# Relation between Lie algebra of conformal Killing vector fields and conformal algebra

I'm new to conformal transformations and I have a question.

Following the book of Barrett O'Neill "Semi-Riemannian geometry with applications to relativity", there is a Lie anti-isomorphism between the Lie algebra of Killing vector fields and the Lie algebra of the isometry group as Lie group. Since isometries are particular cases of conformal transformations, I wonder whether there is something similar.

I know that the conformal Killing vector fields form a Lie algebra. As far as I know, conformal transformations do not form a group since they are not by definition bijective. But we can define the conformal group by keeping only conformal diffeomorphisms and, as far as I know, the "conformal algebra" is its Lie algebra. Is there also a Lie anti-isomorphism between the two Lie algebras? Or are they in fact the same thing?

I followed the calculations for $\mathbb{R}^d_1$ from some physics books and I got confused. The procedure was something like: you find infinitesimal conformal transformations (which was solving the conformal Killing equation), you get the generators (conformal Killing vector fields?), you compute their commutation relations and identify it as the conformal algebra (Lie algebra of conformal Killing vector fields?). In the end, redefine the generators to see the isomorphism with $\mathfrak{so}(d,2)$. As a remark, I saw in many books that the generators of the Lie algebra $\mathfrak{so}(p,q)$ (in representation I think) are usually written like $L_{ij}=x_i\partial_j-x_j\partial_i$ and if I calculate their commutators, the structure constants are - the ones for $\mathfrak{so}(p,q)$ seen as a group of matrices. I think that this anti-isomorphism + an anti-isomorphism between the Lie algebra of conformal Killing vector fields and conformal algebra leads in the end to an isomorphism between conformal algebra and $\mathfrak{so}(4,2)$, but I'm not sure and I'm here for help.

• I'm confused. In what sense are conformal transformations not a group? Also anti-homorphisms (wrong sign of structure constants) are surely trivially equivalent to homomorphisms: simply change the signs of the generators. Am I missing something here? – mike stone Sep 11 '18 at 13:09
• I think they should be invertible in order to be elements of a group. – Silvia G. Sep 11 '18 at 13:16
• As for the signs, I know it doesn't make big difference, but I want to be sure whether everything is the same as in the same of isometries (there I found a rigorous presentation in the book I mentioned) – Silvia G. Sep 11 '18 at 13:18
• But conformal transformations are invertable, surely? If not, can you give an example of a non-invertable conformal transformation? – mike stone Sep 11 '18 at 13:35
• I found the wikipedia article on the conformal group to be useful. The conformal tranformations are a group, but for pseudoeuclidean spaces the null vectors of the metric means that one needs a "projective line over a ring " if one is to realize the conformal group as a bijective set of fractional linear transformations as can be done for the euclidean metric conformal group. – mike stone Sep 11 '18 at 14:01