text:
\section{Introduction}
\label{INTRODUCTION}

Melonic theories \cite{Bonzom:2011zz,Gurau:2011xp,Carrozza:2015adg,Witten:2016iux,Klebanov:2016xxf} are an interesting class of quantum field theories whose essential property is that in an appropriate large $N$ limit, the dominant Feynman diagrams can be generated by iterating on the replacement of a propagator by a melonic insertion, as shown in figure~\ref{MelonicInsertion2} for a melonic version of scalar $\phi^4$ theory.  Melonic theories are interesting for two related reasons: 1) The melonic large $N$ limit is relatively tractable because its Green's functions can be determined through functional techniques including Schwinger-Dyson equations; and 2) The simplest Green's functions are the same as for the Sachdev-Ye-Kitaev (SYK) model \cite{Sachdev:1992fk,Kitaev:2015zz}, widely studied because of its proposed relationship to $AdS_2$.
 \begin{figure}[b]
  \centerline{\includegraphics[width=3in]{MelonicInsertion2.pdf}}
  \caption{A melonic insertion.}\label{MelonicInsertion2}
 \end{figure}

It is well recognized that melonic theories theories exist not just with quartic interactions, but also higher order interactions \cite{Narayan:2017qtw,Choudhury:2017tax,Ferrari:2017jgw,Tarnopolsky:2018env}.  However, in the literature to date there is not much systematic understanding of what sort of higher order interactions are possible.  The aim of this article is to take some steps toward such an understanding in the context of higher order versions of the Klebanov-Tarnopolsky model \cite{Klebanov:2016xxf} where the interaction is of order $q$---meaning that $q$ propagators meet at each interaction vertex.  Inquiries in this direction were initiated in \cite{prethesis}.  Starting at $q=8$, the number of different interaction vertices proliferates quickly.  Some of them are symmetrical under a subgroup of permutations of the propagators leading into them; other have no such symmetry.  Interaction vertices do not mix with one another in the leading melonic limit: The diagrams that would permit this are subleading.  We are therefore content to restrict to theories with only one type of interaction vertex---and each different interaction vertex gives a different theory.  An interaction vertex of order $q$ can be constructed starting from a coloring of the complete graph with $q$ vertices such that each of $q-1$ colors is incident once on each vertex.  The questions we wish to settle regarding these interaction vertices are:
 \begin{enumerate}
  \item What are the possible symmetry groups of these interaction vertices?
  \item How does the number of distinct interaction vertices grow with $q$?
 \end{enumerate}
We claim in sections~\ref{SYMMETRY} and~\ref{CONSTRUCTION} to completely settle question \#1: The possible symmetry groups are $\mathbb{Z}_2^n$, where $0 \leq n \leq v$ if $q = 2^v$ and $0 \leq n < v$ if $q = u 2^v$ where $u$ is odd and larger than~$1$.  (It is easy to show that $q$ must be even.)  Our demonstration is constructive, in that we produce vertices with each possible symmetry.  Question \#2 turns out to be difficult, and it is essentially the problem of counting so-called one-factorizations of complete graphs, where results are generally available up to $q=14$ \cite{Dickson:1906,Gelling:1973,Gelling:1974,Dinitz:1987,Seah,Dinitz:1994,Kaski:2009}: results spanning over a century!  In section~\ref{ONEFACTOR}, we summarize how these results can be combined and modestly extended to give complete results on the number of distinct interaction vertices up to $q=14$.  Our symmetry analysis suggests a new twist on the counting problem: In addition to counting all one-factorizations, one can count one-factorizations with a given symmetry group.  An explicit example of this symmetry-constrained counting is presented in appendix~\ref{AlgorithmAppendix}, and some additional conceptual points are discussed in appendix~\ref{IsomorphismOrdered}.

The symmetry group of the interaction vertex leads to an effective coupling that enters into the Schwinger-Dyson equation for the two-point function and the ladder operator used for computing the four-point function, which we exhibit explicitly in all cases in sections~\ref{TWOPOINT} and~\ref{FOURPOINT}.\footnote{All indications are that every interaction vertex we construct leads to a theory with a melonic limit; however, we do not have a fully rigorous proof of this claim.}  We work out all our results for theories not only over the reals $\mathbb{R}$, but also over the $p$-adic numbers $\mathbb{Q}_p$, where as already shown in \cite{Gubser:2017qed}, there is quite a variety of melonic theories, depending on what sign function one chooses over $\mathbb{Q}_p$.  For approximately half of these melonic theories over $\mathbb{Q}_p$, the Schwinger-Dyson equation can be solved exactly, not just in the infrared, but at all scales, in terms of the solution to a $q$-th order polynomial equation.  Remarkably, as shown in section~\ref{ADELIC}, there is an adelic product formula relating the eigenvalues of the ladder operator integral equation across real and $p$-adic theories.

\section{Structure of higher melonic theories}
\label{STRUCTURE}

The action of the simplest Klebanov-Tarnopolsky model \cite{Klebanov:2016xxf} is
 \eqn{SKT}{
  S = \int_{\mathbb{R}} dt \, \left[
    {i \over 2} \psi^{a_0a_1a_2} \partial_t \psi^{a_0a_1a_2} +
    {g \over 4} \psi^{a_0a_1a_2} \psi^{a_0b_1b_2} \psi^{c_0a_1b_2} \psi^{c_0b_1a_2} \right] \,,
 }
where each index $a_0$, $a_1$, $a_2$, which we can think of as a color index, runs independently from $1$ to $N$, and each $\psi^{a_0a_1a_2}$ is a Majorana fermion.
There is a different $O(N)$ for each index position, so that the whole of $\psi^{a_0a_1a_2}$ transforms in the tensor product of the fundamental representations of three copies of $O(N)$.
We are interested in the color structure for higher rank models with higher degree interactions. We also want to generalize to models defined over the $p$-adic numbers $\mathbb{Q}_p$ and to models with $O(N)$ or $Sp(N)$ indices, as in \cite{Gubser:2017qed}.  Schematically, the lagrangians we will consider take the form
 \eqn{SKTextended}{
  S &= \sigma_\psi
   \int_K d\omega \, {1 \over 2} \psi^A(-\omega) \psi^B(\omega) |\omega|^s (\mop{sgn}\omega)
    \Omega_{AB}  \cr
    &\qquad\qquad{} + (\sigma_\psi)^{\frac{q}{4}}
   \int_K dt \, {g \over |G|} \left( \prod_{i=0}^{q-1} \psi^{A^{(i)}}(t) \right)
    \Omega_{A^{(0)}A^{(1)}\dots A^{(q-1)}} \,,
 }
where $K$ is either $\mathbb{R}$ or $\mathbb{Q}_p$ (or possibly some larger vector space, for instance $\mathbb{C}$ or a field extension of some $\mathbb{Q}_p$) and $G$ is the automorphism group of the interaction vertex, to be discussed further in section~\ref{SYMMETRY}.  The spectral parameter $s$ would usually be chosen to be $1$ for fermionic theories over $\mathbb{R}$ or $2$ for bosonic theories over $\mathbb{R}$, but for theories over $\mathbb{Q}_p$ it is more natural to let it vary continuously over positive real values.  We set $\sigma_\psi = -1$ for fermionic theories, and $\sigma_\psi = +1$ for bosonic theories.  By $\mop{sgn}\omega$ we mean a sign character, which is to say a multiplicative homomorphism of non-zero elements of $K$ to $\{1,-1\}$.  Capital indices are really groups of $q-1$ lowercase indices, each $N$-valued.  For example, to recover \eno{SKT} as a special case, we would set $q=4$, so that $A = a_0a_1a_2$; we would set $\Omega_{AB} = \delta_{a_0b_0} \delta_{a_1b_1} \delta_{a_2b_2}$; we would set
 \eqn{OmegaSimple}{
  \Omega_{ABCD} = \delta_{a_0b_0} \delta_{a_1c_1} \delta_{a_2d_2}
    \delta_{b_1d_1} \delta_{b_2c_2} \delta_{c_0d_0} \,;
 }
and of course we would set $K=\mathbb{R}$ and $s=1$.  It is well recognized (see e.g.~\cite{Gurau:2011xp,Witten:2016iux,Klebanov:2016xxf}) that the structure $\Omega_{ABCD}$ in \eno{OmegaSimple} corresponds to a coloring of the edges of the tetrahedron so that only three colors are used, and opposite edges have the same color.\footnote{It is useful to clarify here one point of terminology: We use the terms ``vertex'' and ``edge'' to describe the inner structure of an interaction vertex like $\psi^{a_0a_1a_2} \psi^{a_0b_1b_2} \psi^{c_0a_1b_2} \psi^{c_0b_1a_2}$.  From this point of view, an interaction vertex is a graph unto itself, with $q$ vertices when the interaction term has $q$ powers of $\psi$.  A full Feynman diagram consists of propagators connecting interaction vertices, and as is familiar from earlier work including \cite{Gurau:2011xp}, the inner structure of a propagator is $q-1$ threads which flow into the edges inside an interaction vertex.}

When considering theories over $\mathbb{Q}_p$, as explained in \cite{Gubser:2017qed}, we must allow $\Omega_{AB}$ to be symmetric ($\sigma_\Omega = 1$) or anti-symmetric ($\sigma_\Omega = -1$); and we must choose $\mop{sgn}\omega$ to be one of the several multiplicative sign characters over $\mathbb{Q}_p$, which are in one-to-one correspondence with the quadratic extensions of $\mathbb{Q}_p$.  To get a real, non-vanishing kinetic term, $\Omega_{AB}$ must be Hermitian, and we must have
 \eqn{sgnConstraint}{
  \sigma_\psi \sigma_\Omega = \mop{sgn}(-1) \,.
 }
(Surprisingly, non-trivial sign characters over $\mathbb{Q}_p$ can have either $\mop{sgn}(-1) = -1$ or $+1$.)

For $q=4$, still following \cite{Gubser:2017qed}, the obvious adaptation of the Klebanov-Tarnopolsky model to a theory over $\mathbb{Q}_p$ is to set $\Omega_{AB} = \Omega_{a_0b_0} \Omega_{a_1b_1} \Omega_{a_2b_2}$ where
 \eqn{OmegaLowercase}{
  \Omega_{ab} = \left\{ \seqalign{\span\TR &\qquad\hbox{for}\qquad \span\TR}{
   {\bf 1}_{N \times N} & \sigma_\Omega = 1  \cr
   \sigma_2 \otimes {\bf 1}_{{N \over 2} \times {N \over 2}} & \sigma_\Omega = -1 \,,} \right.
 }
and to set
 \eqn{OmegaInt}{
  \Omega_{ABCD} =
    \Omega_{a_0b_0} \Omega_{a_1c_1} \Omega_{a_2d_2} \Omega_{b_1d_1} \Omega_{b_2c_2} \Omega_{c_0d_0}
    \,.
 }
(Note that if $\sigma_\Omega = -1$, then because of \eno{OmegaLowercase}, $N$ must be even.)  If indeed $\Omega_{ab}$ is antisymmetric, then one needs a direction on all the edges of the tetrahedral graph in order to decide the order of the indices in each factor on the right hand side of \eno{OmegaInt}.  However, flipping the direction on any one edge flips the sign of $\Omega_{ABCD}$, and so can be compensated for by changing the sign of $g$.

Generalizing the kinetic term to $q>4$ is easy: We need only set
 \eqn{OmegaABgen}{
  \Omega_{AB} = \prod_{i=0}^{q-2} \Omega_{a_ib_i} \,.
 }
Generalizing the interaction tensor turns out to be more subtle, and laying the groundwork for finding suitable generalizations is the focus of the rest of this section.

Up to the minor issue of directedness, constructing a rank $q$ interaction tensor $\Omega_{A^{(0)}A^{(1)}\dots A^{(q-1)}}$ as a product of $q(q-1)/2$ factors $\Omega_{a^{(i)}_r a^{(j)}_r}$ corresponds to a coloring problem on the complete graph of $q$ points (and therefore $q(q-1)/2$ edges), where we use $q-1$ colors (each one labeled by a value of $r$) and require that each of the $q-1$ links incident on a given vertex (each one labeled by a value of $i$) must be a different color.  A special case of Baranyai's theorem guarantees that this can always be done provided $q$ is even.  It is impossible when $q$ is odd.  One can map the problem onto the scheduling of a round-robin tournament, where each link is one game, each vertex is a contestant, and each color is a round, during which each contestant plays exactly one game.  This phrasing makes it obvious that no coloring is possible for $q$ odd, because in a given round one must pair up all $q$ contestants in two-person games.  For $q$ even, there is a canonical solution, which is
 \eqn{rnm}{\seqalign{\span\TL &\qquad\span\TT}{
  r_{ij} \equiv i+j \mod{q-1} & if $0 \leq i < q-1$, $0 \leq j < q-1$, and $i \neq j$ \cr
  r_{i,q-1} \equiv 2i \mod{q-1} & for $0 \leq i < q-1$\,.
 }}
Here, $r_{ij}$ is the color of the edge from vertex $i$ to vertex $j$.  Vertex labels $i$ and $j$ take values from $0$ to $q-1$, while color labels $r$ run from $0$ to $q-2$.  The corresponding interaction tensor is
 \eqn{InteractionTensor}{
  \Omega_{A^{(0)}A^{(1)}\dots A^{(q-1)}} = \prod_{0 \leq i < j \leq q-1}
    \Omega_{a^{(i)}_{r_{ij}} a^{(j)}_{r_{ij}}} \,.
 }
In \eno{InteractionTensor}, we took care of the directedness issue by requiring $i<j$, which is the same as alphabetizing the lowercase indices in \eno{OmegaInt}.

 \begin{figure}
    \centering

\begin{subfigure}[b]{0.3\textwidth}
        \includegraphics[width=\textwidth]{Octagon30}
        \caption{\hspace{12mm}$G=\mathbb{Z}_2^{\,3}$\\30 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 1344}
        \label{fig:oct30}
    \end{subfigure}
    \begin{subfigure}[b]{0.3\textwidth}
        \includegraphics[width=\textwidth]{Octagon630}
        \caption{\hspace{12mm}$G=\mathbb{Z}_2^{\,2}$\\630 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 64}
        \label{fig:oct630}
    \end{subfigure}
    \begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Octagon420}
        \caption{\hspace{12mm}$G=\mathbb{Z}_2^{\,2}$\\420 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 96}
        \label{fig:oct420}
\end{subfigure}

    \begin{subfigure}[b]{0.3\textwidth}
        \includegraphics[width=\textwidth]{Octagon2520}
        \caption{\hspace{12mm}$G=\mathbb{Z}_2$\\2520 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 16}
        \label{fig:oct2520}
    \end{subfigure}
    \begin{subfigure}[b]{0.3\textwidth}
        \includegraphics[width=\textwidth]{Octagon1680}
        \caption{\hspace{12mm}$G=1$\\1680 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 24}
        \label{fig:oct1680}
    \end{subfigure}
    \begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Octagon960}
        \caption{\hspace{12mm}$G=1$\\960 one-factorizations  \\Order of $\mathrm{Aut}(\overline{\mathcal{F}})$ is 42}
        \label{fig:oct960}
\end{subfigure}

\caption{The six inequivalent melonic interactions and their symmetry factors for $q=8$. Each interaction can be identified with an isomorphism class of one-factorizations, and we also list the number of one-factorizations in each class as well as $|\mathrm{Aut}(\overline{\mathcal{F}})|$, the number of permutations in $S_8$ that preserve the one-factorizations in a given isomorphism class (see section \ref{ONEFACTOR} for an explanation of this notation). The canonical coloring is the bottom right.}
    \label{Octagons}
\end{figure}

For $q=2$, $4$, and $6$, the canonical solution \eno{rnm} is the only solution.  Starting at $q=8$, there are multiple solutions: That is, $r_{ij}$ can be chosen differently from \eno{rnm} but still consistent with the requirement that we use only $q-1$ colors and have one edge of each color coming together at each vertex.  For any such $r_{ij}$, we can still use \eno{InteractionTensor} to construct the interaction tensor.  We exhibit the six different solutions for $q=8$ in figure~\ref{Octagons}.  By ``different,'' we mean that there is no way to relabel the colors and/or the vertices to map any of the six $r_{ij}$ into one another.  A striking point is that the solutions have different symmetry groups, composed of up to three factors of $\mathbb{Z}_2$.\footnote{By $\mathbb{Z}_2$ we mean the integers modulo $2$, or equivalently the multiplicative group $\{1,-1\}$---not the $2$-adic integers.}

For $q=10$, there are (we claim) $396$ different interaction vertices, and none of them have any symmetry.  To justify this claim, and to proceed to larger $q$, we need to give a more conceptually organized presentation.  We do so in the next three sections, starting with constraints on the symmetry group in section~\ref{SYMMETRY}, continuing with an explicit construction in section~\ref{CONSTRUCTION}, and concluding with a summary of the problem of counting distinct interaction vertices in section~\ref{ONEFACTOR}.

\section{Symmetry groups of interaction vertices}
\label{SYMMETRY}

In ordinary scalar field theory where the scalar $\phi$ is real-valued, the symmetry group of a $q$-fold interaction vertex $\phi^q$ is the permutation group $S_q$, with order $q!$, because all propagators leading into the interaction vertex are equivalent and can be permuted arbitrarily without changing the structure of the interaction.  In a matrix field theory based on a Hermitian $N \times N$ matrix $\Phi$, the symmetry group of a $\mop{tr} \Phi^q$ interaction vertex is the group $\mathbb{Z}_q$ of cyclic permutations of the propagators.  In the $q=4$ Klebanov-Tarnopolsky model, the symmetry group is $\mathbb{Z}_2 \times \mathbb{Z}_2$ (not $\mathbb{Z}_4$), generated by the permutations $(12)(34)$, and $(13)(24)$ and sometimes referred to as the Klein group.  As should be clear from figure~\ref{VertexGroups}, a permutation in the Klein group reorders propagators leading into the interaction vertex in such a way that we get back to exactly the same diagram that we started with.  This sort of permutation is what we will call a coloring automorphism.  If we look at the inner structure of the interaction vertex, we see that a coloring automorphism permutes the vertices (each one corresponding to an incoming propagator) in such a way as to preserve the colors of each edge.

  \begin{figure}
\centering{
\includegraphics[height=20ex]{ScalarVertex.pdf} \hspace{6em}
\includegraphics[height=20ex]{MatrixVertex.pdf}\hspace{4em}
\includegraphics[height=20ex]{MelonVertex.pdf}
}
\caption{Left: The symmetry group of $\phi^4$ theory is the symmetric group $S_4$.  Middle: The symmetry group of $\mop{tr} \Phi^4$ theory is the cyclic group $\mathbb{Z}_4$.  Right: The symmetry group of the quartic tensor interaction is the Klein group $\mathbb{Z}_2^2$.}\label{VertexGroups}
\end{figure}

To be precise: When we say that $r_{ij}$ is a coloring of the complete graph $K_q$ with $q$ vertices, what we mean is that $ij$ labels an edge (so $i \neq j$), and each $r_{ij}$ is chosen from the set of ``colors'' $\{0,1,\ldots,q-2\}$, with the constraints that $r_{ij} = r_{ji}$ and that for fixed $i$, $r_{ij}$ is a bijection from the $q-1$ vertices that remain after $i$ is omitted to the set of colors.  In other words, each edge leading into a given vertex is a different color.  A coloring automorphism is defined as a map $i \to \pi(i)$ such that $r_{\pi(i)\pi(j)} = r_{ij}$ for all $i$ and $j$.  Let the group of coloring automorphisms be $G$.  The main purpose of the rest of this section is to limit the possibilities for $G$.  Then in section~\ref{CONSTRUCTION} we will show that all groups $G$ not ruled out by the arguments of this section actually can be realized.

Our first claim is that any coloring automorphism is an involution.  Denote the coloring automorphism by $\pi$.  Assume that $\pi$ is not the identity, since otherwise the claim is trivial.  For some vertex $i$, we have $j \equiv \pi(i) \neq i$.  Then $r_{ij} = r_{\pi(i)\pi(j)} = r_{j\pi(j)}$, where in the first equality we remembered that $\pi$ is a coloring automorphism.  From $r_{ji} = r_{j\pi(j)}$ we can conclude that $i = \pi(j)$ because, as noted previously, the coloring $r$ must be a bijection, for fixed $j$, from vertices $i \neq j$ to colors.

Next we remember an elementary result of group theory: Any finite group $G$ consisting only of involutions is isomorphic to $\mathbb{Z}_2^n$ for some $n$.  First let's show that the group is abelian.  Therefore let $g$ and $h$ be group elements.  Because $g$ and $h$ are involutions, we have $(gh)^{-1} = h^{-1} g^{-1} = hg$.  But because $gh$ is also an involution, $(gh)^{-1} = gh$.  So $gh = hg$ as required.  Now the fundamental theorem of finite abelian groups tells us that $G$ must be a direct product of cyclic subgroups of prime-power order.  Because all elements of $G$ are involutions, any cyclic subgroup must be a copy of $\mathbb{Z}_2$, and the result is proven.

Another elementary point to note is that if a coloring automorphism preserves any vertex, then it is necessarily the trivial automorphism that maps all vertices to themselves.  To see this, suppose $\pi(i) = i$ for some vertex $i$, and consider any other vertex $j$.  We have $r_{ij} = r_{\pi(i)\pi(j)} = r_{i\pi(j)}$, and because $r$ is a bijection, for fixed $i$, from vertices $j \neq i$ to colors, we can conclude $\pi(j) = j$. It follows that any permutation that gives a coloring automorphism consists of $q/2$ two-cycles.

We also note that color automorphisms that have a two-cycle in common must be identical. For suppose there are color automorphisms $\pi$ and $\tilde{\pi}$ and a vertex $i$ such that $\pi(i)=\tilde{\pi}(i)$. Then for any $j$ we have that $r_{ij}=r_{\pi(i)\pi(j)}=r_{\tilde{\pi}(i)\pi(j)}$ but also $r_{ij}=r_{\tilde{\pi}(i)\tilde{\pi}(j)}$. Since $r$ is a bijection, it follows that $\pi(j)=\tilde{\pi}(j)$ for all $j$.

It helps our imagination to think of the group $G = \mathbb{Z}_2^n$ of coloring automorphisms as reflections through $n$ orthogonal planes which do not pass through any vertices.  If $q=2^v$ and $n=v$, this line of thinking suggests that we can produce a coloring of $K_q$ whose automorphism group is $\mathbb{Z}_2^v$: See figure~\ref{Reflections} for the first few instances.

  \begin{figure}
\centering{
\includegraphics[height=25ex]{2fold.pdf}
\includegraphics[height=25ex]{4fold.pdf}
\hspace{2mm}
\includegraphics[height=25ex]{8fold.pdf}
}
\caption{The first three instances of maximally symmetrical interaction vertices, with $q=2^v$ and $G = \mathbb{Z}_2^v$.  We omitted one color from the $q=8$ case, namely a seventh color with edges running from each corner to the diametrically opposite corner, for example from $0$ to $6$.}
\label{Reflections}
\end{figure}

With these preliminaries in hand, we now come to the main result of this section: If $q=u 2^v$ where $u$ is odd, then the largest that $G$ can be is $\mathbb{Z}_2^v$ if $u=1$, or $\mathbb{Z}_2^{v-1}$ if $u>1$.  The arguments in the remainder of this section do not demonstrate the existence of interaction vertices with any particular symmetry group $G$; rather, they rule out larger symmetry groups.

The proof of our main result relies on the orbit-stabilizer theorem, which we summarize here for the purposes of a self-contained presentation.  If a group $G$ acts on a set $X$, then the stabilizer $\mop{stab} i$ of an element $i \in X$ is the subgroup of $G$ of elements which preserve $i$.  Meanwhile, the orbit $\mop{orb} i$ is the subset of $X$ consisting of all images of $i$ under the action of elements of $G$.  The theorem says
 \eqn{OrbStab}{
  |\mop{orb} i||\mop{stab} i| = |G| \,.
 }
As a first application, let $X$ be the set of $q$ vertices, assume that $G = \mathbb{Z}_2^n$.  The stabilizer of a vertex $i$ is the trivial group: This is because any $\pi \in G$ that maps $i$ to itself must also map all other vertices to themselves.  Invoking \eno{OrbStab} we see that each orbit contains $2^n$ points.  The union of all vertex orbits is all of $X$, and so there must be
 \eqn{qtildeDef}{
  \tilde{q} \equiv u 2^{v-n}
 }
distinct vertex orbits.  Already, \eno{qtildeDef} shows that $G$ cannot be larger than $\mathbb{Z}_2^v$, because if $n>v$, $\tilde{q}$ is not an integer.

To finish proving the main result, all we need to do is to exclude the possibility that $G = \mathbb{Z}_2^v$ when $u>1$.  This turns out to require somewhat more subtle reasoning than we have used so far, but the essential idea is to consider the quotient of the interaction vertex by $G$ and show that, as a graph with $\tilde{q}=u$ vertices, it leads to an impossible coloring problem.  Clearly, we could just set $v=n$ throughout the following paragraphs, but we refrain from doing so because keeping $n \leq v$ general allows us to see some first hints on how to actually construct graphs with any allowed symmetry.

Let's start with a second application of the orbit-stabilizer theorem.  Consider the set of all edges {\it of a fixed color}.  For any fixed color, there are $q/2$ such edges.  Edges of a fixed color are permuted among themselves by coloring automorphisms, and for each edge we have
 \eqn{OrbStabAgain}{
  |\mop{orb} e||\mop{stab} e| = |G| \,.
 }
Following a standard trick, we divide both sides of \eno{OrbStabAgain} by $|\mop{stab} e|$ and then sum over all distinct orbits to get
 \eqn{OrbSum}{
  {q \over 2} = \sum_{\rm orbits} {|G| \over |\mop{stab} e|} \,.
 }
This can be written more usefully as
 \eqn{OrbSimp}{
  u = 2^{n-v+1} \sum_{\rm orbits} {1 \over |\mop{stab} e|} \,.
 }
To get proper mileage out of \eno{OrbSimp}, we need some knowledge of the values of $|\mop{stab} e|$---which need not be unity!  Consider however the following division of edges (still of any fixed color) into two classes.  There are edges which join two vertices which are in the same vertex orbit; we will call these ``internal'' edges.  All other edges we will refer to as ``external'' edges.  We observe that $\mop{stab} e$ is the trivial group for any external edge, because if it weren't, then there would be some coloring automorphism that exchanges the edge's two ends, and that would make the edge internal.  So $|\mop{stab} e| = 1$ for external edges.

Next we want a count of external edges of a fixed color.  It's easier to start by considering external edges of {\it any} color, i.e.~the disjoint union of external edges of each fixed color.  There are $\binom{\tilde{q}}{2} \cdot 2^{2n}$ such edges, because to specify one we must choose a pair of vertex orbits, and then from each of those two vertex orbits we must choose one vertex.  Because the stabilizer of an external edge is trivial, its orbit must have $2^n$ elements.  So the count of external {\it edge orbits} of any color is
 \eqn{Nedge}{
  N_e = \binom{\tilde{q}}{2} \cdot 2^n = q \left( u 2^{v-n-1} - {1 \over 2} \right) \,.
 }
Suppose now $0 \leq n < v$.  Then $2^{v-n-1}$ is a whole number, and because we have only $q-1$ colors to work with, there must be at least one color---call it red---with at least $u 2^{v-n-1}$ external edge orbits.  Restricting to red edges only, we recall that $|\mop{stab} e| = 1$ for each external edge, and so at least $u 2^{v-n-1}$ terms in the sum on the right hand side of \eno{OrbSimp} must be unity.  Comparing to \eno{OrbSimp}, we see that---for red edges---the sum over orbits works out perfectly with only external edges, implying that there can't be any red internal edges.  This is informative and useful for constructing examples.

Now suppose $n=v$.  If also $u=1$, then from \eno{Nedge} we see that $N_e=0$: There are no external edges at all!  This makes sense because there is only one vertex orbit, and indeed the cubical vertex illustrated in figure~\ref{Reflections} shows that it is entirely consistent to have $n=v$ and $u=1$.  Where things get dangerous is if $n=v$ and $u>1$.  Then, from \eno{Nedge}, $N_e = q(u-1)/2$.  Since we have only $q-1$ colors to work with, there must be at least one color---again call it red---with at least $(u+1)/2$ external edge orbits.  Considering only red edges the sum on the right hand side of \eno{OrbSimp} restricted to external edges gives $u+1$.  This is disastrous, because adding in the contribution of internal edges (if any) results in the absurd inequality $u \geq u+1$.  Another way to put it is that if $n=v$ and $u$ is odd and greater than $1$, then we can't color even the external edges consistently with only $q-1$ colors---let alone the internal edges.  See figure~\ref{SymmetryFailure}.
 \begin{figure}
  \centerline{\includegraphics[width=5in]{SymmetryFailure.pdf}}
  \caption{A partial coloring for $q=6$ which illustrates what goes wrong when one demands too much symmetry.  Here we propose $\mathbb{Z}_2$ symmetry, which according to our general result is too much.  On the left we show a partial coloring with $\mathbb{Z}_2$ symmetry.  On the right, we show the same coloring modded out by the $\mathbb{Z}_2$ symmetry.  The dots on the right are vertex orbits, and the double lines of each color are edge orbits.  The problem is that we need two edge orbits between each pair of vertex orbits, and all of them (in this simple case) must have different colors.  That means we need six colors just for the external edges, and we only have five to work with---leaving us without a second color to use in connecting the $01$ and $34$ orbits.  Internal edges would run within a vertex orbit, for example from $0$ to $1$.}\label{SymmetryFailure}
 \end{figure}

\section{Construction of interaction vertices}
\label{CONSTRUCTION}

Since for any $q$ we can construct an interaction vertex with the canonical ordering, which has no coloring automorphisms, what remains to be shown is that for $q$ equal to twice times an even number, it is always possible to construct an interaction vertex with any allowed non-trivial symmetry group. Figure~\ref{Octagons} explicitly shows that this is true for $q=8$. We now prove inductively that it is also true for $q>8$.
 \begin{figure}
  \centering

\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{ThreeHexagonsA}
        \caption{}
        \label{ThreeHexagonsA}
\end{subfigure}
\hspace{5mm}
  \begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{ThreeHexagonsB}
        \caption{}
        \label{ThreeHexagonsB}
\end{subfigure}
\hspace{5mm}
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{ThreeHexagonsC}
        \caption{}
        \label{ThreeHexagonsC}
\end{subfigure}

  \caption{Constructing an interaction vertex with $q=12$ and $G = \mathbb{Z}_2$ in three steps.  (a) Separate the vertices into two groups of $6$, shown here as inner and outer rings.  We've connected each group only cyclically in order to avoid clutter, but in the full construction, we start with two copies of the complete graph $K_6$, one colored according to $r_{ij}$ and the other according to $\tilde{r}_{ij} = r_{ij} + q-1$.  (b) Reconnect the $12$ vertices according to the prescription shown in the first two lines of \eno{InductiveColoring}.  (c) Add the last color according to the third line of \eno{InductiveColoring}.  In the parlance of section~\ref{SYMMETRY}, these last edges are the internal edges, while the ones colored in the previous step are the external edges.}\label{ThreeSteps}
 \end{figure}

Let $q'=q/2$ and assume that $r_{ij}$ is a coloring of $K_{q'}$ with symmetry group $G'$ (which may be the trivial group). Split the $q$ vertices into two sets of $q'$ vertices; notationally this can be done by identifying a vertex first by indicating which set it's in, say with a Greek index $\alpha=0$ or $1$, and then which vertex within the set it is, say with a Roman index $i \in \{0,1,\dots,q-1\}$. Let $\tilde{r}_{ij}\equiv r_{ij} + q-1$. We now claim that the coloring $r_{\alpha i,\beta j}$ of $K_q$ defined by
\eqn{InductiveColoring}
{
& r_{0i,0j}=r_{1i,1j}=r_{ij}\,, \cr
& r_{0i,1j}=r_{1i,0j}=\tilde{r}_{ij}\,, \cr
& \tilde{r}_{0i,1i}=\tilde{r}_{1i,0i}=q-1\,,
}
has symmetry group $G=G'\times \mathbb{Z}_2$. See figure~\ref{ThreeSteps} for a diagrammatic illustration of how this coloring is generated. That the coloring \eqref{InductiveColoring} inherits the $G'$ symmetry is clear from the fact that, for any $\pi \in G'$, we have that
\eqn{}{
r_{\alpha \pi(i),\beta \pi(j)}=r_{\alpha i,\beta j}\,. }
But by construction the coloring \eqref{InductiveColoring} is also invariant under the permutation $\tau$ that acts on the Greek indices as
\eqn{}{\alpha \rightarrow \alpha +1 \mod{2}\,, \hspace{10mm} \beta \rightarrow \beta +1 \mod{2}\,,}
while leaving the Roman indices unchanged. Furthermore, the coloring \eqref{InductiveColoring} has no other symmetries. For coloring automorphisms that only swap around Roman indices are in one-to-one correspondence with the coloring automorphisms of $r_{ij}$. And if a coloring automorphism $\sigma$ changes the Greek index of any index pair $\alpha i$ such that $\sigma(0i)=1i'$, then for any $j$
\eqn{}
{
r_{\sigma(0i),\sigma(0j)}=r_{1i',\sigma(0j)}\,.
}
But since $r_{0j,0k} \neq r_{1j',0k'}$ for all $j,j',k,k'$, it follows that $\sigma$ must change the Greek index of \emph{all} index pairs and so must be of the form $\sigma = \pi \circ \tau$ for some $\pi \in G'$. This completes the inductive proof.

In general, for a given symmetry, there are multiple interaction vertices different from the one generated by the prescription \eqref{InductiveColoring}. An exception, however, occurs for the maximally symmetric vertex when $q=2^v$. In this case there is only one vertex with $G=\mathbb{Z}_2^{\,v}$. For $G$ consists of the identity element and $q-1$ permutations that commute amongst each other and each consists of $q/2$ two-cycles. And amongst permutations in $S_{q}$ consisting of $q/2$ two-cycles, one can at most form a set of $q-1$ elements that commute amongst each other but don't share a two-cycle; and any two such sets are equivalent by conjugation. But if we consider the edges of a given color, say red, in a maximally symmetric colored graph, then this sub-graph is invariant under the permutation $\pi$ that swaps vertices connected by a red edge. But $\pi$ must commute with all permutations in $G$, and so it follows that $\pi \in G$. Moreover, if we explicitly write $\pi = (a_0a_1)(a_2a_3)\dots(a_{q-1}a_q)$, then we recognize that each two-cycle in $\pi$ corresponds to a red edge; in other words, each non-trivial element of $G$ is precisely associated with a one-factor.  Since there are $q-1$ non-trivial elements of $G$, the one-factors are all specified once $G$ is specified.  And we have argued that $G$ is essentially unique.

\section{One-factorizations and equivalent interaction terms}
\label{ONEFACTOR}
In this section we determine the conditions under which two theories with actions of the form (\ref{SKTextended}) are equivalent. As discussed briefly in section \ref{STRUCTURE}, the choice of an interaction tensor $\Omega_{A^{(0)}A^{(1)}\cdots A^{(q-1)}}$ corresponds to a coloring problem on the complete graph $K_q$ on $q$ vertices.

More precisely, these interaction tensors are in one-to-one correspondence with ordered one-factorizations of $K_q$.  A one-factor of $K_q$ is a set $F$ of edges such that each vertex of $K_q$ belongs to a unique edge in $F$.  A one-factorization of $K_q$ is a partition of the edge-set of $K_q$ into $q-1$ one-factors $F_i$.  We denote a one-factorization as $\overline{\mathcal{F}} = \{F_0,\dots,F_{q-2}\}$, and when speaking of one-factorizations we do not distinguish between different orderings of the $F_i$.  An ordered one-factorization is a $(q-1)$-tuple $\mathcal{F} = (F_0,\dots,F_{q-2})$, i.e.~we have imposed a particular order on a one-factorization.  To see that ordered one-factorizations correspond to interaction vertices, recall that the fields $\psi^{a_0\ldots a_{q-2}}$ don't have any built-in symmetry under interchange of indices $a_i$.  Each such index is associated with a definite color: For example, we could say that red is associated with the first index $a_0$, then green with $a_1$, blue with $a_2$, and so forth.  And each color is associated with a one-factor: red for $F_0$, green for $F_1$, blue for $F_2$, etc.

One can check that any graph automorphism $\pi\colon K_q\to K_q$ carries (ordered) one-factori\-zations into other (ordered) one-factorizations via the actions $\pi \mathcal{F}=\left(\pi(F_0),\ldots,\pi(F_{q-2})\right)$ and $\pi \overline{F} \equiv \overline{\pi\mathcal{F}}$, respectively. Two ordered one-factorizations $\mathcal{F}$ and $\mathcal{G}$ are said to be isomorphic if there exists a graph automorphism $\pi$ such that $\mathcal{F}=\pi\mathcal{G}$. Similarly, $\overline{\mathcal{F}}\simeq\overline{\mathcal{G}}$ means there exists $\pi$ so that $\overline{\mathcal{F}}=\pi\overline{\mathcal{G}}$. We can also consider the natural action of any $\tau\in S_{q-1}$ on ordered one-factorizations $\tau\mathcal{F}=(F_{\tau(0)},\ldots,F_{\tau(q-2)})$. This action simply permutes the colors associated to each one-factor. In this language, $\overline{\mathcal{F}}\simeq\overline{\mathcal{G}}$ if and only if there exists a graph automorphism $\pi$ and a permutation of colors $\tau$ such that $\pi \mathcal{F}=\tau \mathcal{G}$.

Given an ordered one-factorization $\mathcal{F}$, we can form the following interaction tensor:

\eqn{OFTensor}{
\Omega^\mathcal{F}_{A^{(0)}\ldots A^{(q-1)}} \equiv \prod_{r=0}^{q-2}\prod_{\langle i j \rangle \in F_r}\Omega_{a^{(i)}_r a^{(j)}_r} \, ,
}
where $\langle i j \rangle$ is the edge in $K_q$ running between the vertices $i$ and $j$ and $a_r$ is the color index associated to the one-factor $F_r$. In writing $\langle i j \rangle$, we implicitly assume $i<j$. Conversely, any degree $q$ interaction term in which each constituent $\psi^A$ has a single color index contracted with each other $\psi^A$ arises from an interaction tensor of the form (\ref{OFTensor}).

Note that for distinct ordered one-factorizations $\mathcal{F}$ and $\mathcal{G}$ we end up with distinct interaction tensors $\Omega^\mathcal{F}\neq\Omega^\mathcal{G}$. We will however show that if their underlying one-factorizations $\overline{\mathcal{F}}$ and $\overline{\mathcal{G}}$ are isomorphic then $\Omega^\mathcal{F}$ and $\Omega^\mathcal{G}$ give rise to equivalent theories. Our argument proceeds in two parts: First we discuss equivalence under the action of $\tau \in S_{q-1}$ permuting colors, and then we consider the action of a permutation $\pi \in S_q$ of vertices.

Suppose we make the following field redefinition: $\phi^A=\psi^{\tau^{-1} A}$, where $\tau^{-1}\in S_{q-1}$ and $\tau A \equiv a_{\tau^{-1}(0)}\ldots a_{\tau^{-1}(q-2)}$. Then:
\eqn{OFColorEquiv}{
\Omega^\mathcal{F}_{A^{(0)}\ldots A^{(q-1)}}\phi^{A^{(0)}}\cdots\phi^{A^{(q-1)}}=\Omega^{\tau \mathcal{F}}_{A^{(0)}\ldots A^{(q-1)}}\psi^{A^{(0)}}\cdots\psi^{A^{(q-1)}}\,.
}
So, modulo a linear field redefinition implemented by a permutation matrix, the interaction term only depends on the underlying one-factorization $\overline{\mathcal{F}}$.

Now view a graph automorphism as a permutation on $q$ vertices, $\pi\in S_q$. With the change of indices $B^{(i)}=A^{\pi(i)}$ one can show:
\eqn{OGIsoEquiv}{
(\sigma_\psi)^\pi \Omega^\mathcal{F}_{A^{(0)}\ldots A^{(q-1)}}\psi^{A^{(0)}}\cdots\psi^{A^{(q-1)}}&= \Omega^\mathcal{F}_{A^{(0)}\ldots A^{(q-1)}}\psi^{A^{\pi^{-1}(0)}}\cdots\psi^{A^{\pi^{-1}(q-1)}}
\cr
 &=\Omega^{\pi \mathcal{F}}_{B^{(0)}\ldots B^{(q-1)}}\psi^{B^{(0)}}\cdots\psi^{B^{(q-1)}}\,.
}
So any two isomorphic ordered one-factorizations give the same interaction term, up to a sign.\footnote{In the presence of multiple interaction terms in the lagrangian, this argument that a permutation of vertices does not lead to a new theory continues to hold true. But it will no longer be true that an interaction term only depends on the underlying one-factorization $\overline{\mathcal{F}}$, see appendix \ref{IsomorphismOrdered}. }
We can get rid of this sign by sending $g\to -g$.\footnote{See \cite{Bulycheva:2017ilt} and \cite{Klebanov:2018nfp} for a careful discussion, in the context of the $q=4$ fermionic tensor model, of how the Hamiltonian even at the quantum level transforms in the degree 1 sign representation under permutations of indices and of how this affects the spectrum of the theory.}

We should point out here that isomorphic one-factorizations give equivalent interactions to all orders in $N$. As we will show in sections~\ref{TWOPOINT} and~\ref{FOURPOINT}, even non-isomorphic one-factorizations lead to the same two-point and four-point functions in the melonic limit, up to a rescaling of $g$ by a power of the order of the coloring automorphism group $G$.
\begin{table}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|}\hline
\diagbox[innerwidth=6 ex,height=5ex]{$G$}{$q$}& $2$ & $4$ & $6$ & $8$ & $10$ & $12$ & $14$\\\hline
$1$                          &  0 &  0 & 1  &  2 & 396   &   526,910,769 & $\sim 1.13 \times 10^{18}$\\\hline
$\mathbb{Z}_2$     &  1 &  0 &  0 &  1 &  0  &   4851 & 0 \\\hline
$\mathbb{Z}^2_2$ & 0 &  1 &  0 & 2  &  0 &   0 & 0 \\\hline
$\mathbb{Z}^3_2$ & 0 &  0 &  0 &  1 &  0 &   0 & 0 \\\hline
\end{tabular}
\caption{The number of isomorphism classes of one-factorizations of $K_q$ with given symmetry group $G=\mathrm{Aut}(\mathcal{F})$. The values for $q=12$ and $14$ were determined using  \cite{Dinitz:1994} and \cite{Kaski:2009}, respectively; the exact count for $q=14$ is quoted in the main text.}
\label{OFTable}
\end{table}

For future reference, we note that by $\mathrm{Aut}(\mathcal{F})$ we mean all graph automorphisms $\pi$ such that $\pi \mathcal{F}=\mathcal{F}$. $\mathrm{Aut}(\mathcal{F})$ is the exactly the vertex automorphism group described in section \ref{SYMMETRY}. In contrast, $\mathrm{Aut}(\overline{\mathcal{F}})$ consists of all graph automorphisms $\pi$ such that $\pi \overline{\mathcal{F}}=\overline{\mathcal{F}}$, that is all $\pi$ that can be undone by a permutation $\tau\in S_{q-1}$ of the colors $\pi \mathcal{F}=\tau \mathcal{F}$. Clearly $\mathrm{Aut}(\mathcal{F})\leq \mathrm{Aut}(\overline{\mathcal{F}})$. In general $\mathrm{Aut}(\overline{\mathcal{F}})$ has a much richer group structure than $\mathrm{Aut}(\mathcal{F})$. See for instance table~\ref{AutTable} in appendix~\ref{AlgorithmAppendix} where, for $q=12$, we list the ten possible values for $|\mathrm{Aut}(\overline{\mathcal{F}})|$ when $|\mathrm{Aut}(\mathcal{F})|=2$.

 In the case $q=12$, the vast majority of isomorphism classes of one-factorizations have trivial $\mathrm{Aut}(\mathcal{F})$. It is known in the literature \cite{Dinitz:1994} that there are a total of 526,915,620 non-isomorphic one-factorizations (the sum of the two entries in the $q=12$ column in figure \ref{OFTable}) and exactly 252,282,619,805,368,320 distinct one-factorizations of $K_{12}$. Implementing the orderly algorithm used in \cite{Seah} and \cite{Dinitz:1994} on the set of one-factors invariant under a given involution $\pi \in S_{12}$, we find that there are 1,008,649,635,840 one-factorizations of $K_{12}$ with $\mathrm{Aut}(\mathcal{F})=\mathbb{Z}_2$ and that they fall into 4851 isomorphism classes under permutations of vertices, see appendix \ref{AlgorithmAppendix}.

For $q=14$, $\mathrm{Aut}(\mathcal{F})$ can only be trivial and there are 1,132,835,421,602,062,347 nonisomorphic one-factorizations of $K_{14}$ \cite{Kaski:2009}.

For $q \leq 14$, a summary of the number of isomorphism classes of one-factorizations and therefore the number of inequivalent interaction terms is given in table~\ref{OFTable}. It is known that for sufficiently large $q$, the number of non-isomorphic one-factorizations $N(q)$ satisfies $\log N(q) \sim \frac{1}{2}q^2 \log q$ \cite{Cameron:1976}. In particular, $N(q)\to\infty$ as $q\to \infty$.

\section{The two-point function and the Schwinger-Dyson equation}
\label{TWOPOINT}

The kinetic term in the action \eqref{SKTextended} gives rise to a free propagator that in momentum space is given by
\eqn{G0momentum}
{
G_0^{a_0a_1...a_{q-2}b_0b_1...b_{q-2}}(\omega)=\sqrt{\mop{sgn}(-1)}\,\frac{\mop{sgn}(\omega)}{|\omega|^s}\prod_{i=0}^{q-2}\Omega^{a_ib_i}\equiv G_0(\omega)\prod_{i=0}^{q-2}\Omega^{a_ib_i}\, ,
}
where the matrix $\Omega^{ab}$ is the inverse of $\Omega_{ab}$, that is, $\Omega_{ab}\Omega^{bc}=\delta_a^c$.  With the normalization of the kinetic term as given in \eno{SKTextended}, we must choose
 \eqn{sqrtSgn}{
  \sqrt{\mop{sgn}(-1)} = \left\{ \seqalign{\span\TL &\quad \span\TT}{
    1 & if $\mop{sgn}(-1) = 1$  \cr
    i & if $\mop{sgn}(-1) = -1$\,.} \right.
 }

Once we include an interaction term, the propagator picks up loop corrections, but in the limit where $g\rightarrow 0$ and $N\rightarrow \infty$ such that $g^2N^{\frac{(q-1)(q-2)}{2}}$ is kept fixed, only melonic diagrams survive. The melonic contributions to the free propagator can all be obtained by iteratively applying the melonic insertion to the free propagator as shown on figure~\ref{GeneralInsertion}.

\begin{figure}[b]
 \centerline{\includegraphics[width=4.7in]{GeneralInsertion.pdf}}
 \caption{A melonic insertion.}\label{GeneralInsertion}
\end{figure}
\begin{figure}[t]
 \centerline{\includegraphics[width=5in]{SchwingerDyson.pdf}}
 \caption{The Schwinger-Dyson equation for the dressed propagator.}\label{SchwingerDyson}
\end{figure}

Adding together the free propagator and all the melonic corrections yields the dressed propagator
\eqn{}
{
G^{a_0a_1...a_{q-2}b_0b_1...b_{q-2}}(\omega) \equiv G(\omega)\prod_{i=0}^{q-2}\Omega^{a_ib_i}\,.
}
Just as in the cases of the SYK model \cite{Polchinski:2016xgd} and the Klebanov-Tarnopolsky tensor model with $q=4$ \cite{Klebanov:2016xxf}, the dressed propagator satisfies the Schwinger-Dyson equation depicted schematically in figure~\ref{SchwingerDyson}.  It first seems surprising that the right-most propagator in figure~\ref{SchwingerDyson} has no melonic insertion.  This is correct because any Feynman diagram contributing to the dressed propagator must have a right-most melonic insertion, and a free propagator attaches to it from the right.

To work out the exact mathematical expression for the Schwinger-Dyson equation, it is necessary to consider the automorphism group of the interaction as well as the number of sign flips involved in index contraction.

The first melonic correction to the free propagator, depicted on the right-hand side of figure~\ref{GeneralInsertion}, contains two interactions each of which is described by a colored graph with vertices labeled from 0 to $q-1$. For a given Feynman diagram corresponding to the first melonic correction, define a permutation $\sigma \in S_q$ by requiring that if an external propagator enters the left interaction vertex at vertex $i$, then an external propagator enters the right interaction vertex at vertex $\sigma(i)$; and likewise requiring that an internal propagator connected to vertex $j$ in the left interaction vertex connects to vertex $\sigma(j)$ in the right interaction vertex. See figure~\ref{2ptPermutation}.

\begin{figure}[b]
 \centerline{\includegraphics[width=5in]{2ptPermutation.pdf}}
 \caption{To get a melonic diagram, the external legs must be connected to vertices in the two interactions that are related by an automorphism.}\label{2ptPermutation}
\end{figure}

We claim now that such a Feynman diagram is melonic if and only if $\sigma$ belongs to the automorphism group $G$ of the interaction. The claim can be proved by the following reasoning:
\begin{itemize}
\item The two interactions are connected by $q-1$ internal propagators, each of which carry $q-1$ threads, giving a total of $(q-1)^2$ internal threads.
\item The $q-1$ threads of the left external propagator must connect to the $q-1$ threads of the right external propagator, so there are at most $(q-1)(q-2)$ internal threads that can partake in index loops.
\item Since at least two internal threads are needed for an index loop, there can be at most $(q-1)(q-2)/2$ index loops---the number required for a melonic diagram. This maximum number of loops is achieved when and only when the external threads only pass once through the internal propagators and the each index loop contains exactly two internal threads.
\item Labeling the vertex that the left external propagator is incident to by $i$, the fact that the external threads must pass only once through the internal propagators is equivalent to the statement that $r_{ij}=r_{\sigma(i)\sigma(j)}$ for all $j\neq i$.
\item The fact that each index loop must contain exactly two internal threads is equivalent to the statement that for any two indices $j$ and $k$ different from $i$,
$r_{jk}=r_{\sigma(j)\sigma(k)}$.
\end{itemize}
We see then that the melonic diagrams are exactly those for which $r_{lm}=r_{\sigma(l)\sigma(m)}$ for all $l,m$, ie. those for which $\sigma \in G$. Since in writing down the Feynman diagram corresponding to the first melonic correction we may take the left external propagator to be incident to any vertex $i$, while the right external propagator can only be incident to vertices $i'$ for which $i'=\sigma(i)$ for some $\sigma \in G$, the total number of Feynman diagrams contributing is $q|G|$. And for each additional melonic insertion we pick up an extra copy of this factor.

\begin{figure}
 \centerline{\includegraphics[width=5in]{2ptArrows.pdf}}
 \caption{All index loops can be brought to have a uniform orientation by an even number of arrow flips. To get a uniform orientation of any external thread an odd number of arrows flips are requisite.}\label{2ptArrows}
\end{figure}

The other subtlety to consider before writing down the Schwinger-Dyson equation is whether the contraction of index loops causes a sign difference between $O(N)$ and $Sp(N)$ symmetric models. To account diagrammatically for the possibility of $Sp(N)$ symmetry, we replace each thread with an arrow. The orientation of any arrow can be flipped at the cost of a factor of $\sigma_\omega$, and index loops of arrows with uniform orientation can be contracted to give a factor of $N$ without picking up a sign.

For the first melonic correction to the free propagator we may take the threads of all propagators to point from left to right as in figure~\ref{2ptArrows}. For a given Feynman diagram described by a permutation $\sigma \in G$ as explained above, we may for any $l$ and $m$ take the the arrow between vertices $l$ and $m$ to have the same orientation as the arrow between vertices $\sigma(l)$ and $\sigma(m)$ since any automorphism involves an even number of arrow flips. In this case, all index loops consist of two arrows with one orientation and two arrows with the opposite orientation, and so all the loops may be contracted without picking up a sign. But to obtain a uniform orientation of the external threads running from left to right and recover the index structure of the free propagator, it is necessary for each of the $q-1$ colors to flip one thread in one of the two interactions. Hence, we must in total flip an odd number off arrows, and so we conclude that we pick up a factor of $\sigma_\Omega$ with each melonic insertion.  This fact was shown in \cite{Gubser:2017qed} to apply for $q=4$, but we see now that it holds true for all $q$.

Taking into account the symmetry of the interaction and the index structure of the corrections, we arrive finally at the Schwinger-Dyson equation
\eqn{eqSchwingerDyson}{
G(t_1-t_2) &= G_0(t_1-t_2)  \cr
 &\qquad{} +
q|G|\left(\frac{g}{|G|}\right)^2\sigma_\Omega\, N^{\frac{(q-1)(q-2)}{2}}\int dt\,dt'\,G(t_1-t)G(t-t')^{q-1}G_0(t'-t_2)\, .
}
To slightly shorten expressions and remove the explicit dependence on $G$, we find it useful to introduce the definition
\eqn{gtilde}
{
\tilde{g} \equiv \sqrt{\frac{q}{|G|}N^{\frac{(q-1)(q-2)}{2}}}\,g \,.
}

\subsection{The IR solution}
\label{sec:IRsolution}

The Schwinger-Dyson equation can be expressed in term of convolutions as
\eqn{SchwingerDysonConvolution}
{
G(t)=G_0(t) + \sigma_\Omega \tilde{g}^2 (G\ast G^{q-1} \ast G_0)(t)\,.
}
In the infra-red limit, $\tilde{g}$ is large, and to leading order in $1/\tilde{g}$ we can set $G(t)=0$. To solve \eqref{SchwingerDysonConvolution} in the IR, we will adopt the methodology of \cite{Gubser:2017qed}, and so we introduce multiplicative characters defined as
\eqn{multiplicativecharacters}
{
\pi_s(t) \equiv |t|^s, \hspace{10mm} \pi_{s,\mop{sgn}}(t) \equiv |t|^s\mop{sgn}(t)\,.
}
The Fourier transform of these multiplicative characters are given by
\eqn{Fourier}
{
\mathcal{F}[\pi_s](\omega)=\Gamma(\pi_{s+1})\,\pi_{-s-1}(\omega)\, , \hspace{10mm}
\mathcal{F}[\pi_{s,\mop{sgn}}](\omega)=\Gamma(\pi_{s+1,\mop{sgn}})\,\pi_{-s-1,\mop{sgn}}(\omega) \, ,
}
where $\Gamma(\pi_{s})$ and $\Gamma(\pi_{s,\mop{sgn}})$ are instances of Gel'fand-Graev gamma functions, see appendix \ref{GammaAppendix}. Using the above, we may express the bare propagator in momentum and position space as
\eqn{}
{
G_0(\omega)=\sqrt{\mop{sgn}(-1)}\,\pi_{-s,\mop{sgn}}(\omega)\, , \hspace{10mm} G_0(t)=\sqrt{\mop{sgn}(-1)}\,\Gamma(\pi_{1-s,\mop{sgn}})\,\pi_{s-1,\mop{sgn}}(t) \,.
}

As an ansatz for solving \eqref{SchwingerDysonConvolution} with the left-hand side set to zero, we choose
\eqn{ansatz}
{
G(t) = b\,\pi_{-\frac{2}{q},\mop{sgn}}(t)\, ,
}
and so the IR Schwinger-Dyson equation in position space reads
\eqn{}
{
-\pi_{1-s,\mop{sgn}}=\sigma_{\Omega}\,\tilde{g}^2\,b^q\,\pi_{-\frac{2}{q},\mop{sgn}}\ast\pi_{\frac{2(1-q)}{q},\mop{sgn}}\ast\pi_{1-s,\mop{sgn}}\, .
}
Using \eqref{Fourier} to Fourier transform the multiplicative characters, we get the momentum space Schwinger-Dyson equation
\eqn{}
{
-\pi_{s,\mop{sgn}}=\sigma_{\Omega}\,\tilde{g}^2\,b^q\,\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})\,\pi_{\frac{2-q}{q},\mop{sgn}}\ast\pi_{\frac{(q-2)}{q},\mop{sgn}}\ast\pi_{s,\mop{sgn}}\,.
}
The multiplicative characters all cancel, and we read off directly that
\eqn{bsol}
{
b^q = \frac{-1}{\sigma_\Omega\, \tilde{g}^2\, \Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})}\,.
}
The $s$ dependence is seen to cancel out entirely. Insofar as the ansatz \eqref{ansatz} describes the true IR behavior, the dressed propagators of the various non-local theories with different values of $s$ all flow to the same function in the IR. In the case of bosonic theories, however, where the potential can be unbounded from below, the physical significance of the IR solution \eno{ansatz} is uncertain. Nonetheless, we explicitly include bosonic theories under the scope of theories we are subjecting to formal perturbation theory since bosonic theories are required in order to write down an adelic relation connecting Archimedean and $p$-adic theories, as we will see in section \eqref{ADELIC}.

In section \ref{ExactSolution} we will see that for a large subset of the $p$-adic theories, the full Schwinger-Dyson equation \eqref{eqSchwingerDyson} can be solved exactly, and in these cases one can verify by inspection of the full solution that the two-point function flows to the IR solution described by equations \eqref{ansatz} and \eqref{bsol}.

\subsection{The zoo of theories}

As we are working in Euclidean space, the reality of the action \eqref{SKTextended} dictates that $G(t)$ should be real too. The ansatz \eqref{ansatz} can therefore only provide an accurate description of the IR behavior if $b$ is real, that is, if the right-hand side of \eqref{bsol} is positive. In other words, the sign of $\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})$ must be opposite to that of $\sigma_\Omega$. For any choice of number field $\mathbb{R}$ or $\mathbb{Q}_p$ and any choice of sign function in the action, this requirement together with the sign constraint \eqref{sgnConstraint} uniquely specifies whether the theory must be bosonic or fermionic and $O(N)$ and $Sp(N)$ symmetric. Generalizing the table of theories in \cite{Gubser:2017qed} to all values of $q$, table~\ref{ExplicitResults} lists the melonic theories with renormalization group flow from a free theory in the UV to a strongly interacting fixed point in the IR.  The choice of $\tau$ in table~\ref{ExplicitResults} amounts to a choice of the sign function, as explained in appendix~\ref{GammaAppendix}.  Often, the value of $\tau$ involves $\epsilon$, which stands for any integer that is not a square modulo $p$.

\definecolor{Gray}{gray}{0.9}
\def\rowcolor{Gray}{\rowcolor{Gray}}
\renewcommand{\arraystretch}{1.3}
 \begin{table}\begin{center}\begin{tabular}{|c||c|r|c|r|r|l|}
  \hline
  $K$ & condition & $\tau$ &
    $\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})$ &
    $\sigma_\Omega$ & $\sigma_\psi$ & explanation \\[5pt]
    \hline\hline
  $\mathbb{R}$ & & $1$ & $-\frac{2\pi q}{q-2}\tan\left(\frac{\pi}{q}\right)$ & $1$ & $1$ & ${\rm O}(N)$ bosonic \\ [5pt]
  $\mathbb{R}$ & & $-1$ & $-\frac{2\pi q}{q-2}\cot\left(\frac{\pi}{q}\right)$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic   \\ [5pt]\hline
  $\mathbb{C}$ & & $1$ & $-\frac{\pi^2q^2}{(q-2)^2}$ & $1$ & $1$ & ${\rm O}(N)$ bosonic  \\ [5pt] \hline
  $\mathbb{Q}_p$ & $p$ odd & $1$ & $-\frac{(p^{2/q}-1)(p^2-p^{2/q})}{p(p-p^{2/q})^2}$ & $1$ & $1$ & ${\rm O}(N)$ bosonic  \\ [5pt]
  $\mathbb{Q}_p$ & $p$ odd & $\epsilon$ & $\frac{(p^{2/q}+1)(p^2+p^{2/q})}{p(p-p^{2/q})^2}$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\ [5pt]
  \rowcolor{Gray} $\mathbb{Q}_p$ & $p \equiv 1 \mod 4$ & $p$ & $1/p$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_p$ & $p \equiv 1 \mod 4$ & $\epsilon p$ & $1/p$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_p$ & $p \equiv 3 \mod 4$ & $p$ & $-1/p$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_p$ & $p \equiv 3 \mod 4$ & $\epsilon p$ & $-1/p$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\  \hline
  $\mathbb{Q}_2$ & & $1$ & $-\frac{5\cdot 4^{1/q}-4-16^{1/q}}{2(4^{1/q}-2)^2}$ & $1$ & $1$ & ${\rm O}(N)$ bosonic  \\ [5pt]
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $-1$ & $-1/4$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $2$ & $1/8$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $-2$ & $-1/8$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $3$ & $-1/4$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\
  $\mathbb{Q}_2$ & & $-3$ & $\frac{(1+4^{1/q})(4+4^{1/q})}{2(2+4^{1/q})^2}$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\[5pt]
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $6$ & $-1/8$ & $1$ & $-1$ & ${\rm O}(N)$ fermionic  \\
  \rowcolor{Gray} $\mathbb{Q}_2$ & & $-6$ & $1/8$ & $-1$ & $-1$ & ${\rm Sp}(N)$ fermionic  \\ \hline
 \end{tabular}\end{center}\caption{Table of melonic theories, Archimedean or ultrametric. Shaded rows indicate theories for which an exact solution to the Schwinger-Dyson equation is given in subsection \ref{ExactSolution}.}\label{ExplicitResults}\end{table}

\subsection{Full solution to the Schwinger-Dyson equation for direction-dependent theories}
\label{ExactSolution}

Any $p$-adic number $x$ can be represented by a series expansion
\eqn{padicExpansion}
{
x=p^{v(x)}\sum_{m=0}^\infty c_m\, p^{m}
}
where $c_m \in \{0,1,...,p-1\}$ and $c_0 \neq 0$.

It was shown in \cite{Gubser:2017qed} for $q=4$ that the Schwinger-Dyson equation \eqref{SchwingerDysonConvolution} can be solved exactly for $p$-adic theories when the sign function is direction-dependent, that is, when the sign function $\mop{sgn}(x)$ depends not only on the norm $|x|$ but also on the first $p$-adic digit $c_0$ in the expansion \eqref{padicExpansion}. These theories are indicated by shaded rows in table \ref{ExplicitResults}, and the solutions of \cite{Gubser:2017qed} generalize straightforwardly to higher values of $q$.

For each of the theories, the solution can be written as

\eqn{}
{
G(t)=\mathcal{G}(|t|)\mop{sgn}(t) \,,
}
where $\mathcal{G}(|t|)$ is the unique real root of a $q$-th order polynomial.

For the direction-dependent theories with odd $p$, ie. those with a sign function defined by $\tau=p$ or $\tau=\epsilon p$, $\mathcal{G}(|t|)$ is given by the real solution to the equation
\eqn{}
{
\mathcal{G}(|t|)=\sqrt{\mop{sgn}(-1)}\Gamma(\pi_{1-s,\mop{sgn}})|t|^{s-1}\left(1-\frac{\tilde{g}^2}{p}|t|^2 \,\mathcal{G}(|t|)^q \right) .
}
For $p=2$ and $\tau=-1$ or $\tau=3$, $\mathcal{G}(|t|)$ is given by the real solution to
\eqn{}
{
\mathcal{G}(|t|)=\sqrt{\mop{sgn}(-1)}\Gamma(\pi_{1-s,\mop{sgn}})|t|^{s-1}\left(1-\frac{\tilde{g}^2}{2^2}|t|^2 \,\mathcal{G}(|t|)^q \right) .
}
And $p=2$ and $\tau=\pm 2$ or $\tau = \pm 6$, $\mathcal{G}(|t|)$ is given by the real solution to
\eqn{}
{
\mathcal{G}(|t|)=\sqrt{\mop{sgn}(-1)}\Gamma(\pi_{1-s,\mop{sgn}})|t|^{s-1}\left(1-\frac{\tilde{g}^2}{2^3}|t|^2 \,\mathcal{G}(|t|)^q \right) \,.
}
In all cases, we choose $\sqrt{\mop{sgn}(-1)}$ as in \eno{sqrtSgn}.

\section{The four-point function}
\label{FOURPOINT}

As shown in \cite{Klebanov:2016xxf}, the four-point function of the Klebanov-Tarnopolsky tensor model with rank three fermions has the same structure as the SYK model \cite{Polchinski:2016xgd}, and this result generalizes to values of $q>4$ \cite{Narayan:2017qtw}. Working to sub-leading order in the melonic limit, the four-point correlator can be decomposed as
\eqn{4pt}{
&\Omega_{A_1A_2}\Omega_{A_3A_4}\left<T(\psi^{A_1}(t_1)\psi^{A_2}(t_2)\psi^{A_3}(t_3)\psi^{A_4}(t_4))\right>
=
\cr
&N^{2(q-1)}G(t_{12})G(t_{34})+N^{q-1}\bigg(\Gamma(t_1,t_2,t_3,t_4)+\sigma_\psi\Gamma(t_1,t_2,t_4,t_3)\bigg)+\mathcal{O}(N^{q-2}) \,,
}
where $\Gamma$ stands for a sum of so-called ladder Feynman diagrams that can be expanded as $\Gamma= \sum_n \Gamma_n$ according to the number $n$ of sets of $q-2$ rungs in the ladder diagrams. Schematically, for $q=8$:
\eqn{}
{
\Gamma(t_1,t_2,t_3,t_4) =&
\begin{matrix}\includegraphics[height=12ex]{0rungs}\end{matrix}+
\begin{matrix}\includegraphics[height=12ex]{1rung}\end{matrix}+\cr
&\begin{matrix}\includegraphics[height=12ex]{2rungs}\end{matrix}+\,...
}
Here black lines represent propagators while gray lines stand for contraction through the matrix $\Omega_{AB}$. But for a more exact understanding of the four-point function, we must also consider the threads running within each interaction vertex and endow each line with an orientation. For the index contraction in \eqref{4pt}, the arrows of the first ladder diagram are oriented as
\eqn{}
{
\begin{matrix}\includegraphics[height=12ex]{unoriented}\end{matrix}.
}
By flipping two arrows, each corresponding to $q-1$ interchanges of indices in a matrix $\Omega^{ab}$ or $\Omega_{ab}$, we get an oriented diagram,
\eqn{}
{
\begin{matrix}\includegraphics[height=12ex]{oriented}\end{matrix}.
}
Since an even number of arrow flips were performed, there is no $\sigma_\Omega$ dependence so that
\eqn{gamma0}{
\Gamma_0(t_1,t_2,t_3,t_4)=\sigma_\psi G(t_1-t_3)G(t_2-t_4)\,.
}
The absence of a factor of $\sigma_\Omega$ in \eqref{gamma0} owes directly to the choice of contraction with matrices $\Omega_{A_1A_2}\Omega_{A_3A_4}$ rather than say $\Omega_{A_1A_2}\Omega_{A_4A_3}$ in \eqref{4pt}, and so is ultimately the result of a convention. The important question to ask is whether the insertion of each new set of rungs in a ladder diagram leads to a factor of $\sigma_\Omega$. Consider therefore, as in figure~\ref{LadderRecursion}, the diagram obtained by appending an extra set of rungs to some oriented ladder diagram.

\begin{figure}
 \centerline{\includegraphics[width=5in]{LadderRecursion.pdf}}
 \caption{Insertion of an extra set of rungs in an oriented ladder. Not all internal threads in the interactions nor all rungs have been drawn so as to not clutter up the figure.}\label{LadderRecursion}
\end{figure}

The propagator that forms part of the top rail of the ladder immediately to the left of the newly appended set of rungs may be incident to any vertex $i$ of the top interaction vertex. The propagator that forms part of the top rail immediately to the right of the appended sets of rungs may be incident to any other vertex $j\neq i $ of the topmost interaction vertex. But for a given choice of $i$ and $j$, in order to get a ladder diagram that is not suppressed in the melonic limit, the two propagators of the bottom rail immediately to the left and right of the bottom interaction vertex must be given, respectively, by $\sigma(i)$ and $\sigma(j)$ for some $\sigma\in G$. And similarly the appended rungs must connect vertex $l$ of the top interaction vertex to vertex $\sigma(l)$ of the bottom one for all $l\neq i,j$. These facts follow immediately from the above discussion of the two-point function once we note that 1) the threads of the propagators incident to vertices $i$ and $\sigma(i)$ all partake of the same index loops; and 2) from the point of view of index contraction we may look upon the propagators incident to vertices $i$ and $\sigma(i)$, the arrows running to and between $A_3$ and $A_4$, and the two appended interaction vertices and the rungs running between them as all forming part of a melonic insertion into one single propagator. Since there are $q$ choices of vertex $i$, $(q-1)$ choices of vertex $j$, and $|G|$ choices of permutation $\sigma$, the total number of melonic Feynman diagrams that contribute to the ladder diagram in question is $q(q-1)|G|$.

As in our consideration of the two-point function, we can assume that the orientation of an arrow between any two vertices $i'$ and $j'$ of the top interaction vertex is the same as the orientation of the arrow between $\sigma(i')$ and $\sigma(j')$ since automorphisms induce an even number of arrow flips. As to the $q-2$ arrows that make up the rungs, these all have the same orientation prior to flipping any arrows, as one can find by computing the four-point function via functional differentiation. As in figure~\ref{LadderRecursion} we will take these arrows to point downwards rather than upwards, but this is an arbitrary choice that does not affect the parity of arrow flips needed to make all index loops have a uniform orientation of arrows.

To determine whether appending an extra set of rungs gives rise to an overall factor of  $\sigma_\Omega$, we need to consider all the index loops involved. These fall into four types:
\begin{enumerate}
\item The index loops running from $i$ through one of the rungs of the ladder and through $\sigma(i)$ without passing through $A_3$ and $A_4$, as illustrated in blue on figure~\ref{LadderRecursion}. Because there are $q-2$ such loops and they all require the same number of arrow flips to obtain uniform orientation, no net factor of $\sigma_\Omega$ is introduced on account of these index loops.
\item The index loops running from $j$ through $A_3$ and $A_4$ and back to $j$ through one of the rungs in the ladder, illustrated in red on figure~\ref{LadderRecursion}. Again these $q-2$ loops are even in number and require the same number of arrow flips for uniform orientation, so again no net factor of $\sigma_\Omega$ is introduced.
\item The index loops that run between the two interaction vertices and consist of four threads each, two in two rungs and two within the two interaction vertices. There are $(q-2)(q-3)/2$ such loops, an example of which is illustrated in yellow on figure~\ref{LadderRecursion}. Each of these index loops consists of two threads with one orientation and two index loops with the opposite orientation, so once again no net factor of $\sigma_\Omega$ is introduced in bringing about uniform orientation.
\item Lastly, there is the index loop that runs between $i$ and $j$ and $\sigma(i)$ and $\sigma(j)$, illustrated in green on figure~\ref{LadderRecursion}. Since we are assuming that the index loops of the original ladder diagram, prior to insertion of an extra set of rungs, each had a uniform orientation of arrows, two arrows must be flipped in order to give this index loop a uniform orientation (the arrow between $\sigma(i)$ and $\sigma(j)$ and the arrow between $\sigma(j)$ and $A_4$ on figure~\ref{LadderRecursion}) and again we do not pick up any net factor of $\sigma_\Omega$.
\end{enumerate}
In summary, there are no relative sign differences in ladder diagrams between $SO(N)$ and $Sp(N)$ symmetric tensor models.

Having worked out the subtleties relating to $Sp(N)$ symmetry and the automorphism group of the interaction vertex, we are ready to write down the recursive relation describing ladder diagrams:
\eqn{}
{
\Gamma_{n+1}(t_1,t_2,t_3,t_4)= \int dt\,dt'\,K(t_1,t_2,t,t')\Gamma_n(t,t',t_3,t_4)
}
where the integration kernel is given as
\eqn{kernel}
{
K(t_1,t_2,t,t')&=
\sigma_\psi q(q-1)|G|\left(\frac{g}{|G|}\right)^2N^{\frac{(q-1)(q-2)}{2}} G(t_1-t)G(t_2-t')G(t-t')^{q-2}
\cr
&=\sigma_\psi (q-1)\tilde{g}^2G(t_1-t)G(t_2-t')G(t-t')^{q-2}\,.
}

In the IR, we can plug in the expression for the dressed propagator derived in section \ref{sec:IRsolution} to obtain
\eqn{}{
K(t_1,t_2,t_3,t_4)
=&-\frac{\sigma_\psi\sigma_\Omega(q-1)}{\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})}\pi_{-\frac{2}{q},\mop{sgn}}(t_{13})\,\pi_{-\frac{2}{q},\mop{sgn}}(t_{24})\,\pi_{\frac{2(2-q)}{q},\mop{sgn}}(t_{34})\,.
}
Following \cite{Maldacena:2016hyu} and \cite{Klebanov:2016xxf} and defining
\eqn{}{
v(t_1,t_2)\equiv\pi_{h-\frac{2}{q},\mop{sgn}}(t_{12}),}
the integral eigenvalue-equation to solve in order to find the scaling dimensions of two-particle operators is given by
\eqn{}{v(t_1,t_2)=\frac{1}{g(h,q)}\int dt_3 \,dt_4\, K(t_1,t_2,t_3,t_4)\,v(t_3,t_4)\,.
}
Changing variables to
\eqn{}{
u \equiv t_{13}\,, \hspace{20mm} v \equiv t_{42}\,,
}
makes it manifest that the integral in the eigenvalue equation is a convolution since
\eqn{}{
&\int dt_3 \, dt_4 \, \pi_{-\frac{2}{q},\mop{sgn}}(t_{13})\,\pi_{-\frac{2}{q},\mop{sgn}}(t_{24})\,\pi_{h+\frac{2(1-q)}{q}-h,\mop{sgn}}(t_{34})=
\cr
&\mop{sgn}(-1)\int du \, dv\, \pi_{-\frac{2}{q},\mop{sgn}}(u)\,\pi_{-\frac{2}{q},\mop{sgn}}(v)\,\pi_{h+\frac{2(1-q)}{q},\mop{sgn}}(t_{12}-u-v)\,.
}
Invoking the sign constraint \eqref{sgnConstraint}, the eigenvalue equation can therefore be written as
\eqn{}{\pi_{h-\frac{2}{q},\mop{sgn}}=-\frac{(q-1)}{g(h,q)\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})}\,\pi_{-\frac{2}{q},\mop{sgn}}\ast\pi_{-\frac{2}{q},\mop{sgn}}\ast\pi_{h+\frac{2(1-q)}{q},\mop{sgn}}\,.
}
Fourier-transforming this equation using \eqref{Fourier}, the multiplicative characters cancel, as do two of the gamma functions, and we find that
\eqn{}{
g(h,q)=-(q-1)\frac{\Gamma(\pi_{\frac{q-2}{q},\mop{sgn}})\Gamma(\pi_{h+\frac{2-q}{q},\mop{sgn}})}{\Gamma(\pi_{\frac{2-q}{q},\mop{sgn}})\Gamma(\pi_{h+\frac{q-2}{q},\mop{sgn}})}\,.
}
This formula is valid for real as well as $p$-adic numbers, and for an action \eqref{SKTextended} with a kinetic term with any sign function.

On the real numbers, selecting the usual sign function in the action \eqref{SKTextended}, the above equation reproduces the fermionic result of the SYK and melonic tensor models, equations (3.73) in \cite{Maldacena:2016hyu} and (6.8) in \cite{Narayan:2017qtw}. Selecting the trivial sign on the reals, the above equation reproduces the bosonic result, equation (4.14) in \cite{Klebanov:2016xxf} with $d=1$.

On the $p$-adic numbers, there are multiple inequivalent sign functions, each of which can be labeled by a $p$-adic number $\tau$ as explained in appendix \ref{GammaAppendix}. For the sign functions characterized by $\tau=p$ or $\tau=\epsilon p$, $g(h,q)$ reduces to $1-q$. For $\tau=\epsilon$, $g(h,q)$ is a non-constant function in $h$, but the equation $g(h,q)=1$ never has a solution for the $p$-adic theories.

\subsection{Adelic product formula for the integral eigenvalues}
\label{ADELIC}

In \cite{Ruelle:1989dg} it is demonstrated how, by invoking the functional equations of suitably chosen Dirichlet $L$-functions, one can derive an adelic product formula for the signed Gel'fand-Graev gamma functions. By selecting a fixed rational number $\tau$, one picks out a sign function for each of the number fields $\mathbb{R}$ and $\mathbb{Q}_p$. For each of these fields with associated sign function $\mop{sgn}_\tau(x)$ there is a signed character $\pi_{s,\mop{sgn}}(x)$ and an associated gamma function $\Gamma(\pi_{s,\mop{sgn}})$. Taking the product over all these gamma functions for any fixed complex number $s$, one gets the usual sign function of $\tau$:

\eqn{RuelleFormula}
{
\prod_{K=\mathbb{R},\mathbb{Q}_p \forall p}\Gamma(\pi_{s,\mop{sgn}})=\begin{cases}
1 \hspace{8mm} \text{ for }\tau >0 \cr
-1 \hspace{5mm} \text{ for }\tau <0\,.
\end{cases}
}

Note that for any choice of $\tau$ there will be many fields $K$ for which $\mop{sgn}_\tau(x)$ is the trivial sign character, ie. unity, so that $\pi_{s,\mop{sgn}}$ reduces to $\pi_s$. For example, this happens for $K=\mathbb{R}$ when $\tau$ is positive.

As an immediate consequence of \eqref{RuelleFormula}, we find that for fixed rational $\tau$, complex $h$, and even number $q$, the integral eigenvalues $g(h,q)$ satisfy the adelic product formula

\eqn{}
{
\prod_{K=\mathbb{R},\mathbb{Q}_p \forall p}
\frac{g(h,q)}{1-q}=1\,.
}
For most choices of $\tau$, this product mixes bosonic and fermionic, and $O(N)$ and $Sp(N)$ symmetric theories.

\section{Conclusions}

Given the relative uniqueness and simplicity of melonic theories with quartic and sextic vertices, it is surprising that melonic theories proliferate and diversify for larger orders $q$ of the interaction vertex.  Already at $q=8$, there are six different types of vertices, with up to $\mathbb{Z}_2^3$ symmetry.  The situation can be compared with matrix models, where if we restrict to quartic vertices, the most commonly studied interactions are $\mop{tr} \Phi^4$ and $(\mop{tr} \Phi^2)^2$.  The first of these has $\mathbb{Z}_4$ symmetry, and the second has $\mathbb{Z}_2^3$ symmetry.  If we restrict to only one of these two vertices, then we get remarkably different behavior in the large $N$ limit: $\mop{tr} \Phi^4$ leads to a planar limit, while $(\mop{tr} \Phi^2)^2$ leads to bubble diagrams.  Mixing the two gives some interesting modifications of scaling exponents of the pure $\mop{tr} \Phi^4$ theory \cite{Das:1989fq,Sugino:1994zr,Gubser:1994yb,Klebanov:1994kv}.  For melonic theories, our story so far is less featured, in that the treatment of the two-point and four-point functions proceeds almost identically for the myriad theories we can construct.  The one salient difference among the theories we consider is that the effective coupling constant that enters into the self-consistent treatment of two- and four-point functions includes the inverse half power of the order of the symmetry group of the interaction vertex.  An interesting question for future work is to see whether combining different interaction vertices could significantly alter the analysis, for example through some cancellations or modified infrared scaling. Another interesting question for future work is to examine the extent to which the operator counting of \cite{Bulycheva:2017ilt} generalizes to higher melonic theories.

Our construction of $q$-fold interaction vertices is, by necessity, somewhat detailed, amounting to a coloring of the complete graph of $q$ vertices with $q-1$ colors so that all edges meeting at a vertex have different colors.  The combinatorial problem of counting all such colorings is formidable, and it is related to the problem of one-factorizations of the complete graph, for which results are available only for modest values of $q$.  We have however demonstrated the existence of colorings with all possible symmetries groups, namely $\mathbb{Z}_2^n$ with $n$ bounded above by $v$ if $q = 2^v$ and by $v-1$ if $q = u 2^v$ with odd $u$ greater than $1$.  While our construction of interaction vertices is rooted in the natural generalizations of the Klebanov-Tarnopolsky model \cite{Klebanov:2016xxf}, we cannot claim to have exhausted all possible generalizations with melonic limits.  Here is a useful way to frame the question: If we restrict the field content to a single real field $\psi^{a_0a_1\dots a_{q-2}}$ with $q-1$ $N$-valued indices all of which must be contracted amongst each other, is the most general interaction vertex with a melonic limit a linear combination of the interaction vertices that can be described in terms of colorings of a complete graph of $q$ vertices?  We hope to see progress on this and related issues in future work.

\subsection*{Acknowledgments}

We thank Igor Klebanov for useful discussions and suggestions. This work was supported in part by the Department of Energy under Grant No.~DE-FG02-91ER40671, and by the Simons Foundation, Grant 511167 (SSG).  The work of C.J.\ was also supported in part by the US NSF under Grant No.~PHY-1620059.

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Carly Novell and Delaney Tarr told The Opposition host they're hoping for legislation that's a "step in the right direction, something small" to advocate for safe schools without gun violence.
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Since the shooting, they have spent their time becoming "more well-researched" on gun-control issues.
He suggested they instead advocate for armed teachers, metal detectors, and fences.
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Is It Better To Buy a New Entry-Level Camera or a High-End Old Camera?
by Wasim Ahmad April 3, 2021 32 Comments
Video of New Beginner Camera vs Old Pro Camera (M50 Mark ii vs 5D Mark ii)
Here's an interesting thought exercise: Given the same amount of money, is it better to buy a low-end new camera, or a high-end old camera?
Photographer and YouTuber Hyun Ralph Jeong takes the thought out of the exercise by answering that question for viewers, comparing a 2020-era entry level camera, the Canon EOS M50 Mark II, with another Mark II model, the 5D Mark II from 2008.
Right off the bat, it's a bit of an apples-to-oranges comparison. The newer camera is a mirrorless, APS-C sensor camera, while the 5D Mark II is a classic DSLR form factor with a full frame sensor. But at the end of the day, both of these are image-making machines. So, which one offers better bang for the buck? It depends on what kind of shooter you are, Jeong says.
While sheer image quality of the full-frame sensor is "better" in that it has less noise and retains better highlight detail than its newer APS-C counterpart, the shadows are harder to work with and dynamic range isn't up to modern standards. Basically, if you need some margin of error or latitude of editing, you'll get much more of that from a modern sensor than the one in the 5D Mark II.
But beyond that, you'll also get some of the modern creature comforts that are often overlooked when purchasing older cameras. The M50 Mark II nets you a touchscreen, (incredibly fast) autofocus, Wi-Fi capability, 4K video, and an overall smaller form factor.
I really notice these ease-of-use features when I'm teaching, especially with autofocus. Teaching students how to track a subject with non-touch screen older DSLRs, where you set it to servo mode, pick a point, and hang on for dear life is a vastly different experience than say, an EOS R, where you can just tap the subject and let the camera do all the work — same results, just made much easier by modern technology.
So, is it better to buy the old camera and get the higher-end body and pro features from back in the day or get a modern camera that on paper might not have the specs of a higher-end body but in practice will make the job easier?
I guess that depends on whether you like driving the sports car of yesteryear that doesn't have air conditioning or Apple Carplay versus a new compact car that has both of those things and more; both approaches have their own merits.
Which one would you choose? Leave your thoughts in the comments below.
Hyun Ralph Jeong
Wasim Ahmad is an assistant teaching professor teaching journalism at Quinnipiac University. He's worked at newspapers in Minnesota, Florida and upstate New York, and has previously taught multimedia journalism at Stony Brook University and Syracuse University. He's also worked as a technical specialist at Canon USA for Still/Cinema EOS cameras.
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Jerome Brill - April 3, 2021 [Edited]
I think it depends on how old the nighend camera is and what your skill level is. Some of us may take an older camera because we know how to work around it's limitations. A new camera may have a few new bells and whistles but it may also be missing some critical options you may really need to get the most out of it.
If the new camera has a really good autofocus system, I'd say the newer camera. If it doesn't than I'd rather have an older camera. Sensor performance can be worked around on either camera. Not getting the shot can't.
Greg Edwards Jerome Brill - April 4, 2021 [Edited]
This. If you do t need the lastest bleeding edge technology, there's substantial savings to made buying the previous generation. It will still do pretty much everything you need it to do. The same holds true for most other devices such as phones, tablets, TVs etc. Heck, even cars. I hardly buy any tech brand new these days and I'm quite happy with my Eos M5, iPhone 11 and iPad Pro 2018 all of which I bought in the last year.
Nate Jones - April 3, 2021
Spend on lenses.
Tony Clark - April 3, 2021
The highest level that you can afford, put your priorities on lenses since you'll own them the longest.
Kurt Schuster Tony Clark - April 3, 2021
Sage advice to be sure, and while I don't disagree, it may be useful to note that Canon is already starting to discontinue EF-Series lenses. There will always be adapters and a second-hand market but when scarcity makes certain pieces collectors items, be prepared to pay a premium.
Christopher Lloyd - April 3, 2021
I think it depends on the features and what sort of photography you do. I use Olympus gear for travel and landscape/nature while hillwalking. Just moved up from EM-1 to the mk II - 2nd hand for less than any of the newest 5/10 models. From my perspective, build quality and weather sealing are more important than the latest features - but that's me. I'm sure it's different for others.
Rob Davis - April 3, 2021
Would've been a lot cooler with two cameras and the same lens.
Tom Reichner - April 3, 2021
Comparing an M50 to a 5D2 is not a viable comparison. They don't even take the same lenses, for Pete's sake.
There are so many more viable comparisons that Hyun could have used. I mean, at least compare a DSLR with another DSLR, a mirrorless with another mirrorless, an EF/EF-S mount with another EF/EF-S mount, and an M mount with another M mount.
A worthwhile apples-to-apples comparison would be a Canon EOS Rebel SL3 DSLR vs. a Canon 1D Mark 4.
Both are in the $600 range.
Both take EF lenses (note that the Rebel will also accept EF-S lenses).
Both are DSLRs.
OR, we could compare a Canon 5D Mark 4 (current DSLR) with a Canon 1DX Mark 1 (old, discontinued DSLR). Both are roughly in the same ballpark, price wise. Both take EF lenses. Both are full frame.
To compare two completely different cameras that have nothing at all in common is just strange, and doesn't help Hyun to make whatever point he was trying to make in his video.
Morris Erickson - April 4, 2021
The Canon mount adapter EF-EOS M will allow you to use any EF/EF-S lens on an "M" mount camera. I like to use a Zeiss Distagon 2.8/21 ZE lens on an M6 MkII. A tad lighter, a 32 MEG sensor, and manual focus peaking. The Canon EF 70-200mm f/4L IS II USM lens with the M6 makes a good wildlife/hiking combo.
N A - April 4, 2021
If you're patient, wait for a higher end camera to reach end of sale. Not sure about Nikon or Sony but Canon steeply discounts their bodies when the replacements are announced. The 5D4 new was a low as $1899 USD just before the R5 was announced. If you can work with a previous generation body it's good way to get more camera for your money.
Frank Landry - April 4, 2021
Right now its a catch 22. Mirrorless cameras have a limited number of lenses available. Yes you can get the adapter to use your old lenses. But their is no guarantee that they will work properly.
Older cameras have a lot of lenses available. But for how long? Canon has already started cutting back. Nikon is planning on to support their lenses for awhile!
I have an older camera. Im holding onto it. I have all the lenses, i need!
Todd Bryant - April 4, 2021
New camera all the way. I just went out shooting astrophotography with my cousin and the processing time associated her full frame took forever. It was night and day. I also shoot action shots and you can't match the speed of a mirror less camera with a DSLR. Autofocus is also a beast.
J. H. Todd Bryant - April 4, 2021
" you can't match the speed of a mirror less camera with a DSLR" Please take a Nikon D780, D850, D4(S), D5, D6 into consideration.
An older D800(E) is still up to date an it is available in mint state for less than $800.
Tom Reichner Todd Bryant - April 4, 2021
You are mistaken when you say that a DSLR cannot match the focusing speed of a mirrorless camera. Which mirrorless camera? And which DSLR?
If we make a comparison of a new mirrorless camera and a similarly priced old DSLR, I bet the old discontinued DSLR beats the new mirrorless camera when it comes to speed and focusing.
An old, used, discontinued Canon 1D Mark 4 is now about $700. What new $700 mirrorless camera can come anywhere close to the 1D Mark 4 in either frame rate or focusing speed? None can. Absolutely freaking none. They all suck, compared to the 11 year old Canon 1D Mark 4 DSLR.
How about we compare $1200 cameras? A new $1200 mirrorless camera against an old discontinued DSLR that is now around $1200. The old Canon 1DX fits the bill. What new mirrorless camera can you get for $1200 that can keep up with the speed and focusing of the 1DX? I'll tell you - NONO! No $1200 mirrorless can even come close to the speed of the original 1DX, which is now 9 years old.
What you said leads me to believe that you never even shot with a high end DSLR. And if that is the case, then you aren't even qualified to comment about these DSLRs. Don't put something down if you've never even used it.
Wasim Ahmad Tom Reichner - April 4, 2021 [Edited]
I have owned both a 1D Mark IV and a EOS-1D X, both great cameras but most mid-level mirrorless cameras have caught up in terms of speed and focusing, and have certainly surpassed both of those cameras in image quality and ISO performance.
Don't get me wrong, I loved, loved loved both of those cameras but I'd still take an EOS R today which can frequently be found $1500 on the refurb side of the Canon site. What it gives up in terms of sheer FPS, it makes up for in focus accuracy.
Francisc Adamovici - April 4, 2021
For landscape the autofocus performance is useless, better buy an old generation camera and spend more money on a pro lens, filters and a good tripod.
christian thompson - April 4, 2021
Its amazing how spoilt we have become expecting new features every year . Features that so frequently actually never get used.
Back in the old days of silver halide progress was slow, especially in terms of image quality or lens sharpness. The only real path was to get a bigger camera with bigger film in it.
A canon ds mkii or iii offers quality and power us old pros would have given an arm or a leg for. And they are still amazing image making machines today. Forget about mpixel counts. Those older pro cameras deliver in spades and cost €400 ... personally the 5dmkii was a disappointment for me. Maybe too much hype at the time.... what??? Video??mmm.... but i found it too compromised and dysfunctional.
I would recommend any beginner to get a nikon 2d or 3d or canon d2 or 3... and use it in manual for 2 years.. Learn the basics and get amazing quality results with a proper solid machine. I wonder if these older cameras will increase in value one day?
I also love my olympus em5 mkii... amazing camera. Image stabilization is the revolution!
Wasim Ahmad christian thompson - April 4, 2021
My Nikon D2H has not aged well. I'd start at the D3 and go on up from there. I have vague memories of using a Canon 1D Mark IIN in grad school but I can't really remember how I felt about that camera.
Joseph Balson Wasim Ahmad - April 5, 2021
That's funny you mention the D2H.
I actually loved the D2H despite suffering the D2H curse: I owned 3. The first just died for no obvous reason, the second, a D2Hs, drowned, and the last one was stolen. These cameras were my workhorses and took some crazy beating.
When I look back at the photos today up to 16x20 prints, I wouldn't say it didn't age well. The photos are as good as they were 20 years ago.
I especially don't think any modern camera could survive what survived these D2H, desert, mud, monsoon, heat, cold, not being babyed...
considering lots of "photographers" today mostly don't print and only post on social media, 4MP is plenty enough.
I still wouldn't recommend anyone to buy one, but for 150 bucks, why not?the AF is good and it's crazy fast. the trouble would be finding one in good condition, the prices are going up steadily for a couple years.
Wasim Ahmad Joseph Balson - April 7, 2021
I mean, technically you aren't wrong. There's a bus driving around Stony Brook University wrapped with photos I took from a Nikon D2H and I'm still amazed that they enlarged that well.
Stephen Strangways - April 4, 2021
Given the same amount of money? in Toronto, I haven't seen a used M50 Mark II, and the new ones sell for more than twice what a used 5D Mark II sells for. Start considering lens availability and cost, or the cost of an adapter for the M50, and the 5D definitely seems like the better bargain.
Stephen reef - April 4, 2021
Don't buy a Canon DSLR. They've abandoned their DSLR clients.
regan albertson - April 4, 2021
Entry level new vs full framed old for stills? get the old one. For video, yeah, the new would be better. The tanks built in the last dozen years still make great images, when correctly deployed. It's not the brush, it's the artist.
regan albertson regan albertson - April 8, 2021
Some research; A D700 in great shape goes for $300 or so. A medium zoom and a nifty-50 used might add $200. The only thing missing is the dial with AUTO on it. If this newbie is into learning photography, the AUTO can be dispensed with.
Rodolfo Farinas - April 4, 2021
A more advanced camera may push you to better understand and master all the features and advanced settings, on the other hand, a newer entry level may only give you a slight advantage of more speed and or connectivity features that in the end, are not all that necessary to get good pictures. If you get used to work with and handle a more advanced camera, then an entry level camera will be much easier to use. Older advanced verás may give you exceptional Dynamic Range and ISO performance while remaining compatible with older lenses and equipment, And it can also save you some money. But in the end, this is a hobby or profession, that once you get into it, you will always crave the newer stuff, until you reach some level of maturity where you understand that there's a lot more to photography than just the camera body.
Jonathan Ziegler - April 5, 2021
I like both the 5D Mark II & III. I like them because you can get 4k raw video from the mark 3 with magic lantern and raw video (2.5k) from the mark 2. Magic Lantern is one of the main reasons I still buy old t2i and t3i camera bodies (and 5d/6d when available) - I can greatly extend the features and capabilities of these cameras. I first discovered it with the t2i which uses the same sensor as the 7D and has more features than the 7D with ML.
https://magiclantern.fm/
D Smith - April 5, 2021
I am always amused by these "beginner" (their words not mine) recommendations where it involves dropping +$1000 just to get started, let alone adding a few full frame lenses possibly 3x or 4x'ing the price (used) to cover wide to zoom. When I think of a "beginner", I think of someone who has never operated or carried (for hours) an interchangeable lens camera system. Nor have they found their photographic calling (landscape, macro, birds, portrait, etc.)
I would highly recommend to any "beginner" reading this post, go buy the least expensive interchangeable camera system for under $200 with a kit lens. If you cannot take beautiful photos with this camera/lens combo, you have no business dropping thousands, your expensive equipment will not save you. My old T1i could still take beautiful photo's before giving it away to another budding photographer. Everyone needs to remember, beginners use Instagram et.al. yea 1080x1080 resolution. If after spending $200, if you don't like the hobby, you can sell it for what you paid for it.
Once you, the beginner, get to a point where you've discovered your calling and you feel your equipment is holding you back (you'll know when that time is), then and only then do you INVEST into a camera system. My advice will save you thousands while not having this expensive kit sitting in a drawer somewhere.
Shawn W - April 6, 2021
Wrong title and bias comparison. It should be named "13 year old high end full frame DSLR vs 0 year old crop censor mirrorless camera".
I bought a 5D iii in mint condition at $1000 CAD recently. A used 5D ii can be had at $500 CAD. Just checked the price of the M50 mark ii, $899 plus tax. It makes more sense to compare a 5D iii with M50 ii. Then I'd prefer 5D iii.
Wasim Ahmad Shawn W - April 7, 2021
Yeah, there was a huge difference between 5D2 and 5D3 (I owned both for a long time) - enough to swing me in a comparison with the M50 II (which I haven't owned but I had a lot of seat time with its kissing cousin, the M5).
Eriol Brumel - April 10, 2021
With today's technology, photography is probably 80% the skill of the photographer and 20% the performance of his hardware. I have been shooting digitally since 2004 and have the impression that - quite subjectively and only referring to myself - since 2014 the technology of DSLRs has matured quite a bit. So I wouldn't buy an old camera from 2008, since the sensors and image processors in particular have still improved noticeably. But for cameras since 2014, that's totally different ...
I would prefer a "professional" DSLR from 2014/2015 to any top modern entry-level camera, because - quite subjectively - I could not take better photos with the new entry-level camera. And I wouldn't have any more fun with it either. I only take photos through a viewfinder, and I value natural color reproduction (which I can change myself during subsequent photo processing). Professional cameras usually have a better haptics and feel and allow quick settings without having to take the camera off your face.
I don't need a touchscreen or a swiveling 5" display on my DSLR. However, I always have a good photo smartphone with me for special purposes in certain situations.
Ivan Konar - April 10, 2021
Are you hiperactive, want quickly images, are you running to takes photos every where, do you always has the last camera model of the year, are you a easy finger shutter making a thousend pictures per minute ? then buy a mirroless, but if you are a dedicated photographer, then any good dslr of the last 10 years fit perfect with your profile...
Tom Reichner Ivan Konar - May 6, 2021 [Edited]
Ivan Konar said,
"Are you hiperactive, want quickly images, are you running to takes photos every where, do you always has the last camera model of the year, are you a easy finger shutter making a thousend pictures per minute ? then buy a mirroless, but if you are a dedicated photographer, then any good dslr of the last 10 years fit perfect with your profile... "
Ivan,
What you have done in your statement is to create two very separate categories and then put one down in a passive-aggressive way, and build the other category up by using the word "dedicated".
In real life, and in real photography, photographers do not fit so nicely and neatly into your two categories, each if which is so exclusive of the other.
Many dedicated, serious photographers who make wonderfully creative photographs and exercise lots of patience have found that mirrorless cameras suit their needs and their shooting style better than any DSLRs.
And many very casual, impatient, "copycat" photographers who have little patience and care little about creativity and composition prefer to use DSLRs.
Just because you like DSLRs and dislike mirrorless cameras does not mean that you can twist reality around into misinformation, and then use that misinformation to support your likes and bash your dislikes. That is wrong to do, and it is offensive to all of us because it feels like you are trying to manipulate us all into thinking the way you think, by bashing anyone who dares to differ.
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