We use the recently developed framework of the Mellin bootstrap to study
perturbatively free scalar CFTs in arbitrary dimensions. This approach uses the
crossing-symmetric Mellin space formulation of correlation functions to generate algebraic
bootstrap equations by demanding that only physical operators contribute to the OPE.
We find that there are no perturbatively interacting CFTs with only fundamental scalars
in

Article funded by SCOAP3

6$, to the perturbative order we are able to calculate. This provides evidence that there are no interacting CFTs with a single fundamental scalar in $d>6$ dimensions. Our results can also easily be generalized to the case of an arbitrary number of fundamental scalars, again providing evidence that there are no interacting CFTs with \emph{any number of} fundamental scalars in $d>6$ dimensions. Our results are perhaps not surprising: for perturbatively free scalar theories, Lagrangian methods give an easy argument that no non-trivial marginal operator can be constructed in $d>6$ for scalar theories - therefore no interacting scalar theory would be expected to exist. Furthermore, recall that the superconformal algebra does not close in $d>6$, so there are no superconformal field theories in this regime~\cite{Nahm-ml-1977tg}. However, the Mellin bootstrap approach allows us to approach the question of existence of CFTs from a fundamentally different viewpoint not involving Lagrangians at all. This approach also begins to clarify the criteria under which the Mellin bootstrap reduces to a finite set of equations. It would be interesting to expand our results to include fermions, other (fundamental) fields, and to connect these techniques with the analysis of generalized Wilson-Fisher fixed points in~\cite{Gliozzi-ml-2016ysv, Gliozzi-ml-2017hni}. This paper is organized as follows: section~\ref{sec:setup} briefly introduces the Mellin bootstrap and our CFT setup, including our assumptions and the resulting simplifications that occur that reduce the bootstrap equations to a finite number of algebraic equations. Section~\ref{sec:bootstrap} discusses solving these bootstrap equations to show that the theory we obtain is forced to be free. The necessary definitions of functions etc. can be found in appendix~\ref{sec:app:def} while the simplifications are discussed in more detail in appendix~\ref{sec:app:simpl}. ]]>

6$ is the free theory which saturates the unitarity bound in $d_\eps$ dimensions, regardless of the sign of $\eps$. (By contrast, note that $\eps>0$ is necessary in the $d= 6 - \eps$ analysis of~\cite{Gopakumar-ml-2016cpb}; there, when $\eps<0$, the resulting $\Delta_\phi$ can be seen to violate the unitarity bound already~at~$\oeps{1}$.) Our goal is to determine the values of the expansion coefficients in eqs.~(\ref{eq:defdelphi})--(\ref{eq:defCT}) given the assumptions \textbf{A1} and \textbf{A2}. We will do so using the Mellin bootstrap, which we review now. ]]>

0$, we have $c_{\Delta,l'}q^{(a,t)}_{\Delta, l|l'}= \oeps{2}$; i.e.\ to this order in $\eps$, only scalars contribute to the sum $\sum_{l'} c_{\Delta,l'}q^{(a,t)}_{\Delta, l|l'}$. This infinite sum over $l'$ thus reduces~to: \begin{equation} \sum_{l'} c_{\Delta,l'}q^{(a,t)}_{\Delta, l|l'} = c_{\Delta,l'=0} q^{(a,t)}_{\Delta,l|l'=0} + \oeps{2}. \end{equation} \item \textbf{S3}: furthermore, in the $t$-channel, for ``heavier'' scalars of $\Delta > \Delta_0$, we have $c_{\Delta,0}q^{(a,t)}_{\Delta, l|0}= \oeps{2}$. This reduces the infinite sum over $\Delta$ in the $t$-channel to: \be \sum_{\Delta} c_{\Delta,l'=0} q^{(a,t)}_{\Delta,l|l'=0} = c_{\Delta=\Delta_\phi,l'=0} q^{(a,t)}_{\Delta=\Delta_\phi,l|l'=0} + c_{\Delta=\Delta_0,l'=0} q^{(a,t)}_{\Delta=\Delta_0,l|l'=0}+\oeps{2}. \ee \end{itemize} Derivations of these simplifications are in appendix~\ref{sec:app:simpl}. With these simplifications, the bootstrap equations for generic $l$ become \begin{align} \label{eq:ll2} c_{\Delta_{J_l},l} q_{\Delta_{J_l},l}^{(2,s)} + 2 c_{\Delta_\phi,0} q^{(2,t)}_{\Delta_\phi,l|0} +2 c_{\Delta_0,0} q^{(2,t)}_{\Delta_0,l|0} &=\oeps{2}, & (l\neq0,2),\\ \label{eq:ll1} 2q^{(1,t)}_{0,l|0} +c_{\Delta_{J_l},l} q_{\Delta_{J_l},l}^{(1,s)} + 2 c_{\Delta_\phi,0} q^{(1,t)}_{\Delta_\phi,l|0} +2 c_{\Delta_0,0} q^{(1,t)}_{\Delta_0,l|0} &=\oeps{2},& (l\neq0,2), \end{align} and in the special cases $l=0$ and $l=2$ we have \begin{align} \label{eq:l22} c_{d_\eps,2} q_{d_\eps,2}^{(2,s)} + 2 c_{\Delta_\phi,0} q^{(2,t)}_{\Delta_\phi,2|0} +2 c_{\Delta_0,0} q^{(2,t)}_{\Delta_0,2|0} &=\oeps{2},\\ \label{eq:l21} c_{d_\eps,2} q_{d_\eps,2}^{(1,s)} + 2q_{0,2|0}^{(1,t)} + 2 c_{\Delta_\phi,0} q^{(1,t)}_{\Delta_\phi,2|0} + 2 c_{\Delta_0,0} q^{(1,t)}_{\Delta_0,2|0} &=\oeps{2},\\ \label{eq:l02} c_{\Delta_\phi,0}\left(q_{\Delta_\phi,0}^{(2,s)}+2 q^{(2,t)}_{\Delta_\phi,0|0}\right)+c_{\Delta_0,0}\left(q_{\Delta_0,0}^{(2,s)} + 2 q_{\Delta_0,0|0}^{(2,t)}\right) &= \oeps{2},\\ \label{eq:l01} 2q^{(1,t)}_{0,0|0} + c_{\Delta_\phi,0}\left(q_{\Delta_\phi,0}^{(1,s)}+2 q^{(1,t)}_{\Delta_\phi,0|0}\right)+c_{\Delta_0,0}\left(q_{\Delta_0,0}^{(1,s)} + 2 q_{\Delta_0,0|0}^{(1,t)}\right) &= \oeps{2}. \end{align} All of the equations now only involve a finite number of terms. Demanding that the left-hand side of these equations vanish up to the given order in $\eps$ will give us non-trivial constraints on the coefficients appearing in~(\ref{eq:defdelphi})--(\ref{eq:defCJ}). ]]>

2$, and the $t$-channel identity operator only arises in the simple-pole constraints. For $d_0\le6$, all of these terms can make finite, non-zero contributions through $\oeps{1}$ (though note that they do not all contribute in all cases --- for example $\phi_\text{($t$-channel)} = 0$ in the $\mathbb{Z}_2$-symmetric case as $C_{\phi\phi\phi}=0$). The essential story for the $d_0\le6$ regime is that the complicated interplay between non-trivial $s$- and $t$-channel contributions produces a set of linear equations to get cancellations up to $\oeps{2}$. This set of linear equations can then be used to determine operator dimensions, e.g.\ for the Wilson-Fisher fixed point in $d_\eps = 4-\eps$ and the $\phi^3$ theory in $d_\eps = 6-\eps$. We will show that for $d_0>6$, the $\phi$ and $\phi^2$ $t$-channel contributions necessarily begin $\oeps{2}$, so they drop out of the bootstrap equations and all we are left with is the $t$-channel identity to cancel the $s$-channel operator. Without these non-trivial $t$-channel contributions, the only possible solution to the resulting set of equations is the free theory. ]]>

6$ so far, the $\phi^2$ $t$-channel contribution in fact drops out already at $d_0>4$. We can see from a simple series expansion that the normalization coefficient $c_{\Delta_0,0}$ already exhibits a special value at $d_0=4$: \begin{equation} c_{\Delta_0,0} \propto \begin{cases} \oeps{4} & d_0 = 4,\\ \Gamma\left(d_0/2-2\right)^{-2}\eps^{2} & d_0 > 4. \end{cases} \end{equation} A less trivial calculation (see appendices~\ref{sec:app:def} and~\ref{sec:app:simpl} for details) reveals that \begin{equation} q_{\Delta_0,l|0}^{(a,t)}\propto \begin{cases} \oeps{-3} & d_0 = 4,\\ \Gamma\left(d_0/2-2\right)^{3}\eps^{0} & d_0 > 4. \end{cases} \end{equation} Putting these together gives \begin{equation}\label{eq:divergentterm} c_{\Delta_0,0}q_{\Delta_0,l|0}^{(a,t)} \propto \begin{cases} \oeps{1} & d_0 = 4,\\ \Gamma(d_0/2-2)\eps^2 & d_0 > 4. \end{cases} \end{equation} The fact that these poles behave differently in $d_0>4$ is not particularly surprising. We know from~\cite{Gopakumar-ml-2016cpb} that at least in $d_0=6$, the $\phi^2$ operator indeed does only contribute at higher order (otherwise it would compete with the $C_{\phi\phi\phi}$ exchange in the non-$\mathbb{Z}_2$ invariant theory); however, what we find here is that the $\phi^2$ operator cannot contribute to the relevant orders for \emph{any} $d_0>4$ (in particular $d_0=4n$ for $n>1$ is not special in any way). ]]>

6. \end{cases} \end{equation} However, for $d_0>6$, $q^{(a,t)}_{\Delta_0,l|0}$ has an infinite sum of contributions from poles of the integrand of that go as $\oeps{0}$. In particular, this infinite sum is not convergent. This means that in $d_0>6$, we have (schematically): \be c_{\Delta_\phi,0}q^{(a,t)}_{\Delta_0,l|0} = C_\phi^{(0)} \eps^{-1}(\infty),\ee where we have made explicit the factor of $C_{\phi}^{(0)}$ that is included in the normalization $c_{\Delta_\phi,0}$. There are no other infinities that this could cancel with until the infinite tower of heavier operators enters at $\oeps{2}$ in the bootstrap expressions; thus, we are \emph{forced} to set $C_\phi=\oeps{3}$ in order for the bootstrap equations to make any sense to $\oeps{2}$. Note that this suppresses the contribution of $\phi$ from all the bootstrap equations~(\ref{eq:ll2})--(\ref{eq:l01}). It is interesting to compare this with the $\phi^2$ case, as the $\phi$ $t$-channel contributions take an apparently similar form: \begin{equation}\label{eq:divergenttermphi2} c_{\Delta_\phi,0}q_{\Delta_\phi,l|0}^{(a,t)} \propto \begin{cases} \oeps{1} & d_0 = 6,\\[3mm] C_\phi^{(0)}\Gamma\left(\dfrac{d_0-6}{4}\right)^{-2}\eps^{-1}(\infty) & d_0 > 6. \end{cases} \end{equation} However, let us emphasize that the reason $\phi$ must drop out of the bootstrap for $d_0>6$ is different than $\phi^2$: we saw that the contribution of the $\phi^2$ operator is automatically suppressed to $\oeps{2}$ for $d_0>4$; here, the contribution of $\phi$ in $d_0>6$ is actually \emph{infinite}, forcing us to set the OPE coefficient $C_\phi=\oeps{3}$. This could not have happened for the $\phi^2$ operator, as the $\phi\phi\phi^2$ OPE is already non-zero in the free theory. ]]>

4$ for the $\phi^2$ contributions plays a critical role. Using~(\ref{eq:divergentterm}), we find that the $\oeps{1}$ term of~(\ref{eq:l02}) gives us: \begin{equation}\label{eq:delta01} \delta_0^{(1)} = \begin{cases} \frac13 & d_0 = 4\\ 0 & d_0 > 4 \end{cases} \end{equation} The $d_0=4$ value again matches that of~\cite{Gopakumar-ml-2016cpb}. Because $c_{\Delta_0,0}q_{\Delta_0,0|0}^{(2,t)}$ is suppressed by an extra order of $\eps$ in $d_0>4$, we see that $\delta_0^{(1)}$ is forced to vanish in this case, giving the free theory~result. ]]>

6$): \begin{align} \label{eq:delphisum} \Delta_\phi &= \Delta_{\phi,\text{free}} + \oeps{2} ,& \Delta_{\{0,2,l\}}& = \Delta_{\{0,2,l\},\text{free}} + \oeps{2},\\ \label{eq:Cphisum} C_\phi& = 0 + \oeps{3}, & C_{\{0,2,l\}} &= C_{\{0,2,l\},\text{free}} + \oeps{2}. \end{align} Therefore we conclude that any CFT in $d_0>6$ with a conserved stress tensor and a single fundamental scalar must be perturbatively free up to $\oeps{2}$. Furthermore, for $\mathbb{Z}_2$-symmetric theories the range of validity for these results extends to $d_0>4$, since the $\phi^2$ operator drops out already at that dimension. We stress that these results only depend on the value of $d_0$, and in particular not on the sign of $\eps$. These results can also be generalized to higher orders in $\eps$, as well as to include multiple scalars. \paragraph{Higher orders in \texorpdfstring{\boldmath $\eps$}{eps}.} While our results are certainly sugggestive, they are at relatively low orders in $\eps$. To what extent can we push this to deeper orders? We have found two regimes in which we can go further: \begin{align} \label{def:extraassumptions} \begin{split} \bullet ~~d_\eps &= 4n + 2 + \oeps{1}, \text{ or}\\ \bullet ~~d_\eps &= 2n + 2 + \oeps{1} \text{ and the CFT is $\mathbb{Z}_2$-symmetric}, \end{split} \end{align} for $n>1, n \in \mathbb{Z}^+$. As described in appendix~\ref{sec:app:simpl} (see especially~\ref{sec:app:coeffc}), when either of these is true, the simplifications \textbf{S1}-\textbf{S3} actually hold to $\oeps{4}$ (note this also requires $\delta_l^{(1)}=0$, which is guaranteed by~(\ref{eq:delphisum})). This means that the right hand side of~(\ref{eq:ll2})--(\ref{eq:l01}) is $\oeps{4}$ and we can then perform the bootstrap analysis above at orders $\oeps{2}$ and $\oeps{3}$ as well. The analysis of the equations at these orders proceeds entirely analogously as above, and results in determining the quantities in question to higher order in $\eps$:\footnote{Note that we expect the polynomial ambiguities in the Mellin space Witten diagrams in~(\ref{eq:Ms}) and~(\ref{eq:Mt}) to start contributing at $\oeps{4}$~\cite{Dey-ml-2017fab}, which would complicate the analysis beyond this order in $\eps$ even further.} \begin{align} \label{eq:delphisumA3} \Delta_\phi &= \Delta_{\phi,\text{free}} + \oeps{4} ,& \Delta_{\{0,2,l\}}& = \Delta_{\{0,2,l\},\text{free}} + \oeps{4},\\ \label{eq:CphisumA3} C_\phi& = 0 + \oeps{5}, & C_{\{0,2,l\}} &= C_{\{0,2,l\},\text{free}} + \oeps{4}. \end{align} \paragraph{Bootstrap for multiple scalars.} Our analysis was done assuming only one fundamental scalar $\phi$ that saturates the unitary bound when the theory is free at $\eps=0$. However, it is fairly straightforward to see that we can relax this assumption to include an arbitrary number $N$ of fundamental scalars $\phi^{(i)}$, with dimensions: \be \Delta_{\phi^{(i)}} = \frac{d_\eps-2}{2} + \delta_{\phi^{(i)}}^{(1)} \eps + \delta_{\phi^{(i)}}^{(2)} \eps^2 +\oeps{3}.\ee We can perform $N$ versions of the four identical scalar bootstrap that we have performed above in the case of one scalar. The simplifications \textbf{S1}-\textbf{S3} still hold with arbitrary index structures on the heavy operators $\mathcal{O}_{k,l,m}$. The rest of the bootstrap proceeds analogously - for instance, the $l=2$ bootstrap equations will give e.g.\ $\delta_{\phi^{(i)}}^{(1)}=0$ for every $i$. The only possible extra non-triviality that one might worry about is the existence of a conserved spin-1 current (such as in the $O(N)$ model). However, since our results force the scalar theory into a free theory, the relevant spin 1 current is automatically conserved, anyway, without needing any further input.\footnote{Note that the spin-2 conserved current, the stress-energy current, is not automatically conserved. We need to demand its conservation in order to find the condition $\delta_{\phi^{(i)}}^{(1)}=0$.} Our analysis is thus insensitive to any demand of global symmetries the theory could have, as the resulting theory is always free. We thus conclude that our analysis can be extended to the case of $N$ fundamental scalars $\phi^{(i)}$ with arbitrary indices placed on the relevant operators. It is surprising that with very little input or assumptions, we are able to see clear evidence that no perturbative interacting CFTs with \emph{any number of} fundamental scalars exist in $d_0>6$, without needing any notion at all of the Lagrangian. Perhaps by considering more of the infinite class of spurious poles in the Mellin bootstrap one could further relax our assumptions to fully explore the world of CFTs in $d_0>6$. ]]>

2$ vanishes (because e.g.\ the three point function $\phi\phi\phi^k$ vanishes for $k>2$); this implies that: \be \label{eq:app:Ccoeffeps2} C_{\phi\phi\mathcal{O}_{k,2m,l}}= \oeps{2}, \ee (or a higher power) as the $C$ coefficient is the \emph{square} of the $\phi\phi \mathcal{O}_{k,2m,l}$ OPE coefficient (which is reasonable to assume goes as $\oeps{1}$). For $k=2$, we can consider only (primary) operators where the derivatives hit only one of the $\phi$'s, i.e.\ of the schematic form $\phi (\partial^2)^m \partial_{i_1}\cdots \partial_{i_l}\phi$. For $m>0$, these operators identically vanish in the free theory so that we again have~(\ref{eq:app:Ccoeffeps2}). Only the operators with $k=2, m=0$ (i.e.\ the $J_l$ operators) may have: \be \label{eq:app:Ccoeffeps0} C_{\phi\phi\mathcal{O}_{2,0,l}} = \oeps{0}.\ee Next, we note that the normalization factor $\mathcal{N}_{\Delta,l}$ of every heavy operator with $k=2$ goes as:\footnote{We already set $\delta_{\phi}^{(1)}=0$ here for simplicity; this does not change the leading power of $\eps$ in $\mathcal{N}_{\Delta,l}$.} \be \label{eq:norml} \mathcal{N}_{\Delta,l} = \left(\delta^{(1)}_{k=2,2m,l}\right)^2 g(h_0) \oeps{2} + \oeps{4},\ee while when $k>2$, we have: \be \mathcal{N}_{\Delta,l} = \Gamma\left(-\frac12(k-2)(h_0-1)-m\right)^{-2}\tilde{g}(h_0) \oeps{0},\ee where $g(h_0)$ and $\tilde{g}(h_0)$ are unimportant additional factors involving $h_0$. Note that the inverse gamma function has a zero when its argument is a negative integer; the only way to achieve this here is if $k=2n_1,h_0=n_2$ with $n_1,n_2$ integers, or if $h_0=2n+1$ with $n$ integer (since $k$ is always an integer). Remembering that $d=2h$, this is precisely the condition presented in~(\ref{def:extraassumptions}). So, when~(\ref{def:extraassumptions}) holds, we have: \be \label{eq:normA3} \mathcal{N}_{\Delta,l} = \oeps{2}.\ee Putting everything together, we can conclude that for any heavy operator $\mathcal{O}_{k,2m,l}$, we~have: \be \label{eq:app:ceps2} c_{\Delta_{\mathcal{O}},l} = \oeps{2}.\ee When~(\ref{def:extraassumptions}) holds and when $\delta^{(1)}_{2,0,l} = \delta^{(1)}_l = 0$,~(\ref{eq:normA3}) and~(\ref{eq:norml}) imply that we have: \be \label{eq:app:ceps4} c_{\Delta_{\mathcal{O}},l} = \oeps{4}.\ee ]]>

2$ and/or $m>0$, gives \begin{equation} q^{(s)}_{\Delta,l}(s) = \oeps{0}. \end{equation} Since~(\ref{eq:app:ceps2}) holds for these operators, heavier operators in the $s$-channel begin contributing at $\oeps{2}$, i.e.\ we have: \be c_{\Delta_{\mathcal{O}},l} q^{(a,s)}_{\Delta_\mathcal{O},l} = \oeps{2}. \ee When~(\ref{def:extraassumptions}) holds and $\delta^{(1)}_l = 0$, this becomes $c_{\Delta_{\mathcal{O}},l} q^{(a,s)}_{\Delta_\mathcal{O},l} = \oeps{4}$. ]]>

0$ (with $k\geq 2$ and $m\geq 0$) satisfy:
\be \label{eq:S2maininapp} c_{\Delta_{\mathcal{O}}, l'} q_{\Delta_{\mathcal{O}},l|l'}^{(a,t)} = \oeps{2},\ee
for $l=0$ and $l=2$.\footnote{In checking these simplifications for $l\neq2$, it is computationally favorable to already set $\delta_\phi^{(1)}=0$ (as briefly mentioned in~\cite{Gopakumar-ml-2016cpb}). The only place where we need to keep $\delta_\phi^{(1)}$ explicitly non-zero is in checking the simplifications for $q^{(2,t)}$ when $l=2$.}
We first note that the integrand in the integral expression for $q_{\Delta_{\mathcal{O}},l|l'}^{(a,t)}$ (for $a=1,2$) has the following poles:
\begin{itemize}
\item[I.] $\nu = \Delta_\mathcal{O}-h$
\item[II.] $\nu = 2\Delta_\phi + l' - h + 2n_2$
\item[III.] $\nu = h +l'+ 2n_3$
\item[IV.] $\nu = h-1+n_4$; $\qquad$ $(n_4

2$ and/or $m>0$, we have: \be \label{eq:S3maininapp} c_{\Delta_{\mathcal{O}}, l'=0} q_{\Delta_{\mathcal{O}},l|l'=0}^{(a,t)} = \oeps{2}. \ee The analysis proceeds in a similar way to \textbf{S2}. The poles I, II, III (with $l'=0$) are present in the integrand for $q^{(a,t)}$, while the poles IV are now absent. The pole cancellations~(\ref{eq:S2polecan1}) still holds. For $m>0$, the cancellations~(\ref{eq:S2polecan4}) still hold. The final poles present satisfy: \begin{align} \label{eq:S3polecan1} Res_I + Res_{II(n_2=0)} &= \oeps{0} & (m=0),\\ \label{eq:S3polecan2} Res_{II(n_2=0)} &= \oeps{0} & (m>0). \end{align} We again have~(\ref{eq:app:ceps2}), so we can conclude that~(\ref{eq:S3maininapp}) indeed holds. When~(\ref{def:extraassumptions}) holds and $\delta^{(1)}_l = 0$,~(\ref{eq:app:ceps2}) implies that~(\ref{eq:S3maininapp}) becomes suppressed to $\oeps{4}$. ]]>