How to relate a one form to a vector.

  • Thread starter matheinste
  • Start date
1,060
0
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.
 

robphy

Science Advisor
Homework Helper
Insights Author
Gold Member
5,381
661
Last edited by a moderator:
1,060
0
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
 
1,060
0
Thankyou Robphy. The article in your second post looks helpful and I will give it a thorough reading before replying.

Thanks again matheinste.
 
1,060
0
Hi Robphy. Still plodding on and making some progress. Will shout when I need more help.

Cheers Matheinste.
 
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.
 
1,060
0
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.
 

robphy

Science Advisor
Homework Helper
Insights Author
Gold Member
5,381
661
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 https://www.physicsforums.com/showthread.php?t=158698&p=1291561" 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 by a moderator:

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
If you want some unambiguous terminology, try "tangent vector" and "cotangent vector". (a one-form is a cotangent vector field)
 
1,060
0
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
 
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.
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, [tex]\mathbf{a}\cdot\mathbf{b}[/tex], then you can create a one to one mapping between one forms and tangent vectors like so.

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

[tex]\acute{\omega}(\mathbf{x}) = \mathbf{\omega}\cdot\mathbf{x}[/tex]

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.
 
1,060
0
Hi ObsessiveMathsFreak. I'm still confused. Hurkyl defines a one form as a cotangent vector field not as a cotangent vector.

Matheinste.
 
216
1
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
Here's a cleaner way of putting it. Suppose that you have an [itex]m[/itex]-dimensional manifold [itex]M[/itex]. At each point [itex]p\in M[/itex] there exist two natural spaces, the tangent space to [itex]M[/itex] at [itex]p[/itex] (denoted [itex]T_pM[/itex]), and the cotangent space to [itex]M[/itex] at [itex]p[/itex] (denoted [itex]T_p^*M[/itex]). 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 [itex]T_p^*M[/itex] is simply a means of emphasising that these objects act in a special way on elements of [itex]T_pM[/itex] (by "special way" I mean to say that a given covector [itex]\alpha\in T_p^*M[/itex] is to be regarded as an object which takes an element of [itex]T_pM[/itex] as an input and spits out an element of some field [itex]\mathbb{F}[/itex] as an output; for example, in physics we often take [itex]\alpha:T_pM\to\mathbb{R}[/itex], [itex]\alpha:X\mapsto\langle\alpha,X\rangle\in\mathbb{R}[/itex]).

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

[tex]\langle\omega^i,e_j\rangle = \omega^i(e_j) = \delta^i_j[/tex].

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

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

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
ObsessiveMathsFreak said:
The one form is the covector. It doesn't select one. It itself is the covector of which you speak.
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.


matheinste said:
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.
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.
 
1,060
0
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.
 

Chris Hillman

Science Advisor
2,337
8
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.
 

mathwonk

Science Advisor
Homework Helper
10,733
912
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:
1,060
0
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.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
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.
 
1,060
0
Thanks Hurkyl. Will check out bilinear functionals.

I think part of my problem is what makes cotangent and tangent spaces at a point different from each other. This is probably fundamental and I have misunderstood something very basic.

Matheinste.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
Thanks Hurkyl. Will check out bilinear functionals.

I think part of my problem is what makes cotangent and tangent spaces at a point different from each other. This is probably fundamental and I have misunderstood something very basic.

Matheinste.
Well, the tangent and cotangent spaces at a point really aren't very different: they are isomorphic vector spaces. (although the isomorphism is "unnatural" in a certain sense)


What is different here is how the spaces are related to the differential geometry of the manifold. Among other things, tangent vectors were constructed so they capture notions like direction and displacement, while cotangent vectors were constructed to capture notions like a differential.

For example, if you have a scalar function f, the covector w = (df)(P) tells you how f is varying at P. If you have a tangent vector v at P, you can combine them as w(v) to answer "How does f vary in the direction v?".


Another fun thing you can do is to use the notation
P + v
to denote the tangent vector v at the point P. (the + is only a formal symbol; you're not actaully adding anything)

Then, if you have a differentiable function between manifolds, it's literally true that
[tex]f(P + v) = f(P) + f_*(P) v.[/tex]
(I prefer the symbol f* to denote the map f induces on the tangent bundle)
 
Last edited:
39
0
I know this post is a few months old, but I am new to the forum and this is a topic I am very interested in, so I would like to add my two cents worth. I find that many of the properties of vectors and covectors become more transparent when one considers some more intuitive geometric objects that vector and covectors are directly associated with. Vector fields are directly related to parameterized curves, and covectors are closely related to functions. If you understand the similarity and difference between a family of parameterized curves that fill a space and a function defined in the space, you go a long way towards understanding the similarity and difference between vector (fields) and covector (fields).

Covectors (or differential forms) arise natually as the differential of a function. For example, suppose we have the function:
[tex]f(x,y) = x + y^2 + c[/tex]
The covector associated with this is given by:
[tex]df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 1 \cdot dx + 2 \cdot y \cdot dy[/tex]

The most common way to represent a function geometrically is to plot the constant surfaces of the function. The covector represents the rate of change in a function at a point. The geometric symbol that I have seen to represent a covector is two parallel lines with a '^' through one of the lines. The parallel line simply represent the local rate of change of the function [tex]f(x,y)[/tex], and the '^' is used to indicate the direction of increase for the function. As with arrows for a vector field, these parallel line symbols can be scattered throughout space to represent the local change in the value of the function. The parallel lines should approximately line up with the constant surfaces of the function at the point where the covector is represented. I often find it easier to think of the constant surfaces of the function as the geometric representation of the covector FIELD since this is more cohesive and conveys the same information. Note that since the covector is the result of a derivative of the function [tex]f(x,y)[/tex], the constant value [tex]c[/tex] in the [tex]f(x,y)[/tex] equation above is eliminated. This just means that while the constant surfaces of the function provide a geometric characterization of a covector field, the specific value the constant surfaces may differ by a constant value. Not every covector arises as the differential of a function, so this picture is not always valid, but the mental picture of a constant surfaces of a function as the geometric representation of a covector field has some hueristic benefit in my opinion.

A vector field arises by taking the derivative of a set of parameterized curves. For example, suppose we start with the

curves [tex](x(t),y(t))=(t+x_0 , y_0)[/tex]. The associated vector field is found by taking the derivative with respect to t:

[tex] \bar{v} = \frac{dx}{dt} \frac{\partial}{\partial x} + \frac{dy}{dt} \frac{\partial}{\partial y} = 1 \cdot \frac{\partial}{\partial x}[/tex]

Typically the vector field is pictured as as a set of arrows tangent to the curve pointing in the direction that [tex]t[/tex] increases. However, it is just a valid, and sometimes useful to think of the parameterized curve as the geometric representation of the vector field. I tend to think of it as a (one dimensional) curve with tick marks that indicate even increments in [tex]t[/tex]. However, because the vector field arises from taking a derivative, a constant of integration is lost, or in other words, the location of [tex]t=0[/tex] is not defined by the vector field. So while we can picture a curve with tick marks indicating the rate of change of [tex]t[/tex], as a geometric representation of the vector field, no specific value is assigned to the tick marks. Note that the whole (x,y) space is filled with these parameterized curves each with no intrinsic origin. This somtimes reminds me of raw spaghetti when you pour it from the box, it is a bunch of lines packed together, but not tied to each other (i.e. no origin defined for each curve that ties the parameter on each curve to the parameter on the curves nearby).

Now, with the picture of a covector as a function defined in space, and of a vector as a set of curves filling space, the original question may be addressed:

I have trouble in understanding how we get from a given vector to its associated one form.
This is equivalent to saying 'I have trouble understanding how we get from a given set of parameterized curves to THE associated function' or to saying 'how do I get from this function to ITS associated set of parameterized curves'. Functions and parameterized curves are objects that have some similarities and some differences. You SHOULD have a little trouble relating vectors to covectors in a UNIQUE fashion. A function naturally defines a set of constant surfaces in space, but not a one dimensional path through space, like parameterized curves. The parameterized curves, with no fixed origin defined for each curve, aren't tied together to create constant surfaces in the same way as a function. One must nail down one fixed value (the constant of integration or initial condition) for each curve, then since the rate of change of the parameter [tex]t[/tex] is known, the rest of the values for that curve in space are fixed. Once an origin is fixed for a curve and all the nearby curves, the value [tex]t[/tex] can be combined for all the curves to define a function. However, since there are an infinite number curves that fill a space of more than one dimension, there are an infinite number of ways to choose the origins [tex]t=0[/tex] for the set of parameterized curves, so there is no UNIQUE function associated with a vector field.

The way Boothby puts this in 'An Introduction to Differentiable Manifolds and Riemannian Geometry' is that while vectors and covectors are isomorphic (which roughly means that the vector spaces for vectors and covectors are the same size at each point in the domain and can have compatible algebraic operators defined for them), there is no NATURAL isomorphism (i.e. there is no NATURAL way to map a vector (field) to a UNIQUE covector (field)).

While we cannot find a UNIQUE isomorphism between a vector (field) and a covector (field), you can find functions such that the parameter [tex]t[/tex] for a vector field change at the same rate as the constant surfaces of some set of associated functions. This condition is met for vector fields and covector fields which satisfy the following equation:

[tex]\bar{v}(f)=\langle \bar{v} ,df \rangle = \frac{df}{dt} = \frac{dx}{dt} \frac{\partial f}{\partial x} + \frac{dy}{dt} \frac{\partial f}{\partial y} = 1 [/tex]

If [tex] \frac{df}{dt} = 1[/tex] then the function f and the parameter t change at the same rate. The point is that the function f (and the associated covector field) are not UNIQUELY associated with the vector field because functions and parameterized curves are different (but related) kinds of objects.
 
Last edited:
Let V be a vector space over k with basis {e_i}.
The space of one form is a vector space { f:V -> k | linear }.

Define an inner product < , > by <e_i, e_j> := {Kronecker delta ij}

Then given element v of V, we define the associated 1-form as <v, > (or < ,v>).

Remark: Let us denote by w_i the 1-form associated to e_i. Then the 1-form associated to v = {sum}v_i*e_i is {sum}v_i*w_i. (* denotes multiplication.)
 
39
0
I appreciate the clarification. A metric (or inner poduct as shown above) can be used to uniquely relate a vector to a covector. However, it should be noted that the unique map between vectors and covectors is not considered a natural isomophism because a metric is additional structure imposed on a manifold rather than inherent to the manifold.

Sometimes the map induced by the metric is what you want. For example, in physics it is common represent gravity as a potential function [tex]f=mgy[/tex] m-mass, g-accelleration of gravity, y-height of an object or constant potential surfaces. The differential of this is [tex]df=mg dy[/tex]. Using the metric, one can relate the covector field representing the change in the gravitational potential to the direction a force is applied to the object. For newtonian mechanics this is easy since [tex]dy[/tex] maps to the unit vector[tex]\bar j[/tex] giving [tex]mg \bar j[/tex]. However in many cases there is no obvious metric. For example what metric do you use for a space that includes a distance coordinate, a temperature coordinate, and a time coordinate. Vector fields and covector fields (or functions) can be defined on a manifold and a natural inner product exists even when there is no metric defined. However there is not a UNIQUE isomorphism between vector fields amd covector fields based on this natural inner product.
 

Related Threads for: How to relate a one form to a vector.

  • Posted
Replies
16
Views
7K
  • Posted
2 3
Replies
71
Views
26K
Replies
25
Views
3K
Replies
9
Views
929
Replies
4
Views
1K
  • Posted
Replies
4
Views
2K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
11
Views
877

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top