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Covector: A linear map from some set of vectors into real numbers. Also, on its own, it is an element of a vector space. Visualisation: visualize a covector as a stack of hypersurfaces of some density. Operationally a map is described as arrow vector piercing this stack of (hyper)surfaces in some way. Output is a number of (hyper)surfaces pierced. Why this particular visualization? Why not some other?

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  • $\begingroup$ "X or something" or "you know what book, the biggest one" aren't a good way to get help. If you want help from people at least have the courtesy to check the references you give and not expect other people to guess (or do the checking for you). The less helpful you are when asking, the less help you tend to get. The more helpful and precise you are when asking, the more likely people are to try and help. That's been my experience, anyway. $\endgroup$ – StephenG Jul 14 '18 at 11:34
  • $\begingroup$ I believe that the core question is clear...i gave a definiton and an alternative view used in some books.... $\endgroup$ – Žarko Tomičić Jul 15 '18 at 11:03
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    $\begingroup$ If your question is 'why the visualisation' the answer is 'because many people find being able to visualise things helpful'. That's a question about the properties of our minds and brains rather than about physics. $\endgroup$ – tfb Jul 15 '18 at 11:56
  • $\begingroup$ I am sorry about that, I did not realise that the title is not in agreement with the question itself. To say something about visualization, the question aims to that particular type of visualization, and why that type and not some other? Is there some physics in the choice of this visualization or not? $\endgroup$ – Žarko Tomičić Jul 15 '18 at 13:57
  • $\begingroup$ The point of the visualization is that you don't need a concept of distance or angle to calculate the action of a covector. This emphasizes the fact that a covector doesn't need a metric to be defined. $\endgroup$ – Javier Jul 15 '18 at 14:04
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The stacks of surfaces picture is just a helpful way of visualizing a covector which is compatible with the usual picture of vectors as arrows. Since a covector takes a vector and gives back a number, in the stacks of surfaces picture when you place a vector (an arrow), it intersects a few surfaces which corresponds to the number given as output. By changing either the length or the orientation of the arrow, you intersect different number of surfaces meaning you get a different output. Hence its just a helpful picture to have of covectors but isn’t really useful for computation purposes.

Your idea about vector being a covector is kind of right. Precisely for finite dimensional vector spaces, there is an isomorphism between a vector space and the dual vector space (the space of covectors). However there are multiple isomorphisms and no canonical one. Given a metric, you then have a canonical isomorphism. Suppose $g:V \times V \to \mathbb{R}$ is your metric. Then for a vector $v\in V$ , $g(v,-):V \to \mathbb{R}$ is a linear functional and hence is a covector. Call this covector $\omega$. Then for each vector you have a corresponding covector by this identification.

However note a vector and covector are kind of same but not exactly same. The difference lies in the transformations properties. Suppose you have a basis for your vector space and your vector $v=(v_1,v_2,\dots,v_n)$ in this basis and you have a covector $\omega=(\omega_1, \omega_2, \dots, \omega_n)$ in the corresponding dual basis. Then $\omega(v) = \omega_1 v_1+\omega_2 v_2+\dots +\omega_n v_n$ which is a scalar (doesn’t transform when a basis is changed). Suppose you make a change of basis to your vector space. Then the components of your vector transform in a certain way and the components of a covector transform in the opposite way so as to keep the value of $\omega(v)$ the same.

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  • $\begingroup$ So thank you. But, if we call elements of ome vector space vectors, and of some other vector space covectors, than opposite apply, right? $\endgroup$ – Žarko Tomičić Jul 15 '18 at 11:13
  • $\begingroup$ Just calling something a vector or a covector is not the way of defining things. The thing is that a vector is an object which has certain transformation properties. specifically, the components of a vector transform in the opposite way to how coordinates transform under a change of coordinates. On the other hand for a covector, the components transform in the same way as the coordinates transform. This is the property which defines whether something is a vector or a covector $\endgroup$ – ravjotsk Jul 15 '18 at 13:21
  • $\begingroup$ But, lets say there is a vector and there is its dual, a covector. But a dual of this dual is again a vector with which we started. So, if it has these propoerties of transformation, now it is acctually a dual vector. SO, is it important that it is dual of a dual and not just dual? $\endgroup$ – Žarko Tomičić Jul 15 '18 at 13:52
  • $\begingroup$ ANd to add, if something is an element of a vector space (and covectors are elements of a vector space) then it is a vector, regardless of you calling it a covector. $\endgroup$ – Žarko Tomičić Jul 15 '18 at 14:01
  • $\begingroup$ Although it is true in linear algebra that elements of a vector space are vectors, the definition I was referring to earlier is in terms of differential geometry (although the set of real numbers itself is a vector space, geometrically you don’t call them vectors). So just looking at the algebraic properties you are absolutely correct in saying that covectors form a vector space and so must be vectors. However the geometric definition of vectors is based on tangents to a manifold and therein you get a difference between what is a vector and what is a covector. $\endgroup$ – ravjotsk Jul 16 '18 at 11:57

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