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- Thread starter ritzo
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- #2

strangerep

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More accurately, the group SO(4,2) is a double cover of the conformal group of Minkowski space. The Lie algebras of these two groups coincide, as can be seen by direct evaluation.

- #3

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First of all, thank you so much for answer me. Second, I'm sorry, but I'm not familiar with Lie algebra. I don't know if it is possible to solve this in a more elemental way or maybe you can tell me about a book where I can find this. Again, thanks.

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- #4

Bill_K

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Consider a 3-space with orthogonal coordinates T, X, S and signature + - -. Consider the rotations in this 3-space, which form the group SO(2,1), and their actions on the unit hyperboloid T^{2} - X^{2} - S^{2} = 1.

Parametrize points on this hyperboloid with coordinates x, s whose relation to T, X, S is

T - X = e^{x}

T + X = s^{2} e^{x} + e^{-x}

S^{2} = s^{2} e^{2x}

(By substitution, one can verify that for any values of x and s we have T^{2} - X^{2} - S^{2} ≡ 1, which lies on the hyperboloid.)

Now consider the action of SO(2,1) on x and s. One of the rotations, a boost in the T,X plane, amounts to a translation in x and a__rescaling of s__:

x → x + c

s → s e^{-c}

Another, a rotation about the null direction T - X, is a__translation of s__,

s → s + c

Ok, now generalize this to any number of dimensions. Replace the original S^{2} everywhere by Ʃ ε_{i} S_{i}^{2} where ε_{i} = ±1. (This is the line element in an n-dimensional space with signature determined by the choice of ε_{i}'s.) We are now looking at the unit hyperboloid T^{2} - X^{2} - Ʃ ε_{i} S_{i}^{2} = 1. The parameters x and s_{i}

T - X = e^{x}

T + X = Ʃ s_{i}^{2} e^{x} + e^{-x}

Ʃ S_{i}^{2} = Ʃ s_{i}^{2} e^{2x}

and the transformations

x → x + c

s_{i} → s_{i} e^{-c}

and

s_{i} → s_{i} + c_{i}

Thus we have a rescaling of the s_{i}'s and a translation of the s_{i}'s, which, together with rotations among the s_{i}'s, produce the conformal group in n dimensions.

This shows explicitly that the transformations in the rotation group SO(n+1, m+1) map to those in the conformal group C(n,m).

Parametrize points on this hyperboloid with coordinates x, s whose relation to T, X, S is

T - X = e

T + X = s

S

(By substitution, one can verify that for any values of x and s we have T

Now consider the action of SO(2,1) on x and s. One of the rotations, a boost in the T,X plane, amounts to a translation in x and a

x → x + c

s → s e

Another, a rotation about the null direction T - X, is a

s → s + c

Ok, now generalize this to any number of dimensions. Replace the original S

T - X = e

T + X = Ʃ s

Ʃ S

and the transformations

x → x + c

s

and

s

Thus we have a rescaling of the s

This shows explicitly that the transformations in the rotation group SO(n+1, m+1) map to those in the conformal group C(n,m).

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- #5

strangerep

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Hmm. You said you were "reading about the compactification of Minkowski Space", but not knowing anything about Lie group theory is a serious handicap for most subjects in physics.[...] I'm not familiar with Lie algebra. I don't know if it is possible to solve this in a more elemental way or maybe you can tell me about a book where I can find this.

You could try the book by Greiner & Muller: "QM - Symmetries". It contains a lot of introductory material on Lie groups (though of course it's not about compactification of Minkowski Space).

Nice explanation. I hope the OP can appreciate it eventually. (BTW, since it's been ages since I studied this stuff closely, where does the double-cover arise?)Bill_K said:This shows explicitly that the transformations in the rotation group SO(n+1, m+1) map to those in the conformal group C(n,m).

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- #6

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- #7

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And I am like... "What?", Why?. If the linear transformations are of order 2+n, why it is isomorphic to a matrix of SO(2+n,1)?

So, please you guys. I don't know anything about Lie algebra and I have weeks thinking on this. Thank you.

- #8

Bill_K

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This quadric is the "unit hyperboloid" that I referred to above in #4.He then says that every conformal transformation is equivalent, up to a factor c, with a linear transformation that leaves the quadric invariant. This linear transformations is of order n+2.

I didn't mention anything about Lie algebras.So, please you guys. I don't know anything about Lie algebra and I have weeks thinking on this. Thank you.

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