---
abstract: 'The explicit evaluation of the partition function in the Schwarzian theory is presented.'
author:
- 'Vladimir V. Belokurov'
- 'Evgeniy T. Shavgulidze'
title: Exact solution of the Schwarzian theory
---

The Schwarzian theory @(SW) is the basic element of various physical models including the SYK model and the two-dimensional dilaton gravity (see, e.g., @(MS), @(BAK), @(GR), @(MNW), and references therein). The action of the theory is $$\label{Act1}
   I=-\frac{1}{g^{2}}\int \limits _{0}^{2\pi}\,\left[ \mathcal{S}_{\phi}(\tau)+\frac{1}{2}(\phi')^{2}(\tau)\right]d\tau\,.$$ Here, $$\label{Der}
\mathcal{S}_{\phi}(t)\equiv
\left(\frac{\phi''(t)}{\phi'(t)}\right)'
-\frac{1}{2}\left(\frac{\phi''(t)}{\phi'(t)}\right)^2$$ is the Schwarzian derivative, $\phi\in Diff^{3}([0,\, 2\pi]),\ $ and $\phi'(0)=\phi'(2\pi)\,.$

It is convenient to rewrite the action in the form $$\label{Act2}
   I=-\frac{1}{\sigma^{2}}\int \limits _{0}^{1}\,\left[ \mathcal{S}_{\varphi}(t)+2\pi^{2}\dot{\varphi}^{2}(t)\right]dt\,,$$ where $$\sigma=\sqrt{2\pi}g\,,\  \  t=\frac{1}{2\pi}\tau\,,\ \ \varphi(t)=\frac{1}{2\pi}\phi(\tau) \,,$$ $$\varphi\in Diff^{3}([0,\, 1])\,, \ \ \dot{\varphi}(0)=\dot{\varphi}(1)\, .$$

The functional integral for the partition function $$\label{PF1}
   Z(g)=\int \limits _{\dot{\varphi}(0)=\dot{\varphi}(1)}\,\exp\left\{-I\right\}\  d\varphi$$ diverges @(SW). However, as we will see later on (eq. (\[RegPF2\])), the integral $$Z_{\alpha}(g)$$ $$=\int \limits _{  Diff^{3}([0, 1]) }
\exp\left\{-I\right\}\,\exp \left\{\frac{-2\left[ \pi^{2}-\alpha^{2}\right]}{\sigma^{2}}\int \limits _{0}^{1}\dot{\varphi}^{2}(t)dt  \right\}  d\varphi$$ $$\label{RegPF}
=\int \limits _{\dot{\varphi}(0)=\dot{\varphi}(1)}\,\exp\left\{\frac{1}{\sigma^{2}}\int \limits _{0}^{1}\,\left[ \mathcal{S}_{\varphi}(t)+2\alpha^{2}\dot{\varphi}^{2}(t)\right]dt  \right\}  d\varphi$$ converges for $0\leq \alpha < \pi \,. $ Therefore, let us evaluate the integral (\[RegPF\]) first.

The measure $$\label{Measure}
   \mu_{\sigma}(X)=\int \limits _{X}\,\exp\left\{\frac{1}{\sigma^{2}}\int \limits _{0}^{1}\, \mathcal{S}_{\varphi}(t)\,dt  \right\}  d\varphi$$ is quasi-invariant, and the Radon - Nikodym derivative of the measure is @(Sh), @(BSh) $$\frac{d\mu_{\sigma}^{f}}{d\mu_{\sigma}}(\varphi)=\frac{1}{\sqrt{\dot{f}(0)\dot{f}(1)}}$$ $$\label{RadNik}
  \times \exp\left\{ \frac{1}{\sigma^{2}}\left[   \frac{\ddot{f}(0)}{\dot{f}(0)}\dot{\varphi}(0)-  \frac{\ddot{f}(1)}{\dot{f}(1)}\dot{\varphi}(1)\right]   +        \frac{1}{\sigma^{2}}\int \limits _{0}^{1}\, \mathcal{S}_{f}\left(\varphi(t)\right)\,\dot{\varphi}^{2}\,dt  \right\} \,,$$ where $$\mu_{\sigma}^{f}(X)=\mu_{\sigma}(f\circ X)\,.$$ Here, we have used the well known property of the Schwarzian derivative: $$\mathcal{S}_{f\circ \varphi}(t)= \mathcal{S}_{f}(\varphi(t))\dot{\varphi}^{2}(t)+ \mathcal{S}_{\varphi}(t)\,,\ \ \ \ (\,f\circ \varphi\,)(t)=f\left( \varphi(t)\right)\,.$$ Take the function $f$ to be $$\label{f}
f(t)=\frac{1}{2}\left[ \frac{1}{\tan\frac{\alpha}{2}}\tan\left(\alpha(t-\frac{1}{2}) \right)+1 \right]\,.$$ In this case, $$\mathcal{S}_{f}(t)=2\alpha^{2}\,, \ \ \ \dot{f}(0)=\dot{f}(1)=\frac{\alpha}{\sin \alpha}\,,$$ $$\label{tan}
\ \ \ -\frac{\ddot{f}(0)}{\dot{f}(0)}=\frac{\ddot{f}(1)}{\dot{f}(1)}=2\alpha\tan\frac{\alpha}{2}\,.$$ Now we have the following functional integrals equality: $$\frac{\alpha}{\sin \alpha}\,\int \limits _{\dot{\varphi}(0) =\dot{\varphi}(1)}\,F(\varphi)\mu_{\sigma}(d\varphi)$$ $$=\int \limits _{\dot{\varphi}(0)=\dot{\varphi}(1)}\,F(f(\varphi))\,\exp\left\{-\frac{4\alpha}{\sigma^{2}}\tan\frac{\alpha}{2}\,\dot{\varphi}(0)\right\}\,$$ $$\label{FIequality}
\times \exp\left\{\frac{1}{\sigma^{2}}\int \limits _{0}^{1}\,\left[ \mathcal{S}_{\varphi}(t)+2\alpha^{2}\dot{\varphi}^{2}(t)\right]dt  \right\}  d\varphi\,.$$ The next step is the choice of the function $F\,.$ Let it be $$\label{F}
F(f(\varphi))=\exp\left\{\frac{4\alpha}{\sigma^{2}}\tan\frac{\alpha}{2}\,\dot{\varphi}(0)\right\}\,.$$ To find $F(\varphi)$ from the previous equation, note that for $u(t)=f(\varphi(t))$ $$\dot{\varphi}(0)=\frac{1}{\dot{f}(0)}\dot{u}(0)\,.$$ Then $$\label{F(u)}
F(u)=\exp\left\{\frac{8\,\sin ^{2}\frac{\alpha}{2}}{\sigma^{2}}\,\dot{u}(0)\right\}\,.$$ Thus for the regularized partition function we have $$Z_{\alpha}(g)$$ $$\label{RegPF1}
=\frac{\alpha}{\sin \alpha}\,\int \limits _{\dot{\varphi}(0) =\dot{\varphi}(1)}\,\exp\left\{\frac{8\,\sin ^{2}\frac{\alpha}{2}}{\sigma^{2}}\,\dot{\varphi}(0)\right\}\,\mu_{\sigma}(d\varphi)\,.$$ Under the substitution $$\label{subst}
 \varphi(t)=\frac{\int \limits _{0}^{t}\,\exp\{\xi(\eta)\}d\eta  }{\int \limits _{0}^{1}\,\exp\{\xi(\eta)\}d\eta }\,,$$ the measure $\mu_{\sigma}(d\varphi)$ turns into the Wiener measure @(Sh), @(BSh) $$\label{Wiener}
 w_{\sigma}(d\xi)=\exp\left\{ -\frac{1}{2\sigma^{2}}\int \limits _{0}^{1}\,\dot{\xi}^{2}(t)\,dt \right\}\ d\xi\,.$$ In this case, $$\label{xi}
\xi(t)=\ln\dot{\varphi}(t)-\ln\dot{\varphi}(0)\,, \ \ \ \xi\in C([0,\, 1])\,,$$ and $ \xi(0)=\xi(1)=0\,. $

Now $Z_{\alpha}(g)$ is written as $$Z_{\alpha}(g)=\frac{\alpha}{\sin \alpha}$$ $$\label{RegPF2}
\times\int \limits _{\xi(0) =\xi(1)=0}\,\exp\left\{\frac{8\,\sin ^{2}\frac{\alpha}{2}}{\sigma^{2}}\,\frac{1}{\int \limits _{0}^{1}\,\exp\{\xi(\eta)\}d\eta }\right\}\,w_{\sigma}(d\xi)\,.$$ The singularity at $\alpha=\pi $ is canceled out in the ratio $$\frac{Z_{\alpha}(g)}{Z_{\alpha}(1)}\,,$$ and we can remove the regularization there $$\label{Ratio}
   \frac{Z(g)}{Z(1)}=\lim \limits _{\alpha\rightarrow \pi-0}\ \ \frac{Z_{\alpha}(g)}{Z_{\alpha}(1)}\,.$$

To evaluate the integral $$\label{Int}
  \int \limits _{\xi(0) =\xi(1)=0}\,\exp\left\{\frac{8}{\sigma^{2}}\,\frac{1}{\int \limits _{0}^{1}\,\exp\{\xi(\eta)\}d\eta }\right\}\,w_{\sigma}(d\xi)\,,$$ we use the following equation: $$\int \limits _{\xi(0) =\xi(1)=0}\,\exp\left\{\frac{-2\beta^{2}}{\sigma^{2}(\beta +1)}\,\frac{1}{\int \limits _{0}^{1}\,\exp\{\xi(\eta)\}d\eta }\right\}\,w_{\sigma}(d\xi)$$ $$\label{Formula}
  =\frac{1}{\sqrt{2\pi}\sigma}\,\exp \left\{-\frac{4\left( \ln (\beta +1)\right)^{2}}{2\sigma ^{2}} \right\}\,.$$ The proof of the more general formula will be given in the forthcoming paper (also, see @(BSh)). For the integral (\[Int\]) $$(\beta +1)=-1\,.$$

Thus the final result is $$\label{Result}
   \frac{Z(g)}{Z(1)}= \frac{\exp\{-\pi\}}{g}\,\exp\left\{ \frac{\pi}{g^{2}}\right\}\,.$$

It is interesting to compare the one-loop results for $Z(g)$ in @(SW) with the eq. (\[Result\]). Note that the power of the constant $g$ in the denominator is determined by the number of gauge fixing conditions. The one-loop result for the orbit $Diff(S^{1})/U(1)$ (eq. (3.45) in @(SW)) has the same form as our exact result (\[Result\]).

Unlike its compact subgroup $U(1)$, the group $SL(2, \textbf{R})$ is noncompact. Therefore, integrating over the quotient space $Diff^{3}([0, 1])/SL(2,\textbf{R}) $ we get the finite result for the partition function in the Schwarzian theory.

We define the Schwarzian partition function as a limit $$\label{Ratio2}
  Z_{Schw}(g) = \lim \limits_{\alpha\rightarrow \pi - 0}\ \frac{Z_{\alpha}(g)}{V_{\alpha}(g)} \,.$$ Here, $ Z_{\alpha}(g) $ is given by the eqs. (\[RegPF\]), (\[RegPF2\]), and $V_{\alpha}(g) $ is the regularized volume of the group $SL(2,\textbf{R})$ $$\label{V}
V_{\alpha}(g)=\int \limits _{SL(2,\textbf{R})}\exp \left\{\frac{-2\left[ \pi^{2}-\alpha^{2}\right]}{\sigma^{2}}\int \limits _{0}^{1}\dot{\varphi}^{2}(t)dt  \right\}  d\mu _{H}\,.$$ Note that the functional measure in the eq. (\[RegPF\]) and the Haar measure $d\mu _{H}$ on the group $SL(2, \textbf{R})$ in the eq. (\[V\]) are regularized in the same manner.

To perform the integration over the group $SL(2,\textbf{R})$ in the eq. (\[V\]) we choose the representation @(Lang) $$\label{Repres}
  \varphi_{z}(t)=-\frac{i}{2\pi}\ln\frac{e^{i2\pi t}+z}{\bar{z}\,e^{i2\pi t}+1}\,, \ \ \ z=\rho e^{i\theta}\,, \ \ \ \rho < 1  \,.$$ In this case, the Haar measure is @(Lang) $$\label{Haar}
  \mu_{H}(dz)=\frac{4\rho d\rho \,d\theta}{\left(1-\rho^{2} \right)^{2}} \,.$$

The integral $$\label{dt}
 \int \limits _{0}^{1}\dot{\varphi}_{z}^{2}(t)dt =\int \limits _{0}^{1}\frac{\left(1-|z|^{2} \right)^{2} \,dt}{\left( e^{i2\pi t}+z \right)^{2}\left(e^{-i2\pi t}+\bar{z} \right)^{2}}
 =\frac{1+\rho^{2}}{1-\rho^{2}}$$ does not depend on $\theta \,.$ And the regularized volume of the group has the form $$V_{\alpha}(g)
=\int \limits _{0}^{1}\exp \left\{-\frac{\left[\pi^{2}-\alpha^{2}\right]}{\pi g^{2}} \frac{\left(1+\rho^{2}\right)}{\left(1-\rho^{2}\right)} \right\}
\frac{8\pi \rho d\rho }{\left(1-\rho^{2} \right)^{2}}$$ $$\label{Valphag}
=\exp \left\{-\frac{\left[\pi^{2}-\alpha^{2}\right]}{\pi g^{2}}\right\} \frac{4\pi ^{2}g^{2}}{\left[\pi^{2}-\alpha^{2}\right]}\,.$$

Thus we can evaluate the Schwarzian partition function $$\label{Result2}
   Z_{Schw}(g)= \frac{1}{4\pi g^{3}}\,\exp\left\{ \frac{\pi}{g^{2}}\right\}\,.$$

Note that the one-loop result in @(SW), @(MS) has the same form as the exact partition function (\[Result2\]) obtained by the direct functional integration.

<span>7</span>

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V.V. Belokurov and E.T. Shavgulidze, "Extraordinary Properties of Functional Integrals and Groups of Diffeomorphisms", Phys. of Part. and Nuclei, <span>**48**</span>, 267 (2017) \[Fiz. Elem. Chastits At. Yadra <span>**48**</span>, 194 (2017)\]. S. Lang, "$SL_{2}(\textbf{R})$". Addison-Wesley Publishing. 1975.

{'source': 'arXiv_out/1705'}