i am a little bit confused about how to obtain the second line.

i have that $\kappa([A \wedge \delta A] \wedge A) = \kappa([\delta A \wedge A] \wedge A)$ by antisymmetry of $\wedge$ and $[,]$

but i get that $\kappa([A \wedge A] \wedge \delta A) = -\kappa(\delta A \wedge [A \wedge A]) = -\kappa([\delta A \wedge A] \wedge A)$ by cyclicity of $\kappa$ and antisymmetry of $\wedge$ and $[,]$

but this would make the term in question $\kappa(\frac{2}{3} \delta A \wedge A \wedge A)$, which I don't think is correct

in indices i also get this result

okay i think the above expressions are just wrong (the variations i mean)

wait nvm...

ahh i misunderstood the cyclicity proeprty of the killing form

Since earth's gravitational field is independent of object mass all bodies fall at same rate but when we throw a feather and a ball they don't fall at same rate. Is this has to do something with air resistance?

Yes

So what does air resistance depends on I read somewhere it depends on area and in others it depends on velocity. Why does this force does not have a mathematical equation

The problem is that except at very slow speeds the air flow is turbulent.

It does

And turbulence is currently impossible to treat analytically so we have to resort to approximations and numerical methods.

Oh so air flow is different at each point in space in direction and is continuously changing is that why?

Typically you write the equation as $$m\ddot{z} = mg - \rho \dot{z}$$

The friction term is $\rho \dot{z}$

$\rho$ is some friction coefficient and $\dot{z}$ the velocity of the falling object

If you divide by $m$, you get $$\ddot{z} = g - \frac{\rho}{m} \dot{z}$$

@alam The reasons why turbulence is hard to describe are quite involved. Basically it's a chaotic system.

So unlike the frictionless case, it does depend on the mass

Ok thanks

@Slereah That's linear drag, which is easy to describe. Sadly above speeds of a few mm/s the flow becomes turbulent and the linear drag equation no longer applies.

@JohnRennie I tried keeping it simple :p

i have finally completed by study of chern-simons theory to a sufficient degree...for now..!

@SillyGoose Did you follow any particular lecture note?

@Sanjana i only ultimately stationized the non-abelian classical chern-simons action on a trivial bundle and also computed how it changes under arbitrary gauge transformations. so, i didn't actually do a lot of chern-simons theory content-wise. i used nakahara mainly to learn (or perhaps just familiarize) some differential geometry and then looked at D. S. Freed's notes on classical chern-simons part 1

but the Freed notes are a little bit turgid to me now :P maybe to someone more well versed in the mathematics would see it as clear

and then asking questions here to ACM mainly and also some mathoverflow and physics stack answers

i found the answer by josé here particularly helpful

@vengaq Yes. When we talk about Hawking radiation we mean that the radiation escapes to infinity, so it can be observed by an observer far from the horizon.

isnt this inaccurate