A question about the Exterior Product in Yang-Mills theory

In summary: He treats the Lie-algebras as matrix algebras. In summary, Yang-Mills theory involves modifications to the vector potential, resulting in a curvature term that includes an exterior/wedge product. This product may not necessarily be zero, as it depends on the Lie algebra of the internal symmetry group used in the theory. This topic is covered in books such as The Geometry of Physics by Theodore Frankel and Geometry, Topology, and Physics by M. Nakahara.
  • #1
GoldPheonix
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0
Right, so in Yang-Mills theory, the vector potential is modified from:

[tex]F = dA[/tex]​

To:

[tex]F = dA + A\wedge A[/tex]​

However, it is my understanding that the exterior/wedge product is anticommutitive, so that for a given exterior algebra over a vector space, V:

[tex]\omega \wedge \omega = 0, \forall \omega \epsilon \Lambda(V) [/tex]​


Why then is the second term in the curvature, F, not non-zero? I assume I'm missing something, could someone fill me in?


(Sorry, this probably fits better in Topology & Geometry section, but the question technically is a question about multilinear algebra)
 
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  • #2
Argh! I was almost done a long latex post, but I accidentally closed the window.

[itex]A[/itex] is a Lie algebra-valued one-form. not a "normal" one-form. Roughy, think of [itex]A[/itex] as a matrix of "normal" one-forms, and [itex]A \wedge A[/itex] as matrix multiplication, where exterior products of matrix elements (normal one-forms) are used instead of multiplication of numbers.
 
  • #3
Thank you George Jones for your reply.

Is there a particular name for the topic of Lie-algebra valued one-forms? (I assume it's actually a differentiable form, i.e. a section of the cotangent bundle). If there is, I would like to study it a little more.I'm assuming there's a little more to it than this. For instance, the Lie algebra here, is it the Lie algebra of the Lie group of diffeomorphisms/isometries on the differentiable manifold/Riemannian manifold (this is the only straight foreword way, that I can imagine, that they'd be matrices)? Or are we assuming the manifold for which our A-field is defined over is also has smooth group structure?
 
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  • #4
GoldPheonix said:
Thank you George Jones for your reply.

Is there a particular name for the topic of Lie-algebra valued one-forms? (I assume it's actually a differentiable form, i.e. a section of the cotangent bundle). If there is, I would like to study it a little more.

They are treated in The Geometry of Physics: An Introduction by Theodore Frankel, and in Geometry, Topology, and Physics by M. Nakahara.
GoldPheonix said:
I'm assuming there's a little more to it than this. For instance, the Lie algebra here, is it the Lie algebra of the Lie group of diffeomorphisms/isometries on the differentiable manifold/Riemannian manifold (this is the only straight foreword way, that I can imagine, that they'd be matrices)? Or are we assuming the manifold for which our A-field is defined over is also has smooth group structure?

A Yang-Mills theory involves a internal (i.e., not spacetime) symmetry group G, and uses a structure called a principal G bundle. The field strength that you gave in the first post is the curvature (again, internal, not spacetime) of the principal G bundle.
 
  • #5
I own M. Nakahara's book. Is it contained in the section on the theory of connections on fiber bundles?
 
  • #6
GoldPheonix said:
I own M. Nakahara's book. Is it contained in the section on the theory of connections on fiber bundles?

Nakahar covers this stuff briefly (maybe too briefly, I prefer Frankels's treament) in the section on curvature of fiber bundles.
 

1. What is the Exterior Product in Yang-Mills theory?

The Exterior Product in Yang-Mills theory is a mathematical operation that combines two vector fields to create a new vector field. It is commonly used in the study of gauge theories, such as Yang-Mills theory, which is a type of quantum field theory that describes the interactions between elementary particles.

2. How is the Exterior Product used in Yang-Mills theory?

In Yang-Mills theory, the Exterior Product is used to define the field strength tensor, which plays a crucial role in understanding the dynamics of the theory. It is also used to describe the interactions between particles and gauge fields, as well as the symmetries of the theory.

3. What is the significance of the Exterior Product in Yang-Mills theory?

The Exterior Product is a fundamental mathematical operation that helps describe the behavior of gauge fields and particles in Yang-Mills theory. It allows us to understand how these fields interact and how the symmetries of the theory are preserved.

4. How does the Exterior Product relate to other mathematical concepts in Yang-Mills theory?

The Exterior Product is closely related to other mathematical concepts in Yang-Mills theory, such as Lie algebras and differential forms. It is also used in conjunction with other operations, such as the wedge product and covariant derivatives, to fully describe the dynamics of the theory.

5. Are there any practical applications of the Exterior Product in Yang-Mills theory?

While the Exterior Product is primarily used in the theoretical study of Yang-Mills theory, it has also found practical applications in fields such as particle physics and quantum computing. Additionally, the mathematical concepts and techniques used in Yang-Mills theory, including the Exterior Product, have also been applied in other areas of physics and mathematics.

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