- #1

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I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form

Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

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- Thread starter cosmicstring1
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- #1

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I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form

Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

- #2

Chris Hillman

Science Advisor

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- #3

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I'm not aware of anything called the

I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form

Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

Pete

- #4

garrett

Gold Member

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You're going to have to pick a set of orthonormal basis vectors compatible with your metric, then either compute the connection from them, or use them with your Christoffel symbols to get the connection.

This is laid out in detail here:

http://deferentialgeometry.org/#[[Christoffel symbols]]

- #5

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I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form

Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

Firstly, begin by taking [itex]\{\hat{\theta}^i\}[/itex] to be a non-coordinate basis. This non-coordinate basis is related to the coordinate basis [itex]\{dx^i\}[/itex] by

[tex]\hat{\theta}^i = \theta^i_{\phantom{i}j}dx^j[/tex]

for some matrices [itex]\theta^i_{\phantom{i}j}[/itex]. Then the relationship between the connection one-form and the Christoffel symbols is

[tex]\omega^i_{\phantom{i}j} = \Gamma^i_{\phantom{i}jk}\hat{\theta}^k[/tex]

Thus, if you know the Christoffel symbols and the relationship between the non-coordinate and coordinate bases, the answer to your question is to simply read off the value of the components of the connection one form from the above.

pmb said:I'm not aware of anything called the connection 1-forms. All I've heard are Chjristoffel symbols of the first and second kind. Are you sure that is not what you're refering to?

Pete

He was perfectly clear when he said he was interested in the components of the connection one-form. This is a basic idea covered in, for example, the first few lectures of any decent differential geometry course.

- #6

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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Consider the frame bundle, which is a principal SO(n)-bundle. (At least, it is for a Riemannian manifold)I'm not aware of anything called theconnection 1-forms.

If you have any curve, you can take a frame at one endpoint and invoke your connection to parallel transport it to become a frame at the other endpoint. In particular, we get an element of SO(n) detailing how the frame was rotated/twisted/whatever as it was parallel transported along the curve.

If we differentiate (i.e. if we parallel transport by an infinitessimal amount), then to each tangent vector at a point, we get an element of so(n). Since this gives an so(n) value to each tangent vector, and is a linear functional, we call this object a "so(n)-valued one-form".

This so(n)-valued one-form completely determines parallel transport (you integrate it along a curve to get an SO(n) value), thus the term "connection 1-form".

- #7

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your wijk = 1/2( g(ek,[ei,ej]) - g(ei,[ej,ek]) - g(ej, [ek,ei]) )

where ei, ej, ek are the othronormal vectors prescirbed by your metric.

- #8

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Gravitation? Or the textbook by Harley Flanders,Differential forms, with applications to the physical sciences? These books offer clear introductions to computing the connection one-forms, as do many others.

Yes, I read that section from MTW but it does not give the answer. I do'nt have Flanders' book.

- #9

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What is a connection 1-form. Please defined the term for me.

I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form

Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

Thank you

Pete

- #10

- 216

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What is a connection 1-form. Please defined the term for me.

Thank you

Pete

Well, presumably you're familiar with the idea of using a coordinate basis [itex]\{dx^i\}[/itex] in differential geometry, right? Presumably you're also familiar with the idea that in many practical situations it's a far smarter idea to use a

[tex]\omega^i_{\phantom{i}j} \equiv\Gamma^i_{\phantom{i}jk}\hat{\theta}^k[/tex]

That's really all there is to it. Once you have defined the matrix-valued connection one-form like this, it's a simple matter to go on and show that a neat and tidy way to express the curvature and torsion associated with a given metric is to use the Cartan structure equations.

- #11

Chris Hillman

Science Advisor

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Hi, cosmicstring, you asked:

I suggested starting from a frame field, a la Cartan, as explained in MTW. I didn't answer the question you asked, I told you where to find the answer to the question you*should* have asked. You replied:

Sure it does--- look at the sections on computing curvature. This contains what you need to answer the question you asked, as well as showing you a much better way to do all this stuff.

I don't think that even makes sense!

But there is one small point concerning the symmetry properties of the matrix of connection one-forms which MTW slur over. Flanders doesn't cover it either because he only discusses Riemannian manifolds. Namely, the rule for adjoint is a bit different for Lorentzian vice Riemannian. You need to know this when you use the symmetries of the connection one-forms in order to read off the connection one-forms from the exterior derivatives of the basis one-forms, as suggested by Cartan.

By definition, if [itex]A[/itex] is some linear operator and [itex](\cdot , \cdot)[/itex] is some inner product, the adjoint [itex]A^\ast[/itex] is the linear operator satisfying

[tex] (Ax, y) = (x, A^{\ast} y)[/tex]

for all [itex]x,y[/itex]. If you write this out in matrix notation, with the inner product written [itex] (x,y) = x^T L y[/itex], you obtain

[tex] x^T \, A^T \, L \, y = x^T \, L \, A^{\ast} \, y [/tex]

This holds for all [itex]x,y[/itex], so

[tex] A^\ast = L^{-1} \, A^T \, L[/tex]

Now apply this when [itex] L = \operatorname{diag} (-1,1,1,1)[/itex] to obtain the*Lorentzian transpose*. Now apply this to your matrix of one-forms when reading off the connection one-forms as suggested by Cartan.

In the past I have gone through all this stuff in great detail--- you can Google for old posts by myself in sci.physics.*

I have Christoffel symbols for a metric and I want to find the connection 1-forms.

I suggested starting from a frame field, a la Cartan, as explained in MTW. I didn't answer the question you asked, I told you where to find the answer to the question you

Yes, I read that section from MTW but it does not give the answer.

Sure it does--- look at the sections on computing curvature. This contains what you need to answer the question you asked, as well as showing you a much better way to do all this stuff.

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

I don't think that even makes sense!

But there is one small point concerning the symmetry properties of the matrix of connection one-forms which MTW slur over. Flanders doesn't cover it either because he only discusses Riemannian manifolds. Namely, the rule for adjoint is a bit different for Lorentzian vice Riemannian. You need to know this when you use the symmetries of the connection one-forms in order to read off the connection one-forms from the exterior derivatives of the basis one-forms, as suggested by Cartan.

By definition, if [itex]A[/itex] is some linear operator and [itex](\cdot , \cdot)[/itex] is some inner product, the adjoint [itex]A^\ast[/itex] is the linear operator satisfying

[tex] (Ax, y) = (x, A^{\ast} y)[/tex]

for all [itex]x,y[/itex]. If you write this out in matrix notation, with the inner product written [itex] (x,y) = x^T L y[/itex], you obtain

[tex] x^T \, A^T \, L \, y = x^T \, L \, A^{\ast} \, y [/tex]

This holds for all [itex]x,y[/itex], so

[tex] A^\ast = L^{-1} \, A^T \, L[/tex]

Now apply this when [itex] L = \operatorname{diag} (-1,1,1,1)[/itex] to obtain the

In the past I have gone through all this stuff in great detail--- you can Google for old posts by myself in sci.physics.*

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I didn't say he wasn't clear. I mentioned my uncertainty about them once before but it now appears that such things exist, hence the new question, i.e. what are they? That was why I was asking him. No textbook that I have on this subject matter even mentions is. Can you please give me an example? Thank you.He was perfectly clear when he said he was interested in the components of the connection one-form. This is a basic idea covered in, for example, the first few lectures of any decent differential geometry course.

Pete

- #13

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Never mind. I finally found it defined in Lovelock and Rund. Thanks for the help folks.

Pete

Pete

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