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In a general CFT (with no boundary), the one-point correlation function of any primary operator $A_{\Delta}$ is $0$ (unless we are talking about the identity operator): $$ <A_{\Delta}(x)> = 0 $$ The above result is derived using only scaling symmetry and translational symmetry of the theory.

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Now if we consider the upper half plane of $\mathbb{R}^{n}$, which we call $H = \left\{ (x^{1},\ldots,x^{n}) \in \mathbb{R}^{n} | x^{n} \geq 0 \right\}$, certain symmetries of our ordinary CFT are broken.

This forces the one-point correlation function of primary operators to take a certain form, which is non-zero in general. I wanted some confirmation on what symmetries of the theory are now broken? Since in ordinary $\mathbb{R}^{n}$ proving that $<A_{\Delta}(x)>=0$ relied on translational and scaling symmetry, I'm assuming one of these two symmetries is broken.

I have a hunch that scaling symmetry is broken but I don't understand how I can argue this beyond some hand-waving. Is there some way to show this?

Essentially my questions is: does this space break the scaling symmetry of our CFT, and if this is the case how do I show this?

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    $\begingroup$ Your subspace is invariant under scalings but not under translations. If you make a translation in any direction, the description of the subspace in the new coordinates is not of the same form as before. Isn't that quite obvious? I don't know what you actually want to show. $\endgroup$ – Herr_Mitesch Feb 10 '17 at 20:32
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The upper half-plane $H$ is invariant under dilatations because $x^n\geq 0\Rightarrow \lambda x^n\geq 0$ for $\lambda > 0$. It is also invariant under translations $x^i\to x^i + c^i$ for $i<n$, but not under $x^n\to x^n+c^n$. The one-point function must respect the $n-1$ translation invariances and the dilatation invariance, therefore $$ \langle A_\Delta(x) \rangle \propto (x^n)^{-\Delta} $$ This coincides with the $x$-dependence of a two-point function on the plane, with one operator at $x$ and another one at the reflected position $(x^1,\dots ,x^{n-1},-x^n)$, with conformal dimension $\frac{\Delta}{2}$.

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    $\begingroup$ this is exactly the argument I have been looking for...in terms of components of the transformed vector. Is there a way to examine how the one-point correlation function looks like for points $x$ on the inside of a unit sphere? To me it seems like every symmetry in this space is broken - definitely dilations, inversions and translations, but not rotations...? $\endgroup$ – Greg.Paul Feb 12 '17 at 18:07
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    $\begingroup$ In 2d the unit sphere is related to a straight line by a conformal transformation, i.e. $f(z)=\frac{1}{1-z}$ maps the unit sphere $|z|=1$ to a straight line. I think this holds in any dimension, with a hyperplane instead of a straight line. So the inside of the unit sphere is mapped to a half-plane, and you are back to your original problem. $\endgroup$ – Sylvain Ribault Feb 13 '17 at 19:49
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    $\begingroup$ Perfect. One last small question - where does the factor of $2$ come from in $(x^{n})^{-2\Delta}$? When deriving this result myself (using the dilation invariance), I find that I'm only getting $(x^{n})^{-\Delta}$.....This is coming from the fact that the dilation invariance forces the function $f(x^{n})$ (this forced by $n-1$ translation invariances) to satisfy $\lambda^{-\Delta}f(x^{n}) = f(\lambda x^{n})$ - what am I missing here? $\endgroup$ – Greg.Paul Feb 14 '17 at 5:02
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    $\begingroup$ Using the mapping $f(x) = ( \frac{2x_{1}}{1 - 2x_{n}+ x^2},\ldots, \frac{2x_{n-1}}{1 - 2x_{n}+ x^2},\frac{1-x^{2}}{1 - 2x_{n}+ x^2} )$ (which conformally maps the unit disk to the upper-half plane) I've also concluded from what you've said that the correlator inside the unit ball goes like $\propto \frac{(1-2x_{n}+x^{2})^{\Delta+1}}{1 - x^{2}}$. $\endgroup$ – Greg.Paul Feb 14 '17 at 6:27
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    $\begingroup$ I think you are right about the factor of $2$, I corrected the answer to $(x^n)^{-\Delta}$. As for the mapping from the unit ball to the upper half-plane, I think that the resulting correlator should be $(\text{something})^\Delta$. $\endgroup$ – Sylvain Ribault Feb 15 '17 at 7:45

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