# How to relate a one form to a vector.

1. Apr 8, 2007

### matheinste

I asked this question in an SR thread started by someone else but it really belongs in this forum. I received much help there but was naturally told what all the textbooks had already told me and I still could not grasp it.

I understand, or think I do, vector spaces, tangent spaces, reciprocal bases and how to generate them. I also understand how one forms and vectors act linearly upon each other to make a scalar and of the helpful illustrations of how to think of a one forms geometrically. But I have trouble in understanding how we get from a given vector to its associated one form.

One step at a time.

In a book on EM theory some years ago I came accross a diagram showing oblique 2D axes and a vector radiating from the origin. I believe the terminology now not to be the accepted one for co and contra but this can be sorted easily. The coordinates of the vector arrow obtained by dropping perpendiculars to the x and Y axes were called co-variant components. The coordinates obtained by reading the intersection from the vector arrow of the lines parallel to the other axes were called contra-variant components.
This I suppose is just repersenting the vector in a different way.

If we generate the reciprocal axes relative to the original axes and drop perpendiculars to these from the arrowhead of the original vector my thoughts are that this is just the same object referred to a different set of axes and called a different name. I am told that a one form is also called a covariant vector or covector but cannot see how a one form is described by either of these diagrammatic repesentations. I KNOW I am going wrong but WHERE.

If I have not defined the scenario precisely enough I will try to rectify this if you need me to.

Thanks to the SR people for their previous help but it still did not sink in. Perhaps some input from new angles will help.
Thanks Matheinste.

2. Apr 8, 2007

### robphy

Last edited: Apr 8, 2007
3. Apr 8, 2007

### matheinste

Hi Robphy. I understood your constuction showing that one forms and vectors were in a sense reciprocal to each other but I still cannot relate this diagram to the bases or axes. I welcome any input which you can give,

Thank you Matheinste

4. Apr 8, 2007

5. Apr 8, 2007

### matheinste

Thanks again matheinste.

6. Apr 12, 2007

### matheinste

Hi Robphy. Still plodding on and making some progress. Will shout when I need more help.

Cheers Matheinste.

7. Apr 12, 2007

### ObsessiveMathsFreak

One forms aren't vectors and you're not really supposed to relate them to some kind of vector at a point(though a great many people do). Forms in general are like distributions, they only really make sense or have a value after you integrate them over a manifold. Just as you cannot really ask, "What is the value of the Dirac delta function at the point x", so too you cannot really ask the question "What is the vector value of the form at the point (x,y,z)".

OK, yes I know. The coefficients of a one form represent the gradient of a function, and those of a two form are the curl. But those are just the coefficients. The form itself does not take a value until you input the required number of vectors into it.

8. Apr 13, 2007

### matheinste

Hi ObsessiveMathsFreak. Thanks for your input. I think my first contact with co and contravariance where diagrams showed a vector as having either components depending on which axes it is referred to, has brainwashed me into regarding both as vectors because I have been told that one is a contravariant vector and the other a covector ( one form ). I know that what everybody in the forum is telling me is correct and I understand what they say but I am still confused. Should I try to ignore my early experience or try to resolve the problem. It is almost as if, and perhaps this is somehow the answer, that when a vector is referred to it's reciprocal set of axes that it should not be called a one form. As you say it is nothing like one.

There is a diagram showing what I mean in Borisenko and Tarpov, Vector and Tensor Analysis with Applications on page 28, but to be fair they do not mention one forms at this point. If I had never seen this diagram or another like it I would be OK.

I hope I will see the light. Matheinste.

9. Apr 13, 2007

### robphy

I believe one-forms at a point are vectors, in the sense that they are elements of a vector space, and are more specifically described as so-called covariant vectors or tensors of type (0,1). However, these are not the same as the ordinary [so-called contravariant] vectors or tensors of type (1,0).

As I mentioned earlier, to "relate" a one-form to a vector, one needs an additional structure like a metric tensor [or something else, like a symplectic form]. Note again that a vector space alone has no notion of orthogonality [perpendicularity]. So, if the author makes any reference to "perpendicular", there is a metric implicit in the problem. (Note that words like "reciprocal" suggest that there is some kind of additional structure.)

Last edited: Apr 13, 2007
10. Apr 13, 2007

### Hurkyl

Staff Emeritus
If you want some unambiguous terminology, try "tangent vector" and "cotangent vector". (a one-form is a cotangent vector field)

11. Apr 13, 2007

### matheinste

Hi Hurkyl. I think you have helped.

From your definition I think what I now understand, without rigour is :--

At each point on a manifold there is a tangent space. A tangent field ( using some rule ) selects from each tangent space one tangent vector or CONTRAVARIANT vector.

Also at each point there is a cotangent space. A cotangent field OR ONE FORM ( using some rule ) selects from each cotangent space a COVECTOR or COVARIANT vector.

Is this correct. Until I get this right I cannot go any further.

Thanks Matheinste

12. Apr 13, 2007

### ObsessiveMathsFreak

The one form is the covector. It doesn't select one. It itself is the covector of which you speak.

But a covector is not a vector, or at least not a tangent vector. However, if you define the inner product of vectors, $$\mathbf{a}\cdot\mathbf{b}$$, then you can create a one to one mapping between one forms and tangent vectors like so.

Let $$\acute{\omega}$$ be a one form. The tangent vector $$\mathbf{\omega}$$ corressponding to the one form is the tangent vector that for all tangent vectors $$\mathbf{x}$$ in the tangent space satifies:

$$\acute{\omega}(\mathbf{x}) = \mathbf{\omega}\cdot\mathbf{x}$$

For each one form, there is a tangent vector that can in some sense represent, or replace it, but only if you define the inner product of vectors. This may seem a little pedantic, and probably is, but this is in fact the definition. I mostly mention it to emphasise the fact that one forms are not (tangent) vectors, even though they are isomorphic to them.

13. Apr 13, 2007

### matheinste

Hi ObsessiveMathsFreak. I'm still confused. Hurkyl defines a one form as a cotangent vector field not as a cotangent vector.

Matheinste.

14. Apr 13, 2007

### coalquay404

Here's a cleaner way of putting it. Suppose that you have an $m$-dimensional manifold $M$. At each point $p\in M$ there exist two natural spaces, the tangent space to $M$ at $p$ (denoted $T_pM$), and the cotangent space to $M$ at $p$ (denoted $T_p^*M$). An important point to grasp here is that both the tangent space and the cotangent space are vector spaces. Although we call elements of the tangent space "vectors" and elements of the cotangent space "covectors" or "one-forms", they are all to be regarded as vectors since they are elements of vector spaces. The "co-" prefix assigned to elements of $T_p^*M$ is simply a means of emphasising that these objects act in a special way on elements of $T_pM$ (by "special way" I mean to say that a given covector $\alpha\in T_p^*M$ is to be regarded as an object which takes an element of $T_pM$ as an input and spits out an element of some field $\mathbb{F}$ as an output; for example, in physics we often take $\alpha:T_pM\to\mathbb{R}$, $\alpha:X\mapsto\langle\alpha,X\rangle\in\mathbb{R}$).

At any point in $M$ the tangent space and cotangent space are said to be dual to one another. What this means is that if you have some basis $\{e_i\}$ of $T_pM$ then there exists a basis $\{\omega^i\}$ of $T_p^*M$ such that

$$\langle\omega^i,e_j\rangle = \omega^i(e_j) = \delta^i_j$$.

The important point here is that we have made no mention whatsoever of the existence of a metric at this stage. The duality between $T_pM$ and $T_p^*M$ is a completely metric-independent concept and is simply a mild extension of the elementary stuff you already know from (multi-)linear algebra over $\mathbb{R}$.

Now to the concept of "fields". A vector field $X$ over $M$ is a means of smoothly assigning a vector to each point in a given region of $M$. There may be obstacles which prevent us from defining a vector field over all of $M$ (in fact there often are in practice), in which case we say that a vector field is restricted to some $N\subseteq M$.

Last edited: Apr 13, 2007
15. Apr 13, 2007

### Hurkyl

Staff Emeritus
A differential (one-)form is certainly a cotangent vector field.

While "one-form" is certainly an accurate word, albeit old-fashioned, for a linear functional on a vector space, this thread is a very good example of why it's a poor choice of words to use in differential geometry!

In any case, I have been tacitly assuming that we are using the word "one-form" to refer to a differential form on a differentiable manifold.

As such, I would say this is right. Except possibly for the use of contravariant and covariant, because I can never remember which is which.

16. Apr 14, 2007

### matheinste

Hi Hurkyl. Thanks for your confirmation.

From this am I to understand that, as I said, a one form selects a cotangent vector at a point ( on the manifold ) from the cotangent space and that it is this cotangent vector that acts ( linearly ) on a tangent vector to produce a scalar.

Although Kobayashu & Nomizu ( described by Schultz as authoratative ) in Foundatuions of Differential Geometry, Isham in Modern Diff Geom for Physicists, and other books say explicitly that a one form is a selection of a cotangent vector at each point on a manifold, and therefore a field, some books do not. This is confusing and if you only read the wrong book it is also misleading so I still cannot help but feel that somehow I am misreading something somewhere. Others don't seem to have my problem.

Anyway I am with you on this one.

Thanks and I look forward to more help.

Matheinste.

PS I do not know, being a new user op the forum whether posting using mathematical symbology in Word using Mathtype will work.

17. Apr 14, 2007

### Chris Hillman

I have little wish to get into a discussion marred by so much uneccessary confusion, but just thought I'd say that one should avoid the archaic terminology "contravariant" and "covariant" applied to tensors because this directly contradicts the much more important usage in category theory. Speak rather of vector fields and covector fields.

My other comment is that OMF and the others seem to be speaking at cross purposes due to misunderstandings due to archaic/imprecise terminology. Don't forget this little analogy: vector is to covector as vector field is to covector field.

matheinste, you would probably find the classic little book by Flanders helpful. It's a cheap Dover reprint and should be available via amazon.com and through other on-line booksellers.

18. Apr 15, 2007

### mathwonk

bear in mind, it is one thing to understand a concept, and another thing to sort out the various different terminologies people use for that concept.

i think most people agree about what is meant by "tangent vector" "cotangent vector", and fields of same.

i myself intend by the phrase "one form" to describe a field of cotangent vectors, but i think david bachman used "one form" for a single cotangent vector in his nice little book, previously studied here.

this is perhaps because of the old use of the word "linear form" for "linear functional".

the point is to grasp the concepts of tangent vector and cotangent vector, and then families of them.

the next level is bilinear and multilinear functions, and fields of those, and also their duals, bivectors and multivectors.

those are called "tensors" when it is desirous to frighten children by them.

Last edited: Apr 15, 2007
19. Apr 19, 2007

### matheinste

Hello all. I am fairly happy now with most aspects of vectors/covectors having abandoned the terminology of contra and co variance and one-forms. One thing, at the moment still bothers me. Tangent vectors and cotangent vectors inhabit different vector spaces at any point on a manifold so how does the metric map (change) a tangent vector into a cotangent vector.

Matheinste.

20. Apr 19, 2007

### Hurkyl

Staff Emeritus
Let V be a vector space, with an inner product g(_, _).

For any vector v, we can construct the function
g(v, _) : V ---> R : w ---> g(v, w).​

Because g is a bilinear functional, g(v, _) is a linear functional, and therefore a covector.