re-coordinate-sl2c | Parallel Transport

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Yes, pick your favourite coordinates on R3 (or H3) and S3. I really don’t care which ones you pick, I’d like one example expression.

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John Carlos Baez

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I haven’t thought about this. I can tell you formulas for the Haar measure on S^3 and formulas for the Lorentz-invariant measure on H^3 in terms of coordinate. But I don’t know the function F you mentioned.

John Carlos Baez

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This paper claims to have a concrete formula for Haar measure on SL(2,C), in section 3: www-users.cse.umn.edu/~garrett/m/v/S… I got this by typing “Haar measure on sl(2,C)” into Google.

Kartik Prabhu

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That’s as far as I got too! I’ll post a short thread explaining why this is useful for physics later this evening.

Leo C. Stein 🦁

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Should I worry about the fact that the exponential map from the Lie algebra sl(2,c) is not onto SL(2,C)?

John Carlos Baez

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Are you worried that the paper is wrong? I haven’t completely understood it but the author looks like they know what they’re doing.

Kartik Prabhu

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No need to worry about that. The centre is a Z2 i.e. it is the double cover-ness. If you only look at the identity component you get usual Lorentz (no double cover).

Leo C. Stein 🦁

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I agree that they look like they know what they’re doing! My instinct was to construct a map from e->g, and use the differential of the map to translate the measure from T_e to T_g. And I thought that was what the author was doing, too. Except then I remembered that exp is not

Leo C. Stein 🦁

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onto, so I figured I must have a hole in my thinking… or that it doesn’t matter?

Leo C. Stein 🦁

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Anyway, what I was just trying to get at the fact that constructing a top form at the identity is easy; translating it to the rest of G is however not clear in my mind.

Kartik Prabhu

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I have looked at that paper/notes before, but I am not sure I understand it. If you look at the last formula in section 1.3 on pg. 3, the k and k’ are in SU(2) and the r is a real radial coordinate. So that measure looks like it is 7-dimensional which seems wrong.

John Carlos Baez

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Ugh! They use the Cartan decomposition G = KAK where A is a maximal abelian subgroup of G. For G = SL(2,C) you can take A to be diagonal matrices with det = 1, and that’s a 2d Lie group, but their formula for it in Section 1.1 makes it look 1d. “r” should really be complex!!!

John Carlos Baez

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They are trying to get the volume form by writing every element of G as exp(k) exp(a) exp(k’) where k,k’ are in the Lie algebra of a maximal compact subgroup and a is in the Lie algebra of a maximal abelian subgroup. en.wikipedia.org/wiki/Cartan_de…

John Carlos Baez

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This looks like an interesting strategy but I don’t understand it well enough to pass judgement in a definitive way without some real work… and they seem to have screwed up in claiming the maximal abelian of G = SL(2,C) is *real* diagonal matrices!