Can Scalar Fields Be Decomposed into Symmetric and Antisymmetric Parts?

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SUMMARY

A scalar field can indeed be decomposed into symmetric and antisymmetric parts, similar to vector fields. The decomposition is expressed mathematically as $$\phi (q) = \frac{\phi(q) + \phi(-q)}{2} + \frac{\phi(q) - \phi(-q)}{2}$$. This method utilizes the properties of scalar fields as 0-forms on smooth manifolds. The discussion references the Hodge decomposition, which provides a foundational understanding of such decompositions in differential geometry.

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Jhenrique
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If a vector field can be decomposed how a sum of a conservative + solenoidal + harmonic field...
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so, BTW, a scalar field can be decomposed in anothers scalar fields too?
 
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What types of scalar fields are there?
 
How about
$$\phi (q) = \frac{\phi(q) + \lvert \phi(q)\rvert}{2} +\frac{\phi(q) - \lvert \phi(q)\rvert}{2}$$
for a point ##q \in Q## on a smooth manifold and a section ##\phi \in C^{\infty}(Q, Q \times \mathbb R)## of the trivial vector bundle?
This is a decomposition into positive and negative parts.
If the operation ##q \to - q## makes sense, then you can also take
$$\phi (q) = \frac{\phi(q) + \phi(-q)}{2} +\frac{\phi(q) - \phi(-q)}{2}$$
using the same trick. This is a decomposition into symmetric and anti-symmetric parts.
EDIT: You might be interested in this: http://en.wikipedia.org/wiki/Hodge_decomposition
Note that a "scalar field" is a ##0##-form since ##\bigwedge^0 T^*Q \simeq Q \times \mathbb R##.
 
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