Hodge duality and differential forms

In summary, the conversation discusses the calculation of $$d\star A=0$$ where $$\star$$ is the Hodge star operator. The resulting equation is simplified to $$(dt+\lambda)\nabla^2(\bar{\alpha}-\alpha)=0$$ and it is argued that this step can be done on a Lorentzian manifold. The use of the notation $$\nabla^2$$ for the Hodge-Laplacian is questioned, and it is suggested that using ##d^*## for the codifferential may be helpful. However, it is mentioned that this notation may be misleading in the context of Cartan calculus. Ultimately, the final answer is considered correct and the procedure is deemed
  • #1
PhyAmateur
105
2
If we have,$$A=d[(\bar{\alpha}-\alpha)(dt+\lambda)]$$
where $$\alpha$$ is a complex function and $$\lambda$$ is a 1-form. t here represents the time coordinate.

If we want to calculate $$d\star A=0$$ where $$\star$$ is hodge star, we get if I did my calculations correctly $$\nabla^2[(\bar{\alpha}-\alpha)(dt+\lambda)]=0$$

From the product rule, we get $$(\bar{\alpha}-\alpha)\nabla^2(dt+\lambda)+ (dt+\lambda)\nabla^2 (\bar{\alpha}-\alpha) +2 \nabla^2(\bar{\alpha}-\alpha)\nabla^2(dt+\lambda)=0$$
Does it mean here that each term must equal zero and thus since $$(\bar{\alpha}-\alpha)$$ can not be zero in the first term so $$\nabla^2(dt+\lambda)=0$$ thus killing with it the third term in the equation and only leaving the second term?
That is simplifying the above equation, we get, $$(dt+\lambda)\nabla^2 (\bar{\alpha}-\alpha)=0$$ thus leaving us with $$\nabla^2 (\bar{\alpha}-\alpha)=0$$

Can this step be done?
 
Physics news on Phys.org
  • #2
I suppose you're working on a Lorentzian manifold, Minkowski space?
What is ##\nabla^2##, the Hodge-Laplacian?

If you're doing physics, why choose a particular "time coordinate"?
 
  • #3
$$\nabla^2$$ is the normal laplacian of (x,y,z). t is chosen like this such that dt is a 1 form.
 
  • #4
Right, so you're working in a Newtonian spacetime. Maybe using the fact that
$$\Delta = \left( d + d^* \right)^2 = d d^* + d^* d $$
with ##d^*## being the codifferential will help. Btw, if you're doing Cartan calculus ##\nabla^2## is a bad notation for the Hodge-Laplace / Laplace-Beltrami operator, because it suggests that you're applying the connection twice.
 
  • #5
So yes, considering that I used this notation istead of $\nabla^2$ is my final answer correct or at least is the procedure correct that I considered all terms must be set equal to zero?
 

1. What is Hodge duality and why is it important in differential forms?

Hodge duality is a mathematical operation that relates differential forms of different degrees on a manifold. It is important because it allows us to understand the relationship between different types of differential forms and simplifies calculations in differential geometry and physics.

2. How does Hodge duality work?

Hodge duality involves taking the exterior product of a differential form with the Hodge star operator, which is a linear map that preserves the degree of the form. This results in a new differential form of complementary degree.

3. What is the physical significance of Hodge duality?

Hodge duality has important physical applications, particularly in electromagnetism and general relativity. In electromagnetism, it relates the electric and magnetic fields, and in general relativity, it relates the curvature and energy-momentum tensors.

4. Can Hodge duality be extended to higher dimensions?

Yes, Hodge duality can be extended to higher dimensions. In fact, it is a fundamental concept in differential geometry and is used in various fields of mathematics and physics, such as algebraic topology and string theory.

5. How is Hodge duality related to de Rham cohomology?

Hodge duality is closely related to de Rham cohomology, which is a mathematical tool for studying topological spaces. In particular, the Hodge star operator is used to define the Hodge star operator in de Rham cohomology, which allows us to classify differential forms into cohomology classes.

Similar threads

  • Differential Geometry
Replies
29
Views
1K
Replies
9
Views
3K
  • Differential Geometry
Replies
2
Views
511
Replies
2
Views
718
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
535
  • Differential Geometry
Replies
3
Views
3K
  • Differential Geometry
Replies
4
Views
1K
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
3
Views
1K
Back
Top