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In Blumenhagen's book "Introduction to Conformal Field Theory", I found the statement

The algebra of infinitesimal conformal transformations in an Euclidean 2-dimensional space is infinite dimensional.

He concludes this after finding that the generators of the infinitesimal conformal transformations (i.e. a basis for its Lie algebra) are $l_n=-z^{n+1}\partial$ and $\bar l_n= -\bar z^{n+1}\bar\partial$ for $n\in\mathbb N$.
My problem with this is: shouldn't the elements of the Lie algebra be linear combinations (with real coefficients) of the basis elements $\partial,\bar\partial$ of $T_eG$? (Instead of a roduct of the derivatives with polynomials). Furthermore (and related to the previous point), if our Lie group is 2-dimensional (as a manifold), that implies that its Lie algebra is 2-dimensinal (as vector space), so definitely not infinitely dimensional. What is going on here?

soap
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    Related: https://physics.stackexchange.com/q/108472/2451 , https://physics.stackexchange.com/q/163216/2451 and links therein. – Qmechanic May 05 '18 at 19:03
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    Note: The dimension of the conformal algebra in $d$ spacetime dimensions is not $d$. – Qmechanic May 05 '18 at 19:07
  • @Qmechanic I read that, but shouldn't the dimension of the algebra match the dimension of the manifold (Lie group)? – soap May 05 '18 at 19:09
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    Spacetime is one manifold, the Lie group is another manifold. Their dimensions are independent, and do not in general match. – AccidentalFourierTransform May 05 '18 at 19:10
  • @AccidentalFourierTransform Oh of course... So it seems we're in the presence of an infinite dimensional Lie group. Weird (for me!) – soap May 05 '18 at 19:12
  • No, there are two conformal algebras: the local one (infinite dimensional) and the global one (finite dimensional). The global one is the one that corresponds to the conformal group. They both have the same dimensionality. – AccidentalFourierTransform May 05 '18 at 19:14
  • @AccidentalFourierTransform I was not aware of that, and don't really understand that. Blumenhagen didn't really explain that distinction so far. – soap May 05 '18 at 19:46
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    @Soap You may want to have a look at section 1.2 here: https://arxiv.org/abs/hep-th/9108028 – AccidentalFourierTransform May 05 '18 at 20:04
  • @AccidentalFourierTransform Thank you, it was useful to read it to start shedding some light on this. – soap May 05 '18 at 20:46

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You are confusing 2 different things:

A d-dimensional Lie group has a d-dimensional algebra which is the tangent space at the identity.

Here, there is an infinite dimensional group acting on a 2 dimensional space. There is a representation of the algebra as vector fields (derivations) on this space, given by the formulas you stated. (infinitesimal conformal transformations)

The generators should depend on the point z, as each conformal transformations act differently around each point - for example, the map $z$ to $z^2$ acts infinitisimaly as $z+\delta z$ to $z+2z\delta z$. The dependence on the point is easiliy understood - for example under this map, the more far z is away from the origin, it is stretched more.

tsufli
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  • I suppose the infinite dimensional group acting on the 2-dimensional space is the "Infinitesimal conformal group"? (And not the Conformal group itself). This is vague for me since so far I only understood what Blumenhagen did with infinitesimal conformal transformations as transformations of the conformal group near the identity – soap May 05 '18 at 19:49
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    No. It is the conformal group acting on the 2-d space. Upon exponentiation, the generators will become elements of the conformal group itself. The generators themself are not acting on the space, only their exponents do. – tsufli May 05 '18 at 19:53
  • Ok, but how do "infinitesimal conformal transformations" come into play? From what I understood so far, we have: the conformal group $G$ together with a representation of $G$ in a 2D euclidean space. $G$ has a Lie algebra which has a basis given by the generators that I wrote above. What now? – soap May 05 '18 at 20:29