The self-dual connection in LQG

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Main Question or Discussion Point

I'm reading Carlo Rovelli's book "Quantum Gravity". In the second chapter he writes down the Plebanski action by performing a decomposition of the complex Lorentz algebra into self-dual and anti-self dual parts, i.e. [tex]so(3, 1, C)=so(3, C)\oplus so(3, C)[/tex]. I sort of appreciate this fact and what it has to do with the connection, but I don't really understand this in any great depth. For example, why the complex algebras, why into two copies of so(3, C) etc.? I get the rest of the derivation of the action, I just don't get why this decomposition works.

Thanks guys
 

Answers and Replies

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Great book!

It's just a math trick -- shuffling around the degrees of freedom. The obvious thing to do is use a real [itex]so(3,1)[/itex] connection, with 6 degrees of freedom (per direction). But it turns out we can instead use a complex [itex]so(3)[/itex] connection, also with 6 degrees of freedom, and build a map from one connection to the other. So everything we might want to do with one connection we can do with the other, and the map back and forth lets us translate. Letting the [itex]so(3,1)[/itex] connection become complex, and splitting it into self-dual and anti-self-dual parts, one of which can be mapped to a real [itex]so(3,1)[/itex] connection, is how we build the map.

It's not all clear at first why converting to a complex [itex]so(3)[/itex] connection might be a good idea. But this gets developed more in Chapter 4, and the rest of the book.
 
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Oh, ok, thanks. Playing with the degrees of freedom had occurred to me; I just thought there might be some deeper reason for the decomposition that had something to do with representations or whatever.

And yeah, it really is a very good book. I got it for Christmas and I've really enjoyed reading it. Just started reading chapter four.
 
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