A solvable tensor field theory

Abstract

We solve the closed Schwinger–Dyson equation for the 2-point function of a tensor field theory with a quartic melonic interaction, in terms of Lambert’s W function, using a perturbative expansion and Lagrange–Bürmann resummation. Higher-point functions are then obtained recursively.

Appendix A: Recurrence relations

In this section, we will use the recursive equation (11) to determine recurrence relations on the numbers $a_{n,k,m}$. We first perform the integration

$$\begin{aligned} \int \mathrm {d}\mathbf {q}_{\hat{1}} G^{(2)}_p(\mathbf {q}_{\hat{1}}x_1)&= -\frac{\pi }{4}\log (1+x_1^2) \quad \text {if} \,\, p=0, \end{aligned}$$

(42)

$$\begin{aligned}&= \bigg (\frac{\pi }{2}\bigg )^{p+1}\Bigg (\frac{\log ^p(1+x_1^2)}{2p(1+x_1^2)^p}+\frac{(-1)^p}{2(1+x_1^2)^p}\nonumber \\&\qquad \sum \limits _{r=1}^{p-1}(-1)^r\log ^r(1+x_1^2)\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m} \Bigg ) \quad \text {if} \,\, p>0, \end{aligned}$$

(43)

where for $p=1$ the sum on r does not appear. Plugging back the ansatz (19) in the recurrence relation (11) with $c=1$ gives

$$\begin{aligned} G^{(2)}_{n}(\mathbf {x})&= -\frac{2}{|\mathbf {x}|^2+1}\Bigg \{-\frac{\pi }{4}\log (1+x_1^2)G^{(2)}_{n-1}(\mathbf {x})\nonumber \\&\quad + \sum \limits _{p=1}^{n-1}\bigg (\frac{\pi }{2}\bigg )^{p+1}\frac{\log ^p(1+x_1^2)}{2p(1+x_1^2)^p} G^{(2)}_{n-p-1}(\mathbf {x}) \nonumber \\&\quad + \sum \limits _{p=2}^{n-1}\bigg (\frac{\pi }{2}\bigg )^{p+1}\bigg (\frac{(-1)^p}{2(1+x_1^2)^p}\nonumber \\&\quad \sum \limits _{r=1}^{p-1}(-1)^r\log ^r(1+x_1^2)\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m} \Bigg )G^{(2)}_{n-p-1}(\mathbf {x})\Bigg \}. \end{aligned}$$

(44)

The first term of (44) gives

$$\begin{aligned}&\frac{\pi \log (1+x_1^2)}{2(|\mathbf {x}|^2+1)}\bigg (\frac{\pi }{2}\bigg )^{n-1}\Bigg (\frac{\log ^{n-1}(1+x_1^2)}{(1+|\mathbf {x}|^2)^{n}}\nonumber \\&\quad +\frac{(-1)^{n-1}}{(1+x_1^2)^{n-1}}\sum \limits _{k=1}^{n-2}(-1)^k\log ^k(1+x_1^2)\sum \limits _{m=1}^k a_{n-1,k,m}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}}\Bigg ) \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\Bigg (\frac{\log ^{n}(1+x_1^2)}{(1+|\mathbf {x}|^2)^{n+1}}+\frac{(-1)^{n-1}}{(1+x_1^2)^{n-1}}\sum \limits _{k=1}^{n-2}(-1)^k\nonumber \\&\qquad \log ^{k+1}(1+x_1^2)\sum \limits _{m=1}^k a_{n-1,k,m}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+2}}\Bigg ) \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\Bigg (\frac{\log ^{n}(1+x_1^2)}{(1+|\mathbf {x}|^2)^{n+1}}+\frac{(-1)^{n}}{(1+x_1^2)^{n}}\nonumber \\&\qquad \sum \limits _{k=2}^{n-1}(-1)^k\log ^{k}(1+x_1^2)\sum \limits _{m=2}^k a_{n-1,k-1,m-1}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}}\Bigg ), \end{aligned}$$

(45)

where we sent $k \rightarrow k+1$ and $m \rightarrow m+1$ to get to the last line. The second term of (44) gives

$$\begin{aligned}&-\frac{2}{|\mathbf {x}|^2+1}\Bigg (\sum \limits _{p=1}^{n-1}\bigg (\frac{\pi }{2}\bigg )^{p+1}\frac{\log ^p(1+x_1^2)}{2p(1+x_1^2)^p} \bigg (\frac{\pi }{2}\bigg )^{n-p-1}\frac{\log ^{n-p-1}(1+x_1^2)}{(1+|\mathbf {x}|^2)^{n-p}} \nonumber \\&\quad +\sum \limits _{p=1}^{n-3}\bigg (\frac{\pi }{2}\bigg )^{p+1}\frac{\log ^p(1+x_1^2)}{2p(1+x_1^2)^p}\bigg (\frac{\pi }{2}\bigg )^{n-p-1}\frac{(-1)^{n-p-1}}{(1+x_1^2)^{n-p-1}}\nonumber \\&\quad \sum \limits _{k=1}^{n-p-2}(-1)^k\log ^k(1+x_1^2)\sum \limits _{m=1}^k a_{n-p-1,k,m}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}}\Bigg ) \nonumber \\&\quad =-\bigg (\frac{\pi }{2}\bigg )^{n}\frac{\log ^{n-1}(1+x_1^2)}{(1+x_1^2)^{n}} \sum \limits _{p=1}^{n-1}\frac{1}{p}\frac{(1+x_1^2)^{n-p}}{(1+|\mathbf {x}|^2)^{n-p+1}} \nonumber \\&\quad +\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{p=1}^{n-3}\nonumber \\&\quad \sum \limits _{k=1}^{n-p-2}(-1)^{k-p}\log ^{p+k}(1+x_1^2)\sum \limits _{m=1}^k \frac{a_{n-p-1,k,m}}{p}\frac{(1+x_1^2)^{m+1}}{(1+|\mathbf {x}|^2)^{m+2}}. \end{aligned}$$

(46)

Setting $r=p+k$ in the line of the previous equation, let us rewrite the double sum as

$$\begin{aligned} \sum \limits _{k=1}^{n-3}\sum \limits _{r=k+1}^{n-2}\frac{(-1)^{r}}{r-k}\log ^{r}(1+x_1^2)a_{n-r+k-1,k,m} = \sum \limits _{r=2}^{n-2}\sum \limits _{k=1}^{r-1}\frac{(-1)^{r}}{r-k}\log ^{r}(1+x_1^2)a_{n-r+k-1,k,m}. \end{aligned}$$

(47)

Then, we send $m \rightarrow m+1$ and rewrite double sum to get

$$\begin{aligned} \sum \limits _{k=1}^{r-1}\sum \limits _{m=2}^{k+1} \frac{a_{n-r+k-1,k,m-1}}{r-k}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}} = \sum \limits _{m=2}^{r}\sum \limits _{k=m-1}^{r-1} \frac{a_{n-r+k-1,k,m-1}}{r-k}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}}. \end{aligned}$$

(48)

Hence, sending $p \rightarrow n-p$ and collecting the results, we get

$$\begin{aligned}&-\bigg (\frac{\pi }{2}\bigg )^{n}\frac{\log ^{n-1}(1+x_1^2)}{(1+x_1^2)^{n}} \sum \limits _{p=1}^{n-1}\frac{1}{n-p}\frac{(1+x_1^2)^{p}}{(1+|\mathbf {x}|^2)^{p+1}} \nonumber \\&\quad +\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{r=2}^{n-2}(-1)^{r}\log ^{r}(1+x_1^2)\nonumber \\&\quad \sum \limits _{m=2}^{r}\sum \limits _{k=m-1}^{r-1} \frac{a_{n-1+k-r,k,m-1}}{r-k}\frac{(1+x_1^2)^{m}}{(1+|\mathbf {x}|^2)^{m+1}}. \end{aligned}$$

(49)

The third term of (44) gives

$$\begin{aligned}&-\frac{2}{|\mathbf {x}|^2+1}\Bigg \{\sum \limits _{p=2}^{n-1}\bigg (\frac{\pi }{2}\bigg )^{p+1}\Bigg (\frac{(-1)^p}{2(1+x_1^2)^p}\sum \limits _{r=1}^{p-1}(-1)^r\nonumber \\&\quad \log ^r(1+x_1^2)\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m} \Bigg )\bigg (\frac{\pi }{2}\bigg )^{n-p-1}\frac{\log ^{n-p-1}(1+x_1^2)}{(1+|\mathbf {x}|^2)^{n-p}} \nonumber \\&\quad +\sum \limits _{p=2}^{n-3}\bigg (\frac{\pi }{2}\bigg )^{p+1}\Bigg (\frac{(-1)^p}{2(1+x_1^2)^p}\sum \limits _{r=1}^{p-1}(-1)^r\log ^r(1+x_1^2)\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m} \Bigg ) \nonumber \\&\quad \bigg (\frac{\pi }{2}\bigg )^{n-p-1}\frac{(-1)^{n-p-1}}{(1+x_1^2)^{n-p-1}}\nonumber \\&\quad \sum \limits _{k=1}^{n-p-2}(-1)^k\log ^k(1+x_1^2)\sum \limits _{l=1}^k a_{n-p-1,k,l}\frac{(1+x_1^2)^{l}}{(1+|\mathbf {x}|^2)^{l+1}}\Bigg )\Bigg \}. \end{aligned}$$

(50)

The first term of Eq. (50) gives

$$\begin{aligned}&-\bigg (\frac{\pi }{2}\bigg )^{n} \sum \limits _{p=2}^{n-1}\sum \limits _{r=1}^{p-1}(-1)^{p+r}\frac{\log ^{n-p+r-1}(1+x_1^2)}{(1+x_1^2)^n}\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m}\frac{(1+x_1^2)^{n-p}}{(1+|\mathbf {x}|^2)^{n-p+1}} \nonumber \\&\quad =-\bigg (\frac{\pi }{2}\bigg )^{n} \sum \limits _{r=1}^{n-2}\sum \limits _{k=1}^{n-1-r}(-1)^k\frac{\log ^{n-k-1}(1+x_1^2)}{(1+x_1^2)^n}\sum \limits _{m=1}^{r}\frac{a_{r+k,r,m}}{m}\frac{(1+x_1^2)^{n-r-k}}{(1+|\mathbf {x}|^2)^{n-r-k+1}}, \end{aligned}$$

(51)

by setting $k=p-r$. Then, by setting $l=n-1-k$ and rewriting the sums, we get

$$\begin{aligned}&\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{r=1}^{n-2}\sum \limits _{l=r}^{n-2}(-1)^l\log ^{l}(1+x_1^2)\nonumber \\&\quad \sum \limits _{m=1}^{r}\frac{a_{n-1+r-l,r,m}}{m}\frac{(1+x_1^2)^{l-r+1}}{(1+|\mathbf {x}|^2)^{l-r+2}} \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{l=1}^{n-2}(-1)^l\log ^{l}(1+x_1^2)\nonumber \\&\quad \sum \limits _{r=1}^{l}\sum \limits _{m=1}^{r}\frac{a_{n-1+r-l,r,m}}{m}\frac{(1+x_1^2)^{l-r+1}}{(1+|\mathbf {x}|^2)^{l-r+2}}. \end{aligned}$$

(52)

Then, we set $k = l-r+1$ and obtain

$$\begin{aligned} \bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{l=1}^{n-2}(-1)^l\log ^{l}(1+x_1^2)\sum \limits _{k=1}^{l}\sum \limits _{m=1}^{l-k+1}\frac{a_{n-k,l-k+1,m}}{m}\frac{(1+x_1^2)^{k}}{(1+|\mathbf {x}|^2)^{k+1}}. \end{aligned}$$

(53)

The second term of (50) gives, by rewriting the sums,

$$\begin{aligned}&\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{p=2}^{n-3}\sum \limits _{k=1}^{n-p-2}\sum \limits _{r=1}^{p-1}(-1)^{k+r}\log ^{k+r}(1+x_1^2)\nonumber \\&\quad \sum \limits _{l=1}^k\sum \limits _{m=1}^{r}\frac{a_{p,r,m}}{m} a_{n-p-1,k,l}\frac{(1+x_1^2)^{l+1}}{(1+|\mathbf {x}|^2)^{l+2}} \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{r=1}^{n-4}\sum \limits _{k=1}^{n-3-r}(-1)^{k+r}\log ^{r+k}(1+x_1^2)\nonumber \\&\quad \sum \limits _{l=1}^{k}\sum \limits _{m=1}^{r}\sum \limits _{p=r+1}^{n-2-k}\frac{a_{p,r,m}}{m} a_{n-p-1,k,l}\frac{(1+x_1^2)^{l+1}}{(1+|\mathbf {x}|^2)^{l+2}}. \end{aligned}$$

(54)

First by setting $q=k+r$ and by several rewriting of the sums, we get

$$\begin{aligned}&\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{k=1}^{n-4}\sum \limits _{q=k+1}^{n-3}(-1)^q\log ^{q}(1+x_1^2)\sum \limits _{l=1}^{k}\sum \limits _{m=1}^{q-k}\nonumber \\&\quad \sum \limits _{p=q-k+1}^{n-2-k}\frac{a_{p,q-k,m}}{m} a_{n-p-1,k,l}\frac{(1+x_1^2)^{l+1}}{(1+|\mathbf {x}|^2)^{l+2}} \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{q=2}^{n-3}(-1)^q\log ^{q}(1+x_1^2)\sum \limits _{k=1}^{q-1}\sum \limits _{l=1}^{k}\sum \limits _{m=1}^{q-k}\nonumber \\&\quad \sum \limits _{p=q-k+1}^{n-2-k}\frac{a_{p,q-k,m}}{m} a_{n-p-1,k,l}\frac{(1+x_1^2)^{l+1}}{(1+|\mathbf {x}|^2)^{l+2}} \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{q=2}^{n-3}(-1)^q\log ^{q}(1+x_1^2)\sum \limits _{l=1}^{q-1}\sum \limits _{k=l}^{q-1}\sum \limits _{m=1}^{q-k}\nonumber \\&\quad \sum \limits _{p=q-k+1}^{n-2-k}\frac{a_{p,q-k,m}}{m} a_{n-p-1,k,l}\frac{(1+x_1^2)^{l+1}}{(1+|\mathbf {x}|^2)^{l+2}} \nonumber \\&\quad =\bigg (\frac{\pi }{2}\bigg )^{n}\frac{(-1)^n}{(1+x_1^2)^n}\sum \limits _{q=2}^{n-3}(-1)^q\log ^{q}(1+x_1^2)\sum \limits _{l=2}^{q}\sum \limits _{k=l-1}^{q-1}\sum \limits _{m=1}^{q-k}\nonumber \\&\quad \sum \limits _{p=q-k+1}^{n-2-k}\frac{a_{p,q-k,m}}{m} a_{n-p-1,k,l-1}\frac{(1+x_1^2)^{l}}{(1+|\mathbf {x}|^2)^{l+1}}, \end{aligned}$$

(55)

where we send $l \rightarrow l+1$ in the last line.

Now, collecting all the results, we obtain recurrence relations on $a_{n,k,m}$:

$$\begin{aligned}&a_{n,1,1} = a_{n-1,1,1}, \end{aligned}$$

(56)

$$\begin{aligned}&a_{n,n-1,1} = \frac{1}{n-1}, \end{aligned}$$

(57)

$$\begin{aligned}&a_{n,n-1,m} = \frac{1}{n-m} + a_{n-1,n-2,m-1}, \quad \text {for} \,\, m \in \llbracket 2,n-1 \rrbracket , \end{aligned}$$

(58)

$$\begin{aligned}&a_{n,n-2,m} = a_{n-1,n-3,m-1} + \sum \limits _{r=m-1}^{n-3} \frac{a_{r+1,r,m-1}}{n-2-r} + \sum \limits _{l=1}^{n-1-m} \frac{a_{n-m,n-1-m,l}}{l}, \end{aligned}$$

(59)

$$\begin{aligned}&\quad \text {for} \,\, m \in \llbracket 2,n-2 \rrbracket , \nonumber \\&a_{n,k,1} = \sum \limits _{l=1}^{k} \frac{a_{n-1,k,l}}{l}, \quad \text {for} \,\, k \in \llbracket 1,n-3 \rrbracket , \end{aligned}$$

(60)

$$\begin{aligned}&a_{n,k,m} = a_{n-1,k-1,m-1} + \sum \limits _{r=m-1}^{k-1} \frac{a_{n-1+r-k,r,m-1}}{k-r} + \sum \limits _{l=1}^{k-m+1} \frac{a_{n-m,k-m+1,l}}{l} \nonumber \\&+ \sum \limits _{r=m-1}^{k-1}\sum \limits _{l=1}^{k-r}\sum \limits _{p=k-r+1}^{n-2-r}\frac{a_{p,k-r,l}a_{n-p-1,r,m-1}}{l}, \quad \text {for} \,\, k \in \llbracket 2,n-3 \rrbracket \,\, \text {and} \,\, m \in \llbracket 2,k \rrbracket . \end{aligned}$$

(61)

Rewriting these equations gives explicit relations on Stirling numbers of the first kind, harmonic numbers and binomial coefficients. Indeed, from Eq. (60), we recover

$$\begin{aligned} \frac{1}{(n-1)!} \genfrac[]{0.0pt}{}{n-1}{n-k} = \sum _{l=1}^k \frac{1}{(n-l)!} \genfrac[]{0.0pt}{}{n-1-l}{n-1-k}, \quad \text {for} \,\, k \in \llbracket 1,n-3 \rrbracket , \end{aligned}$$

(62)

which corresponds to equation (6.21) in [16].

Setting $l=n-2-r$, $k = n-m-1$ and sending $n-3 \rightarrow n$, Eq. (59) gives

$$\begin{aligned} H_k = \frac{k+1}{2n+3-k}\sum \limits _{l=1}^{k} \frac{n+1-k+l}{l(k+1-l)}, \quad \text {for} \,\, k \in \llbracket 1,n \rrbracket . \end{aligned}$$

(63)

Sending $r \rightarrow k-l$ and in the last term $l \rightarrow r$ of Eq. (61), we get

$$\begin{aligned}&((n-1)m-k(m-1))\frac{(n-2)!}{k!(n-m)!}\genfrac[]{0.0pt}{}{n-m}{n-k} \nonumber \\&\quad = \sum \limits _{l=1}^{k-m+1} \frac{1}{(n-m-l)!}\genfrac[]{0.0pt}{}{n-m-l}{n-k-1}\Bigg (\frac{(n-1-m)!}{(k-m+1)!}+\frac{m-1}{l}\frac{(n-l-2)!}{(k-l)!} \Bigg ) \nonumber \\&\quad + \sum \limits _{l=1}^{k-m+1} \frac{m-1}{l!(k-l)!} \sum \limits _{p=l+1}^{n-2-k+l}\frac{(p-1)!(n-2-p)!}{(n-m-p)!}\nonumber \\&\quad \genfrac[]{0.0pt}{}{n-m-p}{n-k-1-p+l}\sum _{r=1}^l\frac{1}{(p-r)!}\genfrac[]{0.0pt}{}{p-r}{p-l}, \end{aligned}$$

(64)

for $k \in \llbracket 2,n-3 \rrbracket$ and $m \in \llbracket 2,k \rrbracket$.