# Existence of Hodge Dual: obvious or non-trivial?

• nonequilibrium
In summary, the Hodge dual of a p-form \omega \in \Omega^p is defined as the object \omega^\star, where for any p-form \eta \in \Omega^p, their exterior product is equal to the volume form multiplied by the inner product of \eta and \omega. While it is unique, the existence of such a form \omega^\star \in \Omega^{n-p} may not be immediately obvious. However, it can be constructed point-wise using local frames and a smoothness argument.
nonequilibrium
Some sources I have checked define the Hodge dual of a form $\omega \in \Omega^p$ as the object such that $\forall \eta \in \Omega^p: \eta \wedge \omega^\star = g(\eta,\omega) \textrm{ Vol}$ (where "Vol" is a chosen volume form).

I can see that there can be only one form with such a solution (i.e. unicity), but I can't see existence: how do we know there is such a form $\omega^\star \in \Omega^{n-p}$ that satisfies that definition?

(I can kind of see it locally using some basis argument, but how make it global...) The main thing I'm confused about is whether it should be obvious that it exists (since my sources don't give extra arguments), or whether it requires further justification.

Basically just construct it point-wise. Since forms are sections of tensor bundles evaluating at a point reduces the existence problem to vector spaces. Only a smoothness argument remains and for this local frames come to the rescue.

## 1. What is the Hodge dual?

The Hodge dual is a mathematical operation that maps a p-form (a differential form of degree p) to a (n-p)-form, where n is the dimension of the underlying vector space. It is denoted by the symbol * and is used in differential geometry and vector calculus.

## 2. Why is the existence of the Hodge dual non-trivial?

The existence of the Hodge dual is non-trivial because it depends on the existence of certain geometric structures, such as a metric tensor, on the underlying vector space. Additionally, the Hodge dual is only defined in certain dimensions and requires the vector space to have a non-degenerate inner product.

## 3. What is the relationship between the Hodge dual and the exterior derivative?

The Hodge dual and the exterior derivative are closely related. In fact, the exterior derivative of a p-form is equal to the Hodge dual of its (n-p)-form. This relationship is important in differential geometry and allows for the formulation of Maxwell's equations in terms of differential forms.

## 4. Is the Hodge dual unique?

No, the Hodge dual is not unique. In fact, there are infinitely many possible Hodge duals for a given p-form. This is because the Hodge dual depends on the choice of basis for the vector space and different choices of basis will result in different Hodge duals.

## 5. What are some applications of the Hodge dual?

The Hodge dual has many applications in mathematics and physics. It is used in differential geometry, vector calculus, and electromagnetism. It is also important in the study of manifolds and their curvature. In physics, the Hodge dual is used in the formulation of field equations, such as Maxwell's equations and Einstein's equations of general relativity.

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