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

[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.

 

[fisico30]

Thanks mathwonk!

I am used to the idea of differential as an infinitesimal delta of something: dx is the infinitesimal distance, dq the infinitesimal amount of charge...
Infinitesimal intended as something infinitesimally small, but always bigger than zero.

Q1: In differential geometry, the differential changes meaning:
is is true that it represents a unit tangent vector?
is it a functional (operates on a function and outputs a number)?
why its symbol is dx? how is it related to the infinitesimal, differential in calculus?

Q2:p-forms. I am not sure I understand what they are. Could you provide a baisc explanation/example of 1-form and 2 form?

Q3: Manifold: is it correct to say that a manifold is a space, flat or curved?Lethe explains that a surface (a sphere) is a curved 2-manifold. how do we figure out if a space is curved or flat? Is there a curvature function related to the metric distance?
But, what is a space, first of all? For instance, could the electric field, which is a vector field be called a manifold? If so, why?
It seems that a manifold is a general, abstract concept applicable to many things which satisfy a certain criterion of membership But what is that criterion?
(like anything can be a vector, as long as it satisfies those 10 rules that make it be a vector...)

thanks