State operator map, and scalar fields in $R\times S^{D-1}$ to $R^D$

This question is generalized version of my previous questions State operator map in $R \times S^{D-1}$ to $R^D$ , State-operator map, and scalar fields and https://physics.stackexchange.com/q/215060/. (which was wrong, corrected one was states in $R \times S$ to operator in $R^2$)

Due to detail explanation from @ACuriousMind, now i can make conformal map between $R\times S^{D-1}$ to $R^D$. via map \begin{align} t \rightarrow r=e^t \end{align} with angle coordinates, $\Phi$ is in $S^{D-1}$.

For 2d case, From @kau, and Ginsparg's applied conformal field theory i know for conformal map $z \rightarrow f(z)$

\begin{align} \phi(z,\bar{z}) \rightarrow \phi'(z,\bar{z}) = \left(\frac{\partial f}{\partial z} \right)^h \left(\frac{\partial \bar{f}}{\partial \bar{z}} \right)^{\bar{h}}\phi(f(z), \bar{f}(\bar{z})) \end{align} where we call such $\phi$ as a primary field of conformal dimension $(h, \bar{h})$.

How about general case, $i.e$, $R\times S^{D-1}$ to $R^D$ case?

• Just a remark (since I'm not sure what exactly happens in $d>2$ to the fields): You have to be careful with the notion of "primary field". Since the conformal algebra is smaller in $d>2$ (it's just $\mathfrak{so}(p+1,q+1)$ for $p+q=d$ and $(p,q)$ the signature of the metric), the notion of "primary field" reduces to being invariant under special conformal transformations. Nov 12 '15 at 13:47

For the transformation of field \begin{align} \phi_{ R \times S^{D-1}}(t, \Phi) \rightarrow \phi_{R^D}(r, \Phi) = r^{-\Delta}\phi_{ R \times S^{D-1}}(t, \Phi) \end{align} where $\Delta$ is scaling dimension.