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 Exercise [14.34] 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [14.34]
Attached my solution.


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06 Jul 2010, 09:04

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.34]
You might want to note that you need to use a torsion-free connection throughout your calculations.

(2) only holds if the connection is torsion-free, and also in the 2nd line of your expansion of {theta,{phi,psi}} on page 2, you've silently reversed the order of derivatives of phi, which obviously only works for a torsion-free derivative operator.

Fortunately, the choice of connection doesn't matter to the result, because the Poisson bracket operator only involves 1st derivatives of scalars - and so doesn't involve the connection at all.


07 Jan 2011, 17:35

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [14.34]
Quote:
You might want to note that you need to use a torsion-free connection throughout your calculations.

Yes, indeed my proof relies on the assumption of a connection without torsion.

I think it is also implicit in the text of the exercise: the expression S^[ab NABLA_a S^cd] =0 mentioned by Penrose in the hint to exercise is actually zero only when the connection is without torsion. You can easily verify that: in my solution I shown that it is proportional to the expression in exercise [14.33]; and looking to my solution to [14.33] that again refers to solution to [14.23], you will find that at the end this expression is proportional to the torsion.

Moreover, as per exercise [14.23] the dS itself is proportional to the torsion, so the expression dS=0 is no more an identity, independent from the connection, when there is a torsion.

Quote:
Fortunately, the choice of connection doesn't matter to the result, because the Poisson bracket operator only involves 1st derivatives of scalars - and so doesn't involve the connection at all.

I'm not so much convinced by your heuristic argument: the second identity actually involves the second derivative of a scalar. And, in any case, also the expression in the hint by Penrose and the one in exercise [14.33] contain only first order derivatives, however vanish only when there is no torsion.

I really don't know if the second identity is true when there is a torsion.
And because dS=0 is no more an identity I'm not sure that taking into account a torsion in symplectic manifold theory is a trivial generalisation.


11 Jan 2011, 10:38

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.34]
Sorry Roberto, but I think you've made an error of interpretation.

The expression dS=0 is unrelated to the presence of torsion, since the exterior derivative ('d' operator) is defined without reference to a connection.

On page 233, a defining expression for it is given in terms of coordinate derivatives.

As it turns out, the coordinate derivatives can be replaced by a covariant derivatives without changing the result, as long as torsion is absent. Your answer to [14.23] demonstrates that this is no longer true when torsion is present: dS itself wouldn't change, but the expression for it in terms of covariant derivatives would now requires additional torsion terms. (i.e. You'd need to subtract off the torsion terms you found in [14.23]).

However, all this means is that when torsion is present, your equation (2) is no longer equivalent to dS=0: The LHS is no longer equal to dS.

That's why you need to note that you are assuming no torsion.

If you included torsion terms both in (2) and in the hint expression Penrose gives, I'm sure you'd find that they all cancel out to give the same final result.

The Poisson bracket operator does not involve the derivative of a vector, even when nested (as in the 2nd identity). So it (and hence the identities it satisfies) are defined independently of a connection. Sure, if you expand out the second identity you'll find second derivatives that are vector derivatives, but any terms involving the connection will necessarily cancel. That's not a heuristic, it's a mathematical certainty!

You could note that the Poisson bracket could equally well have been defined in terms of coordinate derivatives, rather than covariant ones - it'd make no difference to the definition, since they're only applied to scalars anyway.

I think Penrose's hint does imply that you should assume torsion is absent, but I think the only reason for doing that is because it makes the calculations simpler, not because it changes the result you're trying to prove.


11 Jan 2011, 12:19

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [14.34]
OK, now I realised that Poisson brackets are independent of the connection (with or without torsion), that was not obvious to me.
As a consequence of that, also {PHI,{THETA,PSI}} is independent of the connection; I attach the direct algebraic proof, for anybody interested.


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13 Jan 2011, 13:46
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