The Exact Solution of the Schwarzian Theory

TL;DR

This paper provides an exact analytical evaluation of the partition function within the Schwarzian theory, offering precise insights into its mathematical structure and physical implications.

Contribution

It introduces an exact solution for the Schwarzian theory's partition function, advancing beyond previous approximate or numerical methods.

Findings

Exact partition function derived analytically

Enhanced understanding of Schwarzian theory's structure

Potential applications in quantum gravity and holography

Abstract

The explicit evaluation of the partition function in the Schwarzian theory is presented.

Equations

$$I=-\frac{1}{g^2}\underset{0}{\overset{2\pi }{\int }}\left[{\mathcal{S}}_{\varphi }(\tau )+\frac{1}{2}({\varphi }^{\prime }{)}^2(\tau )\right]d\tau .$$

$${\mathcal{S}}_{\varphi }(t)\equiv {\left(\frac{{\varphi }^{\prime \prime }(t)}{{\varphi }^{\prime }(t)}\right)}^{\prime }-\frac{1}{2}{\left(\frac{{\varphi }^{\prime \prime }(t)}{{\varphi }^{\prime }(t)}\right)}^2$$

$$I=-\frac{1}{{\sigma }^2}\underset{0}{\overset{1}{\int }}\left[{\mathcal{S}}_{\phi }(t)+2{\pi }^2{\dot{\phi }}^2(t)\right]dt,$$

$$\sigma =\sqrt{2\pi }g,t=\frac{1}{2\pi }\tau ,\phi (t)=\frac{1}{2\pi }\varphi (\tau ),$$

$$\phi \in Dif{f}^3([0,1]),\dot{\phi }(0)=\dot{\phi }(1).$$

$$Z(g)=\underset{\dot{\phi }(0)=\dot{\phi }(1)}{\int }\exp \left\{-I\right\}d\phi$$

$$Z_{\alpha }(g)$$

$$=\underset{Dif{f}^3([0,1])}{\int }\exp \left\{-I\right\}\exp \left\{\frac{-2\left[{\pi }^2-{\alpha }^2\right]}{{\sigma }^2}\underset{0}{\overset{1}{\int }}{\dot{\phi }}^2(t)dt\right\}d\phi$$

$$=\underset{\dot{\phi }(0)=\dot{\phi }(1)}{\int }\exp \left\{\frac{1}{{\sigma }^2}\underset{0}{\overset{1}{\int }}\left[{\mathcal{S}}_{\phi }(t)+2{\alpha }^2{\dot{\phi }}^2(t)\right]dt\right\}d\phi$$

$${\mu }_{\sigma }(X)=\underset{X}{\int }\exp \left\{\frac{1}{{\sigma }^2}\underset{0}{\overset{1}{\int }}{\mathcal{S}}_{\phi }(t)dt\right\}d\phi$$

$$\frac{d{\mu }_{\sigma }^f}{d{\mu }_{\sigma }}(\phi )=\frac{1}{\sqrt{\dot{f}(0)\dot{f}(1)}}$$

$$\times \exp \left\{\frac{1}{{\sigma }^2}\left[\frac{\ddot{f}(0)}{\dot{f}(0)}\dot{\phi }(0)-\frac{\ddot{f}(1)}{\dot{f}(1)}\dot{\phi }(1)\right]+\frac{1}{{\sigma }^2}\underset{0}{\overset{1}{\int }}{\mathcal{S}}_f\left(\phi (t)\right){\dot{\phi }}^{2}dt\right\},$$

$${\mu }_{\sigma }^f(X)={\mu }_{\sigma }(f\circ X).$$

$${\mathcal{S}}_{f\circ \phi }(t)={\mathcal{S}}_f(\phi (t)){\dot{\phi }}^2(t)+{\mathcal{S}}_{\phi }(t),(f\circ \phi )(t)=f\left(\phi (t)\right).$$

$$f(t)=\frac{1}{2}\left[\frac{1}{\tan \frac{\alpha }{2}}\tan \left(\alpha (t-\frac{1}{2})\right)+1\right].$$

$${\mathcal{S}}_f(t)=2{\alpha }^2,\dot{f}(0)=\dot{f}(1)=\frac{\alpha }{\sin \alpha },$$

$$-\frac{\ddot{f}(0)}{\dot{f}(0)}=\frac{\ddot{f}(1)}{\dot{f}(1)}=2\alpha \tan \frac{\alpha }{2}.$$

$$\frac{\alpha }{\sin \alpha }\underset{\dot{\phi }(0)=\dot{\phi }(1)}{\int }F(\phi ){\mu }_{\sigma }(d\phi )$$

$$=\underset{\dot{\phi }(0)=\dot{\phi }(1)}{\int }F(f(\phi ))\exp \left\{-\frac{4\alpha }{{\sigma }^2}\tan \frac{\alpha }{2}\dot{\phi }(0)\right\}$$

$$\times \exp \left\{\frac{1}{{\sigma }^2}\underset{0}{\overset{1}{\int }}\left[{\mathcal{S}}_{\phi }(t)+2{\alpha }^2{\dot{\phi }}^2(t)\right]dt\right\}d\phi .$$

$$F(f(\phi ))=\exp \left\{\frac{4\alpha }{{\sigma }^2}\tan \frac{\alpha }{2}\dot{\phi }(0)\right\}.$$

$$\dot{\phi }(0)=\frac{1}{\dot{f}(0)}\dot{u}(0).$$

$$F(u)=\exp \left\{\frac{8{\sin }^2\frac{\alpha }{2}}{{\sigma }^2}\dot{u}(0)\right\}.$$

$$Z_{\alpha }(g)$$

$$=\frac{\alpha }{\sin \alpha }\underset{\dot{\phi }(0)=\dot{\phi }(1)}{\int }\exp \left\{\frac{8{\sin }^2\frac{\alpha }{2}}{{\sigma }^2}\dot{\phi }(0)\right\}{\mu }_{\sigma }(d\phi ).$$

$$\phi (t)=\frac{\underset{0}{\overset{t}{\int }}\exp \{\xi (\eta )\}d\eta }{\underset{0}{\overset{1}{\int }}\exp \{\xi (\eta )\}d\eta },$$

$$w_{\sigma }(d\xi )=\exp \left\{-\frac{1}{2{\sigma }^2}\underset{0}{\overset{1}{\int }}{\dot{\xi }}^2(t)dt\right\}d\xi .$$

$$\xi (t)=\ln \dot{\phi }(t)-\ln \dot{\phi }(0),\xi \in C([0,1]),$$

$$Z_{\alpha }(g)=\frac{\alpha }{\sin \alpha }$$

$$\times \underset{\xi (0)=\xi (1)=0}{\int }\exp \left\{\frac{8{\sin }^2\frac{\alpha }{2}}{{\sigma }^2}\frac{1}{\underset{0}{\overset{1}{\int }}\exp \{\xi (\eta )\}d\eta }\right\}{w}_{\sigma }(d\xi ).$$

Full text

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

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Here,

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is the Schwarzian derivative, \phi\in Diff^{3}([0,\,2\pi]),\ and $\phi^{\prime}(0)=\phi^{\prime}(2\pi)\,.$

It is convenient to rewrite the action in the form

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where

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The functional integral for the partition function

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diverges (SW) . However, as we will see later on (eq. (17)), the integral

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converges for $0\leq\alpha<\pi\,.$ Therefore, let us evaluate the integral (5) first.

The measure

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is quasi-invariant, and the Radon - Nikodym derivative of the measure is (Sh) , (BSh)

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where

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Here, we have used the well known property of the Schwarzian derivative:

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Take the function $f$ to be

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In this case,

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Now we have the following functional integrals equality:

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The next step is the choice of the function $F\,.$ Let it be

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To find $F(\varphi)$ from the previous equation, note that for $u(t)=f(\varphi(t))$

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Then

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Thus for the regularized partition function we have

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Under the substitution

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the measure $\mu_{\sigma}(d\varphi)$ turns into the Wiener measure (Sh) , (BSh)

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In this case,

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and $\xi(0)=\xi(1)=0\,.$

Now $Z_{\alpha}(g)$ is written as

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The singularity at $\alpha=\pi$ is canceled out in the ratio

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and we can remove the regularization there

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To evaluate the integral

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we use the following equation:

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The proof of the more general formula will be given in the forthcoming paper (also, see (BSh) ). For the integral (19)

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Thus the final result is

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It is interesting to compare the one-loop results for $Z(g)$ in (SW) with the eq. (21). 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 (21).

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

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Here, $Z_{\alpha}(g)$ is given by the eqs. (5), (17), and $V_{\alpha}(g)$ is the regularized volume of the group $SL(2,\textbf{R})$

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Note that the functional measure in the eq. (5) and the Haar measure $d\mu_{H}$ on the group $SL(2,\textbf{R})$ in the eq. (23) are regularized in the same manner.

To perform the integration over the group $SL(2,\textbf{R})$ in the eq. (23) we choose the representation (Lang)

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In this case, the Haar measure is (Lang)

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The integral

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does not depend on $\theta\,.$ And the regularized volume of the group has the form

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Thus we can evaluate the Schwarzian partition function

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Note that the one-loop result in (SW) , (MS) has the same form as the exact partition function (28) obtained by the direct functional integration.