1

If we have, expression (1) with the $\star$ sign used for Hodge star $$\star(d(\alpha))$$ where $\alpha$ is a complex function. We are speaking in 3 dimensions (x,y,z) that is expression(1) can be written as $$\star (\partial_x(\alpha)dx + \partial_y(\alpha)dy+\partial_z(\alpha)dz)$$ this can be written more explicitly as $$\partial_x(\alpha)(dy\wedge dz) + \partial_y(\alpha)(dz\wedge dx) + \partial_z(\alpha) (dx\wedge dy)$$

Okay, my first intention was to take the exterior derivative of expression (1) that is $$d(\star(d(\alpha))$$, now after I expanded it, it seems to me that this is equivalent to: $$d[\partial_x(\alpha)(dy\wedge dz) + \partial_y(\alpha)(dz\wedge dx) + \partial_z(\alpha) (dx\wedge dy)]$$ which pretty much looks like zero to me. Am I thinking in the correct way here?

PhilosophicalPhysics
  • 755

2 Answers2

2

What are you getting is the Laplacian of $\alpha$ multiplied by the volume element $\text dx\wedge\text dy\wedge\text dz$, i.e. a 3-form, which is dual to the 0-form $\nabla^2\alpha$. Observe that $\delta\text d+\text d\delta = \nabla^2$, where $\delta$ is the codifferential $\star\text d\star$. On 0-forms $\delta$ vanishes, hence $\nabla^2\alpha = \delta\text d\alpha$ in this case.

Phoenix87
  • 678
  • Sorry what is exactly a codifferential? –  Feb 23 '15 at 16:54
  • And where exactly should I observe $\delta d+d\delta=\nabla^2$? –  Feb 23 '15 at 16:54
  • The codifferential is defined in the answer: $\delta := \star\text d\star$. What you are trying to compute is the Hodge dual of $\delta\text d + \text d\delta$ applied to $\alpha$, which is then the Hodge dual of $\nabla^2\alpha$, i.e. $\nabla^2\alpha\text dx\wedge\text dy\wedge\text dz$. – Phoenix87 Feb 23 '15 at 16:57
  • So you are telling me the exterior derivative of $\star (d(\alpha))$ is $\star \nabla ^2 \alpha$? –  Feb 23 '15 at 17:00
  • @PhilosophicalPhysics yes. – Phoenix87 Feb 23 '15 at 17:03
1

Elaborating on my comment, since I think this is what you are misunderstanding: $$ d[\partial_x(a)(dy\wedge dz)]=d(\partial_x(a))\wedge dy\wedge dz + \partial_x(a) d( dy\wedge dz) $$ Yes the second term vanishes, but the first term is $$ \partial^2_x (a)dx \wedge dy\wedge dz $$ If you work out the other terms you get the full Laplacian multiplying the three form $ dx \wedge dy\wedge dz $

octonion
  • 381
  • Thank you, yes I obviously missed a very important pt. Thanks for pointing it out! –  Feb 23 '15 at 17:02