The Sachdev-Ye-Kitaev (SYK) model and conformal $SL(2,\mathbb{R})$ symmetry in a 1D spacetime

Question

Two-point function in SYK model is given by $$\begin{align} G_{ij}(\tau,\tau')=\frac{b}{|\tau-\tau'|^{1/2}}{\rm sgn}(\tau-\tau')\delta_{ij} \end{align} \tag{1}$$ where $i$ and $j$ are the indices of the fermion $\chi_i, \chi_j$ and $b$ is just coefficient. SYK model has the following reparametrization invariants in IR limit. $$\begin{align} G_{ij}(\tau,\tau')=|f'(\tau)f'(\tau')|^{1/4}G_{ij}(f(\tau),f(\tau')) \end{align} \tag{2}$$ where $f(\tau)$ is reparametrization of time $\tau$. Several papers have argued that it has only $SL(2,\mathbb{R})$ symmetry. I would like to prove this, but I'm not sure how to do it. The $SL(2,\mathbb{R})$ transformation can be written as the following transformation. $$\begin{align} f(\tau)=\frac{a\tau+b}{c\tau+d},\ a,b,c,d\in\mathbb{R},\ ad-bc=1. \end{align} \tag{3}$$ If we consider this based on the reparametrization invariance mentioned earlier, $$\begin{align} G_{ij}(\tau,0)&=|f'(\tau)f'(0)|^{1/4}\frac{b}{|f(\tau)-f(0)|^{1/2}}{\rm sgn}(f(\tau)-f(0))\delta_{ij}\nonumber \\ &=\left|\frac{1}{d(c\tau+d)}\right|^{1/2}\frac{b}{\left|\frac{a\tau+b}{c\tau+d}-\frac{b}{d}\right|^{1/2}}{\rm sgn}\left(\frac{a\tau+b}{c\tau+d}-\frac{b}{d}\right)\delta_{ij}\nonumber \\ &=\left|\frac{1}{d(c\tau+d)}\right|^{1/2}\frac{b}{\left|\frac{\tau(ad-bc)}{d(c\tau+d)}\right|^{1/2}}{\rm sgn}\left(\frac{\tau(ad-bc)}{d(c\tau+d)}\right)\delta_{ij}\nonumber \\ &=\frac{b}{\left|\tau\right|^{1/2}}{\rm sgn}\left(\frac{\tau}{d(c\tau+d)}\right)\delta_{ij} \end{align}\tag{4}$$ where I used the fact $$f'(\tau)=\frac{1}{(c\tau+d)^2}>0.\tag{5}$$ But here the sgn function does not become sgn$(\tau)$. Depending on the sign of $d(c\tau+d)$, it can be positive or negative in general. However, many papers claim that it is invariant under $SL(2,\mathbb{R})$. What am I doing wrong? Please tell me.

Answer 1

1. One particular simple solution to the connected zero-temperature 2-point function in the SYK model in the infrared (IR) $ |\tau|,|\tau^{\prime}| \gg J^{-1}$ is $$G(\tau,\tau^{\prime})~=~C\frac{\color{red}{{\rm sgn}(\tau-\tau^{\prime})}}{|\tau-\tau^{\prime}|^{2\Delta}}, \tag{A}$$ cf. Refs. 1-4.

   The absolute value $|G(\tau,\tau^{\prime})|$ has the standard power-law of a CFT 2-point function of primary fields.

2. The full solution to the connected 2-point function $G(\tau,\tau^{\prime})$ in the infrared (IR) is an infinite family of solutions weaved together by local 1D conformal transformations, i.e. (Euclidean) time reparametrizations/diffeomophisms $f$, cf. Ref. 3 and e.g. this Phys.SE post. They transform covariantly $$|G(\tau,\tau^{\prime})|~=~|f^{\prime}(\tau)|^{\Delta}~|f^{\prime}(\tau^{\prime})|^{\Delta}~|G(f(\tau),f(\tau^{\prime}))|.\tag{B}$$

3. OP wants to show that eq. (A) is invariant under global 1D conformal transformations, i.e. Moebius/$PSL(2,\mathbb{R})$ transformations, but a sign factor $\color{red}{{\rm sgn}(\tau-\tau^{\prime})}$ gets in their way.

4. Note that eq. (A) is odd (and hence not invariant) under (Euclidean) time reversal $f(\tau)=-\tau.$ For this (and some other) reasons we restrict from now on to (Euclidean) time reparametrizations/diffeomophisms $f\in{\rm Diff}^+(\mathbb{S^1})$ that are monotonically increasing: $f^{\prime}\geq 0$. We furthermore assume that all zeros and poles of $f^{\prime}$ are of even order, so that we can extract a smooth square root.

5. Proposal for monotonically increasing diffeomophisms. Let us modify OP's conformal transformation rule (2) to $$ G(\tau,\tau^{\prime})~=~\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau)})}|f^{\prime}(\tau)|^{\Delta}~\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau^{\prime})})}|f^{\prime}(\tau^{\prime})|^{\Delta}~G(f(\tau),f(\tau^{\prime}))\tag{C}$$ where we, say, choose the smooth square root branch with$^1$ $$\sqrt{f^{\prime}(0^+)}~>~0.\tag{D}$$

6. Proposal for Moebius transformations. The 2-point function (A) becomes invariant under Moebius transformations $f:\mathbb{R}\to\mathbb{R}\cup\{\infty\}$ $$\begin{align} f(\tau)~=~&\frac{a\tau+b}{c\tau+d},\tag{E} \cr f^{\prime}(\tau)~\stackrel{(C)}{=}~&\frac{1}{(c\tau+d)^2}~>~0, \tag{F} \cr a,b,c,d~\in~\mathbb{R}, \end{align} $$ with the group $PSL(2,\mathbb{R})\cong SL(2,\mathbb{R})/\{\pm {\bf 1}_{2\times 2}\}$ where $$ SL(2,\mathbb{R})~=~\left\{ M=\begin{pmatrix}a&b \cr c&d\end{pmatrix}\in GL(2,\mathbb{R}) ~\mid~ \det(M)= 1 \right\}. \tag{G}$$ if we modify OP's conformal transformation rule (2) to$^2$ $$\begin{align} G(\tau,\tau^{\prime})~=~&\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau)})}|f^{\prime}(\tau)|^{\Delta}~\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau^{\prime})})}|f^{\prime}(\tau^{\prime})|^{\Delta}~G(f(\tau),f(\tau^{\prime}))\tag{H}\cr ~=~&\color{red}{{\rm sgn}(c\tau+d)}|c\tau+d|^{-2\Delta}~\color{red}{{\rm sgn}(c\tau^{\prime}+d)}|c\tau^{\prime}+d|^{-2\Delta}~G(f(\tau),f(\tau^{\prime})),\tag{I}\end{align} $$ and if we, say, pick a branch of the square root$^1$ $$ \sqrt{f^{\prime}(\tau)}~:=~\frac{1}{c\tau+d}, \tag{J} $$ cf. modular forms. To prove eq. (H) we have used that $$ f(\tau)-f(\tau^{\prime})~\stackrel{(E)}{=}~\frac{\tau-\tau^{\prime}}{(c\tau+d)(c\tau^{\prime}+d)}. \tag{K} $$

7. Proposal for finite temperature $\beta<\infty$ in the infrared (IR) $|\tau|\gg J^{-1}$. Consider a diffeomorphism $f:\mathbb{S}^1\to\mathbb{R}$ from the circle to the line $$ f(\tau)~=~\tan\frac{\pi\tau}{\beta}~=~f(\tau+\beta),\tag{L}$$ $$f^{\prime}(\tau)~=~\frac{\pi}{\beta}\frac{1}{\cos^2\frac{\pi\tau}{\beta}}~>~0.\tag{M}$$ If we modify OP's reparametrization rule (2) to $$\begin{align} G(\tau,\tau^{\prime})~\to~&\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau)})}|f^{\prime}(\tau)|^{\Delta}~\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau^{\prime})})}|f^{\prime}(\tau^{\prime})|^{\Delta}~G(f(\tau),f(\tau^{\prime}))\tag{N}\cr ~=~&\left(\frac{\pi}{\beta}\right)^{2\Delta}\color{red}{{\rm sgn}(\cos\frac{\pi\tau}{\beta})}\left|\cos\frac{\pi\tau}{\beta}\right|^{-2\Delta}~\color{red}{{\rm sgn}(\cos\frac{\pi\tau^{\prime}}{\beta})}\left|\cos\frac{\pi\tau^{\prime}}{\beta}\right|^{-2\Delta}~G(f(\tau),f(\tau^{\prime}))\tag{O}\cr ~=~&C\left(\frac{\pi}{\beta}\right)^{2\Delta}\frac{\color{red}{{\rm sgn}(\sin\frac{\pi}{\beta}(\tau-\tau^{\prime}))}}{|\sin\frac{\pi}{\beta}(\tau-\tau^{\prime})|^{2\Delta}}, \tag{P} \end{align} $$ and if we pick a branch of the square root$^1$ $$ \sqrt{f^{\prime}(\tau)}~:=~\sqrt{\frac{\pi}{\beta}}\frac{1}{\cos\frac{\pi\tau}{\beta}}, \tag{Q} $$ we get the connected finite-temperature 2-point function (P), cf. Ref. 4. To prove eq. (N) we have used the subtraction formula for tangent $$ f(\tau)-f(\tau^{\prime})~\stackrel{(L)}{=}~ \left(1+\tan\frac{\pi\tau}{\beta}\tan\frac{\pi\tau^{\prime}}{\beta}\right)\tan\frac{\pi}{\beta}(\tau-\tau^{\prime}) ~=~\frac{\sin\frac{\pi}{\beta}(\tau-\tau^{\prime})}{\cos\frac{\pi\tau}{\beta}\cos\frac{\pi\tau^{\prime}}{\beta}}. \tag{R} $$

8. The above proposal solves OP's sign problem. It can be additionally defended by pointing out that

   • the notion of 1D conformal/angle-preserving transformations is artificial in the first place, and

   • that similar sign factors are implicit in the notion of 2D conformal/angle-preserving transformations in the complex plane.

9. $sl(2,\mathbb{R})$ Lie algebra. Conversely, let us consider an arbitrary infinitesimal transformation $$ \epsilon(\tau)~=~f(\tau)-\tau ~=~\sum_{n=0}^{\infty}\epsilon_n\tau^n, \qquad \epsilon_n~\in~\mathbb{R},\tag{S} $$ that leaves the 2-point function (A) invariant. We may assume that $$ f^{\prime}(\tau)~=~1 +\epsilon^{\prime}(\tau)~>~0 \tag{T}$$ and its square root$^1$ $$ \sqrt{f^{\prime}(\tau)}~=~1 +\frac{\epsilon^{\prime}(\tau)}{2}~>~0 \tag{U}$$ are positive. The linearized invariance condition (C) becomes $$ \epsilon^{\prime}(\tau)+\epsilon^{\prime}(\tau^{\prime}) ~=~2\frac{\epsilon(\tau)-\epsilon(\tau^{\prime})}{\tau-\tau^{\prime}}.\tag{V} $$ Now set $\tau^{\prime}=0$ in eq. (V). The complete solution to eq. (V) is an arbitrary 2nd-order polynomial $$ \epsilon(\tau)~=~\sum_{n=0}^2\epsilon_n\tau^n~=~\epsilon_0+\epsilon_1\tau+\epsilon_2\tau^2,\tag{W} $$ which precisely corresponds to the infinitesimal form of the $sl(2,\mathbb{R})$ symmetry transformation (E). This shows that the full stabilizer subgroup is $PSL(2,\mathbb{R})$.

References:

1. J. Maldacena & D. Stanford, Comments on the SYK model, arXiv:1604.07818; eq. (2.9).

2. G. Sarosi, $AdS_2$ holography and the SYK model, arXiv:1711.08482; eqs. (121) + (126).

3. V. Rosenhaus, An intro to SYK, arXiv:1807.03334; eq. (3.3).

4. D.A. Trunin, Pedagogical introduction to SYK model and 2D Dilaton Gravity, arXiv:2002.12187; eqs. (3.31) + (3.33) + (3.34).

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$^1$ The other branch choice works just as well for $n$-point functions with $n$ even. The main point is that we should use the same choice of branch everywhere.

$^2$ Note that eq. (H) is invariant under $M\to -M$ in eq. (G). More generally, this proposal is only single-valued for $n$-point functions with $n$ even.