What Are Differential Forms and How Do They Apply in Differential Geometry?

[lethe]

Originally posted by turin
Holy crap!

Ya, uh, question. What is "symplectic form?"

a symplectic form is a nondegenerate closed 2-form. in the context of classical mechanics, it is related to the Poisson bracket (which is, as you know, antisymmetric)

This seems to contradict the definition of ∂_μf dx^μ as a vector. Is a vector the same thing as a scalar in math land?

no, in math land, vectors and scalars are, of course, different.

i think where we are getting confused is this: dx^\mu are the basis 1-forms. they are dual to the basis vectors \mathbf{\partial}_\mu which means that you feed a basis vector to a basis dual vector and get a number dx^\mu(\mathbf{\partial}_\nu)=\delta^\mu_\nu. make special note of the fact that i made those basis vectors bold. so here, i contract a dual vector with a vector and get a scalar.

now let's recall how i defined vectors in the first place: they are derivations on the algebra of functions. in other words, they are differential operators (this is why i use the symbol \mathbf{\partial} for the basis vectors. it makes them look like differential operators.

so a bold \partial represents a tangent (basis) vector.

the exterior derivative of a function was defined by df(\mathbf{v})=\mathbf{v}(f), from which we derived that the components of df are \partial_\mu f. notice that i didn t put my \partial in bold. because \partial by itself is a tangent vector, it should be bold. but \partial_\mu f is just a number, a component of a 1-form. a scalar (in the math sense of the word, not the physics sense; it is not invariant under coordinate transformations)

so when you contract \partial_\mu f with dx^\mu, you just taking a linear combination of 1-forms, and so you end up with a 1-form (not a scalar)

when you contract dx^\mu with \mathbf{\partial}_\mu, you are letting your 1-form eat your vector, and since 1-forms (by definition) eat vectors and spit out scalars (here these are scalars in the math sense and the physics sense)

note: i am not very strict or consistent about requiring my vectors to be bold, but i m hoping in this case, it will help.

OK, so a (1,0) tensor is not synonymous with a contravariant vector, nor is a (0,1) tensor synonymous with a covariant vector?

well, that depends. are you using the words contravariant and covariant according to the physics convention? if so, then a (1,0) is a contravariant tensor, and a (0,1) tensor is a covariant tensor, as you say.

I have also been told rather emphatically that the metric is a scalar because it does not get transformed by a Lorentz transformation. Not true?

this is simply not true. not even for a physicist. whoever told you that was just wrong.

according to the math convention, the metric is a coordinate independent object, and it does not transform under coordinate transformations. but in the math convention, this does not make something a scalar. so the statement is not correct.

in the physics convention, we say something is a scalar if it is invariant under coordinate transformations, and i just finished saying the metric is coordinate independent, right?

no, not right. in this physics convention, when we say metric, we don t mean the coordinate independent (0,2) tensor, we mean its coordinate components g_{\mu\nu} and these certainly do transform under coordinate transformations.

so i can t think of any interpretation in which that statement makes any sense.

By writing one of the indices as a subscript, contraction with the metric tensor is already implied, and the dx_μ is supposed to be the covariant form of dx^μ. Is this just a matter of confusing termonology?

in math land, there is no object dx_\mu. that is physics shorthand for g_{\mu\nu}dx^\nu. there is also no object dx^\mu dx^\nu, that is physics shorthand for dx^\mu\otimes dx^\nu. putting the two statements together, i have g(\cdot,\cdot)=g_{\mu\nu}dx^\mu\otimes dx^\nu. this is the metric, written in coordinate components, in terms of the dx^\mu\otimes dx^\nu basis for the (0,2) tensors.

I think it may be deeper than termonology and notation, because I would have sworn yesterday that dτ² was a scalar, and that g_μν was a second rank tensor. You're starting to scare me.

but i suppose if you mean ds^2 to be the invariant distance between two very close points, then this is indeed a scalar. but here, in math land, dx^\mu does not mean distance between neighboring points (such a concept has no meaning; there is always a finite seperation, and to find the distance, you must integrate)

I understand that dx^μdx^ν is a second rank tensor (and so, I guess that means a (0,2) tensor?). But dx_βdx^β?

like i said above, that is just physics shorthand for something else.

I wish I was in the math camp right now. I am starting to think that physics is teaching me bad habits.




How do you know that they are 1-forms and not vectors? What is wrong with saying that df is a vector, and in the ∂_μf basis, it has components dx^μ?

well... i guess you can say that if you want, but in this thread, we are going to use my definitions, not yours. but let me point out why i think your choice of notation sucks: the notation \partial_\mu f suggests that this object depends on some function f. but basis vectors should only depend on your choice of coordinates, not on some function. also, the components of the vector associated with f should depend on f, and your notation doesn t show that.

 

[turin]

Originally posted by lethe
a symplectic form is a nondegenerate closed 2-form.

I guess I will need to know what degenerate means in this context. Does it mean that it gives a unique value for every unique pair of vectors that you feed it?

Originally posted by lethe
make special note of the fact that i made those basis vectors bold.

so a bold \partial represents a tangent (basis) vector.

notice that i didn t put my \partial in bold. because \partial by itself is a tangent vector, it should be bold. but \partial_\mu f is just a number, a component of a 1-form. a scalar (in the math sense of the word, not the physics sense; it is not invariant under coordinate transformations)

so when you contract \partial_\mu f with dx^\mu, you just taking a linear combination of 1-forms, and so you end up with a 1-form (not a scalar)

when you contract dx^\mu with \mathbf{\partial}_\mu, you are letting your 1-form eat your vector, and since 1-forms (by definition) eat vectors and spit out scalars (here these are scalars in the math sense and the physics sense)

I'm not quite following the distinction between bold and not-bold. Is the difference whether or not the partial derivative operates on a function?

Originally posted by lethe
... i use the symbol \mathbf{\partial} for the basis vectors. it makes them look like differential operators.

I don't understand what you mean by this. If they are in fact differential operators, then of course you would use the differential operator symbol? Are they not really differential operators, but they just act a lot like them?

Originally posted by lethe
in this physics convention, when we say metric, we don t mean the coordinate independent (0,2) tensor, we mean its coordinate components g_{\mu\nu} and these certainly do transform under coordinate transformations.

I thought that "metric" meant "ds²" in physics. At any rate, that's what I meant by it. So, is ds² a (0,2) tensor? I think I'm starting to see how it could be. It returns a math_scalar value when you feed it two 1-forms, but if you don't feed it anything, then it doesn't even make sense.

Originally posted by lethe
according to the math convention, the metric is a coordinate independent object, and it does not transform under coordinate transformations. but in the math convention, this does not make something a scalar. so the statement is not correct.

Can you give the math definition of scalar?

Originally posted by lethe
in math land, there is no object dx_\mu. that is physics shorthand for g_{\mu\nu}dx^\nu.

Is not the g_μνdx^ν a 1-form? Are you saying that g_μνdx^ν is also mathematically meaningless?

Originally posted by lethe
... in this thread, we are going to use my definitions, not yours.

but let me point out why i think your choice of notation sucks:

Hey, whoa. Sorry, there, partner. I didn't mean to offend you or stomp on your territory unwarranted. I just want to figure out how the notation and stuff that I have learned fits here, because some of the things that have been said seemed contradictory. I in no way intended to imply that you were contradicting yourself, and I appologize if I have admitted such interpretation.

Originally posted by lethe
the notation \partial_\mu f suggests that this object depends on some function f. but basis vectors should only depend on your choice of coordinates, not on some function.

I seem to remember the definition of coordinates as functions. Are they a different kind of function or what? By choosing coordinates, have you not chosen functions?

Originally posted by lethe
also, the components of the vector associated with f should depend on f, and your notation doesn t show that.

What does it mean for a vector to be associated with f? Is this related to the selection of the dual space?

 

[lethe]

Originally posted by turin
I guess I will need to know what degenerate means in this context. Does it mean that it gives a unique value for every unique pair of vectors that you feed it?

a bilinear form B(\cdot,\cdot) is degenerate if there exists some vector \mathbf{v}\neq 0 such that B(\mathbf{v},\mathbf{w})=0 for all \mathbf{w}\in V. if you think of the bilinear form as a matrix (if V is finite dimensional, then you can always do this), then this is equivalent to saying that its determinant is zero.

the existence of a nondegenerate bilinear form on a vector space gives you an isomorphism between the vector space and its dual space. it is this isomorphism that allows you to pretend that vectors with raised and lowered indices are just different names for the same thing. in the absence of this bilinear form, you cannot do this. in Riemannian geometry, the bilinear form is the metric.

I'm not quite following the distinction between bold and not-bold. Is the difference whether or not the partial derivative operates on a function?

the boldness is supposed to draw your attention away from the fact that these guys are differential operators, and remind you that they are tangent vectors. and tangent vectors can really be thought of as arrows. bold face is somehow supposed to remind you of that.

but they are still differential operators (by definition).

I don't understand what you mean by this. If they are in fact differential operators, then of course you would use the differential operator symbol? Are they not really differential operators, but they just act a lot like them?

tangent vectors are differential operators, and to remind you of this, i use the symbol \partial

to recap: \mathbf{\partial}_\mu is a vector. if you feed it to a 1-form, you get a scalar. if you feed a function to this vector, you get a scalar.

\partial_\mu f is a component of a 1-form. if you contract it with basis 1-forms, you get a linear combination of those basis 1-forms, which is of course still a 1-form, not a scalar.

I thought that "metric" meant "ds²" in physics. At any rate, that's what I meant by it. So, is ds² a (0,2) tensor? I think I'm starting to see how it could be. It returns a math_scalar value when you feed it two 1-forms, but if you don't feed it anything, then it doesn't even make sense.

feed it 2 tangent vectors, not 1-forms. things that eat vectors are type (0,s) tensors, things that eat 1-forms are type (r,0) tensors.

Can you give the math definition of scalar?

a vector space is defined to be the pair (F,V) where F is a field, and V is an abelian group, and which satisfies a few axioms (which you probably know)

in math, i call things that live in F scalars, and things that live in V vectors. since we are doing differential geometry, F will probably always be R, the real numbers. sometimes the complexes. those guys are math scalars.

Is not the g_μνdx^ν a 1-form? Are you saying that g_μνdx^ν is also mathematically meaningless

yes, it is a 1-form, and no it is not mathematically meaningless. recall where we said this:

Originally posted by turin

Originally posted by lethe

but you could have figured out that it was a (0,2) rank tensor just by looking at its coordinate components g_{\mu\nu}. 2 lowered indices on the coordinate components = (0,2) rank tensor.

I understand that dx^μdx^ν is a second rank tensor (and so, I guess that means a (0,2) tensor?). But dx_βdx^β?

now since you yourself have told me that dx_\mu is a 1-form (which is a (0,1) rank tensor), and we know that dx^\mu is also a 1-form, then it should be clear that dx_\mu\otimes dx^\mu must be a (0,2) rank tensor. i only said that bit about it mathematically meaningless because i thought you were getting hung up on the fact that one of those 1-forms had a lowered index. that is just physics short hand for the components of a metric. nothing more. it doesn t change the fact that you are tensoring 2 (0,1) tensors, and therefore get a (0,2) tensor. just because you are contracting an index, it does not change the fact that there are 2 tensored 1-forms up there.

Hey, whoa. Sorry, there, partner. I didn't mean to offend you or stomp on your territory unwarranted. I just want to figure out how the notation and stuff that I have learned fits here, because some of the things that have been said seemed contradictory. I in no way intended to imply that you were contradicting yourself, and I appologize if I have admitted such interpretation.

hey man, no need to apologize. you didn t offend me. i just thought your suggestion for a change of notation was silly, and i was ribbing you for it.

I seem to remember the definition of coordinates as functions. Are they a different kind of function or what? By choosing coordinates, have you not chosen functions?

coordinates are indeed a different kind of function. coordinates are mappings from the manifold to Rⁿ (for an n dimensional manifold), whereas functions that we are dealing with here are mappings from the manifold to R.

you can consider a single coordinate as a single valued function. then i guess you could call this a choice of a function. but f is a different (arbitrary) function, and your coordinates shouldn t depend on it.

What does it mean for a vector to be associated with f? Is this related to the selection of the dual space?

well, in this context, it means that there is a unique differential form df for each function f on the manifold that satisfies the following equation:
<br /> df(\mathbf{v})=\mathbf{v}(f)<br />

 

[turin]

It's "soak" time for me. I think you've answered all of my questions. As always, thanks.

Please proceed with your original exposition.

 

[Peterdevis]

Excuse me for my rather bad english (I'm dutch speaking)

first:
In many replies people are making a distinguis between physicists and maths views about tensors ( p-forms and vectors).
I think that both agree that tensors are coordinate independent objects (because that's the whole idea behind differential geometry)
Only when you have to do some real calculation you have to choose a coördinate system and base vectors and p forms. The tensor components you get then are of cours coordinate dependent.

second:
It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)

 

[lethe]

Originally posted by Peterdevis
Excuse me for my rather bad english (I'm dutch speaking)

first:
In many replies people are making a distinguis between physicists and maths views about tensors ( p-forms and vectors).
I think that both agree that tensors are coordinate independent objects (because that's the whole idea behind differential geometry)
Only when you have to do some real calculation you have to choose a coördinate system and base vectors and p forms. The tensor components you get then are of cours coordinate dependent.

there are two reasons why i outline the distinction between the maths convention and the physics convention.

1. while, as you say, both the mathematician and the physicist know that a tensor is a coordinate independent object, i think this fact is completely obscure for the physics student who sees only the coordinate transformations of the components which follow a prescribed but somewhat mysterious rule.

2. math and physics have the use of the words covariant and contravariant exactly switched, so there is a tangible difference that needs to be made clear.

second:
It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)

yeah, sure. i don t know who said the coordinates of the metric are coordinate independent. they are not. the components of any tensor (except a (0,0) tensor, otherwise known as a scalar) are not coordinate independent. i agree with you, but i don t know why you say "it seems that way". in fact, it doesn t even seem like the coordinates of the metric are invariant under coordinate transformations. the transformation looks like this:

g_{\mu\nu}&#039;=\frac{\partial x^\rho}{\partial x&#039;^\mu}\frac{\partial x^\sigma}{\partial x&#039;^\nu}g_{\rho\sigma}

so i don t know why anyone would think that the components are invariant, or even seem invariant.

although... now that i think of it, what did turin actually say before?

Originally posted by turin
I have also been told rather emphatically that the metric is a scalar because it does not get transformed by a Lorentz transformation. Not true?

you know, when i answered this before, i answered it talking about general coordinate transformations, for which there is no sense in which this statement can me made to even resemble the truth. For some reason, i didn t even notice that you were asking about Lorentz transformations, instead of coordinate transformations. i guess it is because when i am in the differential geometry subforum, i am never thinking about flat manifolds.

when the manifold in question has a large number of symmetries, we may consider those coordinate transformations that respect those symmetries. in the case of your manifold being flat Minkowski space, the set of coordinate transformations that respect its symmetries are Lorentz transformations.

in other words, by construction, a Lorentz transformation leaves the metric invariant. even the components.

i guess you would say that the components of the metric are scalars, but i think only a sick person would do this. i think it is much better to keep the metric in your head as a (0,2) tensor, whose components are therefore neither tensors or scalars, but rather components of a tensor, and just view the Lorentz transformation as a specific coordinate transformation that leaves those components invariant. not by luck, but because this is what we want, this is how we define the Lorentz transformation.

 

[Peterdevis]

It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)

I wrote a mistake independent must be dependent. So the components of the metric (and any tensor) can be seen as coördinate independent.

The whole transformation rule of the metric is:

g= g_{\mu^\prime\upsilon^\prime}(dx^{\mu^\prime}\otimes dx^{\upsilon^\prime}) =\frac{\partial x^\mu}{\partial x^{\mu^\prime}}\frac{\partial x^\upsilon}{\partial x^{\upsilon^\prime}}g_{\mu \upsilon }(\frac{\partial x^{\mu^\prime}}{\partial x^\mu}dx^{\mu }\otimes \frac{\partial x^{\upsilon^\prime}}{\partial x^\upsilon}dx^{\upsilon })
whow it takes me 20 minutes with latex

So the componenets are changing not because of te coordinate transformation but because when we change of coordinates, we automaticaly change the basic tensors of the vector space. By changing the basic tensors, the components of a tensor must change.
But it is not necessary to change basic tensors when you change coordinates!

 

[maddy]

Exterior differentiation of a 1-form

...but before we leave the 2-forms, let s find a basis for them, and look at their coordinate representation. it should be obvious how to do that, right? we built our 2-form from two 1-forms, so we should be able to build a basis for our 2-forms from the basis for our 1-forms. let s recall what that was (6):

df=\partial_\nu f dx^\nu

let s take two of those, and wedge them together:

df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu<br />

lethe, are you going to continue with this thread further later i.e. more on exterior derivatives and Hodge star operators? Thanks to your thread, am getting a lot of confusions cleared up esp the difference between a vector and the vector components and the basis vectors.

I'd like to ask some questions related to the above. Can we write d\omega^\mu=df\wedge dg?

Let the {\omega^\mu} be the basis of 1-forms dual to a basis {X_\mu}.

Can we relate the structure coefficients (\partial_\mu f)(\partial_\nu g) in df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu<br /> with the structure coefficients of the expansion of the Lie bracket of 2 basis vectors? (am trying to work out an equation in the section dealing with exterior differentiation in Ryan and Shepley's Homogeneous Relativistic Cosmologies)

 

[lethe]

lethe, are you going to continue with this thread further later i.e. more on exterior derivatives and Hodge star operators?

well, when i started this thread, i had planned on going all the way to the Yang-Mills lagrangian, which means i would include more on exterior derivatives, covariant derivatives, and exterior covariant derivatives, as well as the Hodge dual. and i had ideas on where i could go after that. higher form gauge fields, Palatini formalism of GR, Chern-Simmons forms, i dunno, a lot of things, some of which i am still learning myself, but the best way to learn a subject is to try to teach it, right?

however, i haven't written a new addition to the main exposition of this thread in almost a year. i seem to have run out of steam. i need something to motivate me to write some more of these, but they are a lot of work.

i have gotten a lot of encouraging private messages about this thread, and i like to think that i have made a difficult subject accesible to people. but i have abandoned it halfway through.

i will make you a deal, maddy. throughout my exposition (this thread is long, it has almost 100 replies. but there are only about 10 or so posts, written by me only, that i consider to be the main kernel of the exposition. i wouldn't mind if every post other than those that make up my exposition were deleted), so throughout these 10 posts or so by me, there have been a couple of homework exercises. no one ever did my homework exercises. i think they should have been very easy for anyone following along. so my deal for you is: do my homework exercises, and, in exchange, i will write 5 more posts.

what do you think?

Thanks to your thread, am getting a lot of confusions cleared up esp the difference between a vector and the vector components and the basis vectors.

I am glad you liked it! i think it is an exciting subject, I really liked learning it, and this thread is a way for me to communicate that excitement to others. i am glad someone is listening

I'd like to ask some questions related to the above. Can we write d\omega^\mu=df\wedge dg?

hmm... well, you can write whatever you want, and if you define things appropriately, it may even make sense. i am not sure what \omega is in your equation above, so i am having trouble making sense of it. usually \omega^\mu is a function, in which case, the left-hand side of your equation is a 1-form, and the right-hand side is a 2-form. this would make the equation meaningless. furthermore, it violates the einstein summation convention, where the index content of both sides of the equation should match. your right-hand side contains a raised \mu index, whereas your left-hand side contains no index at all. in short, i would say this is an invalid equation.

Let the {\omega^\mu} be the basis of 1-forms dual to a basis {X_\mu}.

these symbols don't agree with the spirit of my thread here, in that you are using coefficients of vectors with respect to some basis as the basis itself. but, OK, i can live with that. it is the convention that most physicists follow.

Can we relate the structure coefficients (\partial_\mu f)(\partial_\nu g) in df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu<br /> with the structure coefficients of the expansion of the Lie bracket of 2 basis vectors? (am trying to work out an equation in the section dealing with exterior differentiation in Ryan and Shepley's Homogeneous Relativistic Cosmologies)

the equation above has pretty much nothing to do with the structure constants of my basis.

why don't you post the equation here that you are trying to make sense of (i don't have the book you refer to), and let's see if we can help you

 

[rdt2]

well, when i started this thread, i had planned on going all the way to the Yang-Mills lagrangian, which means i would include more on exterior derivatives, covariant derivatives, and exterior covariant derivatives, as well as the Hodge dual...

however, i haven't written a new addition to the main exposition of this thread in almost a year. i seem to have run out of steam. i need something to motivate me to write some more of these, but they are a lot of work...

i will make you a deal, maddy. throughout my exposition (this thread is long, it has almost 100 replies. but there are only about 10 or so posts, written by me only, that i consider to be the main kernel of the exposition...

The original thread on http://www.sciforums.com/showthread.php?t=20843&page=2&pp=20 is still fairly clean. Over there I started a new thread for comments/questions/side-issues about Lethe's notes, leaving the main thread for the notes themselves. Then I was off-line for some time and lost the thread, so to speak. But perhaps that's what's needed here.

RDT2.

Still around - and still trying to integrate this into mech eng teaching. So many notations, so little time!

 

[maddy]

i will make you a deal, maddy...my deal for you is: do my homework exercises, and, in exchange, i will write 5 more posts.

what do you think?

It's a deal.

Homework no 1 is:-

i will ask for a volunteer to show that the set of linear functionals on a given vector space is itself a vector space. it s not to hard, just check the vector space axioms given above.

Let \sigma and \omega are members of a set of linear functionals, p^\alpha be the basis of the linear functional \sigma, \nu be a vector, {e_\alpha} be its basis, and a and b be arbitrary members of R.

(\sigma+\omega)\nu
=(\sigma+\omega)(\nu^\alpha e_\alpha)
=\sigma(\nu^\alpha e_\alpha)+\omega(\nu^\alpha e_\alpha)
=\sigma(\nu)+\omega(\nu)
=(\omega+\sigma)\nu
So, the addition of linear functionals is commutative.

Because \sigma=\sigma_\alpha p^\alpha,
-\sigma=-(\sigma_\alpha p^\alpha)
=(-\sigma_\alpha)p^\alpha
So, given a linear functional \sigma, -\sigma is also a linear functional.

(a+b)\sigma(\nu)=a\sigma(\nu)+b\sigma(\nu)
So, the addition of the real numbers is distributive.

a(\sigma+\omega)(\nu)
=a(\sigma(\nu)+\omega(\nu))
=a\sigma(\nu)+a\omega(\nu)
So, the addition of linear functionals is distributive.

a(b\sigma(\nu))
=a(b\sigma(\nu^\alpha e_\alpha))
=a(b \nu^\alpha \sigma_\alpha)
=(ab)\nu^\alpha \sigma_\alpha
=(ab)\sigma(\nu^\alpha e_\alpha)
=(ab)\sigma(\nu)
So, scalar multiplication of the linear functionals is associative.

The set of linear functionals thus form a vector space.

 

[maddy]

Homework no 2 is:-

OK, it should be easy to show that the set of tangent vectors, thusly defined, satisy the axioms of the vector space. i will call this vector space TMp. that is, the tangent space to the manifold M at the point p is TMp. for an n dimensional manifold, the tangent space is always an n dimensional vector space.

Let f and g be functions, and e_\alpha=\partial_\alpha.

Likewise,
(\partial_\alpha+\partial_\mu)f
=(\partial_\alpha)f+(\partial_\mu)f
=(\partial_\mu+\partial_\alpha)f
Addition of tangent vectors is commutative.

(a+b)\partial_\alpha f
=a\partial_\alpha f+b\partial_\alpha f
Addition of the real numbers is distributive.

a(\partial_\alpha+\partial_\mu)f
=a((\partial_\alpha f)+(\partial_\mu f))
=a(\partial_\alpha f)+a(\partial_\mu f)
Addition of tangent vectors is distributive.

a(b\partial_\alpha f)
=ab(\partial_\alpha f)
=(ab)(\partial_\alpha f)
Scalar multiplication of the tangent vectors is associative.

The set of tangent vectors thus form a vector space.

 

[maddy]

i am not sure what \omega is in your equation above, so i am having trouble making sense of it. usually \omega^\mu is a function, in which case, the left-hand side of your equation is a 1-form, and the right-hand side is a 2-form. this would make the equation meaningless. furthermore, it violates the einstein summation convention, where the index content of both sides of the equation should match. your right-hand side contains a raised \mu index, whereas your left-hand side contains no index at all. in short, i would say this is an invalid equation.

Oops, sorry, yes, I made a lethal careless mistake!
The equation I was referring to is
d\omega^\mu=D^\mu_{\alpha\beta} \omega^\alpha \wedge \omega^\beta.
(which at a glance, I saw the right-hand side wrongly as the wedge product of d\omega^\alpha and d\omega^\beta)

Ok, so the above equation shows that the exterior derivative operator turns the 1-form \omega^\muinto a 2-form \omega^\alpha \wedge \omega^\beta.

these symbols don't agree with the spirit of my thread here, in that you are using coefficients of vectors with respect to some basis as the basis itself. but, OK, i can live with that. it is the convention that most physicists follow.

The {\omega^\mu} is the basis of a 1-form dual to a basis {X_\mu} of a vector.

the equation above has pretty much nothing to do with the structure constants of my basis.

why don't you post the equation here that you are trying to make sense of (i don't have the book you refer to), and let's see if we can help you

Here is the Lie bracket of the basis of a vector.
\left[ X_\mu , X_\nu \right] = C^\lambda_{\mu\nu} X_\lambda

I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.

I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.

Is there any concrete way that I can prove this?

 

[meteor]

Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?

 

[maddy]

Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?

Mmm, as far as I learnt, exterior calculus applies only to functions and forms to generate forms of higher rank. An arbitrary tensor is a product of an arbitrary number of forms and vectors. We use another type of calculus for tensors i.e. Lie derivatives, covariant derivatives. (I stand corrected)

 

[maddy]

I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.

I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.

Is there any concrete way that I can prove this?

Oops, didn't know that the Maurer-Cartan machinery does the trick.
Some discussions on Maurer-Cartan forms at SPR

Any other recommendations for an excellent book on Lie algebras that's not too hard for a physics student? (sorry if veering out of topic here)

 

[lethe]

The set of linear functionals thus form a vector space.

OK, this is all correct. one comment though, as Matt Grime would say "gentlemen never choose a basis". none of these proofs require you to specify a basis, and you can just delete those steps where you expand in terms of a basis, and the proofs still go through correctly

 

[lethe]

The set of tangent vectors thus form a vector space.

yes, very good. there is a basis independent way to prove this too, but since the definition of a tangent vector was in terms of a basis, the proof must be as well. thus proving that i am not a gentleman.

i think these are too easy for you!

 

[lethe]

Oops, sorry, yes, I made a lethal careless mistake!
The equation I was referring to is
d\omega^\mu=D^\mu_{\alpha\beta} \omega^\alpha \wedge \omega^\beta.
(which at a glance, I saw the right-hand side wrongly as the wedge product of d\omega^\alpha and d\omega^\beta)

OK, now i see. this is Cartan's first structure equation for a torsion free connection. if there is torsion, then you have T=d\omega+D\wedge\omega

and that D is your connection coefficient, right? not a differential operator?

Ok, so the above equation shows that the exterior derivative operator turns the 1-form \omega^\muinto a 2-form \omega^\alpha \wedge \omega^\beta.

yes. but of course that is no surprise. an exterior derivative, by construction, always turns a p-form into a (p+1)-form.

The {\omega^\mu} is the basis of a 1-form dual to a basis {X_\mu} of a vector.

OK, i apologize. I saw X_\mu and just assumed that meant coefficients of a vector. which is what i usually use that symbol for, with raised index. but i do use \omega^\mu for basis 1-forms, so you are all good.

Here is the Lie bracket of the basis of a vector.
\left[ X_\mu , X_\nu \right] = C^\lambda_{\mu\nu} X_\lambda

I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.

I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.

Is there any concrete way that I can prove this?

sure:

by definition, the torsion tensor acts on a pair of vectors like this:

<br /> T(v,w)=D_vw-D_wv-[v,w]<br />
since we are dealing with a torsion-free connection, we can set that equal to zero, and this gives you your required result.

 

[lethe]

Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?

you can take the exterior covariant derivative of a tensor, but not a plain old exterior derivative.

 

[mathwonk]

How fascinating. The popularity of this thread shows that many people would prefer to exchange posts than to just read a book on the topic, where all of these questions would be answered. That is not shocking, as science too is a social activity, and it is fun to share it. Still if these questions are really of interest, one could just read a book about it and learn most of the answers, I am guessing. Sometimes it seems as if the questions here are asked by people who do not read, and the answers are provided by those who do. No disrespect intended, just curious about the learning habits of the younger generation. I also like to read these posts to see what I can learn. Books are so dry. Keep it going.

 

[chroot]

mathwonk,

Too bad you didn't get to see this thread when it was active.. the starter of this thread, lethe, decided to leave PF after our membership voted to keep our profanity filter in place (he felt it was unfair censorship: https://www.physicsforums.com/showthread.php?t=24373). It's a shame he decided to delete all his posts before leaving out of some sort of vindictiveness. The posts are still physically present in our database, however, and I might be legally able to reinstate them -- I'll have to look into it.

- Warren

 

[mathwonk]

a riemannian manifold is locally flat if and only if the curvature tensor is zero. this is what spivak class the "test case" for the relevance of riemann's curvature tensor.

 

[mathwonk]

trying again after the post did not appear:

a riemannian manifold is locally eucldean in the sense of isometric, if and only if the riemann curvature tensor is zero. that is the whole point of the curvature tensor.

 

[VASILSIS]

i am going to assume that you are a little familiar with euclidean vectors. a euclidean vector is an arrow between two points. it has direction and magnitude¹. mathematically, we can specify a vector in euclidean space with a pair of points in the space, and let the vector be the arrow directed from one point to the other. or you can assume that the first point is always the origin, and specify the vector with just a single point. by doing this, you are essentially moving the vector from it s basepoint, to the origin. this is possible because euclidean space is both a manifold and a vector space.
this won t be true when we move to noneuclidean manifolds. for example, there is no sensible way to make points on a sphere into a vector space. there is no sensible way to define addition on these points.[/size]
¹Well, the vectors don t have magnitude or direction until we endow the space with a metric. almost everything we are going to talk about here is independent of metric, and we will not need to specify a metric on this space. when using metric dependent quantities, this is the differential geometry, and when dealing with the more general metric independent quantities, this is differential topology. if you don t know what any of this means, ignore it.[/size][/QUO

I may say that all familiar curced shapes can be expressed in a Eucledian space using one extra dimension.
A line can be expressed in a 2D eucl.. space .A sphere or other 2D shape can be expressed in an 3D eucl. space .
a) Is this correct that all shapes no matter how many dimension have can be expressed in D+1 eucledian space?
b) If we want to express them using a space with different metric can we always manage it with a tranformation equation ?

 

[Mulder]

I may say that all familiar curced shapes can be expressed in a Eucledian space using one extra dimension.
A line can be expressed in a 2D eucl.. space .A sphere or other 2D shape can be expressed in an 3D eucl. space .
a) Is this correct that all shapes no matter how many dimension have can be expressed in D+1 eucledian space?

I'm probably misunderstanding your question, but doesn't a (flat) 2-torus (for example) only stay flat in 4D (not in 3D)? Sure someone with more knowledge will answer fully.
The differential forms webbook is great by the way for someone just learning differential geometry (e.g. me). Clears up 4 weeks lectures straightaway.

 

[mathwonk]

even smooth manifolds without metric amy not be embeddable as hypersurfaces - e.g. klein bottle in 4 space.

the general theorem (whitney embedding) says it takes 2n dimensions to embed a smooth n manifold, and this is best possible in general. e.g. the 8 dimensional projective spacedoes not embed in 15 dimensional euclidean space projective space cannot be smoothl;y embedded in 15 dimensional

 

[Ratzinger]

Reading this thread I must say very sad that Lethe left PF.

PLEASE LETHE, COME BACK!

 

[fisico30]

Hi Lethe,

can I bug you with some super elementary questions about differentials, in order to catch up with the rest of you?
thanks

 

[mathwonk]

sure, anyone can ask anything about them at all.