text:
\section{Introduction}
In string theory, the sphere partition function without operator insertions is a fundamental but confusing quantity. In principle, it should give minus the classical value of the on-shell action of the string background. In the simplest string backgrounds, like empty flat spacetime, this action vanishes. However, there are backgrounds of critical string theory where the on-shell action is nonzero and physically important, like for thermal AdS${}_3\times S^3\times X$.

It isn't known how to compute the sphere partition function in such cases. Part of the puzzle is that one has to divide by the volume of the conformal Killing group $\text{PSL}(2,\mathbb{C})$, which has infinite volume and no sensible finite regularized value \cite{Liu:1987nz}\cite{Eberhardt:2021ynh}. Another aspect is that in cases where the partition function is expected to be nonzero, the target space is noncompact, and there is a divergent integral over the location of the string worldsheet.

Some proposals exist in the literature. Tseytlin \cite{Tseytlin:1988tv}\cite{Tseytlin:1988rr} has proposed to replace the division by the divergent volume of $\text{PSL}(2,\mathbb{C})$ by a derivative wrt the worldsheet UV cutoff. See also \cite{Callan:1986jb} and \cite{Susskind:1994sm}. Another proposal \cite{Kutasov:1999xu} was that at least for the case of AdS$_3$, the noncompactness and the $\text{PSL}(2,\mathbb{C})$ could cancel each other directly. This proposal was questioned in \cite{Kraus:2002cb}.

It seems likely that the on-shell sphere partition function really is zero up to effects having to do with the noncompactness of the target space. This is consistent with the fact that in the gravity theory one gets from the low-energy limit of string theory, the on-shell action vanishes up to boundary terms \cite{Tseytlin:1988tv}\cite{Gutperle:2002ki}. In that theory, to compute the on-shell action of a noncompact spacetime, one has to put some kind of radial cutoff, add the GHY boundary term together with additional counterterms, and take a limit. Perhaps the resolution of the sphere puzzle from the worldsheet perspective will involve a similar procedure. This seems like a technical challenge: how does one put a spacetime radial cutoff in the worldsheet path integral?

We don't know how to do this. But as a (possibly irrelevant) warmup, in this paper we will discuss an example from noncritical string theory, first studied by Zamolodchikov \cite{Zamolodchikov:1982vx}, where the answer for the sphere partition function is finite and nonzero. Specifically, we study the noncritical string theory consisting of Liouville theory and the $(2,p)$ minimal model, and we do the Liouville path integral directly in the semiclassical limit $c_\text{Liouville} \to +\infty$,  which is relevant for large values of $p$. As shown in \cite{Zamolodchikov:1982vx}, in this example the conformal symmetry of the Liouville theory is spontaneously broken by a semiclassical saddle point, leading to Goldstone zero modes. These zero modes are noncompact, and the integral over them cancels against the divergent volume of $\text{PSL}(2,\mathbb{C})$ that we are supposed to divide by, giving a finite and nonzero answer. So indeed, the noncompactness and the $\text{PSL}(2,\mathbb{C})$ cancel each other neatly.

We do the calculation from \cite{Zamolodchikov:1982vx} in a bit more detail, and match the answer to the predictions of the matrix integral. To do this matching, it is important to compare not the sphere partition function itself, but the well-defined quantity sphere/disk$^2$, where we consider the specific case of the disk with FZZT boundary conditions. So one also has to compute this disk partition function. This is structurally similar to the sphere, with $\text{PSL}(2,\mathbb{R})$ playing the role of $\text{PSL}(2,\mathbb{C})$.

In fact, this comparison with the matrix integral was already done (with ZZ disk instead of FZZT) in the work of Alexandrov, Kazakov and Kutasov \cite{Alexandrov:2003nn}, using exact Liouville methods \cite{Dorn:1994xn}\cite{Zamolodchikov:1995aa}\cite{Fateev:2000ik}\cite{Zamolodchikov:2001ah}. Specifically, \cite{Alexandrov:2003nn} gets a formula for the sphere partition function by integrating the the DOZZ formula for the three point function of cosmological constant operators, which represent derivatives with respect to $\mu$ of the sphere partition function.\footnote{This procedure would also work for all the $(q,p)$ minimal string theories and also the $c=1$ string theory.} Our direct semiclassical calculation is a less complete match to the matrix integral, because it is only valid for large $p$. However, it has the advantage that the role of $\text{PSL}(2,\mathbb{C})$ and $\text{PSL}(2,\mathbb{R})$ are more obvious. Also, we are able to determine the overall numerical coefficient, which was fitted in \cite{Alexandrov:2003nn}.

On the matrix integral side, defining the analog of the sphere and disk partition functions requires some care, and we explain this in detail. We find that in the matrix integral dual to the $(2,p)$ minimal string, which has the leading density of eigenvalues
\begin{equation}
\rho(E) = \frac{e^{S_0}}{2\pi^2}\sinh\left[\frac{p}{2}\text{arccosh}\left(1 + \frac{8\pi^2}{p^2}E\right)\right],
\end{equation}
the universal part of the matrix integral free energy is
\begin{equation}\label{sphereansintro}
\log(\mathfrak{Z}) \supset -\frac{e^{2S_0}}{2^{10}\pi^6}\frac{p^5}{p^2-4}.
\end{equation}
In the string theory, the density of states is related to the disk with FZZT boundary conditions, and the free energy is the sphere partition function. Our Liouville formulas for these quantities agree with (\ref{sphereansintro}) in the large $p$ limit where one ignores the ``$-4$'' in the denominator. Note that in the strict large $p$ limit, in which this system approaches JT gravity, the sphere partition function diverges, as suggested in \cite{Maldacena:2019cbz}.

While our work was nearing completion, \cite{Anninos:2021ene}\cite{Muhlmann:2021clm} appeared which also study the semiclassical limit of the sphere partition function in Liouville theory.

\section{Liouville computations}
In this section we will compute Liouville path integrals on the sphere, and on the disk (hemisphere) with FZZT boundary conditions, in a semiclassical approximation at large positive Liouville central charge. In this limit, Liouville theory is weakly coupled, and one can compute the path integral by summing over saddle points and including a one-loop determinant. There are several subtleties involved in getting a well-defined answer from these partition functions, and we will start by explaining these.

First, a problem that one runs into is that the one-loop determinants are infinite, due to the existence of noncompact zero modes. The origin of these zero modes is that the saddle point configurations of the Liouville field spontaneously break the conformal symmetry of the theory, leading to a finite number of Goldstone modes. On the sphere, the globally defined conformal symmetry group is $\text{PSL}(2,\mathbb{C})$, and on the disk it is $\text{PSL}(2,\mathbb{R})$, and the zero modes parametrize the quotient space $G/H$ where $G = \text{PSL}(2,\mathbb{C})$ or $\text{PSL}(2,\mathbb{R})$ and $H = $ PSU(2) or U(1) is the subgroup of $G$ that preserves the saddle point solutions.

One can get a finite and well-defined answer by computing the ratios
\begin{equation}\label{divideBy}
\frac{Z_{\text{sphere}}}{\text{vol($\text{PSL}(2,\mathbb{C})$)}}, \hspace{20pt} \text{and} \hspace{20pt} \frac{Z_{\text{disk}}}{\text{vol($\text{PSL}(2,\mathbb{R})$)}}.
\end{equation}
Concretely, the division by the infinite volumes in the denominator is accomplished by omitting the zero modes from the one-loop determinants, and then dividing by the volume of the stabilizer subgroup $H$ that leaves the saddle point invariant (note that $H$ has finite volume).

A second subtlety is that the overall normalization of the path integral is ambiguous, due to (i) the conformal anomaly, (ii) the existence of a finite counterterm proportional to the Euler characteristic, and (iii) an arbitrary choice of measure on the group $G$ whose volume we divide by in (\ref{divideBy}). These ambiguities really exist, but they can be made to cancel out in the ratio
\begin{equation}\label{ratio}
\frac{Z_{\text{sphere}}}{\text{vol($\text{PSL}(2,\mathbb{C})$)}} \cdot\left(\frac{\text{vol($\text{PSL}(2,\mathbb{R})$)}}{Z_{\text{disk}}}\right)^2.
\end{equation}
In order to make the ambiguities (i) and (ii) cancel out, we will use the same metric for the two problems, taking the disk to the be the hemisphere.
We will also use the same cutoff procedure for the computation of the sphere and the disk one loop determinants.

In order to address (iii), which is the ambiguity in the measure on $G$, we need a principle which chooses related measures on $\text{PSL}(2,\mathbb{C})$ and $\text{PSL}(2,\mathbb{R})$.
One might be tempted to use the fact that $\text{PSL}(2,\mathbb{R})$ is a subgroup of $\text{PSL}(2,\mathbb{C})$ and (up to a normalization that cancels out in the ratio) there is a preferred metric on $\text{PSL}(2,\mathbb{C})$ that induces a measure on both spaces. In fact, for our purposes, this is actually not the right answer: instead the metric in the two spaces should be multiplied by a further factor of the volume of the sphere or hemisphere that the theory is defined on. This factor introduces some factors of two relative to the naive guess just described.

This prescription is the correct one for the application of Liouville theory to noncritical string theory. There one is interested in $Z/\text{vol}(G)$ because in string theory, the conformal symmetry is treated as part of the diffeomorphism and Weyl gauge symmetry. From this perspective, the factors of $1/\text{vol}(G)$ arise from zero mode integrals in the path integral over the $bc$ ghosts, and the factor of the volume of the sphere or hemisphere in the $G$ metric described above arises from the normalization of these zero modes on the two spaces.

Having explained these subtleties, let's now give an overview of the computation and set conventions for Liouville theory. We will define the Liouville field $\sigma$ so that the physical metric is
\begin{equation}
\d s^2 = e^{2\sigma}\hat{\d s^2}.
\end{equation}
In the explicit computations, we will use the sphere or hemisphere as the reference metric
\begin{equation}
\hat{\d s^2} = \d\theta^2 + \sin^2(\theta)\d\phi^2, \hspace{20pt} \hat{R} = 2, \hspace{20pt} \hat{K}|_{\text{equator}} = 0.
\end{equation}
It is conventional to write the central charge of Liouville theory as $c = 1 + 6(1/b + b)^2$, and to approach the limit of large $c$ by taking $b$ small. Then the Liouville action is
\begin{align}
I &= \frac{1}{b^2}\left\{\frac{1}{4\pi}\int \sqrt{\hat{g}}\left[(\hat\partial\sigma)^2 + \hat{R}\sigma  + 4\pi\mu e^{2\sigma}\right] + \frac{1}{2\pi}\int \sqrt{\hat{h}}\left[\hat{K} + 2\pi \mu_B e^\sigma\right]\right\}\label{firstLine}\\
&\hspace{20pt}+\frac{1}{4\pi}\int \sqrt{\hat{g}}\hat{R}\sigma + \frac{1}{2\pi}\int \sqrt{\hat{h}}\hat{K}\sigma.\label{secondLine}
\end{align}
The parameter $\mu$ is called the cosmological constant, and the parameter $\mu_B$ is called the boundary cosmological constant. Our conventions for these parameters differ by a factor of $b^2$ from the standard ones in the literature.

As written, the terms on the first line are proportional to $b^{-2}$, and the terms on the second line are of order one. We will treat the theory to one loop order in the small $b$ expansion, which means that we want to compute the order $b^{-2}$ and order one terms in $\log(Z)$. This means that we will need to retain the terms on the second line (\ref{secondLine}). However, to the order that we work, these terms can be treated by first-order perturbation theory, simply evaluating them on the classical solution that is obtained from the leading $b^{-2}$ terms. As long as we remember to do this, we only need to take the first line (\ref{firstLine}) into account in determining the classical solutions and one-loop determinants.

The goal is to compute two different partition functions: (1) the partition function on the sphere, and (2) the FZZT partition function on the disk (hemisphere), with $\mu_B$ fixed.
More precisely, as explained above, we will compute the partition functions divided by the volumes of the respective conformal groups.
For both cases, the answer for small $b$ is
\begin{equation}
\frac{Z}{\text{vol}(G)} = \sum_{\text{saddles}}e^{-I_{\text{classical}}} \cdot (\text{one-loop det}')\cdot(\text{gauge-fixing factor}).
\end{equation}
Here the one-loop determinant is computed with the zero modes omitted, and the gauge-fixing factor will convert this prescription into a properly normalized division by $\text{vol}(G)$.

In the rest of the computation, we will go through and evaluate each of these three factors for the sphere and for the disk.

\subsection{Classical solutions and action}
\subsubsection{Sphere}
On the sphere, there is a simple family of classical solutions given by constant configurations of $\sigma$. Restricting to such configurations, the equation of motion (obtained by varying the $b^{-2}$ part of the action with respect to $2\sigma$) is
\begin{equation}\label{sphereSolutions}
1 + 4\pi \mu e^{2\sigma} = 0, \hspace{20pt} \implies \hspace{20pt} 2\sigma = \log(\frac{1}{4\pi\mu}) + \i\pi(1 + 2n).
\end{equation}
We see that there are actually an integer-indexed family of solutions, in which the Liouville field $\sigma$ differs by $2\pi \i n$. Associated to each constant solution is a family of position-dependent solutions with the same action, obtained by acting with $\text{PSL}(2,\mathbb{C})$ on the constant solutions; we will address these later.

One might be surprised by the fact that there are any classical solutions for Liouville theory on a spherical topology, given that the equations of motion impose that the physical metric $e^{2\sigma}\hat{\d s^2}$ should have constant negative curvature, and that no everywhere-negative-curvature metric is possible on a spherical topology. In fact, for the solutions (\ref{sphereSolutions}), the metric $e^{2\sigma}\hat{\d s^2}$ is a round sphere with an overall negative sign in front. Formally, these solutions have negative curvature $R <0$ and count as valid complex solutions to the equations of motion (this point was explained in JT gravity in \cite{Maldacena:2019cbz}).
Plugging them into the action, we find
\begin{equation}
e^{-I_{\text{classical}}} = -e^{-\frac{\i\pi}{b^2}(1+2n)}e^{\frac{1}{b^2}}(4\pi\mu)^{\frac{1}{b^2}+1}.
\end{equation}

Which, if any, of these solutions are we supposed to sum over? If the theory is defined by analytic continuation in $b$, starting from the region where $b$ has a positive imaginary part, then the correct answer is to sum the solutions with $n = 0,1,2,\dots$ \cite{Harlow:2011ny}. This gives the result
\begin{equation}\label{sumovern}
\sum_{n = 0}^\infty e^{-I_{\text{classical}}} = \frac{\i}{2\sin(\frac{\pi}{b^2})}e^{\frac{1}{b^2}}(4\pi\mu)^{\frac{1}{b^2}+1}.
\end{equation}
In principle, one should sum over the saddle points at the end, after including the one-loop determinants and gauge fixing factors, but these are independent of $n$, so it is allowable to sum over the saddles at this early stage.

To motivate this prescription, one can consider a toy integral
\begin{equation}
\int \d \sigma e^{a \sigma - e^{2\sigma}}.
\label{toyint}
\end{equation}
After setting $a = -2(b^{-2} + 1)$ and shifting $\sigma$ by a constant, this corresponds to the truncation of the Liouville path integral to the constant mode of $\sigma$. If the real part of $a$ is positive, then the integral converges along the real axis. For our problem, $a$ is negative, and the integral does not converge, but we can imagine defining it by analytic continuation, starting from positive values of $a$, and gradually adjusting the defining contour as we vary $a$ in order to make the integral remain convergent. If we vary $a$ through the upper half plane from positive values almost all the way to the negative real axis, then one acceptable defining contour is the one shown below:
\begin{equation}\label{contourDiagram}
\includegraphics[width = .5\textwidth, valign = c]{figures/contours.pdf}
\end{equation}
Also shown on the diagram are the locations of the saddle points and the steepest-descent contours for each of the saddles (the real part of the locations of these saddles depends on $\mu$). The defining contour is equivalent to a sum of all of the steepest-descent contours for $n = 0,1,2,\dots$, justifying the sum in (\ref{sumovern}).

\subsubsection{FZZT disk}
On the disk (hemisphere), a constant $\sigma$ is not a solution to the equations of motion even with a complex value of $\sigma$. The next simplest thing is to find solutions $\sigma(\theta)$ that are independent of the angular coordinate $\phi$. The Liouville action for such a configuration is
\begin{align}
I &= \frac{1}{b^2}\left\{\frac{1}{2}\int_0^{\frac{\pi}{2}}\d\theta\sin(\theta)\left[\sigma'^2 + 2\sigma  + 4\pi\mu e^{2\sigma}\right] + 2\pi \mu_B e^{\sigma(\pi/2)}\right\}+\frac{1}{2}\int_0^{\frac{\pi}{2}}\d\theta\sin(\theta)2\sigma.
\end{align}
The equations of motion are obtained by varying the first term, of order $b^{-2}$, with respect to $\sigma$. We find the equations
\begin{equation}
1 + 4\pi \mu e^{2\sigma} = \frac{1}{\sin(\theta)}(\sigma' \sin(\theta))', \hspace{20pt} 2\pi\mu_B e^{\sigma(\pi/2)} + \sigma'(\pi/2) = 0.
\end{equation}
One can check that the following is a solution
\begin{align}
\label{hemisphereSol}
\sigma &= 2\pi \i n-\frac{1}{2}\log(4\pi\mu) + \log\left[\frac{2\alpha}{(1+\alpha^2)\cos(\theta) + (1-\alpha^2)}\right], \\
1 &+ \alpha^2 + 2\sqrt{\frac{\pi}{\mu}}\, \mu_B \alpha = 0. \label{alphamub}
\end{align}
This solution corresponds to the metric $e^{2\sigma}\hat{\d s^2}$ being a piece of the hyperbolic disk, written in a way that is conformal to the hemisphere. The action is given by plugging in and integrating:
\begin{align}
I &= \frac{1}{b^2}\left[2\pi \i n -\frac{1}{2}+\log(\alpha) - \frac{1}{2}\log(4\pi \mu)\right] + \left[-\frac{1}{2}\log(4\pi \mu) +  1 + \frac{2\log\alpha - (1-\alpha^2)\log\frac{2\alpha}{1-\alpha^2}}{1+\alpha^2}\right].
\end{align}

In this expression, $\alpha$ is a parameter of the solution, and is determined by the boundary cosmological constant $\mu_B$ in (\ref{alphamub}). When we compute the one-loop determinants later, we will actually only do the computation in the limit of small positive $\alpha$, which corresponds to large negative $\mu_B$. Physically, this is a high energy limit in the boundary theory (matrix integral). In this limit, we have the leading behavior
\begin{equation}
e^{-I_{\text{classical}}} = \frac{2}{e}e^{-2\pi \i n/b^2} e^{\frac{1}{2b^2}}\left(\frac{\sqrt{4\pi\mu}}{\alpha}\right)^{\frac{1}{b^2}+1}.
\end{equation}

Again, there is an integer-indexed family of solutions, and one has to decide which solutions should be included. We will assume that it is correct to imitate the case of the sphere, and sum over $n = 0,1,2,\dots$, which leads to the answer
\begin{equation}
\sum_{n = 0}^\infty e^{-I_{\text{classical}}} = -\frac{\i e^{\i \pi/b^2}}{2\sin(\frac{\pi}{b^2})}\cdot \frac{2}{e}e^{\frac{1}{2b^2}}\left(\frac{\sqrt{4\pi\mu}}{\alpha}\right)^{\frac{1}{b^2}+1}.
\end{equation}
Our understanding of the contour is not as good for this case as for the sphere, but one piece of evidence for this formula is that exact Liouville formulas \cite{Fateev:2000ik}\cite{Kutasov:2004fg} do contain a factor of $1/\sin(\pi/b^2)$, which arises in this expression from the sum over saddles.

\subsection{One loop determinants}
To compute the one-loop determinant, we expand around a classical solution
\begin{equation}
\sigma = \sigma_{\text{cl}} + \chi
\end{equation}
and integrate over the fluctuation $\chi$ with an appropriate action and measure. The measure is derived from an ultralocal metric in field space
\begin{equation}\label{metric}
\d s^2 = C^2\cdot (\d\chi,\d\chi)
\end{equation}
where we introduced an arbitrary constant $C$ to parametrize the normalization ambiguity in the metric, and we defined
\begin{equation}\label{innerProd}
(f,g) = \frac{1}{4\pi}\int \d^2x\, \sqrt{\hat{g}}\, f(x)g(x).
\end{equation}
The action for the fluctuations is just the quadratic approximation to the full action near the saddle point. This can be written
\begin{equation}
I \supset \frac{1}{b^2}(\chi,\mathcal{O}\,\chi)
\end{equation}
for a particular differential operator $\mathcal{O}$ that depends on the solution we are expanding around.

To compute the integral, it is convenient to work in a basis of eigenfunctions of the quadratic action, $\chi(x) = \sum_i \chi_i Y_i(x)$ where
\begin{equation}
[\mathcal{O} Y_i](x) = \lambda_i Y_i(x), \hspace{20pt} (Y_i,Y_j) = \delta_{ij}.
\end{equation}
In this basis, the field-space metric (\ref{metric}) is
\begin{equation}
\d s^2 = C^2 \sum_i \d \chi_i^2
\end{equation}
and the path integral is formally
\begin{equation}\label{oneLoopFormal}
\int \mathcal{D}\chi e^{-\frac{1}{b^2}(\chi,\mathcal{O}\,\chi)} = \prod_i C\int \d \chi_i e^{-\lambda_i \chi_i^2/b^2} = \prod_i \frac{\sqrt{\pi}\,bC}{\sqrt{\lambda_i}}.
\end{equation}

\subsubsection{Sphere}

Expanding around any of the classical solutions (\ref{sphereSolutions}), the quadratic part of the action (\ref{firstLine}) is
\begin{equation}
I \supset \frac{1}{4\pi b^2}\int \sqrt{\hat{g}}\left[(\partial\chi)^2 -2\chi^2\right].
\end{equation}
The eigenfunctions of this problem are the spherical harmonics, and the eigenvalues are
\begin{equation}
\lambda = \ell(\ell+1) - 2, \hspace{20pt} \text{degeneracy $2\ell+1$}.
\end{equation}
We have to deal separately with the $\ell = 0$ eigenfunction, the $\ell = 1$ eigenfunctions, and all of the others.

First, note that the $\ell = 0$ eigenfunction is a negative mode, with eigenvalue $\lambda = -2$. This is to be expected based on the diagram (\ref{contourDiagram}). In that diagram, the steepest-descent contours pass vertically through the saddle points, which means that the action is unstable with respect to real perturbations in the constant mode of $\chi$. In the quadratic approximation, the steepest descent contour is just the imaginary axis, and the integral is
\begin{equation}
C\int_{+\i \infty}^{-\i \infty} \d \chi_0 e^{2 \chi_0^2/b^2} = -\i \frac{\sqrt{\pi}\,bC}{\sqrt{2}}.
\end{equation}

Next, the $\ell = 1$ modes are the zero modes that we promised. These correspond to the Goldstone modes of the $\text{PSL}(2,\mathbb{C})$ symmetry that is spontaneously broken by the constant solutions (\ref{sphereSolutions}). We will take these properly into account in the gauge-fixing part of the computation; for now we simply insert delta functions, so the contribution of these modes is
\begin{equation}
\prod_{i = 1}^3 C\int \d \chi_i \delta(\chi_i) = C^3.
\end{equation}

Finally, we have the product over all of the other modes with $\ell \ge 2$:
\begin{equation}
\prod_{\ell = 2}^\infty \left[ \frac{\sqrt{\pi}b C}{\sqrt{\ell(\ell+1)-2}}\right]^{2\ell+1}.
\end{equation}
This is a divergent product, but we can compute a regularized version of its logarithm using the following sums:
\begin{align}\label{sum111}
\sum_{\ell = 0}^\infty (2\ell+1) e^{-\epsilon^2\ell(\ell+1)} &= \frac{1}{\epsilon^2} + \frac{1}{3} + O(\epsilon^2)\\
\sum_{\ell=2}^\infty(2\ell+1)\log\Big[\ell(\ell+1)-2\Big]e^{-\epsilon^2 \ell(\ell+1)} &= \frac{\log(\frac{1}{\epsilon^2}) - \gamma}{\epsilon^2}-2\log(\frac{1}{\epsilon^2}) + 2.32713 + O(\epsilon^2),\label{sum222}
\end{align}
where $\gamma$ is the Euler-Mascheroni constant. We determined the divergent terms by approximating the sums as integrals, and we determined the constant terms numerically.
Using these formulas, one can compute
\begin{align}
\sum_{\lambda > 0} \log\left[\frac{\sqrt{\pi}b C}{\sqrt{\lambda}}\right]e^{-\epsilon^2 \lambda} &= \sum_{\ell  = 2}^\infty (2\ell+1)\log\left[\frac{\sqrt{\pi}b C}{\sqrt{\ell(\ell+1)-2}}\right]e^{-\epsilon^2 \ell(\ell+1)}\\
&= \log(\sqrt{\pi}bC)\times (\ref{sum111}) - \frac{1}{2}\times (\ref{sum222}).\label{sphereSum}
\end{align}

\subsubsection{Hemisphere}
Expanding around the solution (\ref{hemisphereSol}) to quadratic order, one finds
\begin{align}
I &\supset \frac{1}{b^2}\left\{\frac{1}{4\pi}\int_{\text{hemisphere}}\hspace{-20pt}\sqrt{\hat{g}}\left[(\partial \chi)^2 + \frac{8\alpha^2\chi^2}{((1{+}\alpha^2)\cos(\theta) + (1{-}\alpha^2))^2}\right] - \frac{1}{4\pi}\frac{1+\alpha^2}{1-\alpha^2}\int_{\text{equator}}\hspace{-13pt}\chi^2\right\}\\
&\approx \frac{1}{b^2}\left\{\frac{1}{4\pi}\int_{\text{hemisphere}}\hspace{-20pt}\sqrt{\hat{g}}(\partial \chi)^2  - \frac{1}{4\pi}\int_{\text{equator}}\hspace{-13pt}\chi^2\right\}.
\end{align}
In the second line, we gave an approximate formula for small $\alpha$. The eigenfunctions are determined by solving
\begin{equation}
-\partial^2\chi = \lambda \chi,
\end{equation}
where $\partial^2$ is the Laplacian on the sphere, and by imposing the boundary condition at the equator
\begin{equation}
\chi'(\pi/2) = \chi(\pi/2).
\end{equation}
The solutions are
\begin{equation}
P^{m}_\ell(\cos(\theta))e^{\i m\phi}, \hspace{20pt} \lambda = \ell(\ell+1)
\end{equation}
where $P$ is the generalized Legendre function and $\ell$ is a (non-integer!) parameter that is determined by solving the boundary condition equation.

The spectrum of this operator is qualitatively similar to the one-loop spectrum on the sphere. There is one negative eigenvalue, there are two zero modes, and there are an infinite number of other eigenvalues for which the product requires regularization. The negative mode is in the $m = 0$ sector, and numerically, its eigenvalue is $\lambda_0\approx -1.51095$. Its contribution to the one-loop determinant is
\begin{equation}
C\int_{+\i \infty}^{-\i \infty} \d \chi_0 e^{-\lambda_0 \chi_0^2/b^2} \approx -\i \frac{\sqrt{\pi}\,bC}{\sqrt{1.51095}}.
\end{equation}
The two zero modes are the lowest eigenvalues in the $m = \pm 1$ sectors, with $\ell = 0$. These correspond to the two spontaneously broken generators of $\text{PSL}(2,\mathbb{R})$.
We will treat these in the gauge-fixing step, but for now we insert delta functions, so they contribute
\begin{equation}
\prod_{i = 1}^2 C\int \d \chi_i \delta(\chi_i) = C^2.
\end{equation}

Next we discuss the product over all of the other modes. A regularized version of the product (\ref{oneLoopFormal}) can be computed using the following sums
\begin{align}\label{sum1}
\sum_{\lambda>0}e^{-\epsilon^2\lambda} &= \frac{1}{2\epsilon^2} + \frac{\sqrt{\pi}}{4\epsilon} - \frac{11}{6} + O(\epsilon)\\
\sum_{\lambda>0}\log(\lambda) e^{-\epsilon^2\lambda} &= \frac{\log\frac{1}{\epsilon^2} - \gamma}{2\epsilon^2} + \sqrt{\pi}\frac{\log\frac{1}{\epsilon^2} - \log(4) - \gamma}{4\epsilon} +0.57136 + O(\epsilon).\label{sum2}
\end{align}
Here, the sum runs over the positive eigenvalues, omitting the two zero eigenvalues and the one negative eigenvalue. The divergent terms are the same as with Neumann boundary conditions, where the eigenfunctions are a subset of the ordinary spherical harmonics with integer $\ell$. This allowed us to compute the divergent terms analytically by approximating them as integrals. The finite terms in the sums were computed numerically. To get good precision, it was necessary to use the package \url{https://github.com/JamesCBremerJr/ALegendreEval} \cite{bremer2018algorithm} to compute generalized Legendre functions with large parameters.

These formulas can be used to compute the regularized log-determinant
\begin{equation}
\sum_{\lambda > 0} \log\left[\frac{\sqrt{\pi}b C}{\sqrt{\lambda}}\right]e^{-\epsilon^2 \lambda}.\label{hemisphereSum}
\end{equation}
But after doing so, one finds a problem. For the ratio $Z_{\text{sphere}} / Z_{\text{disk}}^2$, we need the log determinant for the sphere (\ref{sphereSum}) minus two times the log determinant for the disk, (\ref{hemisphereSum}). In this combination, our formulas imply that the leading quadratic divergence cancels out (they can also be individually absorbed using an area counterterm), but the linear and the log divergences remain. The linear divergence can be absorbed into a boundary length counterterm for the disk, but the mismatch in the log term would mean that the ratio is not well-defined.

This problem can be avoided if we regularize the disk in a different way, by inserting in the sum a slightly different convergence factor $e^{-\epsilon^2\tilde{\lambda}_i}$ where
\begin{align}
\tilde{\lambda}_i &= \frac{1}{4\pi}\int \sqrt{\hat{g}} (\partial Y_i)^2\\
&= \lambda_i + \frac{1}{4\pi}\int_{\text{bdy}}Y_i^2.\label{bdyTERM}
\end{align}
One can now recompute the sums (\ref{sum1}) and (\ref{sum2}) using the new regulator. In fact, this isn't much work: the difference $\tilde{\lambda} - \lambda$ is of order one, even for very large eigenvalues, so the change makes a multiplicative correction of order $\epsilon^2$. The only way a small correction like this can affect the order-one terms in the answer is by correcting the leading quadratic divergences. These are dominated by large eigenvalues, so it is enough to know that on average the boundary term in (\ref{bdyTERM}) approaches two for large eigenvalues. (This can be shown by using the fact that for large eigenvalues, the eigenfunctions approach those of the hemisphere with Neumann boundary conditions.)

The upshot is that one finds the revised answers for the sums
\begin{align}\label{sum3}
\sum_{\lambda>0}e^{-\epsilon^2\tilde{\lambda}} &= \frac{1}{2\epsilon^2} + \frac{\sqrt{\pi}}{4\epsilon} - \frac{17}{6}+ O(\epsilon)\\
\sum_{\lambda>0}\log(\lambda) e^{-\epsilon^2\tilde{\lambda}} &= \frac{\log\frac{1}{\epsilon^2} - \gamma}{2\epsilon^2} + \sqrt{\pi}\frac{\log\frac{1}{\epsilon^2} - \log(4) - \gamma}{4\epsilon} - \log(\frac{1}{\epsilon^2})+1.14858 + O(\epsilon).\notag
\end{align}
Now when we subtract twice the disk answer from the sphere answer, the log term cancels, and we can proceed to analyze the finite parts. Note that a priori, both $\lambda$ and $\tilde{\lambda}$ seem like reasonable quantities to use in the regularization, and to  be honest, we would not have known which was right. However, cancellation of the log term seems to require using $\tilde{\lambda}$, and once this choice is made, the finite parts are determined.

\subsection{Dividing by the volume of the conformal group}
Liouville theory on the sphere has an exact $\text{PSL}(2,\mathbb{C})$ conformal symmetry.
A PSU(2) subgroup of this corresponds to ordinary rotations of the sphere; these symmetries are preserved by the constant saddle points (\ref{sphereSolutions}). However, as we will see, the remaining three directions in $\text{PSL}(2,\mathbb{C})$ are spontaneously broken, which means that if we act with an $\text{PSL}(2,\mathbb{C})$ generator in this subspace, it changes the saddle point nontrivially to a new saddle point with shifted values of the zero modes $\chi_1,\chi_2,\chi_3$ that we found in the one-loop determinant.

We will use coordinates $s_1,s_2,s_3$ for the stabilizer subgroup that is preserved, and $b_1,b_2,b_3$ for the directions that are broken by the classical solution. The $\chi_1,\chi_2,\chi_3$ zero modes can be considered functions of the $b_j$ coordinates.

One can define $Z_{\text{sphere}}/\text{vol}(\text{PSL}(2,\mathbb{C}))$ using the Fadeev-Popov procedure. Starting with the measure for the field zero modes, divided by the measure on $\text{PSL}(2,\mathbb{C})$, we replace it as follows
\begin{align}\label{sphereFP}
\frac{\d(\text{sphere zero modes})}{\d(\text{PSL}(2,\mathbb{C}))} &= \frac{\d \chi_1\d \chi_2\d \chi_3}{\underbrace{\d s_1 \d s_2 \d s_3}_{\text{PSU}(2)}\underbrace{\d b_1\d b_2\d b_3}_{\text{PSL}(2,\mathbb{C})/\text{PSU}(2)} } = \frac{1}{\d s_1\d s_2\d s_3} \text{det}(\frac{\partial \chi_i}{\partial b_j}).
\end{align}
In the final expression, we have an inverse measure on PSU(2) and a Fadeev-Popov determinant. The integral gives the determinant divided by the volume of PSU(2).

The situation for the disk is very similar to that of the sphere, except that we only have a $\text{PSL}(2,\mathbb{R})$ subgroup of the conformal symmetry. The analogous Fadeev-Popov procedure is
\begin{align}\label{diskFP}
\frac{\d(\text{disk zero modes})}{\d(\text{PSL}(2,\mathbb{R}))} &=\frac{ 2^{3/2}\d \chi_1\d \chi_2}{\underbrace{\d s_1}_{\text{U}(1)}\underbrace{\d b_1\d b_2}_{\text{PSL}(2,\mathbb{R})/\text{U}(1)} } =  \frac{2^{3/2}}{\d s_1} \text{det}(\frac{\partial \chi_i}{\partial b_j}).
\end{align}
In this case, we end up with a Fadeev-Popov determinant and an inverse measure on U(1). Note that we inserted an important factor of $2^{3/2}$ in this expression relative to (\ref{sphereFP}). This factor will be explained below.

Let's now work out the details explicitly. We write the reference sphere or hemisphere in stereographic coordinates
\begin{equation}
\hat{\d s^2} = \frac{4 \d z \d \bar z}{(1 + z\bar{z})^2}.
\end{equation}
The infinitesimal $\text{PSL}(2,\mathbb{C})$ or $\text{PSL}(2,\mathbb{R})$ transformations correspond to the following set of six holomorphic vector fields ($c$ ghost zero modes)
\begin{equation}\label{transf}
\begin{tabular}{c|c|c|c}
$a$ & $C_{0,a}^z = \delta_a z$ & $C_{0,a}^{\bar{z}} = \delta_a \bar{z}$ & \text{coordinate} \\
\hline
1 & $\i z$ & $-\i \bar{z}$ & $s_1$\\
2 & $\tfrac{1}{2}(1-z^2)$ & $\tfrac{1}{2}(1-\bar{z}^2)$ & $b_1$\\
3 & $\tfrac{\i}{2}(1+z^2)$ & $-\tfrac{\i}{2}(1+\bar{z}^2)$ & $b_2$\\
4 & $z$ & $\bar{z}$ & $b_3$\\
5 & $\tfrac{\i}{2} (1-z^2)$ & $-\tfrac{\i}{2}(1-\bar{z}^2)$ & $s_2$\\
6 & $\tfrac{1}{2}(1+z^2)$ & $\tfrac{1}{2}(1+\bar{z}^2)$ & $s_3$
\end{tabular}
\end{equation}
The first three of these vector fields preserve the hemisphere $|z|\le 1$, and these correspond to the $\text{PSL}(2,\mathbb{R})$ subgroup of $\text{PSL}(2,\mathbb{C})$.
The last three make sense only on the full sphere. In the final column, we have anticipated results below and labeled the transformations according to whether they are preserved ($s$) or broken ($b$) by the classical solution.

One way to get the right meausure in the $s_i$ and $b_i$ coordinates is to use the perspective of the $bc$ ghost path integral. The determinants of the nonzero modes cancel between the sphere and the disk$^2$ (see e.g.~\cite{Douglas:1986eu}) and the zero modes are integrated with a measure that is given by the square root of the determinant of the field-space inner product of the $c$-ghost zero modes:
\begin{equation}
M_{ab} = \frac{3}{8\pi}\int \sqrt{\hat{g}} C_{0,a}^\alpha C_{0,b}^{\beta}\hat{g}_{\alpha\beta}.
\label{Mabmatrix}
\end{equation}
The constant $3/8\pi$ out front will cancel out in the ratio sphere$/($disk$)^2$, and we chose it to so that the answer is simply that $M$ is the $6\times 6$ identity matrix for the sphere, and one-half of the $3\times 3$ identity matrix for the disk. So the measure is one for the sphere, and $2^{-3/2}$ for the disk, justifying the numerical factor in (\ref{diskFP}).

It remains to compute the determinants of $\partial \chi_i/\partial q_j$. Conformal transformations are defined to act on the Liouville field in such a way that the physical metric remains invariant. So, under a general
\begin{equation}
z\rightarrow \tilde{z}(z)
\end{equation}
we require that
\begin{equation}
e^{2\tilde{\sigma}(\tilde{z},\bar{\tilde{z}})}\frac{\d \tilde{z} \d \bar{\tilde{z}}}{(1+\tilde{z}\bar{\tilde{z}})^2} = e^{2\sigma(z,\bar{z})}\frac{\d z\d \bar{z}}{(1+z\bar{z})^2}.
\end{equation}
Infinitesimally, for $\tilde{z}(z) = z + \delta z$, this implies that $\delta \sigma (z) = \tilde{\sigma}(z) - \sigma(z)$ is given by
\begin{equation}
\delta\sigma = -\left[\delta z \left(\partial_z\sigma - \frac{\bar z}{1+z\bar{z}}\right) + \delta\bar{z} \left(\partial_{\bar{z}}\sigma - \frac{z}{1 + z\bar{z}}\right) + \frac{1}{2}\left(\partial_z\delta z + \partial_{\bar{z}}\delta\bar{z}\right)\right].
\end{equation}
We see that the transformation depends on the classical solution $\sigma$ that we start with. The classical solutions for the sphere and the disk are
\begin{align}
\text{sphere:} \hspace{20pt} \sigma &= \text{const.}\\
\text{disk:} \hspace{20pt} \sigma &= \text{const.} + \log\frac{\alpha(1+z\bar{z})}{1-\alpha^2z\bar{z}}\approx \text{const.}' + \log(1+z\bar{z})
\end{align}
where we gave the small $\alpha$ limit in the final expression. Plugging in, one finds that for the transformations (\ref{transf}), the corresponding perturbations to $\sigma$ are
\begin{align}
\text{sphere:} \hspace{20pt} \delta_a\sigma &= \left\{0,\frac{z+\bar{z}}{1+z\bar{z}},-\i\frac{z-\bar{z}}{1+z\bar{z}},\frac{-1+z\bar{z}}{1+z\bar{z}},0,0\right\}\\
\text{disk:} \hspace{20pt} \delta_a\sigma &= \frac{1}{2}\left\{0,z+\bar{z},-\i(z-\bar{z})\right\}.
\end{align}
We see that the $a = 1,5,6$ directions are the PSU(2) symmetry directions that stabilize the classical solution, justifying the labeling in (\ref{transf}).

The nonzero $\delta_a\sigma$ functions correspond precisely to the zero modes of the one-loop determinants, but with an arbitrary normalization. The $\chi_i$ coordinates are the coefficients of {\it normalized} zero modes. To see the discrepancy, we can evaluate the matrix $m_{ab}=(\delta_a \sigma,\delta_b\sigma)$  where the inner product is defined in (\ref{innerProd}). One finds
\begin{align}
\text{sphere:} \hspace{20pt} m  &= \frac{1}{3}\text{diag}(0,1,1,1,0,0)\\
\text{disk:} \hspace{20pt} m &= \frac{\log(4)-1}{4}\,\text{diag}(0,1,1).
\end{align}
The determinants $\text{det}(\partial \chi_i/\partial q_j)$ are just the square roots of the determinants of the nonzero submatrices here,
\begin{align}
\text{sphere:} \hspace{20pt} \text{det}(\frac{\partial \chi_i}{\partial q_j})  &= \frac{1}{3^{3/2}}\\
\text{disk:} \hspace{20pt} \text{det}(\frac{\partial \chi_i}{\partial q_j}) &= \frac{\log(4)-1}{4}.
\end{align}

So the gauge-fixing factors should be in the two cases
\begin{align}
\frac{1}{\text{vol}(\text{PSL}(2,\mathbb{C}))}= \frac{1}{\text{vol}(\text{PSU}(2))}\frac{1}{3^{3/2}} \delta(\chi_1)\delta(\chi_2)\delta(\chi_3)\\
\frac{1}{\text{vol}(\text{PSL}(2,\mathbb{R}))}= \frac{2^{3/2}}{\text{vol}(\text{U}(1))}\frac{\log(4)-1}{4}\delta(\chi_1)\delta(\chi_2).
\end{align}
We normalized the original transformations (\ref{transf}) so that with a unit measure, a full rotation has length $2\pi$. For the case of $U(1)$, this means simply $\text{vol}(\text{U}(1)) = 2\pi$. For the case of PSU(2), we can use the fact $\text{vol}(\text{PSU}(2)) = 2\pi \, \text{vol}(S^2) = 8\pi^2$.

\subsection{Putting the pieces together}
Putting the pieces together and dropping the divergent terms in the one-loop determinants (including the dangerous logarithmic piece, which will cancel between the two expressions), we find the following formulas
\begin{align}
\frac{Z_{\text{sphere}}}{\text{vol}(\text{PSL}(2,\mathbb{C}))} &= \left(\frac{\i e^{\frac{1}{b^2}}}{2\sin(\frac{\pi}{b^2})}(4\pi\mu)^{\frac{1}{b^2}+1}\right)\left( -\i \frac{\sqrt{\pi}\,bC}{\sqrt{2}}\cdot C^3 \cdot\frac{(\sqrt{\pi}b C)^{-\frac{11}{3}}}{e^{\frac{1}{2}\cdot 2.32713}}\right) \left(\frac{1}{8\pi^23^{3/2}}\right) \label{zsphereliouv} \\
\frac{Z_{\text{disk}}}{\text{vol}(\text{PSL}(2,\mathbb{R}))} &=
\left(\frac{-\i e^{\frac{1}{2b^2}(1 +2\pi \i)}}{e\sin(\frac{\pi}{b^2})}\left(\frac{\sqrt{4\pi\mu}}{\alpha}\right)^{\frac{1}{b^2}+1}\right)\left(  \frac{-\i\sqrt{\pi}\,bC}{\sqrt{1.51095}}\cdot C^2 \cdot\frac{(\sqrt{\pi}b C)^{-\frac{17}{6}}}{e^{\frac{1}{2}\cdot 1.14858}}\right) \left(\frac{2^{\frac{3}{2}}(\log(4)-1)}{2\pi\cdot 4}\right)\notag
\end{align}
We remind the reader that these formulas are valid in the semiclassical small $b$ limit, and further (for the disk) in the high-energy limit of small positive $\alpha$ (or large negative $\mu_B$).
In these expressions, the first term is the classical action, the second term is the one-loop determinant, and the third term comes from the gauge fixing. The invariant ratio is
\begin{equation}\label{ratio2}
\frac{Z_{\text{sphere}}}{\text{vol}(\text{PSL}(2,\mathbb{C}))} \cdot\left(\frac{\text{vol}(\text{PSL}(2,\mathbb{R}))}{Z_{\text{disk}}}\right)^2 = 8.889\, e^{-\frac{2\pi \i}{b^2}} b \sin(\frac{\pi}{b^2})\, \alpha^{2 + \frac{2}{b^2}}.
\end{equation}

Let's now apply this to the minimal string. To compute the partition functions of the minimal string, we set $b = \sqrt{2/p}$ where $p$ is an odd integer (and which must be large for our semiclassical approximation to be valid) and multiply by the partition function of the matter sector, which is the $(2,p)$ minimal model:
\begin{equation}
\mathcal{Z}_{\text{sphere}} = Z_{\text{sphere}}^{\text{minimal model}}\frac{Z_{\text{sphere}}}{\text{vol}(\text{PSL}(2,\mathbb{C}))}, \hspace{20pt} \mathcal{Z}_{\text{disk}} = Z_{\text{disk $(1,1)$}}^{\text{minimal model}}\frac{Z_{\text{disk}}}{\text{vol}(\text{PSL}(2,\mathbb{R}))}.
\end{equation}
Using the formula (here $S_{(1,1),(1,1)}$ is an element of the modular S-matrix relating the identity characters in the two channels, see e.g.~\cite{DiFrancesco:1997nk},  Chapter 10)
\begin{equation}
\frac{(Z_{\text{disk $(1,1)$}}^{\text{minimal model}})^2}{Z_{\text{sphere}}^{\text{minimal model}}} = S_{(1,1),(1,1)} = -\frac{2}{\sqrt{p}}\sin(\frac{\pi p}{2})\sin(\frac{2\pi}{p}),
\label{minimalmodelS11}
\end{equation}
and approximating $\sin(\frac{2\pi}{p}) = \frac{2\pi}{p}$, we find that for large $p$ and small $\alpha$
\begin{equation}\label{liouvillefinalanswer}
\frac{\mathcal{Z}_{\text{sphere}}}{(\mathcal{Z}_{\text{disk}})^2} = 1.000\, p \,\alpha^{p+2}.
\end{equation}
We remind the reader that this formula is accurate in the limit of large $p$ and small $\alpha$. Here $p$ is an odd integer that labels the $(2,p)$ minimal string, and $\alpha$ is a parameter that determines the energy of the FZZT boundary condition. Small $\alpha$ corresponds to high energy. In the next section we will compute the same thing in the matrix integral language, and we will find that the numerical constant in (\ref{liouvillefinalanswer}) should be exactly one.

\section{Matrix integral computations}
A Hermitian matrix integral (see \cite{DiFrancesco:1993cyw}\cite{Eynard:2015aea} for reviews) is an integral of the form
\begin{align}
\mathfrak{Z} &= \int \d H \, e^{-L \ \text{Tr}V(H)},
\end{align}
where $L$ is the rank of the matrix and $V$ is the ``potential.'' This can be written as an integral over the eigenvalues,
\begin{align}\label{matrixEigs}
\mathfrak{Z} =C_L\int \mathrm{d}^L\lambda\, e^{-L\sum_{j = 1}^L V(\lambda_j)}\,\prod_{i<j}(\lambda_i - \lambda_j)^2.
\end{align}
Here the constant $C_L$ and the final term (Vandermonde determinant) both arise from integrating out the non-eigenvalue parts of the matrix.

In the leading order at large $L$, one can formally ignore the discreteness of the eigenvalues and trade in the $L$ eigenvalues for a smooth density $\rho(\lambda)$, normalized so that $\int \d \lambda \rho(\lambda) = 1$:
\begin{align}
\mathfrak{Z} &\sim \int \mathcal{D}\rho  \, e^{-L^2 I[\rho]}\\
I[\rho] &= \int \d \lambda \rho(\lambda) V(\lambda) - \frac{1}{2}\int\hspace{-7pt}\int \d \lambda_1 \d \lambda_2 \log[(\lambda_1-\lambda_2)^2] \rho(\lambda_1)\rho(\lambda_2).
\label{iofrho}
\end{align}
In particular, for large $L$, we can think of the matrix integral as being dominated by a single saddle point $\rho_0(\lambda)$ which stationarizes this action, subject to the constraint $\int \d \lambda \rho_0(\lambda) = 1$. Because there are a total of $L$ eigenvalues, the total or ``physical'' density of eigenvalues is $L$ times this normalized density, so $L \rho_0(\lambda)$. This function is supported on an interval or a union of intervals, and generically it vanishes like a square root at the ends of each interval.

The $(2,p)$ minimal string theory is conjectured to be related to a type of matrix integral where $\rho_0$ is supported on the entire positive real axis, and with \cite{Moore:1991ir}
\begin{equation}\label{rhoMS}
\rho_0(x) = \sinh\Big(\frac{p}{2}\text{arccosh}(1+2x)\Big).
\end{equation}
This does not fit the definition of a standard matrix integral, because the density cannot be normalized so that its integral is one. However, it makes sense as an example of what is called a ``double scaled'' matrix integral. This can be defined as a limiting procedure applied to an ordinary matrix integral, where a family of potentials parametrized by $\epsilon$ are arranged so that very near the endpoint, the density locally approximates a rescaled version of (\ref{rhoMS}):
\begin{equation}\label{confbackground}
\rho_0(\lambda) = \epsilon^{p/2}\sinh\Big(\frac{p}{2}\text{arccosh}(1 + 2E)\Big) + O(\epsilon^{p/2+1}), \hspace{20pt} \lambda = \lambda_{\text{endpoint}} + \epsilon E.
\end{equation}
So in the limit $\epsilon \to 0$, we recover the full density (\ref{rhoMS}) in a ``zoomed-in'' view of a small neighborhood of one of the endpoints. ``Double scaling'' refers to following this limiting procedure, while also adjusting $L$ so that the total density of eigenvalues in the $E$ coordinate, which is proportional to
\begin{equation}
\epsilon^{p/2+1} L = e^{S_0},
\end{equation}
is fixed.
The result is a region near the edge of the spectrum that resembles (\ref{rhoMS}), attached to a larger ``garbage'' region at higher energies that depends on the details of the limiting procedure that was used.

What are we supposed to compute in this double scaled matrix integral? In the minimal string, we computed the sphere partition function, and we normalized it using the (FZZT) disk partition function in the high-energy limit. Both the sphere and disk  quantities have duals in the matrix integral picture. First, the sphere partition function is related to the leading term $L^2\mathcal{F}_0$ in the logarithm of the full matrix partition function:
\begin{equation}
\log(\mathfrak{Z}) = L^2 \mathcal{F}_0 + \mathcal{F}_1 + L^{-2} \mathcal{F}_2 + \dots.
\end{equation}
One can get this term by simply evaluating the action $I$ on the stationary configuration $\rho_0$:
\begin{equation}\label{FREEEN}
\mathcal{F}_0 = -I[\rho_0].
\end{equation}
Second, the (FZZT) disk partition function is given by a similar leading term $L\mathcal{G}_0$ in the expectation value
\begin{equation}
\langle \text{Tr}\log(H-x)\rangle = L\mathcal{G}_0(x) + L^{-1} \mathcal{G}_1(x) + L^{-3} \mathcal{G}_2(x) + \dots
\end{equation}
Again, this is given simply in terms of the stationary configuration $\rho_0$:
\begin{equation}
\mathcal{G}_0(x) = \int \d\lambda \rho_0(\lambda) \log(\lambda-x).
\end{equation}

The terms in the expansion that are proportional to negative powers of $L$ are well-defined in the double-scaled limit, in the sense that they do not depend on the ``garbage'' region that (\ref{rhoMS}) is attached to at high energies. However, the leading terms do depend on the garbage region. In fact, they are numerically dominated by it! There is a good analog of this in the Liouville path integral. The terms proportional to inverse powers of $L$ correspond to Liouville partition functions on surfaces with negative Euler characteristic. For such surfaces, the integral over the Liouville field converges along the real axis. But for the sphere or the disk (or, marginally, the torus), the integral is divergent in the large negative $\phi$ region. This corresponds to very small surfaces, and the ambiguity in how this part of the path integral is regulated corresponds to the ambiguity in the nonuniversal garbage that is used to construct the double-scaled limit of the matrix integral.

In Liouville, the nonuniversal pieces and the universal pieces can be distinguished by their dependence on the cosmological constant $\mu$. The contribution of the nonuniversal small $\sigma$ region is analytic in $\mu$, because for large negative $\sigma$, one can expand down in powers of $\mu e^{2\sigma}$, giving a power series in $\mu$.\footnote{The contour prescription we used for Liouville throws out these nonuniversal analytic parts automatically, but they would be there if for example we had defined the integral over $\sigma$ to be on a contour on the real axis that ended at some finite but large negative value, corresponding to a UV cutoff on the physical metric.} But as we saw above, the universal part depends on the $\mu e^{2\sigma}$ term in an essential way, and the result is proportional to a nontrivial power of $\mu$. So the interesting part of the answer can be selected by keeping the part that is nonanalytic in $\mu$.

In the matrix integral, $\epsilon^2$ plays the role of $\mu$, and the nonanalytic terms correspond to odd powers of $\epsilon$. As we will see, these terms are numerically highly subleading, but they are distinguished by their nonanalyticity as functions of $\epsilon^2$. For this to work, it is important that we do not accidentally introduce any direct nonanalyticity in $\epsilon^2$ through the matrix integral potential. So we will make sure that the potential is analytic in $\epsilon^2$, and we will then take the leading nonanalytic part of the free energy. Schematically, if the potential is chosen to be analytic in $\epsilon^2$, then we will have
\begin{equation}
L^2 \mathcal{F}_0 = L^2(1 + \epsilon^2 + \epsilon^4 + \dots) + L^2 \epsilon^{p + 2}(1 + \epsilon^2 + \epsilon^4+\dots).
\end{equation}
The leading nonanalytic piece is proportional to $L^2 \epsilon^{p+2} = e^{2S_0}$, and this piece can be identified as the universal part that can be compared to the universal part of the Liouville answer. There is a similar procedure for extracting the universal part of the FZZT partition function $\mathcal{G}_0(x)$, which we will describe below.

\subsection{The conformal background}
Let's now carry this out in detail. The first step will be to construct a family of potentials that gives the desired double-scaled limit. To do so it will be helpful to define a special set of functions that we will add together to get the desired density of states
\begin{align}
\rho_j(\lambda|a) = \frac{2^{2j}}{\binom{2j-1}{j-1}}\frac{(a^2-\lambda^2)^{j - \frac{1}{2}}}{2\pi a^{2j}}.
\end{align}
For each value of $j$, this is a normalized and symmetric density of states that extends between endpoints $\pm a$. Near one of these endpoints, the density of states behaves as
\begin{equation}
\rho_j(\lambda|a) = \frac{2^{3j-\frac{1}{2}}}{\binom{2j-1}{j-1}}\frac{(\epsilon E)^{j-\frac{1}{2}}}{2\pi a^{j+\frac{1}{2}}} + \dots, \hspace{20pt} \lambda = -a+\epsilon E
\end{equation}
where the dots are higher order in $\epsilon$. Because we can get different powers with different values of $j$, an appropriate linear combination of such functions with different coefficients can be used to approximate the conformal background near the edge.

The density of states $\rho_j(\lambda|a)$ is the large $L$ stationary configuration for a matrix integral with a particular potential $V_j(\lambda|a)$. Explicitly, this potential (or rather its first derivative) is
\begin{equation}
V'_j(\lambda|a) = (-1)^{j+1}\frac{2^{2j}}{\binom{2j-1}{j-1}}\frac{(\lambda^2 - a^2)^{j-\frac{1}{2}}_+}{a^{2j}}
\end{equation}
where the subscript $(\cdot)_+$ means the terms proportional to non-negative powers of $\lambda$, when the expression is expanded around $\lambda = \infty$ (see,  for example,  section 2.2 of \cite{DiFrancesco:1993cyw}).
For example
\begin{equation}
(\lambda^2-a^2)^{\frac{1}{2}}_+ = \lambda, \hspace{20pt} (\lambda^2 - a^2)^{\frac{3}{2}}_+ = \lambda^3 - \frac{3 a^2}{2}\lambda, \hspace{20pt} \dots
\end{equation}
If we choose one of these potentials $V'_j(\lambda|a)$, then the resulting large $L$ density of states will be $\rho_j(\lambda|a)$. More generally, if we take a superposition with coefficients $c_i$ such that $\sum_{i}c_i = 1$, then the density of states will be the corresponding superposition:
\begin{equation}\label{correspondence}
\rho_0(\lambda) = \sum_{j = 1}^\infty c_j \rho_j(\lambda|a) \hspace{20pt} \leftrightarrow  \hspace{20pt} V'(\lambda) = \sum_{j = 1}^\infty  c_jV_j'(\lambda|a).
\end{equation}

In principle, we could use this to get the desired double scaled background (\ref{confbackground}), by setting $a$ to some value and then adjusting the coefficients $c_j$ as a function of $\epsilon$ so that near $\lambda = -a$, the density of states approximates (\ref{confbackground}). However, if we do this in the most straightforward way, the potential will not be analytic in $\epsilon^2$, and it will not be simple to isolate the desired term in the free energy.

To avoid this problem, we need to be more careful, adjusting the coefficients in the potential in a manifestly analytic way. As we will see, this will lead to an endpoint $a$ that is not analytic in $\epsilon^2$, contaminating the functions $V_k(\lambda|a)$ and making it difficult to impose analyticity. So rather than working with $V_j(\lambda|a)$, we will expand the potential in terms of $V_j(\lambda|1)$:
\begin{equation}\label{potentialtk}
V'(\lambda) = \sum_{k = 1}^\infty t_k V'_k(\lambda|1).
\end{equation}
This makes it easier to be sure that the potential is analytic: all we need to do is choose $t_k$ coefficients that are analytic in $\epsilon^2$. One can determine the density of states associated to this potential by the following procedure: (i) write $(\ref{potentialtk})$ in terms of $V'_j(\lambda|a)$ for an arbitrary $a$ to be fixed later, (ii) use (\ref{correspondence}) to get $\rho$, assuming that value of $a$ (iii) determine the correct value of $a$ by imposing the normalization constraint. For the first step, one can expand $V_k(\lambda|1)$ in terms of $V_j(\lambda|a)$ using
\begin{equation}
(\lambda^2-1)^{k-\frac{1}{2}}_+ = \sum_{j = 0}^{k-1} \binom{k-\frac{1}{2}}{j}(a^2-1)^j (\lambda^2-a^2)^{k-j-\frac{1}{2}}_+
\end{equation}
which follows from linearity of the $(\cdot)_+$ operator. Substituting into (\ref{potentialtk}) leads directly to
\begin{align}\label{cjfromtk}
c_j &= a^{2j}\sum_{k = j}^\infty \binom{k}{j}(1-a^2)^{k-j}t_k.
\end{align}
These $c_j$ coefficients give the density of states, from (\ref{correspondence}). However, the result depends on an as-yet-undetermined endpoint $a$. This can be fixed by imposing the normalization condition
\begin{equation}\label{normcond}
1 = \sum_{j = 1}^\infty c_j = \sum_{k = 1}^\infty \Big(1-(1-a^2)^{k}\Big)t_k.
\end{equation}

Our goal is find a set of $t_k$ coefficients that are analytic in $\epsilon^2$, such that the procedure just outlined leads to a density of states that behaves as (\ref{confbackground}) near the endpoints. We will parametrize the odd integer $p$ in terms of an unrestricted integer $m$, so the relevant minimal model is
\begin{equation}
(2,p) = (2,2m-1).
\end{equation}
We claim that the solution for the $t_k$ parameters $t_1,\dots,t_m$ is\footnote{Up to a choice of convention, the terms that are explicitly written here are the nonzero KdV times of the conformal background \cite{Moore:1991ir}\cite{Turiaci:2020fjj}.}
\begin{equation}\label{tksol}
t_k = \begin{cases} (-1)^{\frac{m-k}{2}}\frac{\binom{m-1}{\frac{m-k}{2}}\binom{k+m-2}{k}}{\binom{2m-2}{m}}\epsilon^{m-k}   + O(\epsilon^{m-k+2})& m-k \text{ even} \\ O(\epsilon^{m-k+1}) & m-k\text{ odd}\end{cases}
\end{equation}
There is some freedom in choosing the subleading terms, but to get the correct density of states, we will have to choose them so that the solution to (\ref{normcond}) satisfies
\begin{equation}
a^2 = 1-\epsilon + O(\epsilon^2).
\end{equation}
For example, one valid choice is to set to zero the $O(\epsilon^{m-k+2})$ correction on the first line of (\ref{tksol}), and then to set $t_{k}$ for odd $k$ to be equal to minus $t_{k-1}$. Explicitly, this can be written as
\begin{equation}
t_{m-2\ell} = -t_{m-2\ell+1} = (-1)^\ell\frac{\binom{m-1}{\ell}\binom{2m-2-2\ell}{m-2\ell}}{\binom{2m-2}{m}}\epsilon^{2\ell}.\label{texact}
\end{equation}
A nice property of this particular choice is that the sum of the $t_k$ parameters telescopes, so that
\begin{equation}\label{telescopes}
\sum_{k=0}^\infty t_k = 1
\end{equation}
where we have formally defined
\begin{equation}
t_0 = \begin{cases} (-1)^{\frac{m}{2}+1} \frac{\binom{m-1}{\frac{m}{2}}}{\binom{2m-2}{m}}\epsilon^{m} & m \text{ even} \\ 0 & m \text{ odd}\end{cases}.
\end{equation}
This simplifies the normalization condition (\ref{normcond}) that determines $a$ to
\begin{equation}
\sum_{k = 0}^\infty (1-a^2)^k t_k = 0,
\end{equation}
which one can check is indeed solved by $1 - a^2 = \epsilon  + O(\epsilon^2)$.

It remains to check that near the endpoint, we get the desired density of states (\ref{confbackground}). For this one can neglect the higher order terms and use only the terms written explicitly in (\ref{tksol}). Using (\ref{cjfromtk}), we find
\begin{align}
c_j &= \frac{\binom{m}{j}\binom{m+j-2}{j-1}}{2^{j-m}\binom{2m-2}{m-1}}\epsilon^{m-j} +O(\epsilon^{m-j+1}).
\end{align}
This implies that near the endpoint at $a \approx -1$, we have
\begin{align}
\rho_0(\lambda)  &=\frac{2^{m+\frac{1}{2}}\epsilon^{m-\frac{1}{2}}}{2\pi} \sum_{j = 1}^m \frac{\binom{m}{j}\binom{m+j-2}{j-1}}{\binom{2j-1}{j-1}\binom{2m-2}{m-1}}(4E)^{j-\frac{1}{2}}  + O(\epsilon^{m+\frac{1}{2}})\\&= \frac{2^{m+\frac{1}{2}}\epsilon^{m-\frac{1}{2}}}{\pi\binom{2m-1}{m-1}}\sinh\left[\frac{2m-1}{2}\text{arccosh}(1+2E)\right] + O(\epsilon^{m+\frac{1}{2}}).\label{matrixDensity}
\end{align}
Up to an overall constant that can be absorbed into the definition of $e^{S_0}$, this is indeed the desired density of states.

\subsection{The free energy}
The explicit values (\ref{texact}) determine a potential exactly, which in turn determines $\rho_0$ exactly, and together these quantities determine the free energy $\mathcal{F}_0$ via for example (\ref{FREEEN}) or a somewhat more efficient version of this formula in \cite{Eynard:2015aea}. Using Mathematica, we found the first few cases:
\begin{align}
2\mathcal{F}_0^{(2,3)} &= -\log2 - \frac{25}{24} - \frac{\epsilon^2}{3} + \frac{\epsilon^4}{4} \boxed{- \frac{4\epsilon^5}{15}} + \dots \\
2\mathcal{F}_0^{(2,5)} &= -\log2 - \frac{49}{40}-\frac{\epsilon ^2}{5}+\frac{\epsilon ^4}{24}\boxed{-\frac{4\epsilon ^7}{105}} + \dots\\
2\mathcal{F}_0^{(2,7)} &= -\log2 - \frac{761}{560}-\frac{6 \epsilon ^2}{35}+\frac{8\epsilon ^4}{125}-\frac{\epsilon ^6}{50}+\frac{\epsilon ^8}{100}\boxed{-\frac{64\epsilon ^9}{7875}} + \dots\\
2\mathcal{F}_0^{(2,9)} &= -\log2 - \frac{7381}{5040}-\frac{10 \epsilon ^2}{63}+\frac{109 \epsilon ^4}{1372}-\frac{\epsilon ^6}{49}+\frac{3 \epsilon ^8}{392}\boxed{-\frac{64\epsilon ^{11}}{33957}} + \dots
\end{align}
Here we have boxed the first nonanalytic term in each case. This one term is the universal part of the answer and everything else is ``nonuniversal garbage,'' depending on specific decisions we made in constructing the double-scaled theory.

Note that in each case we have written $2\mathcal{F}_0^{(2,p)}$ on the LHS. This is because with the symmetric potential we have chosen, there is an identical copy of the same double-scaled theory at both edges of the spectrum. The matrix integral free energy includes contributions from both, but we intend $\mathcal{F}_0^{(2,p)}$ to mean the free energy associated to just a single copy, so $2\mathcal{F}_0^{(2,p)} = \mathcal{F}_0$.

It rapidly becomes impractical to calculate the answer this way, but in fact there is a simple general answer for arbitrary $p = 2m-1$:
\begin{equation}\label{generalAns}
2\mathcal{F}_0^{(2,p)} = (\text{analytic in }\epsilon^2) -\frac{2^{p-1}p}{(p^2-4)\binom{p}{\frac{p-1}{2}}^2}\epsilon^{p+2} + O(\epsilon^{p+3}).
\end{equation}
To derive this in an efficient way, one can use a formula, reviewed in appendix \ref{APPORTHOG}, that computes the free energy from the function $u(\xi) = 1-a^2(\xi)$ that solves the ``genus zero string equation''
\begin{align}
1-\xi &= \sum_{k = 0}^\infty t_k \Big(1-a^2\Big)^k\label{stringEQ}\\
&= \sum_{k = 0}^\infty t_k u^k\\
&\approx \frac{\epsilon^m}{2^m \binom{m-\frac{1}{2}}{m}}\Big(P_m(u/\epsilon) - P_{m-2}(u/\epsilon)\Big)\label{legendreP}\\
&=f(u).
\end{align}
In the second-to-last line, we used an approximate form for the $t_k$ that resums to a combination of Legendre polynomials, as pointed out in \cite{Turiaci:2020fjj}.
Concretely, this form neglects all of the unspecified higher-order terms in (\ref{tksol}), and it is a sufficiently good approximation to compute the leading nonanalytic term in the free energy. In the final line, we introduced a temporary notation $f(u)$. In solving this equation to get $u(\xi)$, we choose the branch of the solution that is equal to $\epsilon$ when $\xi = 1$.

In terms of this function $u(\xi)$, the formula for the free energy is
\begin{align}
\mathcal{F}_0 &= \int_0^1 \d \xi (1-\xi)\log\left(\frac{1-u(\xi)}{4}\right)\label{ORTHOG}\\
&=\int_1^\epsilon (-\d u f'(u)) f(u) \log(\frac{1-u}{4}).
\end{align}
The function $f(u)$ is analytic in $\epsilon^2$, so for generic $\xi$, the solution $u(\xi)$ will also be analytic in $\epsilon^2$. However, this breaks down near $\xi = 1$, where the solution is $u = \epsilon$. So, to accurately compute the nonanalytic terms, we only need to do the integral in the vicinity of $\xi = 1$, which corresponds to $u$ in the vicinity of $\epsilon$, where $\log(1-u)\approx -u$. Using $\sim$ to denote equality of the leading nonanalytic terms, we therefore have
\begin{align}
\mathcal{F}_0& \sim  \int_?^\epsilon (-\d u f'(u))f(u)(-u)\\
&\sim \int_0^\epsilon (-\d u f'(u))f(u)(-u).
\end{align}
In the first line, we introduced an arbitrary lower limit of integration. As long as this point is chosen to be analytic in $\epsilon^2$, it will not affect the nonanalytic terms in the answer, and in the second line we chose a convenient value of zero. As we will see, this has the nice effect of removing completely the analytic terms, leaving only the nonanalytic part that we are seeking. After integrating by parts, we continue
\begin{align}
 &= -\int_0^\epsilon \d u \frac{f^2(u)}{2}\\
&=-\frac{\epsilon}{2}\left(\frac{\epsilon^m}{2^m\binom{m-\frac{1}{2}}{m}}\right)^2\left(\frac{1}{2m+1} + \frac{1}{2m-3}\right).
\end{align}
To get the final line, we inserted the expression for $f(u)$ in (\ref{legendreP}) and used
\begin{equation}
\int_0^1 \d y P_m(y)P_m(y) = \frac{1}{2m+1}, \hspace{20pt} \int_0^1 \d y P_m(y)P_{m-2}(y) = 0
\end{equation}
which follow from the orthogonality relation for the $P_m$ functions, and the fact that $P_m$ and $P_{m-2}$ are either both even functions or both odd functions. After substituting in $p = 2m-1$, one finds the term in (\ref{generalAns}).

\subsection{FZZT disk}
We start by reminding the reader of the definition
\begin{equation}
\mathcal{G}_0(x) = \int_{-a}^a \d \lambda \rho_0(\lambda)\log(\lambda - x).
\end{equation}
To define a function appropriate for the double-scaled limit, we consider this function at an argument $x$ that is close to the lower endpoint. For example, in the case of the $(2,3)$ model, we have the explicit formula (after giving $E$ a small negative imaginary part)
\begin{equation}\label{disk23}
\mathcal{G}_0^{(2,3)}(-a + \epsilon E) = \frac{7}{12}-\frac{2(1+2E)}{3}\epsilon +\left(\frac{2}{3}+\frac{(1+2E)^2}{2}\right) \epsilon ^2+\boxed{\i\frac{2^{7/2}}{15} E^{3/2} (5+4 E) \epsilon^{5/2}}+O(\epsilon^3).
\end{equation}
Here we have boxed the universal term, which will be compared to the Liouville computations below. We expect that everything else in this expression is ``nonuniversal garbage'' which depends on the way in which we take the double-scaled limit.

At first, this might seem puzzling, because the ``nonuniversal garbage'' contains a term $\epsilon^1$, which would seem to be a nonanalytic function of $\mu \sim \epsilon^2$. However, from the Liouville perspective, precisely the combination $(1+2E)\epsilon$ is proportional to $\mu_B$, the boundary cosmological constant, and we should expect nonuniversal analytic terms in both $\mu$ and $\mu_B$, associated to the divergence of the path integral in the large negative $\sigma$ region.\footnote{Again, the contour we used for Liouville effectively set all such terms to zero, but with a different contour prescription, such terms would be present.} In Liouville language, the term at order $\epsilon^2$ is a linear combination of $\mu$ and $\mu_B^2$. By contrast, the boxed term is genuinely nonanalytic in $\mu_B$ and $\mu$, and corresponds to a contribution from the universal continuum region of the Liouville path integral.

In the Liouville computation, the disk path integral was pure imaginary, which suggests that the first nonanalytic term (in this sense) will be pure imaginary. This is true of (\ref{disk23}), and also true for the $(2,5)$ and $(2,7)$ cases, which we checked explicitly. We don't have a general proof of this from the matrix side, although we suspect it is possible to show this. However, what we can do easily is compute the imaginary part:
\begin{align}
\text{Im}\,\mathcal{G}_0(-a+\epsilon E) &= \text{Im} \int_{-a}^a \d \lambda \rho_0(\lambda)\log(\lambda+a-\epsilon E) \\
&= \pi \int_{-a}^{-a+\epsilon E} \d\lambda \rho_0(\lambda).
\end{align}
This only depends on the density of states near the edge, where it is constrained by the double-scaled limit. Putting in (\ref{matrixDensity}) and taking high energies $E \gg 1$, this is
\begin{align}\label{diskmatrixanswer}
\text{Im}\,\mathcal{G}_0(-a+\epsilon E) \to \frac{(2\epsilon)^{\frac{p}{2}+1}}{4\binom{p}{\frac{p-1}{2}}}\frac{(4E)^{\frac{p}{2}+1}}{p+2}.
\end{align}

\subsection{Comparing the ratio to the Liouville answer}
As a final step, we need to relate the $\alpha$ parameter of the Liouville theory to the energy $E$. We can compare $Z_{\text{disk}}'(\mu_B)$ with $\mathcal{G}_0'(x)$, which is proportional to $\rho_0(E)$. For this we need to study the theory at finite energy, where we did not compute the one-loop determinant. Fortunately, the classical answer will be enough. In the classical approximation (which means only keeping terms of order $\frac{1}{b^2} = \frac{p}{2}$ in the exponential), the $\alpha$ dependence of $Z_{\text{disk}}'(\mu_B)$ is proportional to
\begin{equation}
e^{-\frac{1}{b^2}\log(\alpha)} = e^{\frac{p}{2}\log(\frac{1}{\alpha})}.
\end{equation}
On the other hand, for large $p$, the density $\rho_0(E)$ is proportional to
\begin{equation}
e^{\frac{p}{2}\text{arccosh}(1+2E)}.
\end{equation}
Comparing the two, we conclude that\footnote{With (\ref{hemisphereSol}), this implies $
-\sqrt{\frac{\pi}{\mu}}\mu_B = 1 + 2E$,
which justifies a statement made above that $(1+2E)\epsilon \propto \mu_B$.}
\begin{equation}
\label{equivto}
\log\frac{1}{\alpha} = \text{arccosh}(1+2E).
\end{equation}
Or, at high energies,
\begin{equation}
\frac{1}{\alpha} \approx 4E.
\end{equation}
Substituting this into (\ref{diskmatrixanswer}), and also using (\ref{generalAns}), we find
\begin{equation}
\frac{\mathcal{F}_0^{(2,p)}|_{\text{universal}}}{(\mathcal{G}_0^{(2,p)}|_{\text{universal}})^2} = \frac{p+2}{p-2}p\, \alpha^{p+2}, \hspace{20pt} \alpha \ll 1.
\end{equation}
This exact answer agrees with (\ref{liouvillefinalanswer}) at large $p$.

\section*{Acknowledgements}
RM is supported in part by Simons Investigator Award \#620869.
DS was supported in part by DOE grant DE-SC0021085.

redpajama_set_name: RedPajamaArXiv


text:
Ecru Royalty Outside Envelope
100 envelopes
/ box
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The tradition of inner and outer envelopes dates back to the time of white gloved butler and muddy streets. The outer envelope was to protect the inner envelope and its contents for clean delivery within the home. This is the stylish way to deliver an invitation for your special event. The address goes on the outer envelope.
Light weight paper
Pearl White Royalty Outside Envelope
Crane is widely considered the standard bearer of impeccable stationery. For over 200 years we have earned a legacy of excellence. From the generations of Crane family members to the master artisans who continue to fulfill that legacy today, we pride ourselves on setting the standard of craftsmanship that all others aspire to. Discover more about Crane's legendary craftsmanship here.
For personalized envelopes, the sender's name and address appear in the top left corner of the front, or on the back flap

redpajama_set_name: RedPajamaCommonCrawl


The Breg compact X2K OA knee brace offers people under 5'6" the same support and comfort as the longer Breg X2K knee brace models, with a custom fit just for them. Active men and women who need extra stability and support for ACL, PCL and other knee ligament problems appreciate the hard-frame construction without bulk and the internal strap design for more comfort and a secure fit.
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Pasta salads are perfect for picnics and barbecues. This one has been upgraded with wholemeal pasta and a generous quantity of vegetables. Reduced-fat mayo and low-fat yogurt make a creamy dressing with a fraction of the saturated fat of typical mayonnaise dressings.
Bring a large pot of water to the boil. Cook pasta according to the packet directions. Drain and refresh under cold running water.
Whisk mayonnaise, yogurt, oil, vinegar, garlic and pepper in a large bowl until smooth. Add the pasta and toss to coat. Add the tomatoes, capsicum, carrot, spring onions, olives and basil. Toss to coat well. The salad will keep, covered, in the refrigerator for up to 1 day.


text:
Pennsylvania Becomes Ground Zero for Trump Referendum
By: Brent Sailhamer, Director of Government Affairs, ABC Keystone
Historically, midterm elections have not gone well for sitting Presidents. Only two years into a term, Presidents have lost seats in the U.S. House of Representatives in 9 out of the last 10 elections. In the Senate, the party of the President has given up seats in 6 out of the last 10. While critics point to sagging poll numbers and an out-of-control, bombastic tempo from President Trump, the 2018 election could buck trends.
Despite recent unpopularity among organized polling, Trump remains popular among his base. So popular, in fact, that 70% of supporters claim that President Trump could do "literally nothing" to lose their support. Because Trump won in places that were unexpected, that voracious support could present a problem to Democrats looking to push back on an aggressive agenda this year. While Republicans control the 100-seat Senate by a 52-member majority, Democrats will be defending 25 of their current seats in 2018, including U.S. Senator Bob Casey. Trump shook the pundits when he picked up Pennsylvania in 2016, a feat that no one saw coming. Despite solid, historic Democrat majorities in Pennsylvania and states like West Virginia, North Dakota, and Montana, Trump won all of those states handily, making Senate elections in each of those states less certain for Democrat incumbents. In 2018, Casey will likely face Congressman Lou Barletta, who has remained a steadfast Trump loyalist since day one. Not only will that loyalty pay off in heavy support from the White House and Trump allies, but Barletta is banking on it paying off among voters as well.
With his candidacy for U.S. Senate, Barletta also gives up his seat in Congress. The 11th Congressional district, which stretches from Hazleton to Shippensburg – more than 240 miles – will be up for grabs. The seat is mildly favored for Republicans by voter registration, but is heavily weighted toward the northern end of the district. Current state House member Stephen Bloom was early to announce his candidacy for the 11th Congressional seat, but will face stiff opposition in former Revenue Secretary Dan Meuser, who hails from northeast Pennsylvania, and Andrew Lewis, who narrowly lost in the 2016 Republican primary to state Senator John DiSanto.
Making things more complex is the recent nomination of U.S. Congressman Tom Marino as the head of the Office of National Drug Control Policy (ONDCP). While Marino was nominated earlier in the year for the same position, he backed out shortly after his name was floated, citing personal health concerns. If Marino is cleared by the Senate for the position, the 10th Congressional district, which stretches from the New York border in Pike County to western Perry County, will also be up for grabs. Marino was also a strong supporter of Trump throughout the 2016 campaign, and because of their strong cooperative approach in campaigning Pennsylvania for Trump, Barletta and Marino earned the nickname "Thunder and Lightning."
Advocacy, Government Affairs, Midterm Election


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"It's PredPol, and it's going to reduce crime": Agencies take algorithmic effectiveness on faith, with few checks in place
Out of over 50 agencies surveyed, none indicated they had examined effectiveness of "data-driven" approach
Written by Beryl Lipton
Edited by Michael Morisy
Predictive policing software is being used by law enforcement agencies nationwide, yet a review suggests that almost none of its users, past or present, have clear ways to measure the effectiveness or accuracy of the tool, a lack of oversight many researchers consider irresponsible.
"I remember the only heads up I really got about PredPol was from my chief coming to our briefing and telling us, 'We have this great new tool. We're going to support it. It's PredPol, and it's going to reduce crime,'" said Captain Brian Bubar of the San Pablo Police Department, whose 2.4 square-mile California town has fewer than 40,000 people.
Bubar was a San Pablo patrol officer when the department acquired the program in 2013; the department's subscription ended on July 31, 2015. PredPol, one of the first predictive systems, was created by Los Angeles Police Department and University of California, Los Angeles, and it has been used in dozens of departments now, employing law enforcement data from the previous days and months to generate recommended areas for patrolling.
MuckRock has submitted requests to more than 50 agencies known to have used the technology, asking for information on their predictive policing training and use. Of those that have responded, a few have been able to send their contracts, input data, scientific papers written by PredPol's makers, annual mayoral presentations, and even some training materials, but none have been able to provide validation studies.
"I was never able to explain how it was identifying these boxes, how it was telling us to get to these points, which I think was a huge component [of its failure], because it did not give an opportunity to give the system validity to the officers," said Bubar. "And that could happen in a department our size. At the time, we didn't have a dedicated IT infrastructure. We didn't have a committee to learn how to incorporate our training staff, incorporate the sergeants, to educate them on these new systems, who were able to get the buy-in from the staff to give it a chance. It was just kind of delivered in our laps, and we were told to use it."
Though it is sold as "crime prediction" and officer allocation tools for under-resourced agencies, one of the reasons predictive policing tools are so controversial is the contention that they are built on and reinforce racial bias. Using data generated by unfair or questionable policing strategies to train a computer system can simply result in automation of that bias.
"You have to understand that every bias that goes into an instrument's development is going to be manifest in the tool's ability. It's a great concept, but it doesn't recognize that human behavior is negatively impacted when we use algorithms to predict it. You look at the research: blacks are twice as likely as whites to be targeted for certain behaviors when those behaviors are evenly distributed," said Howard Henderson, founding director of the Center for Justice Research at Texas Southern University. "You can't adopt an instrument without validating. That's unacceptable."
A report from the AI Now Institute found that some police departments cited by the Department of Justice for unfair policing practices use data from those eras to inform predictive policing systems.
"These tools are not sufficiently well-tested. They're biased premises that, in some cases, are faulty. They're based on a process of learning that I think are misapplied. And in the case of person-based predictive policing, they're based on assumptions and structures that we've already seen have huge numbers of problems. I don't see any reason to keep using them until we know, what is it exactly that they're trying to do," said Suresh Venkatasubramanian, professor at the University of Utah.
Many academics argue that, as they're currently built and used, predictive policing systems shouldn't be employed by police at all. In early October, more than 400 academics signed onto a letter to the LAPD Commission challenging claims that scholarship supports use of the tool.
"We keep telling ourselves, 'We're not stupid. We know correlation does not equal causation.' But we're going to use it anyway," Venkatasubramanian said. "Why are you going to use it anyway?"
In addition to the concerns around encoding biased policies into police equipment, local validation and accuracy checks are important for seeing whether a system can realistically help an agency. Multiple police departments have said they stopped using PredPol because it simply did not work for them, suggesting areas they already knew to be problematic or not offering suggestions that could work with the existing demand created by daily urgent needs for service.
"It became almost offensive to patrol officers when the city commits funding to a new analytical tool that's supposed to reduce crime. We need to be out in high visibility," Bubar explained. "These were things that we were already identifying."
Changes to standard procedure
LAPD, one of the longest users of PredPol, announced only last month that it would begin measuring the effectiveness of its data-driven policing techniques—nearly eight years after it started using them. An April report from the LAPD's Inspector General found the department needs a lot of improvement around how it captures data and evaluates the fairness of the tool's application. At a meeting in mid-October of the Los Angeles Police Commission, representatives from the LAPD acknowledged the shortcomings.
To address them, the department will be creating a data-driven policing unit and a reference manual, scheduled for release in January, detailing how it is used.
Dozens of other departments still use similar technology, but community pressure is starting to limit deployments.
"I think that's where the change has occurred. LA and Philly are places where you can see good examples of this, but that's because of the enterprising nature of the local community that they were able to do this," said Venkatasubramanian. "To expect or require communities to be the source of pushback is unfair, although, yes, that's probably the only way it's going to happen."
One point at which oversight might be implemented is during the initial procurement or subsequent justification meetings with city councils.
The use of a tool making predictions or decisions about force deployment is akin to implementing a policy stance, says Deirdre K. Mulligan, faculty director of the Berkeley Center for Law and Technology, and as with any policy enacted by law enforcement, there needs to be thought and intention behind its adoption.
"When it comes to predicting crime the underlying data is so fraught that it seems it can't do anything but play racism forward," wrote Mulligan. "I've been focused on all the policies that are embedded in the 'tools,' and the need for agencies to understand that adopting a tool may be akin to adopting a policy and, therefore, requires expertise, reasoned decision making, and public participation. How do we make sure the public is informed and in control of key normative decisions when processes are offloaded to an algorithm?"
Image by Chris Yarzab licensed under CC-BY-2.0
Algorithmic Control is part of a joint research and reporting project from MuckRock and the Rutgers Institute for Information Policy and Law. Support for this project is provided by Rutgers, but the reporting is editorially independent. View the full database of requests, learn more about the project, or get in touch with the reporter.
Algorithmic Control by MuckRock Foundation is licensed under a Creative Commons Attribution 4.0 International License.
Based on a work at https://www.muckrock.com/project/algorithmic-control-automated-decisionmaking-in-americas-cities-84/.
Algorithmic Control: Automated Decisionmaking in America's Cities
police oversight
predictive policing
predpol
validation study
With task force finished, New York City looks to next steps for AI regulation
by Beryl Lipton
The U.S. needs to "get AI right" — and fast — says government group
In case after case, courts reshape the rules around AI
As facial recognition begins to face regulatory scrutiny, industry and privacy advocates spar over who owns your data


Monitoring disease activity in individuals with multiple sclerosis, either to predict flare-ups or to check treatment response, might be done with a simple blood test that measures levels of a nerve protein, according to a new study from Norway.
A simple, low-cost blood test could offer a less invasive alternative to conventional MS monitoring methods.
The findings, which have now been published in the journal Neurology: Neuroimmunology & Neuroinflammation, are the work of Dr. Kristin N. Varhaug, of the University of Bergen in Norway, and her colleagues.
They came to their conclusions after studying more than 80 people with relapsing-remitting multiple sclerosis and comparing the results of blood tests with results of MRI scans and other assessments.
"Since MS varies so much from person to person," says Dr. Varhaug, "and is so unpredictable in how the disease will progress and how people will respond to treatment, identifying a biomarker like this that can help us make predictions would be very helpful."
MS develops when the immune system attacks healthy tissue in the central nervous system, which is a part of the body that includes the brain, spinal cord, and optic nerves.
In MS, immune attacks destroy the myelin sheath that protects the nerves, resulting in disruption to the electrical signals that they carry to and from the brain and the rest of the body.
Depending on which parts of the central nervous system come under attack, the symptoms can vary from person to person and vary at different times in the same person. In relapsing-remitting MS, the symptoms flare up unexpectedly and then go away. In progressive MS, they remain and gradually get worse.
Symptoms of MS may include: blurred or double vision; blindness; fatigue; muscle weakness; problems with balance and coordination; numbness and prickling sensations; dizziness; pain; tremors; speech impediments; hearing loss; paralysis; and difficulty remembering and concentrating.
Because doctors in the United States do not have to report new cases of MS to the Centers for Disease Control and Prevention (CDC), it is not easy to arrive at an accurate figure of how many people may be living with the disease nationwide.
Some estimates, based on research that was done before MRI scans were introduced to diagnose MS, suggest that there might be around 2.5 million people worldwide with MS, including an estimated 300,000–400,000 in the U.S.
Preliminary findings of a recent study, however, suggest that the numbers are much higher, instead proposing that there are around 1 million people in the U.S. living with MS.
The new study looks at a nerve protein called "neurofilament light chain" (NFL), which is shed into the cerebrospinal fluid that surrounds the brain and spinal cord when nerve cells and fibers are damaged.
Higher levels of NFL are found in cerebrospinal fluid in various diseases that damage nerve cells and fibers, including MS, Alzheimer's disease, and motor neuron disease.
Learn about two ways in which white blood cells overcome the blood-brain barrier to cause nerve damage in MS.
It is, however, possible to detect even low concentrations of NFL in the bloodstream. Therefore, Dr. Varhaug and her colleagues wanted to investigate whether using blood levels of NFL might offer a suitable clinical way to monitor disease activity in relapsing-remitting MS.
Testing for an MS biomarker in blood samples would be preferable to taking samples of spinal fluid "as lumbar puncture is invasive and not appropriate for repetitive sampling and long-time follow-up," they note.
They recruited 85 people with the relapsing-remitting form of MS and followed them for 2 years. During this time, they did not receive any treatment for 6 months and then received 18 months of treatment with interferon-beta 1a, which is given to MS patients to reduce flare-ups and slow buildup of brain lesions.
Over the 2 years of the study, the subjects also underwent various assessments of their condition, including disability status (scored at baseline and then every 6 months) and MRI scans (at baseline and then every month for 9 months, and then at the end of the first and second years).
The researchers collected blood samples from the participants at baseline and after 3, 6, 12, and 24 months. They measured levels of NFL in the samples and then used various statistical tools to compare them with the other assessments of disease activity.
The results showed that blood NFL levels were higher when the MRI scans showed new areas of damage in the brain known as T1 and T2 lesions.
The blood level of NFL was 37.3 picograms per milliliter in people with new T1 lesions compared with only 28 picograms per milliliter in people without new T1 lesions.
People with new T2 lesions had 37.3 picograms per milliliter of NFL in their blood compared with 27.7 picograms per milliliter in those without new T2 lesions.
The raised levels of NFL persisted for 3 months as new lesions developed. They then fell once the subjects began their treatment with interferon-beta 1a.
The statistical analysis calculated that for every 10-picogram per milliliter increase in blood NFL, there was a 48 percent raised risk of a new T1 and a 62 percent raised risk of a new T2 lesion.
The team also checked whether or not another protein, called chitinase 3-like 1 (CHI3L1), might also serve as a blood biomarker for MS. Raised levels of CHI3L1 have been found in several inflammatory conditions and studies have proposed it as a biomarker for MS.
However, in this study, the results showed that changes in "CHI3L1 were not associated with clinical or MRI disease activity or interferon-beta 1a treatment."
Summarizing the findings, Dr. Varhaug says that they found that levels of NFL "were higher when people had new disease activity and lower when they took medication to reduce the number of symptom flare-ups."
"These blood tests could provide a low-cost alternative to MRI for monitoring disease activity."


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THE LATE HENRY "TATANG" SY, SR. AND HIS EMPIRE
Henry "Tatang" Sy, Sr. was considered the richest Filipino, the King of Retail, and an iconic business figure in the Philippines.
Tatang, as Henry Sy was fondly called, was a migrant from Amoy (now Xiamen) in China.
Henry Sy's social ascent story is awe-inspiring and rousing. His vision, mission, determination, persistence, hardwork, and benevolence brought him fortune and fame and his famous empire.
Photo by Google
Henry Sy was naturally born with an entrepreneurial spirit. He dreamt big but worked hard.
After World War II, he established small shoe outlets in Manila. Then, he tied up with Senen Mendiola, a commerce graduate of San Beda College, who came from a political clan from Mindoro, whom Henry Sy supposedly derived the initials, SM.
Nonetheless, Sy still finished high school education in now Chiang Kai Shek College. Likewise, he pursued his Associate of Arts (AA) degree in commercial studies at Far Eastern University.
Henry Sy built his first Shoemart along with Senen Mendiola and Lao Kang at the corner of Avenida Rizal and Plaza Goiti.
During that time, Escolta was the hub of shopping along with Good Earth Emporium which was well known in Manila previously for having the first escalator, which was very uncommon at that time.
Later on, Henry Sy's built his second shoe store, the Makati Commercial Center, known today as SM Makati, situated beside the famous Hotel Intercontinental.
Since the establishment of Makati Commercial Center (now SM Makati) in 1975, SM has become a household name.
Sy's future plans of building other stores in Mandaluyong and Quezon City were shared and told to then business editor and now Manila Standard contributor and columnist, Emil P. Jurado, who was then a resident of Philam Homes, at Elysee Coffee Shop, after asking the former of his business plans.
"If I can sell a pair of shoes (for) every person, then I consider myself successful." — Henry Sy
Subsequently, Sy set his sights on acquiring a tract of land in a swampy location filled with watercress on North Avenue in Quezon, where SM North EDSA, the first SM mall, is presently located.
Despite the economic crisis and high-interest rates the Philippines experienced during that time, Sy remained unfazed acquiring that piece of land. He rationalized and compared the strategic area from a fishnet catching people from the North and South. Further, he visualized the location as a future regional hub.
Thus, SM North EDSA came to existence in 1985, becoming the first SM Supermall.
SM City North EDSA in the 1980s
It was Henry Sy's dream of allowing the Filipinos to own a pair of shoes and establishing SM in every corner of the Philippines—the idea is apparently materializing these days—not to mention Savemore and Hypermarket.
Footbridge on North Avenue in Quezon City in front of SM North EDSA in 2008
In 1993, the SM patriarch founded the SM Foundation Scholarship program, whose scholars are entitled to free tuition fees, matriculation, monthly allowance, and part-time jobs. Now, each scholar is given a personalized wristwatch with their names inscribed onto it on their graduation.
As of 2019, Henry Sy's SM has approximately 79 branches nationwide and 6 branches in China and is still counting.
As time went on, Henry Sy diversified his company. Later, Sy endeavored in other profitable investments like Realty, Banking, Hotel, Gaming, Tourism and most importantly Education—as Chairman of Pacific Colleges and National University—which he envisioned for the Filipinos as early as he got into a shoe store business.
Recently, SM Investments chairman emeritus Henry Sy Sr., who has a net worth of 20 Billion dollars, was ranked 52nd wealthiest people in the world by Forbes Magazine.
GENERAL INFORMATION, HISTORY
Henry Sy Sr., Senen Mendiola, Shoemart, SM, SM Supermalls
REMINISCENT OF MANNY PACQUIAO'S ACHIEVEMENTS
KUYA RAYMOND's KAMBINGAN HOUSE
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In Reception, we follow the 'Statutory Framework for the Early Years Foundation Stage'. We provide fun and enjoyable learning experiences, through lots of play-based learning both indoors and out. We provide children with daily focused activities aimed to develop understanding and skills in reading (RWI), writing and mathematics. All our activities and learning experiences are aimed to develop independent and confident learners.
We aim to develop children's curiosity by encouraging them to use their senses to explore the world around them. We encourage the children to be playful, and be willing to have a go and take risks by engaging in new experiences.
Children will be encouraged to develop their concentration and attention. Children will be encourage to persist when they find something tricky, and take pride when they achieve their goals.
In the early years we aim to provide children with problems to solve, and encourage them to find new ways of doing things. Children are supported in giving predictions and testing their ideas, as well as teaching them different methods to approach problems with.
-Learning about relationships with both adults and children, making friendships and learning to share, to take turns and to negotiate with others.
-Be able to talk about what we want and how we feel and will have opportunities to express this through role play.
-Be able to enjoy stories, poems, songs and rhymes, and will develop the ability to use language to describe past and present experiences and a sequence of events.
-Become good at moving, climbing, controlling and coordinating as we play.
-Learn to use our body, feet and hands with more and more skill.
-Recognise that words tell stories and that the words we see in the world around us tell people something.
-See letters and words written down and to try and write them and use them when we play.
-Learn the alphabetical sounds and through a mixture of speaking and listening.
-Through Letters and Sounds in Nursery, and Read Write Inc. in Reception, children will become familiar with early reading and writing.
-Sort and group objects or materials and explain why we have done this.
-Play with lots of resources indoors and outdoors and make comparisons in things like height, weight, size and texture.
-Use words like under, on top, between and in front to tell where things are and to put things in order when we play.
-Have opportunities to build and make things with shapes and talk about what these shapes look like.
-We will be able to look at and talk about objects, materials, living creatures and plant life, noticing lots of details and things that are the same or different.
-Be able to take part in explorations and investigations at all different times of the year in the classroom and in the outdoor environment giving us the chance to notice changes as time passes.
-Make predictions, try things out and come to some conclusions and even give some reasons as to why we think things happen.
-Express our own thoughts, ideas and feelings through drawing, painting, model making and by using clay.
-Use natural and manmade materials to design and make our own creations, models and structures.
All of our work in these areas will be documented by photographs, snapshot observations, mark making and other pieces of evidence in our Learning Journeys.
As the children come to the end of their time in the Early Years, they will be supported with their transition into Year 1 to ensure the best possible start in Key Stage 1.