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If we have expression (1) $$\star (F \wedge d\alpha)$$ where $$ (F \wedge d\alpha)$$ is a $2$-form field strength ($F$ and $d\alpha$ are 1 forms)and $\star$ represents Hodge star.

How can we simplify this rather more?

I want to exterior derive expression (1) and was wondering if I can simplify it before applying exterior derivative on it?

To elaborate my question: The Hodge dual on, let's say, $d(\alpha)$ where $\alpha$is any complex function can easily be deducted as one applies the Hodge Duality rule. Why? Because $d(\alpha)$ can be written as $\partial _x \alpha dx + \partial _y \alpha dy + \partial _z dz$ and we know how to go from this expression to its dual because we know that $\star dx$ let's say is $dy \wedge dz$ in 3 spatial coordinates, but here we don't know how to apply the rule..

PhilosophicalPhysics
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2 Answers2

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There is nothing that you can simplify in this, except if some further simplification comes from the explicit form of $F$ and $\alpha$. Also the action of the Hodge operator on basis element is well defined (cf. wikipedia).

Phoenix87
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Suppose $\omega = \star(F\wedge d\alpha)$. Then $\star\omega = \star ^2 F\wedge d\alpha \Rightarrow$ $\star\omega = (-1)^{n(n-p)}F\wedge d\alpha$, where $n$ is the dimension of manifold and $p$ is the degree of $\star(F\wedge d\alpha)$. Then applying the exterior derivative at both sides we obtain $d\star\omega = (-1)^{n(n-p)}dF\wedge d\alpha$ Applying $\star$ again we gain $\delta \omega = (-1)^{n(n-p)}\star(dF\wedge d\alpha) \Rightarrow$ $\delta \star (F\wedge d\alpha) = (-1)^{n(n-p)}\star(dF\wedge d\alpha)$

L.F. Cavenaghi
  • 4,651