Vectors, one-forms and gradients

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  • #1
Neoma
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I'm currently working through Schutz's "A first course in general relativity" as a preparation for a graduate course in General Relativity based on Carroll's notes. I'm a little confused about vectors, one-forms and gradients.

Schutz says the gradient is not a vector but a one-form, because it maps vectors into the reals in a linear way and explains that you should represent a gradient (and a one-form in general) by a series of surfaces. When you contract a one-form with a vector, the number you get is the number of surfaces the vector crosses.

If this was all I knew about the subject, it would be quite clear, however, my advanced calculus says the gradient is a vector pointing in the direction of fastest increase and it's a vector that's, when evaluated at a point p, is perpendicular to the tangent plane of the surface at point p. Schutz says there's nothing really wrong with this definition, because in normal Euclidean space, vectors and one-forms are the same, but how can something be both a vector (an arrow) and a one-form (a series of surfaces)?

Later on, Schutz says that a vector can also be seen as a linear map from one-forms into the reals (Carroll also says the dual of a dual vector space is the original vector space), so this way it seems whether to call something a vector or a one-form is totally arbitrary (as long as you do so in a consistent way), whether you're dealing with Euclidean space or not... :grumpy:

When I turn to other texts I get confused only more, for example, in some text on Differential Geometry a vector is defined as a linear operator on a function space that produces a real number...? I find this a strange definition, wouldn't this imply that a function is actually a one-form (and a vector, since it's totally up to you what to call the original vector space and what it's dual).

I think it's all equivalent in some peculiar way, but I really don't see how...
 
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  • #2
Berislav
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Schutz says there's nothing really wrong with this definition, because in normal Euclidean space, vectors and one-forms are the same, but how can something be both a vector (an arrow) and a one-form (a series of surfaces)?
Think about the size of the surfaces. By Schutz's definition the number you get when you contract is the number of surfaces that the vector passes. This would imply that you would only get rational numbers, unless the surfaces were infintesmal. Now how can you define a surface in Euclidian space? By a vector peripendicular to it! Then you take the product of the two vectors and get a number.

Schutz says that a vector can also be seen as a linear map from one-forms into the reals (Carroll also says the dual of a dual vector space is the original vector space),
Yes. Duals of tensors are those that contract the tensors into the original numbers.

so this way it seems whether to call something a vector or a one-form is totally arbitrary (as long as you do so in a consistent way),
Read on, when you get to tensors everything will be more clear.
 
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  • #3
quetzalcoatl9
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the gradient, [tex] \vec{\nabla f}[/tex] is a vector. it has vector components [tex]\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}[/tex]

whereas [tex]df[/tex] is a one-form with covector components [tex]\frac {\partial f}{\partial x^i}[/tex]

df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]:

[tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]

the gradient is a vector in the tangent space with the basis [itex]\frac{\partial}{\partial x^i}[/itex], defined as:

[tex]df(\vec{v}) = <{\nabla f}, \vec{v}>[/tex]

so the gradient vector is the vector whose inner product with [itex]\vec{v}[/itex] maps to the same real value as acting the one-form [itex]df[/itex] on the vector [itex]\vec{v}[/itex]

it is an important distinction because they live in entirely different spaces, being related by the metric tensor.
 
  • #4
quetzalcoatl9
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Neoma said:
Later on, Schutz says that a vector can also be seen as a linear map from one-forms into the reals (Carroll also says the dual of a dual vector space is the original vector space), so this way it seems whether to call something a vector or a one-form is totally arbitrary (as long as you do so in a consistent way), whether you're dealing with Euclidean space or not... :grumpy:

think of it this way.

[tex]\alpha: V \rightarrow \mathbb{R}[/tex]

so the one-form (a dual-vector) alpha takes a vector and maps it to a real. it is a linear functional.

[tex]\alpha(\vec{v}) = a[/tex]

so what would be my vector acting on the one-form be? simple, let's define it this way:

[tex]\vec{v}(\alpha) = \alpha(\vec{v}) = a[/tex]
 
  • #5
quetzalcoatl9
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Neoma said:
When I turn to other texts I get confused only more, for example, in some text on Differential Geometry a vector is defined as a linear operator on a function space that produces a real number...? I find this a strange definition, wouldn't this imply that a function is actually a one-form (and a vector, since it's totally up to you what to call the original vector space and what it's dual).

these are very good questions that you are asking.

yes, the vector can be viewed as a differential operator. what operator? how 'bout the directional derivative:

[tex]\vec{v}(f) = v^i \frac{\partial f}{\partial x^i}[/tex]

which is a real number. this is different than how you may normally think of the directional derivative. before, you probably thought about the function being fixed, and varying the vector to get the derivative. here we are saying that the tangent vector is fixed and that we can vary the function.

if we can vary the function, then we can define our vector as:

[tex]\vec{v} = v^i \frac{\partial}{\partial x^i}[/tex]

in order to see this, consider constructing a map from a tangent vector on one manifold to another, by way of a function that maps points from one manifold to the other. you will see that neither the coordinate system nor the function f matter in constructing the tangent vector, and that the two tangent vectors will be mapped by the Jacobian matrix, in essence a linear transformation between tangent vectors.
 
  • #6
Berislav
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df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]
[tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]
Schutz's book is basic text on General Relativity. It doesn't cover differential forms in detail. His text was the first I studied on the subject and I found it helpful to read it parallel with a book on differential geometry.
 
  • #7
matt grime
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there are many things in maths that are the same but which we wish to keep separate, and this is just one of them. one of the reasons is that the isomorphism from the tangent space at a point on a manifold to its cotangent space (tangent vectors and 1 forms resp) is that it is contravariant: duality reverses the direction of the arrows. these directions are important! so a map from manifolds M to N induces a map on the tangent vectors at each point from M to N but induces a map from the one forms of N to the one forms on M (hope i got that the right way round). if maps pushfoward on tangents and pull back on cotangents (one forms). years since i did this, apologies for any gross inaccuracies.

or another way of thinking is that whilst they may have the same underlying structure they do different things.
 
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  • #8
javanse
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quetzalcoatl9 said:
the gradient, [tex] \vec{\nabla f}[/tex] is a vector. it has vector components [tex]\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}[/tex]

whereas [tex]df[/tex] is a one-form with covector components [tex]\frac {\partial f}{\partial x^i}[/tex]

df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]:

[tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]

the gradient is a vector in the tangent space with the basis [itex]\frac{\partial}{\partial x^i}[/itex], defined as:

[tex]df(\vec{v}) = <{\nabla f}, \vec{v}>[/tex]

so the gradient vector is the vector whose inner product with [itex]\vec{v}[/itex] maps to the same real value as acting the one-form [itex]df[/itex] on the vector [itex]\vec{v}[/itex]

it is an important distinction because they live in entirely different spaces, being related by the metric tensor.
The problem is that the components of the gradient transform like a covector, and till now i have not seen a satisfy solution for this fact, to interprete the gradient as a vector or why he could exist at all as a vector. :yuck:
 
  • #9
nrqed
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quetzalcoatl9 said:
the gradient, [tex] \vec{\nabla f}[/tex] is a vector. it has vector components [tex]\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}[/tex]

whereas [tex]df[/tex] is a one-form with covector components [tex]\frac {\partial f}{\partial x^i}[/tex]

df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]:

[tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]

the gradient is a vector in the tangent space with the basis [itex]\frac{\partial}{\partial x^i}[/itex], defined as:

[tex]df(\vec{v}) = <{\nabla f}, \vec{v}>[/tex]

so the gradient vector is the vector whose inner product with [itex]\vec{v}[/itex] maps to the same real value as acting the one-form [itex]df[/itex] on the vector [itex]\vec{v}[/itex]

it is an important distinction because they live in entirely different spaces, being related by the metric tensor.


I am grateful for this post....

I usually hear saying that the gradient is a one-form!!!:yuck: Which is very confusing and it disagreement with what quetzalcoatl9 wrote here.

If I understand correctly, the components of df (in spherical coordiantes, say) would be [itex] { \partial f \over \partial r}, {\partial f \over \partial \theta}, {\partial f \over \partial \phi} [/itex].

However, I would never have called these the components of the gradient in spherical coordinates, of course! These are [itex] {\partial f\over \partial r}, {1 \over r } {\partial f \over \partial \theta}, {1 \over r \sin(\theta} {\partial f \over \partial \phi} [/itex]!

So the gradient (in the sense of what is being taught in introductory physics classes) *is* a vector.

Now I am sure someone will come along to tell me I have it completely wrong:surprised

But that's fine, my goal is to get it right!
 
  • #10
mathwonk
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this is a problem of duality, of having two ways to picture something.

imagine a heat source like a radiator. do you see it as a concentric family of spheres of constant temperature with the radiator at the center? or do you see it as stream of heated particles flowing radially away from the radiator?

the two points of view are dual and equivalent to each other. either you prefer to represent ypour phenomenon by the family of surfaces of constant temperature, or by the flow lines of greatest increase (or decrease) of temperature.
trictly speaking the word "gradient" seems to correctly refer to the directions of greatest increase, but for some reason this name has been taken over also by some people to refer to the level surfaces of the temperature function.

i.e. some people including myself, carelessly refer to the one form df as the "gradient of f", when really the gradient is the object dual to df. once one becomes familiar with the stuff, it becomes more tempting to be careless in ones use of language.
 
  • #11
nrqed
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quetzalcoatl9 said:
the gradient, [tex] \vec{\nabla f}[/tex] is a vector. it has vector components [tex]\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}[/tex]

whereas [tex]df[/tex] is a one-form with covector components [tex]\frac {\partial f}{\partial x^i}[/tex]

df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]:

[tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]

the gradient is a vector in the tangent space with the basis [itex]\frac{\partial}{\partial x^i}[/itex], defined as:

[tex]df(\vec{v}) = <{\nabla f}, \vec{v}>[/tex]

so the gradient vector is the vector whose inner product with [itex]\vec{v}[/itex] maps to the same real value as acting the one-form [itex]df[/itex] on the vector [itex]\vec{v}[/itex]

it is an important distinction because they live in entirely different spaces, being related by the metric tensor.


I know this is all old stuff and knowledgeable people here find this type of questions tedious and incredibly boring, but I really want to understand diff geometry (I will go completely insane if I don't sort out all the notation and terminology!!!!) and I will be posting tons of *really* stupid questions, mostly about notation. I apologize ahead of time for that and if that kind of questions make you cranky, just ignore me:shy:


Firstset of stupid questions: In the book by Baez and Muniain, they write (p.41)

[tex] df(v) = vf [/tex]
(where vf is there notation for the directional derivative of the scalar field f along the vector v)

and then they write below Think of this as a slick way to write [itex] \nabla f \cdot v = vf[/itex]).

Ok. Here are my stupid questions.

What is df, explicitly? Let's work in spherical coordinates (so that the metric is not trivial!!)

I thought that df was simply [tex] {\partial f \over \partial r } dr + {\partial f \over \partial \theta} d \theta + {\partial f \over \partial \phi} d \phi [/tex]. Is that correct or not?

Let's say v is [tex] v_r {\partial \over \partial r} + v_\theta {\partial \over \partial \theta} + v_\phi {\partial \over \partial \phi} [/tex].

Then I would think that feeding v to df would simply give [tex] v_r {\partial f \over \partial r }+ v_\theta {\partial f \over \partial \theta} + v_\phi {\partial f \over \partial \phi} [/tex]

However, this is (at first sight) not at all what we learned in introductory maths the directional derivative of a scalar field was, viz. [tex]v_r {\partial f \over \partial r }+ v_\theta {1 \over r} {\partial f \over \partial \theta} + v_\phi {1 \over r sin \theta} {\partial f \over \partial \phi} [/tex]

So I am missing something surely obvious :mad: .

One thing that bothers me is that the components of a vector field using the "diff geometry language" uses for basis the partial derivatives. But that means that these components cannot be the same as the components of a vector field using the }. "intro maths approach" with basis [itex] {\hat e_r}, {\hat e_{\theta} }, {\hat e_{\phi}} [/itex], just by looking at the dimensions ([itex] {\hat e_r} [/itex] and [itex] {\partial \over \partial r}[/itex] don'thave the same dimensions).
So what am I missing?

Another thing that bothers me is that the expression df(v), thought as feeding a vector to a differential form is supposed to be completely metric independent (right?). But looking at the standard (i.e. "intro maths") expression for the directional derivative of a scalar field, it does look (at least at first sight) as if it does depend on the metric, through the presence of the 1/r and 1/(r sin theta) terms. What am I missing?


I know this is all extremely basic and just a question of notation, but it is impossible to focus on the maths and the physics if the notation is not clear.

Thanks to anyone who will be willing to help!!!

Patrick
 
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  • #12
George Jones
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One thing that bothers me is that the components of a vector field using the "diff geometry language" uses for basis the partial derivatives.

It doesn't have to.

But that means that these components cannot be the same as the components of a vector field using the }. "intro maths approach" with basis [itex] {\hat e_r}, {\hat e_{\theta} }, {\hat e_{\phi}} [/itex], just by looking at the dimensions ([itex] {\hat e_r} [/itex] and [itex] {\partial \over \partial r}[/itex] don'thave the same dimensions).

"Diff geometry language" is not restricted to coordinate bases, so it also can use this basis. If [itex]\left\{v_{r}, v_{\theta}, v_{phi}\right\}[/itex] are the components of [itex]v[/itex] with respect to an orthonormal basis, and [itex]\left\{v'_{r}, v'_{\theta}, v'_{phi}\right\}[/itex] with respect to a coordinate basis, then everything should work out OK.
 
  • #13
nrqed
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George Jones said:
It doesn't have to.



"Diff geometry language" is not restricted to coordinate bases, so it also can use this basis. If [itex]\left\{v_{r}, v_{\theta}, v_{phi}\right\}[/itex] are the components of [itex]v[/itex] with respect to an orthonormal basis, and [itex]\left\{v'_{r}, v'_{\theta}, v'_{phi}\right\}[/itex] with respect to a coordinate basis, then everything should work out OK.
Thanks.

Then I have to understand "coordinate basis" vs "orthonormal basis". I have always seen the partial derivatives used as basis for vectors in differential geometry, I thought that it was the only thing people worked with, at a fundamental level (i.e. if a metric is not introduced).

I am confused because it seems to me that in order to define an orthonormal basis, one needs to introduce a scalar product, which means a metric must be introduced.

And then what is the explicit relation between the components of the vectors in the different basis?

Thanks!
 
  • #14
nrqed
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George Jones said:
It doesn't have to.



"Diff geometry language" is not restricted to coordinate bases, so it also can use this basis. If [itex]\left\{v_{r}, v_{\theta}, v_{phi}\right\}[/itex] are the components of [itex]v[/itex] with respect to an orthonormal basis, and [itex]\left\{v'_{r}, v'_{\theta}, v'_{phi}\right\}[/itex] with respect to a coordinate basis, then everything should work out OK.

Oh...now it is slowly starting to sink in. Now I *think* I understand a bit better. My mistake is that I was assuming that the components of a vector field in the basis [itex] {\partial \over \partial_i} [/itex] were the same as the components in the basis of the usual unit vectors.
Of course it makes sense that they will not be (:blushing: ). Now, it must be that the relation between the two involves the metric (or its inverse). How does one establish the relation between the components in the two basis?

Patrick
 
  • #15
George Jones
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nrqed said:
Then I have to understand "coordinate basis" vs "orthonormal basis". I have always seen the partial derivatives used as basis for vectors in differential geometry, I thought that it was the only thing people worked with, at a fundamental level (i.e. if a metric is not introduced).

I am confused because it seems to me that in order to define an orthonormal basis, one needs to introduce a scalar product, which means a metric must be introduced.

Yes, a metric is needed. Differential geometry sometimes deals with manifolds that don't have metrics, but it is not restricted to such manifolds.

And then what is the explicit relation between the components of the vectors in the different basis?

Let's continue on with this example, i.e., the [itex]\mathbb{R}^3[/itex] with metric for which [itex]\right\{\partial / \partial x, \partial / \partial y, \partial / \partial x \right\}[/itex] is an orthonormal basis. Then, e.g.,

[tex]\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial}{\partial z}[/tex].

Dot this [itex]\partial / \partial r[/itex] with itself, and use the orthonormality of the coordinate basis to find the (squared) length of [itex]\partial / \partial r[/itex]. Divide [itex]\partial / \partial r[/itex] by its length to find [itex]\hat{e}_r[/itex]. Do the same for the other two coordiante.

In this case we're a bit lucky in that the coodinate vectors are othogonal, but it doesn't always have to be this way.

Now,

[tex]
v = v_{r} \hat{e}_{r} + v_{\theta} \hat{e}_{\theta} + v_{\phi} \hat{e}_{\phi} = v'_{r} \frac{\partial}{\partial r} + v'_{\theta} \frac{\partial}{\partial \theta} + v'_{\phi} \frac{\partial}{\partial \phi}.[/tex]

Writing either the hatted basis vectors or the unhatted basis vectors in terms of the other set gives the relationship between the compoments.
 
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  • #16
nrqed
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George Jones said:
Yes, a metric is needed. Differential geometry sometimes deals with manifolds that don't have metrics, but it is not restricted to such manifolds.
Thanks by the way for being patient withme:redface: I appreciate.

I understand this.
I guess that I was saying this because it seemed to me (from what I was reading in Baez and Muniain) that one important lesson they wanted to make was that the directional derivative was defined before introducing any metric (which they discuss for the first time 32 pages later than the above quote). So I wanted to see how get to the usual expression of the directional derivative without invoking the metric.

I am guessing that it will cancel out when I see the derivation in full (in the sense that there will be the metric contracted with its inverse).


Let's continue on with this example, i.e., the [itex]\mathbb{R}^3[/itex] with metric for which [itex]\right\{\partial / \partial x, \partial / \partial y, \partial / \partial x \right\}[/itex] is an orthonormal basis. Then, e.g.,

[tex]\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial}{\partial z}[/tex].

Dot this [itex]\partial / \partial r[/itex] with itself, and use the orthonormality of the coordinate basis to find the (squared) length of [itex]\partial / \partial r[/itex]. Divide [itex]\partial / \partial r[/itex] by its length to find [itex]\hat{e}_r[/itex]. Do the same for the other two coordiante.

Ok, but my dilemma is the following. I don't know how to take the scalar product between a pair of [itex] {\partial \over \partial_i} [/itex]. I can guess what the result should be (<partial_x,partial_x> =1 and so on) but I don't want to make any guess.

so that becomes my question: how is the scalar product between partial derivatives (I guess these are what you call coordinate bases?) is defined?

That's my stumbling block for now.

Thanks a lot!!
 
  • #17
nrqed
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nrqed said:
Thanks by the way for being patient withme:redface: I appreciate.

I understand this.
I guess that I was saying this because it seemed to me (from what I was reading in Baez and Muniain) that one important lesson they wanted to make was that the directional derivative was defined before introducing any metric (which they discuss for the first time 32 pages later than the above quote). So I wanted to see how get to the usual expression of the directional derivative without invoking the metric.

I am guessing that it will cancel out when I see the derivation in full (in the sense that there will be the metric contracted with its inverse).




Ok, but my dilemma is the following. I don't know how to take the scalar product between a pair of [itex] {\partial \over \partial_i} [/itex]. I can guess what the result should be (<partial_x,partial_x> =1 and so on) but I don't want to make any guess.

so that becomes my question: how is the scalar product between partial derivatives (I guess these are what you call coordinate bases?) is defined?

That's my stumbling block for now.

Thanks a lot!!

Just to add to my previous post.

First let me emphasize again that I am not trying to be difficult. I am just trying to *really* understand (as opposed to doing calculations and getting sometimes the correct results without feeling that I am really understanding what I am doing).

Just to add a comment: I can understand how to get from the [itex] \partial_r, \partial_\theta, \partial_\phi [/itex] to the [itex] \partial_x, \partial_y,\partial_z [/itex] of course.

What bothers me is that once I have the [itex] \partial_x, \partial_y,\partial_z [/itex], I have to switch back to thinking of them as the usual i,j,k and to do scalar products and finding magnitudes. It bothers me for two reasons.

First, I know that we are in R^3 and that there *is* a metric around, but I thought that the point of working with the "partial derivative basis" was that they did not require a metric to be define. So I feel strange about calculating scalar products directly in terms of these.

Second, if I am allowed to simply make the correspondence [itex] \partial_x \rightarrow {\hat i} [/itex] and so on , why was it possible to simply make the correspondence [itex] \partial_r \rightarrow {\hat r} [/itex] and so on?

The answer is surely that there is a non trivial step in going from the partial derivative as a basis to unit vectors....But that step becomes probably so trivial in the case of the x,y,z basis that it's not even mentioned. But to really understand what is going on I would need to see that step clearly explained, I am afraid.

I hope this helps clarify my confusion!

Thanks!

Patrick
 
  • #18
George Jones
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First, I know that we are in R^3 and that there *is* a metric around, but I thought that the point of working with the "partial derivative basis" was that they did not require a metric to be define. So I feel strange about calculating scalar products directly in terms of these.

A bit of care has to taken when using the term "scalar product." Modern differential geometry emphasizes that 1-forms exist without metrics, and that a 1-form acts on a vector to give a scalar. All of this can be done interms of coordinate bases. If a metric does happen to exist, then there is a natural correspondence between getting a scalar by letting a 1-form act on a vector, and by letting the metric act on two vectors.

Second, if I am allowed to simply make the correspondence [itex] \partial_x \rightarrow {\hat i} [/itex] and so on , why was it possible to simply make the correspondence [itex] \partial_r \rightarrow {\hat r} [/itex] and so on?

It is possible!

In order to have explicitly a manifold with metric, the metric must be defined in some way. The manifold R^3 is turned into normal Euclidean space by defining a metric for which the basis vectors of a Cartesian coordinate are orthonormal. This is the definition of the metric.

Defining the basis vectors associated with spherical coordinates to be orthonormal results in a completely different metric. In this case, the relationship between the two metrics is a (variable) change of scale. For this metric, e.g., a line of constant r and theta and changing phi is a geodesic.

Usually, metrics that are useful, either mathematically or physically, are chosen. In the case of R^3, the first metric above is almost always chosen, and this is the metric that Baez and Munian chooses implicitly on the top of page 74.

The answer is surely that there is a non trivial step in going from the partial derivative as a basis to unit vectors....

Just divide a vector by its metric-given length.

But that step becomes probably so trivial in the case of the x,y,z basis that it's not even mentioned. But to really understand what is going on I would need to see that step clearly explained, I am afraid.

You might not like my answer to this. The basis vectors associated with the {x, y, z} coordinate system are orthonormal ... by definition.

This defines the usual metric.
 
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  • #19
nrqed
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George Jones said:
A bit of care has to taken when using the term "scalar product." Modern differential geometry emphasizes that 1-forms exist without metrics, and that a 1-form acts on a vector to give a scalar. All of this can be done interms of coordinate bases. If a metric does happen to exist, then there is a natural correspondence between getting a scalar by letting a 1-form act on a vector, and by letting the metric act on two vectors.
Thanks again George. I have to leave shortly unfortunately but I hope you will keep helping me out!

I understand the above, yes.
And, unles I am mistaken, vector fields also exist without a metric. This is my main source of confusion. We have a vector field that we write using [itex] \partial_r...[/itex] and then we want to write it using [itex] {\hat e_r}...[/itex].

If we go from the [itex] \partial_r...[/itex] to [/itex] \partial_x ...[/itex], there is no need to invoke the metric (unless I am mistaken).

What confuses me is the role of the [itex] {\hat e_r}...[/itex] basis (or the i,j,k or any other unit vector basis. In *that* case, a metric is needed to define them, am I right? (I guess the fact that they are of unit length is a giveaway!). So they are fundamentally different than the partial derivative basis in that regard. This is the first point I want to clarify because I used to see all those basis as just corresponding to a differnt choice of basis without invoking a metric.


My second point is that I am still wondering about scalar product between partial_x and partial_y, say. Again, these basis are defined independently of a metric, right? But the scalar product involves a metric. So it seems to me that there is a nontrivial step there that involves introducing a metric before the scalar product is taken. Am I correct? What I mean is that I could have a 3-D manifold with coordinates x,y,z on which no metric is defined and I still would work with partial_x, etc. But then I would never be able to get to a basis with the usual i,j,k or e_r, e_theta, r_phi, is that right??

I guess that I want to see the "big picture".

I hope this makes sense. I really appreciate yoru help. I have to run but I am looking forward to reading your comments!!
 
  • #20
mathwonk
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sighhh... i try to explain these things in down to Earth terms, and yet the students seem to prefer computational details and technical symbols to my plain explanations of what is really going on. [voice in wilderness]
 
  • #21
nrqed
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mathwonk said:
sighhh... i try to explain these things in down to Earth terms, and yet the students seem to prefer computational details and technical symbols to my plain explanations of what is really going on. [voice in wilderness]

Maybe it's because I am too stupid but I need to see both explicit calculations and discussions of the general concepts when I learn a new framework.

I don't know why but it seems as if asking questions about differential geometry always bring comments of the type "this is obvious, you are wasting my time" :mad:

EDIT: I want to add that I have appreciated immensely your comments in the past and that it's quite possible that you have already explained this to me before. I would not be surprised if that has happened. But I find myself spending several months at a time without having the opportunity to think about this subject (unfortunately) because I sometimes have over 20 contact hours of teaching per semester. This, in combination with the fact that I never reached a point where everything seemed crystal clear to me and the fact that I also spend my free time time learning other stuff (I spent the last month learning Python and deciphering code written in Fortran 90 and Python that performs calculations in lattice QCD) explains why I may find myself asking the same or similar questions a few months apart. Maybe that makes me stupid, I don't know. But it does not mean that I was not paying attention the previous time.

Regards
 
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  • #22
George Jones
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mathwonk said:
sighhh... i try to explain these things in down to Earth terms, and yet the students seem to prefer computational details and technical symbols to my plain explanations of what is really going on. [voice in wilderness]

I agree that post #10 by a certain mathwonk was very nice, and we will eventually get back to it; in the meantime, cover your eyes.
 
  • #23
George Jones
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nrqed said:
If we go from the [itex] \partial_r...[/itex] to [itex] \partial_x ...[/itex], there is no need to invoke the metric (unless I am mistaken).

Right - no metric required.

What confuses me is the role of the [itex] {\hat e_r}...[/itex] basis (or the i,j,k or any other unit vector basis. In *that* case, a metric is needed to define them, am I right? (I guess the fact that they are of unit length is a giveaway!).

Right.

So they are fundamentally different than the partial derivative basis in that regard. This is the first point I want to clarify because I used to see all those basis as just corresponding to a differnt choice of basis without invoking a metric.

Right, saying, e.g., which bases are orthonormal requires the use of a metric. In general relativity, a tetrad is an orthonormal basis, and, in many important spacetimes, tetrads are not coordinate bases.

My second point is that I am still wondering about scalar product between partial_x and partial_y, say. Again, these basis are defined independently of a metric, right?

Right.

But the scalar product involves a metric.

Right.

So it seems to me that there is a nontrivial step there that involves introducing a metric before the scalar product is taken. Am I correct? What I mean is that I could have a 3-D manifold with coordinates x,y,z on which no metric is defined and I still would work with partial_x, etc.

Right.

But then I would never be able to get to a basis with the usual i,j,k or e_r, e_theta, r_phi, is that right??

Since a vector space is closed under scalar multiplication, you, without a metric, could work with, e.g., the vector

[tex]\frac{1}{r} \partial_\theta,[/tex]

but you couldn't assign a length to this vector, and you coudn't say which vectors are orthogonal to this vector.

In our example, Cartesian coordinates are used to define the metric. In other words, define [itex]g[/itex] by:

1) [itex]g[/itex] is bilinear;

2) [itex]g \left( \partial_{i} , \partial_{j} \right) = \delta_{ij}.[/itex]

This defines the scalar product of any two vectors.
 
  • #24
nrqed
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George Jones said:
Right - no metric required.



Right.



Right, saying, e.g., which bases are orthonormal requires the use of a metric. In general relativity, a tetrad is an orthonormal basis, and, in many important spacetimes, tetrads are not coordinate bases.



Right.



Right.



Right.



Since a vector space is closed under scalar multiplication, you, without a metric, could work with, e.g., the vector

[tex]\frac{1}{r} \partial_\theta,[/tex]

but you couldn't assign a length to this vector, and you coudn't say which vectors are orthogonal to this vector.

In our example, Cartesian coordinates are used to define the metric. In other words, define [itex]g[/itex] by:

1) [itex]g[/itex] is bilinear;

2) [itex]g \left( \partial_{i} , \partial_{j} \right) = \delta_{ij}.[/itex]

This defines the scalar product of any two vectors.

Ok, then it all fits perfectly together. Thank you George!

Now I realize what the source of my confusion was. What I had not realized is that some bases for vectors require the introduction of a metric and some other bases don't. *This* is really the reason for my confusion. In my mind, all bases were metric independent . Of course, in hindsight, the very name "orthonormal basis" is a dead giveaway. But my problem was that I knew that vector fields exist independently of a metric and I was (subconsciously) assuming that any representation of a vector field (in the sense of giving its components) had to be independent of a metric.

Now this is much clearer!

In regard to Mathwonk's comment of post # 10. Yes, that WAS *very* instructive and useful to me!!!! I finally understood that the gradient of introductory math was in fact a vector and I understood, because of Mathwonk, that more knowledgeable people are used to switch between the form df and its "associated" (through the metric) vector nabla f. And that cleared up a major source of confusion (and frustration) for me so I highly appreciate that post! (my sincere apologies for not replying with a thank you post, I really should have :frown: :frown: ).

But this then gave me the impetus to actually go beyond discussing the concepts and actually sitting down and working out all the details in a specific case (which is the true test of understanding a new formalism, imho). And this is when I ran into a problem which led me to my post #11.

I hope that this did not give the impression that I had ignored Mathwonk's post!!!!
Notice that my question of post #11 is not the same as the question that led to Mathwonk's post #10! Mathwonk clarified for me the disticntion between the one-form df and the vector nabla f. In post #11 I started to use that knowledge to explicitly calculate a *directional derivative*. This was no longer an issue of distinguishing df and nabla f. It was an attempt at working out explictly df(v) and understanding why this comes out independent of a metric.

Again, if Mathwonk felt that I had ignored his very appreciated post (and there were other posts in other threads that were illuminating), I apologize sincerely!

Patrick
 
  • #25
mathwonk
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thank you george and nrqed, for the feedback and encouragement. i am very insecure and appreciate it a lot.

best,

mwonk.
 
  • #26
nrqed
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mathwonk said:
thank you george and nrqed, for the feedback and encouragement. i am very insecure and appreciate it a lot.

best,

mwonk.
:frown: Then I reiterate my apologies! I have to say that I feel completely out of depth in pure maths stuff and I often feel dumb for all the questions that I keep asking myself! So I guess I feel insecure myself.

I am a bit surprised by your feelings because you are among the most knowledgeable and helpful persons to have helped me understand better diff. geometry!! There was another thread (which I can't seem to find anymore...are threads in the diff geometry forum regularly deleted? Are they archived somewhere?) in which you posted a few replies to my questions which were extremely clear and illuminating to me. You were explaining things in a very clear way, just at the level that I needed. I was really grateful for your input!

Best regards

Patrick
 
  • #27
mathwonk
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silly remark about terminology: to my knowledge, actually the subject of differential geometry does require a metric (or something equivalent that allows one to define curvature). this is what distinguishes it from differential topology.

i.e. the two differential manifolds, the unit circle, and the circle of radius 2, are diffeomorphic, but have different curvature, hence are not the same in differential geometry.


this is just language, but i think it is pretty standard usage.

happy 4th, or 7th, as the case may be now.
 
  • #28
nrqed
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mathwonk said:
silly remark about terminology: to my knowledge, actually the subject of differential geometry does require a metric (or something equivalent that allows one to define curvature). this is what distinguishes it from differential topology.
oh! But there are things that can be defined without a metric. I mean, one can introduce differential forms and vector fields without a metric, am I worng? One can then "feed" vector fields to the one-forms without introducing a metric, no? One can even calculate a directional derivative of a scalar function df(v) without introducing a metric, no?
I thought this was one of the main points of differential geometry, that there were quantities that can be defined without introducing a metric and that one of th eimportant lessons was to distinguish what requires the introduction of a metric and what doesn't. Or maybe you are saying that when one computes the directional derivative, say, one is not really doing differantial geometry but differential topology? Maybe that's what I was missing...I had not realized that there was something called differential topology !

That sounds like an important thing for me to realize!!
 
  • #29
nrqed
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mathwonk said:
this is a problem of duality, of having two ways to picture something.

imagine a heat source like a radiator. do you see it as a concentric family of spheres of constant temperature with the radiator at the center? or do you see it as stream of heated particles flowing radially away from the radiator?

the two points of view are dual and equivalent to each other. either you prefer to represent ypour phenomenon by the family of surfaces of constant temperature, or by the flow lines of greatest increase (or decrease) of temperature.
trictly speaking the word "gradient" seems to correctly refer to the directions of greatest increase, but for some reason this name has been taken over also by some people to refer to the level surfaces of the temperature function.

i.e. some people including myself, carelessly refer to the one form df as the "gradient of f", when really the gradient is the object dual to df. once one becomes familiar with the stuff, it becomes more tempting to be careless in ones use of language.
It was very good for me to see this expressed out loud!! Because I would read Schutz too (like the OP) or other sources and I would also see "df" referred to as the divergence. And this is extremely confusing when learning the subject.

If I am used to think of something as a vector and one tells me that it's actually a one-form, then there are two possible meanings:

Do they mean this literally, i.e. that what I am used to think of as [itex] A_1 {\hat e_1} + A_2 {\hat e_2} + A_3 {\hat e_3} [/itex] is actually [itex] A_1 dx_1 + A_2 dx_2 + A_3 dx_3 [/itex], that its the components are actually the components of a one-form

OR do they mean that what I am used to is a true vector and using the metric I can obtain a one-form corresponding to it and then think in terms of this dual picture.

The two possibilities are inequivalent so it's extremely confusing.

From what you wrote (and what Quetzalcoatl9) wrote, when people say that the divergence is a one-form, it's in the second sense above (that the divergence we learned about in elementary maths is really a vector but that one can use the metric to get a one-form out of it). Whereas for a while I thought that people meant it literally (in the first sense above), that really the components of the divergence we are used to were truly components of a one-form. This is very confusing.

I understand what you are saying about passing from one view to the other but I still find it confusing to refer to df as the divergence. (instead of, maybe, calling it "the one-form dual to the divergence vector"). I would expect physicists to be careless with terminology like this but not mathematicians :shy:

And I think that the distinction is important given that one could even work with a space in which no metric would be defined, in which case df would still be defined, but one could not use the metric to associate df to a vector. Then would one still call df the divergence?

Thanks for the comments! You can see why newbies like me keep having those annoying questions and are confused when learning the subject!

Regards

Patrick
 
  • #30
mathwonk
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you are mistakenly using the term differential geometry for those concepts that are part of pure differential manifolds and differential topology.


people who do not know the subject link them all together, they think anything involving a manifold is differential geometry. just look at some of the books on the topic. i admit it is a bit confusing to beginners.

but i8 think sopivak e.g. amkres it clear that his first volume is mere backgound material on differential manifolds and that the second volume is where the geometry, i.e. the metrics, begin.


i.e. forms and tangent and cotangent bundles have nothing to do specifically with differenmtial geometry. that is pure basic theory of differential calculus on manifolds. there is no metric hence no curvature hence no geometry.


without a metric teo diff manifolds are isomorphic iff they are diffeomorphic, that is by definition differential topological equivalence.


as is aid beforew it is just language, but it is standard language.

if you cannopt tell a circle of radius one froma circle of radius 2, as you cannot from their differential structure, then you are not doing differential geometry.

differential geometry is the subject mostly advanced by gauss and riemann, dealing primarily with curvature, theorema egregium, and so on...

just look at the names of the thoperems in any course on diff geom, you will see they cannot be stated or proved usually withouta metric.
 
  • #31
mathwonk
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similarly galois theory is not field theory, but is discussed in the language of field theory.

galois theory means the study of symmetries of algebraic field extensions.
 
  • #32
mathwonk
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let me be more rpecise. calculus on, manifolds is not differential geometry but is the language in which differential geometry is discussed.


since most moidern books on differential geometry do not wish toa ssume he reader ahs studied calculus on manifolds, they tend to begin with a preliminary chapter on differential manifolds so mthe reader will understand the geometry when they get to it.

e.g. spivak has an entrie first volume just on differnetial manifolds and starts the geometry in volume 2. noel j hicks has a one short chapter on differential manifolds and starts the geometry subtly in chapter 2. it is hard to notice but if you read carefully and think about it, his definition of covariant derivative of a vector field wrt another one, uses the existence of a standard basis for the tangent bundle of R^n, hence they define a metric in which they are unit vectors.

he does not mention it anywhere but there it is.

in the same way some books may not assume we know any topology or any set theory or linear algebra, and begin with preparatory chapters on those topics, but they are still not differential geometry.

so the chapters on calculus on manifolds, tangent bundles, forms, etc... in books on diff geom are really prerequisites for diff geom, which they think the reader does not have.


to see this more clearly, do not ask why books on diff geom contain treatents of calc on manifodls, but why books not on diff geom also have them. i.e. if this topic appears also in other books on other subjects then it is a common prereq not a part of diff geom rpoper.


e.g. the book by guillimen and pollack on diff top also has a thorough treatment of calc on diff manifolds, as does spivak's little book calculus on manifolds,

in bott - tu e.g. on diff forms in alg top, there does occur a use of metrics but diff geom is merely used to prove a theorem in topology. so the matter is further complicated since diff geom is also helpful in proving facts that in the end do not depend on the metric.

in the same way a metric on R^n can be used to prove facts that are purely topological.

but bott - tu make it clear what they are doing by saying: "the proof of this theorem will use a little diff geom" and they promptly introduce a metric, which after the proof they promptly throw out again.

milnors beautiful book on morse theory combines the two topics and shows how to use diff geom techniques to prove more subtle things about topology. e.g spaces which admit certain types of metrics have strong restrictions on their homotopy groups.

am i making it clear what the difference is? if there is no metric or curvature, or covariant derivative, or connexion, then by definition there is no diff geom.:biggrin:
 
  • #33
nrqed
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mathwonk said:
let me be more rpecise. calculus on, manifolds is not differential geometry but is the language in which differential geometry is discussed.


since most moidern books on differential geometry do not wish toa ssume he reader ahs studied calculus on manifolds, they tend to begin with a preliminary chapter on differential manifolds so mthe reader will understand the geometry when they get to it.

e.g. spivak has an entrie first volume just on differnetial manifolds and starts the geometry in volume 2. noel j hicks has a one short chapter on differential manifolds and starts the geometry subtly in chapter 2. it is hard to notice but if you read carefully and think about it, his definition of covariant derivative of a vector field wrt another one, uses the existence of a standard basis for the tangent bundle of R^n, hence they define a metric in which they are unit vectors.

he does not mention it anywhere but there it is.

in the same way some books may not assume we know any topology or any set theory or linear algebra, and begin with preparatory chapters on those topics, but they are still not differential geometry.

so the chapters on calculus on manifolds, tangent bundles, forms, etc... in books on diff geom are really prerequisites for diff geom, which they think the reader does not have.

Hi mathwonk...

wow, this really is an eye opener for me!

In my mind, I was making the distinction between "calculus involving a metric" and "calculus not involving any metric", but I did not know th ecorrect terminology and I thought that all of this was part of differential geometry!

What confused me, I guess, is the term "differential" in "differential geometry". That confused me because I though that the term "differential" implied "calculus", whether a metric was involved or not.
And I had never heard of "differential topology".

Please correct me if I am still wrong but if I understand correctly now, the key word is "geometry" which does imply a metric (right?). One can introduce differential forms and vector fields and calculate metric independent quantities like a directional derivative as part of differential topology (all concepts which I used to think, mistakenly, were part of differential geometry). But if one talks about differential geometry then a metric is necessarily involved.

Is this all correct?

to see this more clearly, do not ask why books on diff geom contain treatents of calc on manifodls, but why books not on diff geom also have them. i.e. if this topic appears also in other books on other subjects then it is a common prereq not a part of diff geom rpoper.
Ok. The few books I have looked at do not make the distinction clear, i.e. they do not say "so far the concepts introduced were concepts of differential topology. Starting from here, we are doing differential geometry". At least, if they do I had not been paying close enough attention.
e.g. the book by guillimen and pollack on diff top also has a thorough treatment of calc on diff manifolds, as does spivak's little book calculus on manifolds,

in bott - tu e.g. on diff forms in alg top, there does occur a use of metrics but diff geom is merely used to prove a theorem in topology. so the matter is further complicated since diff geom is also helpful in proving facts that in the end do not depend on the metric.
ok! That's an interesting point!!
in the same way a metric on R^n can be used to prove facts that are purely topological.
ok! Interesting
but bott - tu make it clear what they are doing by saying: "the proof of this theorem will use a little diff geom" and they promptly introduce a metric, which after the proof they promptly throw out again.
Ok! It's good to know because I am sure I would not have understood the subtlety if I had looked at this and not be warned.
milnors beautiful book on morse theory combines the two topics and shows how to use diff geom techniques to prove more subtle things about topology. e.g spaces which admit certain types of metrics have strong restrictions on their homotopy groups.
Sounds fascinating. I hope to be able one day to understand that level of maths.
am i making it clear what the difference is? if there is no metric or curvature, or covariant derivative, or connexion, then by definition there is no diff geom.:biggrin:
Yes, I get it now:redface: :redface:

I would appreciate it if you could tell me if everything I wrote above is correct or if I am still misunderstanding some things.

I appreciate these explanations very much!

Regards

Patrick
 
  • #34
masudr
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The problem is that the components of the gradient transform like a covector, and till now i have not seen a satisfy solution for this fact, to interprete the gradient as a vector or why he could exist at all as a vector. :yuck:

The satisfactory solution you are looking for is the chain rule.

So, if a basis (the co-ordinate basis) for our tangent space is

[tex]\partial_\mu = \frac{\partial}{\partial x^\mu}[/tex]

then how do we transform to another co-ordinate system? From ordinary calculus

[tex]\frac{d}{dy} = \frac{d}{dx}\frac{dx}{dy}[/tex]

and the corresponding rule for partial derivatives is

[tex]\frac{\partial}{\partial x^{\mu'}}=\sum_\mu \frac{\partial}{\partial x^\mu}\frac{\partial x^\mu}{\partial x^{\mu'}}[/tex]

So the basis vectors transform like this. The components must transform in the opposite way so that the actual vector itself

[tex]V=\sum_\mu V^\mu \partial_\mu[/tex]

does not change under a co-ordinate transformation.


To get the opposite result for covectors (or one-forms) we first define the co-ordinate basis for the cotangent space by requiring that our cotangent basis [itex]dx^\mu[/itex] satisfies

[tex]\partial_\mu dx^\nu = \delta^\mu_\nu[/tex]

and working out the above calculation.


I hope that answers your problem (which I gathered to be why vector/covector components transform as they should).
 
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  • #35
mathwonk
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I know you believe me, but just to illustrate how easy it is to find these definitions and explanations on the web, i searched for a few seconds under diff geom and found this:

(of course it helps if you already know the answer):
http://www.math.niu.edu/~rusin/known-math/index/53-XX.html [Broken]

Introduction

Differential geometry is the language of modern physics as well as an area of mathematical delight. Typically, one considers sets which are manifolds (that is, locally resemble Euclidean space) and which come equipped with a measure of distances. In particular, this includes classical studies of the curvature of curves and surfaces. Local questions both apply and help study differential equations; global questions often invoke algebraic topology.


also:
http://www.math.niu.edu/~rusin/known-math/96/diff.geom [Broken]

# What is Differential Geometry; how does it differ from differential topology?

The study of manifolds is Differential Topology; the study of
Riemannian manifolds is Differential Geometry. Each has become
incredibly robust, with many directions of research, fascinating
examples, and significant applications. The material is not easy, and
requires a good background in topology and analysis. Spivak's
"Comprehensive Introduction to Differential Geometry", a mere 3000
pages, is an excellent resource starting at the undergraduate level.

dave :blushing:
 
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