Is every conformal transformation given by a conformal Killing field?

179 Views Asked by At

I am reading A Mathematical Introduction to Conformal Field Theory, Second Edition by Schottenloher. In the first chapter, he classifies the conformal transformations on $\Bbb R^{p,q}$ by classifying those transformations that arise as the flows of conformal Killing fields. He states that

Every conformal transformation $\varphi$ on a connected open subset of $\Bbb R ^{p,q}$ which is an element $\varphi = \varphi_{t_0}$ of a one-parameter group $(\varphi_t)$ of conformal transformations, is a composition of maps of the form $$ q \mapsto \left\{q + c, \Lambda q \ (\text{with } \Lambda \in O(p,q)), e^\lambda q, \frac{q - \langle q, q\rangle b}{1 - 2 \langle q, b\rangle + \langle q, q \rangle \langle b, b \rangle}\right\}. $$

Now, I know that all sufficiently smooth conformal transformations are of this form, so I am wondering if there is a way to derive the more general classification (all sufficiently smooth conformal transformations) from that for the transformations that arise as flows of conformal Killing fields.

On the Wikipedia page for conformal groups it states that all conformal groups are Lie groups. I wonder if using the fundamental vector field, I can show that all the elements of the conformal group are given by conformal Killing fields. I imagine these results are probably in some introductory textbook, but I am not sure where to look. Can anybody recommend a comprehensive reference?

1

There are 1 best solutions below

0
On BEST ANSWER

Let $(M, g)$ be a Riemannian manifold and let $G$ be the group of conformal diffeomorphisms from $M$ onto itself. Assume $G$ is a connected Lie group. Any $X \in \frak{g} = \text{Lie}(G)$ generates a one-parameter subgroup of $G$ given by $\phi_t = \exp(t X)$. Denote the action of $\phi_t$ on $M$ by $p \cdot \phi_t$. Because $\phi_t$ is a one-parameter group in $G$, the map $p \mapsto p \cdot \phi_t$ defines a global flow on $M$ with infinitesimal generator $\hat X$. We want to show that $\hat X$ is a conformal Killing vector field. \begin{align*} (\mathcal L_{\hat X} g)_p(Y, Z) &= \lim_{h \to 0} \frac{1}{h} \left( (\phi_h ^*g)_p(Y, Z) - g_p(Y, Z) \right) \\ &= \lim_{h \to 0} \frac{1}{h} \left( \lambda_h(p) g_p(Y, Z) - g_p(Y, Z) \right) \\ &= \lim_{h \to 0} \frac{1}{h} \left( \lambda_h(p) - 1 \right) g_p(Y, Z) = \frac{\text d }{\text d t} \Big|_{t = 0} \lambda_t(p) g_p(Y, Z) \end{align*} In particular, if we let $\kappa(p) = \frac{1}{2}\frac{\text d }{\text d t} \Big|_{t = 0} \lambda_t(p)$, and expand the Lie derivative in terms of the Levi-Civita connection to get \begin{equation} \text{Sym}(\nabla \hat X ^{\flat}) = \kappa g. \label{eq:conf-killing-eq} \end{equation} Solving this system of linear, first order partial differential equations for $\hat X$ and then constructing its flow furnishes every $1$-parameter subgroup of $G$. If $G ^*$ is the subgroup generated by all $1$-parameter subgroups, then $G ^*$ is path-connected because each of its points can be reached from the identity by following a finite sequence of flows. It follows that $\bar{G ^*}$ is also connected, and by Cartan's theorem it is a Lie subgroup of $G$ which contains all of its $1$-parameter subgroups, whence $\text{Lie}(G) = \text{Lie}(\bar{G ^*}) \Rightarrow G = \bar{G ^*}$ by the Lie Correspondence (Theorem 20.21 in Lee SM). We have hence shown that when $\text{Conf}(M)$ is a connected Lie group, then all conformal transformations of $M$ can be found by solving the conformal Killing equation