conformal field theory - State operator map in $R times S^{D-1}$ to $R^D$

This question is relevant in my former question 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 $R^2$)

First what i know from Ginsparg's applied conformal field theory the state in $R\times S^1$, cylinder and operator in $R^2$, plane, there is a one-to-one map. $i.e$, following conformal map we can make one to one map between them.

$$\begin{align} \xi = t+ix, \quad z = \exp[\xi]=\exp[t+ix] \end{align}$$ here $\xi$ is a cylinder's complex coordinate, and $z$ is a plane's complex coordinate.

---

How about in $R \times S^{D-1}$ to $R^D$? (From Tong's lecture notes on string theory, section 4.6, i know there is a State operator map between them.) Is it okay just set

$$\begin{align} \xi = t + i( x_1 + x_2 + \cdots x_{D-1}), \quad z=\exp[\xi] \end{align}$$

Answer

This is not a state-operator correspondence map. The 2D state-operator correspondence is given by a map between the Hilbert space of states $\mathcal{H}$ with a $\mathrm{PSL}(2,\mathbb{C})$-invariant vacuum $\Omega$ and fields $\phi: \mathbb{C}\to\mathrm{U}(\mathcal{H})$ explicitly given by

$$\{\text{fields}\}\to\mathcal{H},\ \phi \mapsto \lim_{z\to 0}\phi(z,\bar z)\Omega$$ with inverse $$\mathcal{H}\to \{\text{fields}\}, v \mapsto \phi_v \,\text{with}\,\phi_v(z,\bar z) = \mathrm{e}^{zL_{-1} + \bar z \bar L_{-1}}$$

By this correspondence, the state-operator correspondence holds for every 2D conformal field theory that possesses a conformal map to the plane. This is one of the motivating reasons for using radial quantization for the cylinder $\mathbb{R}\times S^1$, where it is mapped to the plane by the map you mention, which makes the radial coordinate the time.

This state-operator correspondence crucially relies on the existence of a conformal map between the cylinder and the plane, but it is not the same. We need the plane to be able to talk about $\lim_{z\to 0}$, but we need the cylinder to have a reason to have the radial coordinate be the time direction, and hence the limit point $z\to 0$ be a spatial slice in the infinite past instead of just some point in the plane.

Now, in higher dimensions for CFTs, this still works - we keep a conformal map between the "cylinder" $\mathbb{R}\times S^{d-1}$ and the $\mathbb{R}^d$ that is given by

$$\mathbb{R}\times S^{d-1} \to \mathbb{R}^d,\, (t,\Phi)\mapsto (e^t,\Phi)_\text{pol}$$ where I mean to say that the map is given by observing that $\mathbb{R}^d$ is the union of the spheres $S^{d-1}$ with radius $r$, and then sending the point $(t,\Phi)\in \mathbb{R}\times S^{d-1}$ to the point $\Phi\in S^{d-1}$ on the sphere with radius $e^t$ in $\mathbb{R}^d$.

One can show that this suffices to give a higher-dimensional analogue of the 2D state-operator correspondence. However, let me stress again that is it not the case that states live on the cylinder and operators on the plane. The states at a time are associated to the spatial slice at that time of the spacetime, i.e. to the $S^{d-1}$-slices, regardless of whether we are on the cylinder or the plane, and the state-operator correspondence map is more than just the conformal mapping of the cylinder to the plane.

Also, one has to note that the maps $\mathbb{R}\times S^{d-1}\to\mathbb{R}^d$ are not bijections - they are precisely not surjective onto $0\in\mathbb{R}^d$, which should not be surprising, since that corresponds to the infinite past, and $-\infty$ is not in $\mathbb{R}$, either. The conformal map is precisely what allows us to make the limit towards the infinite past well defined as a limit towards a single point - the origin - in $\mathbb{R}^d$.