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 Exercise [14.17] 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [14.17]
Attached my solution.


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25 Jun 2010, 08:15

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.17]
I think it may be simpler than that.

Curvature is (or can be) defined as the change in a vector as it is parallel-transported around a loop (an infinitesimal parallelogram in particular, but for my purposes here it doesn't matter).

For there to be curvature, the parallel transport of a vector V from a point p to a point q must depend on the path from p to q.

Contrast this with the case of Euclidean space: there is a global notion of 'parallel', such that given a vector V at any point p, you can immediately construct a parallel vector V at any other point q, without consideration of the path between the points.

Euclidean space has no curvature, as an immediate consequence of this -
i.e. 'parallel transporting' any vector around any loop just leaves you with the same vector you started with.

Now the text says that in a Lie group:
Quote:
The infinitesimal group elements are to be pictured as particular vector fields on G (or, indeed, H). That is, we think of ‘moving G’ infinitesimally along the relevant vector field V on G, in order to express the transformation that corresponds to pre-multiplying each element of the group by the infinitesimal element represented by V.

and
Quote:
The relevant notion of ‘parallelism’ comes from the group action, supplying the needed notion of ‘parallel transport’...

(pages 312 and 313 respectively)

The first quote means that (just as in the Euclidean case), given a vector V (say, at the origin), it can immediately be transferred to any other point on the manifold, without consideration of the path taken.

The second quote confirms that this is indeed how 'parallel' (and hence 'parallel transport') is defined.

Thus, 'parallel transport' is path-independent in Lie groups, just like in Euclidean space, and it follows immediately from this that Lie groups have no curvature.


23 Oct 2010, 18:57

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [14.17]
You have explained why Lie groups generate a connection without curvature; but in my understanding the exercise asks also to show that this connection has a torsion.


03 Nov 2010, 14:18

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [14.17]
Well, figures 14.9 and 14.15 basically shows what the relationship is between Lie group elements and torsion, and the text spells it out explicitly as well:

There is no torsion if 'parallelograms' close (to O(epsilon^2)). Geometrically, this is the same as saying that commutators (and hence Lie derivatives) are always zero, or that the Lie group elements commute.

If that's the case, then the "parallel vectors" of each group element could never change size or angle anywhere... and the resulting group is just the abelian (i.e. commutative) group of vector additions in Euclidean n-space. (i.e. dimension = n).



I'm not arguing with your math here, I just think that in the case of this "try to explain why" question, Penrose is probably asking for a geometrical insight, rather than an algebraically calculated solution... Namely, that torsion means pairs of parallel vectors don't 'close' as their size approaches zero; and that curvature means that the notion of "parallel" is path dependent. My interpretation could be wrong, of course!


04 Nov 2010, 02:56
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