basis $e_{i}$) let say $\vec{v}$=$v^{i} e_{i}$ be expressed in terms of the basis of V* (with basis $e^{i}$) as $\vec{v}$=$v_{i} e^{i}$ ???!!! The basis of V* span another vector space, namely the dual space V* and not V. So how is it that vector $\vec{v}$ can be written in both ways! $\endgroup$ – Ben Mar 2 '15 at 18:18
  • $\begingroup$ @Ben: it's abuse of notation, and I do not believe I've seen any literature going quite that far; the basic idea is that given a non-degenerate bilinear form (resp. a metric tensor in case of (pseudo-)Riemannian geometry), $V$ and $V^*$ become canonically isomorphic and can be 'identified'; if you prefer, you could introduce a new space $V^\blacktriangle$ isomorphic to both, and a single geometric object from $V^\blacktriangle$ can be represented by elements of $V$ and $V^*$ both; the culmination of that idea is that there are only (4-)vectors with co- and contravariant components $\endgroup$ – Christoph Mar 2 '15 at 18:51
  • $\begingroup$ @Ben: personally, I thinks this is misleading, in particular the Riemannian case, both in terms of differential geometry (vectors are equivalence classes of curves $\mathbb R\to M$, whereas covectors are equivalence classes of functions $M\to \mathbb R$) and physics (velocities vs momenta) $\endgroup$ – Christoph Mar 2 '15 at 18:53
  • $\begingroup$ @Christoph: I think it might be true in the case of a vector space equipped with a dot product. en.wikipedia.org/wiki/Dual_basis claims that for Cartesian coordinates $e_{i}$=$e^{i}$. Please leek at the example sections. In fact it provides formula in order to find $e^{i}$ in terms of $e_{i}$ in Euclidean space. It started to make sense to me, since the action of the vectors in the dual space V* can be defined by the help of the dot product, we can assume that these two spaces are the same. However, the tangent basis and cotangent basis are different. $\endgroup$ – Ben Mar 2 '15 at 19:02
  • $\begingroup$ @Ben: well, if you do not make the distinction between row- and column vectors, you can do that, but I'd advise against it (vectors are traditionally represented as column vectors, and covectors as row vectors, ie $n\times 1$ and $1\times n$ matrices) $\endgroup$ – Christoph Mar 2 '15 at 19:10

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