# Conformal group in two dimensions [closed]

How can one show in a group-theoretical way that each of SO(d,2) and SO(d+1,1) is isomorphic to two-copies of Virasoro algebra for d=2?

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## closed as not a real question by Qmechanic♦Jun 17 '13 at 21:48

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

You want a proof that (the Lie algebras of) SO(2,2) and SO(3,1) are isomorphic to two copies of the Virasoro algebra? That will be a hard proof, since it would be wrong. For example, the first two are finite-dimensional while the latter is infinite-dimensional. Can you elaborate why you think there is such an isomorphism? The Virasoro algebra has a sl(2) subalgebra, is that related to what you want to see? –  Heidar Jun 15 '13 at 5:55
@Heidar Yah..exactly...can so(2,2) be written as sl(2,C)$\otimes$ sl(2,C) and from there can we get some idea about this? –  layman Jun 15 '13 at 6:10
$so(2,2)$ has $6$ generators, each $sl(2,C)$ has $6$ generators too, don't you see a problem ? –  Trimok Jun 15 '13 at 6:30
As pointed out in comments, the question (v1) asks for a proof of something that is not true, so I'm closing it as not a real question. –  Qmechanic Jun 17 '13 at 21:47
I guess here is some confusion. The Lie algebra of SO(2,2) or SO(3,1) is just a sub Lie algebra of Vir $\oplus$ Vir, namely it is generated by the generators $L_{0,\pm 1}$ of the Virasoro algebra. Some people call SO(2,2) or SO(3,1) the conformal group but the conformal group in 2D is actually infinite dimensional. –  Marcel Jun 18 '13 at 17:17