Pondering basis vectors and one forms

Click For Summary
SUMMARY

This discussion centers on the relationship between basis vectors and one forms in the context of differential geometry. It establishes that the property \(\tilde{\omega}^j(\vec{e}_i)=\delta_i^j\) does not necessitate a dual relationship defined by a metric tensor, as duality can exist independently of a metric. The conversation also touches on the tetrad method, highlighting that when using orthonormal basis vectors, the metric simplifies to diag(1,1,1,...). References to Wald's notation and Andrew Hamilton's notes provide additional context for understanding these concepts.

PREREQUISITES
  • Understanding of basis vectors and one forms in differential geometry
  • Familiarity with the concept of duality in linear algebra
  • Knowledge of metric tensors and their role in raising and lowering indices
  • Basic comprehension of the tetrad method in general relativity
NEXT STEPS
  • Study the properties of basis vectors and one forms in differential geometry
  • Learn about the implications of duality without a metric
  • Research the tetrad method and its applications in general relativity
  • Examine Andrew Hamilton's notes on non-orthonormal tetrads for deeper insights
USEFUL FOR

Mathematicians, physicists, and students of general relativity seeking to deepen their understanding of the relationship between vectors, one forms, and metric tensors.

Matterwave
Science Advisor
Homework Helper
Gold Member
Messages
3,971
Reaction score
329
So, I've been thinking about this for a while...and I can't seem to resolve it in my head. In this thread I will use a tilde when referring to one forms and a vector sign when referring to vectors and boldface for tensors. It seems to me that if we require the basis vectors and one forms to obey the property that:

\tilde{\omega}^j(\vec{e}_i)=\delta_i^j

Then, we cannot require the basis vectors and one forms to be "dual" to each other in the sense that we raise and lower their indices with the metric tensor. I.e.:

\tilde{\omega}^i=\bf{g}(\vec{e}_i,\quad)

Since, unless my metric is the Cartesian metric, I cannot get the first property from the second quantity.

Is this true, or have I messed up somewhere?
 
Last edited:
Physics news on Phys.org
The "duality" between vector and one forms that you cite does not require a metric. When the vector basis is changed, the dual forms are also changed.

The "duality" between vectors and one forms that can be defined with a metric is different, as the duals do not change if you change basis.

So you are right. The former has nothing to do with raising and lowering indices with a metric (it's defined even without a metric), the latter involves raising and lowering indices with a metric.
 
Last edited:
The natural set of one form basis vectors then, is defined by the first property and not the second one then right?
 
Yes.
 
I have another question. If we choose to use the tetrad method to describe our manifold, then the metric is automatically diag(1,1,1,...)? Since we are choosing ortho-normal basis vectors, then it would make sense that that would have to be the case.

I could never figure out Wald's take on the tetrad method...His abstract index notation there makes me really confused what his equations mean...>.>
 
Just reading wikipedia, the metric is the Minkowski metric times the "square" of the vierbein field. http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

I remember Andrew Hamilton had a good set of notes about this (including using non-orthonormal tetrads). http://casa.colorado.edu/~ajsh/phys5770_08/grtetrad.pdf
 
Last edited by a moderator:

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
7K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 146 ·
5
Replies
146
Views
10K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K