The Sachdev-Ye-Kitaev (SYK) model and conformal $SL(2,\mathbb{R})$ symmetry in 0+1D

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. The connected 2-point function in the SYK model is$^1$ $$G(\tau,\tau^{\prime})~\propto~\frac{{\rm sgn}(\tau\!-\!\tau^{\prime})}{|\tau\!-\!\tau^{\prime}|^{2\Delta}}, \tag{1'}$$ cf. Refs. 1-3.

2. Proposal. The 2-point function (1') becomes invariant under Moebius transformations $$\begin{align} f(\tau)~=~&\frac{a\tau+b}{c\tau+d},\tag{3'} \cr f^{\prime}(\tau)~\stackrel{(3')}{=}~&\frac{1}{(c\tau+d)^2}~>~0, \tag{5'} \cr a,b,c,d~\in~\mathbb{R}, \end{align} $$ with the group $PSL(2,\mathbb{R})\cong SL(2,\mathbb{R})/\{\pm {\bf 1}_{2\times 2}\}$ where $$ SL(2,\mathbb{R})~=~\left\{ M=\begin{pmatrix}a&b \cr c&d\end{pmatrix}\in GL(2,\mathbb{R}) ~\mid~ \det(M)= 1 \right\}. \tag{6'}$$ if we modify the 1D conformal transformation rule (2) to$^2$ $$\begin{align} G(\tau,\tau^{\prime})~\to~&\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau)})}|f^{\prime}(\tau)|^{\Delta}~\color{red}{{\rm sgn}(\sqrt{f^{\prime}(\tau^{\prime})})}|f^{\prime}(\tau^{\prime})|^{\Delta}~G(f(\tau),f(\tau^{\prime}))\tag{2'}\cr ~=~&\color{red}{{\rm sgn}(c\tau+d)}|c\tau+d|^{-2\Delta}~\color{red}{{\rm sgn}(c\tau^{\prime}+d)}|c\tau^{\prime}+d|^{-2\Delta}~G(f(\tau),f(\tau^{\prime})),\tag{4'}\end{align} $$ where we have picked the following branch of the square root: $$ \sqrt{f^{\prime}(\tau)}~:=~\frac{1}{c\tau+d}, \tag{7'} $$ cf. modular forms. To prove eq. (4') we have used that $$ f(\tau)-f(\tau^{\prime})~\stackrel{(3')}{=}~\frac{\tau-\tau^{\prime}}{(c\tau+d)(c\tau^{\prime}+d)}. \tag{8'} $$

3. The proposal (2') solves OP's sign problem. Eq. (2') can be somewhat 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.

4. Conversely, let us consider an arbitrary infinitesimal transformation $$ \epsilon(\tau)~=~f(\tau)-\tau ~=~\sum_{n=0}^{\infty}\epsilon_n\tau^n, \qquad \epsilon_n~\in~\mathbb{R},\tag{9'} $$ that leaves the 2-point function (1') invariant. We may assume that $$ f^{\prime}(\tau)~=~1 +\epsilon^{\prime}(\tau)~>~0 \tag{10'}$$ (and its square root) are positive. The linearized invariance condition (2') becomes $$ \epsilon^{\prime}(\tau)+\epsilon^{\prime}(\tau^{\prime}) ~=~2\frac{\epsilon(\tau)-\epsilon(\tau^{\prime})}{\tau-\tau^{\prime}}.\tag{11'} $$ The complete solution to eq. (11') is arbitrary 2nd-order polynomials $$ \epsilon(\tau)~=~\sum_{n=0}^2\epsilon_n\tau^n~=~\epsilon_0+\epsilon_1\tau+\epsilon_2\tau^2,\tag{12'} $$ which precisely corresponds to the infinitesimal form of the $sl(2,\mathbb{R})$ symmetry transformation (3'). This shows that the full symmetry group is $PSL(2,\mathbb{R})$.

References:

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

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

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

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$^1$ Note that eq. (1') is odd (and hence not invariant) under time reversal $f(\tau)=-\tau.$ For this reason we only consider time reparametrizations $f:\mathbb{R}\to \mathbb{R}$ with $f^{\prime}>0$.

$^2$ Note that eq. (4') is invariant under $M\to -M$ in eq. (6). More generally, this proposal is only single-valued for $n$-point functions with $n$ even.