# Three Dimensional Black Hole in the Low Energy Heterotic String Theory

Byon N. Jayawiguna<EMAIL_ADDRESS>Departemen Fisika, FMIPA,
Universitas Indonesia, Depok 16424, Indonesia

###### Abstract

We study the BTZ black holes (2+1 dimensional space-time) in the low energy
heterotic string theory (BTZ-Sen BH). This concept requires us to include a
non-trivial dilaton $\phi$ and a 3-form $H_{\mu\nu\rho}$ field. By using the
Hassan-Sen transformation and BTZ black hole as a seed solution, we obtain the
solution in the string frame. Some properties of the black hole solutions are
discussed.

###### pacs:

97.60.Lf, 04.20.Jb

##  I Introduction

Banados, Teitelboim, and Zanelli (BTZ) in 1992 were the first to obtain exact
solution of a vacuum rotating black hole in three-dimensional space-time with
non-zero cosmological constant Banados:1992wn ; Banados:1992gq . This proposal
also supports the existence of three-dimensional space-time in general
relativity as the foundation of the classical and quantum gravity aspects that
have been conceived by Deser Deser:1983tn , Jackiw Deser:1988qn , ’t Hooft
tHooft:1988qqn , and Witten Witten:1988hc ; Witten:1989sx  circa 1984-1989.
Because the three-dimensional space-time has no Newtonian limit, BTZ black
holes are different from Kerr black holes Kerr:1963ud . The fundamental
difference is that the BTZ black holes are asymptotically anti-de-Sitter and
do not have a curvature singularity, whereas the Kerr black holes are
asymptotically flat and have a ring singularity. Nevertheless, both objects
are still black holes; have the inner horizon as well as the outer horizon,
ergopshere, static limit region, and frame dragging effects.

From the vacuum BTZ case, it took eight years for Martinez, Teitelboim, and
Zanelli to obtain the charged and rotating black holes in three-dimensional
space-time Martinez:1999qi . The method used are the same with BTZ. They do
not neglect the boundary action, with the consequence that the scalar Ricci
evaluated on the hypersurface $\Sigma$ must be taken into account. With the
Hamiltonian formulation and $(2+1)$ decomposition, they choose stationary
conditions and axial symmetry for the consequences of rotating object. Under
these conditions, the lapse and shift functions, $N^{\perp}$ and $N^{i}$,
could be obtained using the Hamiltonian and momentum constraints
$\mathcal{H}_{\perp}=0$ and $\mathcal{H}_{i}=0$.

Although the BTZ black hole solution is a rotating object, it is without doubt
that there is a vast literature on the static BTZ black hole $(\Lambda\neq
0)$. BTZ with radial and azimuthal electric fields has been studied by the
authors in Martinez:1999qi ; Cataldo:2002fh ; Hirschmann:1995he  while BTZ
with pure magnetic solutions has also been studied in Koikawa:1997am ;
Cataldo:2000ns . Interestingly, studies on static BTZ black holes have been
widely discussed in the case of nonlinear electrodynamics as well as the
phenomenological aspects Myung:2008kd ; Hendi:2017oka ; Dehghani:2019noj ;
Fernando:2017qrq ; Balart:2019uok ; Fernando:2019cez ; Gurtug:2018bxu .

In the realm of low energy string theory, Sen Sen:1992ua  obtained an exact
four-dimensional, rotating and charged black hole solution, commonly referred
to as a Kerr-Sen black hole. This solution is obtained by the Hassan-Sen
transformation Hassan:1991mq  which requires a seed solution (in the
literature, Sen uses a Kerr black hole solution) that satisfies the stationary
and axial symmetry conditions. The black hole solution obtained by Sen is
analogous to the Kerr-Newman black hole (rotating charged black hole in four
dimensional space-time). Despite the similarities between the Kerr-Sen and
Kerr-Newman black hole solutions, there are several features that could
distinguish between the two solutions. For example, the appearence of dilaton
scalar and 3-form tensor field on Kerr-Sen. It can be inferred that the Kerr-
Sen and Kerr-Newman black hole is the same but different families. On the
other hand, a study on the merger estimates Siahaan:2019oik  and the energy
requirements needed to destroy the horizon of the Kerr-Sen black hole
Siahaan:2015ljs  shows that the study of Kerr-Sen black holes is still of
particular interest for researchers. Moreover, since the black hole in string
theory is consistent with the descriptions of classical and quantum gravity, a
vast literature on AdS/CFT correspondence to the Kerr-Sen black hole are
discussed in the reference Ghezelbash:2012qn ; Siahaan:2015xia ;
Siahaan:2018wvh . Recently, accelerating black hole in the low energy string
theory and several aspects such as horizons, conical singularities, angular
velocity, and temperatures have been obtained Siahaan:2018qcw .

For the case of static black holes in string theory, Gibbons and Maeda have
obtained black holes in extra dimensions Gibbons:1987ps  (also Horowitz and
Strominger Horowitz:1991cd ) while in the four dimension Garfinkle, Horowitz,
and Strominger have studied the charged black hole in string theory (Einstein-
Maxwell-Dilaton) Garfinkle:1990qj . On the other hand, the Kerr-Sen black hole
is parameterized with a mass $M$, charge $Q$, and angular momentum $J$. If we
set $Q=0$, the solution will be reduced to Kerr black hole solution. If we
switch off the contribution of the rotation parameter $a$ followed by the
transformation of $r\rightarrow r-\frac{Q^{2}}{M}$ Siahaan:2014ihe , we can
find that the solution is nothing but a static charged black hole solution in
string theory Garfinkle:1990qj . It can be inferred that in order to arrange
the static and charged string object, a coordinate transformation must be
performed.

In this paper, we devote our effort to constructing what Sen had already
obtained on Kerr-Sen black hole to the BTZ black hole object. The output of
this paper is to obtain the BTZ black hole in the low energy heterotic string
theory. We also include the contribution of cosmological constant
$\Lambda=-\frac{1}{l^{2}}$ since the BTZ black hole will exist if it has a
negative cosmological constant (anti-de-Sitter spacetime). BTZ black holes in
both rotating and static in low energy string field have been studied
extensively in some literature Horowitz:1993jc ; Horowitz:1994ei ; Chan:1994qa
; Chan:1995wj ; Chan:1996rd ; Sa:1995vs ; Garcia-Diaz:2017cpv ;
Aniceto:2017gtx ; Fernando:2009tv ; Fernando:2004ay ; Fernando:2003ai .
However, construction of a rotating, charged black hole with a complete low
energy string term $(R,~{}\Lambda,~{}F_{\mu\nu},~{}\phi,~{}H_{\mu\nu\rho})$
has never been studied.

The paper is structured as follows. In section II we will do a quick review on
BTZ vacuum black hole. This is used because the incoming seed solution for
Hassan-Sen transformation is a BTZ black hole solution. This section will be
filled by the metric solution, event horizons, and static limit radius. In
section III, we construct the BTZ black hole on the low energy heterotic
strings (BTZ-Sen). From the solution we obtained in the string frame, we
transform conformally into Einstein frame. After we obtain the solution, we
also discuss in detail this solution to study the geometrical aspects of the
BTZ-Sen black hole (event horizon, static limit radius, and ergosphere) and
the test particles inside and outside of the black hole. The angular velocity
of the particle and black hole perimeter are also discussed.

##  II BTZ Black Hole: A Review

Three dimensional (2 + 1) black holes or commonly known as BTZ black holes
were first obtained in 1992 Banados:1992gq ; Banados:1992wn  (see also
Carlip:1995qv ). This black hole is quite easy to imagine compared to the
black hole that we are familiar with (Schwarzschild, Kerr, etc.) since the BTZ
black hole has a lower dimension than the ordinary four dimension (3+1). For
more details on the concept above, the author begins with the action form of
the BTZ black hole. The BTZ black hole action is given by

$I=\frac{1}{16\pi G}\int\sqrt{-g}\left(R+2l^{-2}\right)d^{3}x,$ (1)

from the action above, the symbol $g$ is stands for determinant of the metric
tensor $g_{\mu\nu}$ whereas $R$ and $l$ is the Ricci scalar and the radius
associated with the cosmological constant $-\Lambda=l^{-2}$, respectively. The
corresponding Einstein’s equation is

$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\Lambda g_{\mu\nu}=0.$ (2)

The equation above can be made in the nice way by contracting it with
$g^{\mu\nu}$. Then we found the equation to be

$R_{\mu\nu}=2\Lambda g_{\mu\nu}.$ (3)

In general, if $T_{\mu\nu}\neq 0$, the Einstein field equation can be written
as

$R_{\mu\nu}=2\Lambda g_{\mu\nu}+8\pi G(T_{\mu\nu}-g_{\mu\nu}T),$ (4)

In addition, the selection of the momentum energy tensor $T_{\mu\nu}$
completely determines the Ricci $R_{\mu\nu}$ tensor, but generally does not
determine the Riemann tensor $R_{\mu\nu\alpha\sigma}$. However, in the
3-dimensional gravity concept, the traceless part of $R_{\mu\nu\alpha\sigma}$
vanishes so that the Riemann tensor only depends linearly with its Ricci
scalar and Ricci tensors

$\displaystyle R_{\mu\nu\gamma\sigma}$ $\displaystyle=$ $\displaystyle
g_{\mu\gamma}R_{\nu\sigma}+g_{\nu\sigma}R_{\mu\gamma}-g_{\nu\gamma}R_{\mu\sigma}-g_{\mu\sigma}R_{\nu\gamma}$
(5)
$\displaystyle-\frac{1}{2}(g_{\mu\gamma}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\gamma})R.$

By inserting Eq. 4, we get a relationship between the Riemann tensor and the
momentum energy tensor. For vacuum cases $(T_{\mu\nu}=0)$, the Riemann tensor
and Ricci scalar can be written as
$R_{\mu\nu\gamma\sigma}=(g_{\mu\gamma}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\gamma})\Lambda$,
and $R=6\Lambda$. Eq. 3 above can be solved to obtain the black hole solution
with ansatz

$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}=-N^{2}dt^{2}+N^{-2}dr^{2}+r^{2}\left(N^{\phi}dt+d\phi\right)^{2},$
(6)

Where $N(r)$ and $N^{\phi}(r)$ are the lapse and the angular shift function,
respectively. The solution is written as

$\displaystyle
N(r)=-M+\frac{r^{2}}{l^{2}}+\frac{J^{2}}{4r^{2}},~{}N^{\phi}(r)=-\frac{J}{2r^{2}},$
(7)

with $-\infty<t<\infty$, $~{}0<r<\infty$, and $0\leq\phi\leq 2\pi$. The metric
solution could be written in a form

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-\left(\frac{r^{2}}{l^{2}}-M\right)dt^{2}-Jdtd\phi+\left(\frac{r^{2}}{l^{2}}-M+\frac{J^{2}}{4r^{2}}\right)^{-1}dr^{2}$
(8) $\displaystyle+r^{2}d\phi^{2}.$

The two constants $M$ and $J$ contained in Eq. (7) are the consequences of
stationary conditions (invariant with respect to time) and axial symmetry
(invariant with respect to rotation), mass and angular momentum. In addition,
the metric $g^{rr}$ will be zero if its obey the value of $r$ as follows

$r_{\pm}=l\sqrt{\frac{M}{2}\left[1\pm\sqrt{1-\left(\frac{J}{Ml}\right)^{2}}\right]}.$
(9)

The value of $r$ indicates that there are two event horizon in the case of the
BTZ black hole. The three variables $l$, $M$, and $J$ play a major role in
this case. To keep the event horizon on this BTZ black hole exist, the
variables $l$, $M$, and $J$ must satisfy the relation below

$\displaystyle M>0,~{}|J|\leq Ml.$ (10)

In extreme conditions $|J|=Ml$, event horizon $r_{+}$ $r_{-}$ merges into one
with a radius $r_{e}=l\sqrt{\frac{M}{2}}$. Another aspect that can be obtained
is the emergence of ergosphere

$r_{sl}=l\sqrt{M}.$ (11)

With all these $r$ values obtained, it can be said that the characteristics of
BTZ black holes are almost the same as those of Kerr black holes, $r_{+}$ is
an event horizon, $r_{sl}$ is a static limit, and the range of
$r_{+}<r<r_{sl}$ is the ergosphere area.

##  III BTZ Black Hole in the low energy Heterotic String Theory

Black hole in string field (such as black p-brane) certainly has different
characteristics compared to black holes that often appear in Einstein’s theory
of gravity. Most of these solutions contain more than one charge such as the
Yang-Mills field or the antisymmetric tensor field, and the dilaton scalar
field. When these charges are switched off, the solution will be reduced to an
ordinary Schwarzschild solution. In this paper, we construct the classical and
exact solution of the BTZ black hole in the low energy string that describe
black hole carrying some charge and angular momentum. The method used to
obtain this solution is the twisting procedure which requires a stationary and
axial symmetry solution as an input. This twisting procedure which will later
be referred to as Hassan-Sen transformation. The explanation above starts from
the action of string theory in the 3d string frame

$S=\int
d^{3}x\sqrt{-g}~{}e^{-\Phi}\left(\mathcal{R}-2\Lambda+(\nabla\Phi)^{2}-\frac{F^{2}}{8}-\frac{H^{2}}{12}\right),$
(12)

where $g$ is a determinant of metric tensor $g_{\mu\nu}$, $R$ is a Ricci
scalar, $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is a Maxwell
field, $\phi$ as a scalar dilaton, and

$H_{\kappa\mu\nu}=\partial_{\kappa}B_{\mu\nu}+cyclic~{}permutation-\left[\Omega_{3}(A)\right]_{\kappa\mu\nu},$
(13)

where $B_{\mu\nu}$ is antisymmetry tensor field, and

$\left[\Omega_{3}(A)\right]_{\kappa\mu\nu}=\frac{1}{4}\left(A_{\kappa}F_{\mu\nu}~{}+~{}cyclic~{}permutation\right)$
(14)

is a Chern-Simons term. The $3$-form field tensor $H_{\kappa\mu\nu}$ above can
be made in a compact form into

$\displaystyle H_{\kappa\mu\nu}$ $\displaystyle=$
$\displaystyle\partial_{\kappa}B_{\mu\nu}+\partial_{\nu}B_{\kappa\mu}+\partial_{\mu}B_{\nu\kappa}$
(15)
$\displaystyle-\frac{1}{4}\left(A_{\kappa}F_{\mu\nu}+A_{\nu}F_{\kappa\mu}+A_{\mu}F_{\nu\kappa}\right).$

One can obtain the equation of motion by varying the action (12) with respect
to $\\{g_{\mu\nu},~{}\Phi,~{}A_{\mu},~{}B_{\mu\nu}\\}$

$\displaystyle R_{\mu\nu}$ $\displaystyle=$ $\displaystyle
g_{\mu\nu}\left[2\Lambda-\nabla^{2}\Phi+(\nabla\Phi)^{2}-\frac{1}{8}F^{2}-\frac{1}{6}H^{2}\right]$
(16)
$\displaystyle-\nabla_{\mu}\Phi\nabla_{\nu}\Phi+\frac{1}{4}g^{\lambda\rho}F_{\mu\lambda}F_{\nu\rho}+\frac{1}{4}H_{\mu\kappa\sigma}H_{\nu}^{\kappa\sigma},$
$\displaystyle\nabla^{2}\Phi-(\nabla\Phi)^{2}-2\Lambda+\frac{F^{2}}{8}+\frac{H^{2}}{6}=0,$
$\displaystyle\nabla_{\mu}\left(e^{-\Phi}F^{\mu\nu}\right)=\frac{1}{2}F_{\alpha\beta}H^{\nu\alpha\beta},$
(17)

and

$\nabla_{\mu}\left(e^{-\Phi}H^{\kappa\mu\nu}\right)=0.$ (18)

It is shown that the ordinary BTZ equation in (3) can be obtained by switched
off the non-gravitational field. Thus, taking the BTZ black hole as a seed
solutions are allowed in string frame. However, rather than considering the
solutions only in the string frame, we will make a conformal transformation to
the Einstein frame. We therefore define the conformal transformation

$ds_{E}^{2}=e^{-\Phi}\tilde{ds}^{2},$ (19)

The solution in the Einstein frame will be used when we discuss some aspects
BTZ black holes in heterotic string theory. Hassan and Sen show that the new
field
$\\{g^{\prime}_{\mu\nu},~{}B^{\prime}_{\mu\nu},~{}A^{\prime}_{\mu},~{}\phi^{\prime}\\}$
will satisfies the equation of motion of 12 if the relationship of the new
field with the old is

$\displaystyle\mathcal{M}^{\prime}=\Omega\mathcal{M}\Omega^{T},~{}\Phi^{\prime}=\Phi+\ln\sqrt{\frac{g^{\prime}}{g}}.$
(20)

The definition for matrix $\mathcal{M}$ and $\Omega$ in the (20) are as
follows

$\footnotesize\mathcal{M}=\begin{pmatrix}(K^{T}-\eta)g^{-1}(K-\eta)&(K^{T}-\eta)g^{-1}(K+\eta)&-(K^{T}-\eta)g^{-1}A\\\
(K^{T}+\eta)g^{-1}(K-\eta)&(K^{T}+\eta)g^{-1}(K+\eta)&-(K^{T}+\eta)g^{-1}A\\\
-A^{T}g^{-1}(K-\eta)&-A^{T}g^{-1}(K+\eta)&A^{T}g^{-1}A\end{pmatrix},$ (21)

and

$\Omega=\begin{pmatrix}I_{5x5}&...&...\\\ ...&\cosh\alpha&\sinh\alpha\\\
...&\sinh\alpha&\cosh\alpha\end{pmatrix}.$ (22)

The dot sign in the matrix $\Omega$ indicate that the contents of the matrix
are zero whereas $I_{5x5}$ is an identity matrix 5x5. The definition inside
the matrix $M$ could be written as

$\displaystyle K_{\mu\nu}=\begin{pmatrix}K_{rr}&K_{r\phi}&K_{rt}\\\ K_{\phi
r}&K_{\phi\phi}&K_{\phi t}\\\
K_{tr}&K_{t\phi}&K_{tt}\end{pmatrix},~{}\eta_{\mu\nu}=\begin{pmatrix}1&0&0\\\
0&1&0\\\ 0&0&-1\\\ \end{pmatrix},$ (23)

and

$\displaystyle g^{\mu\nu}=\begin{pmatrix}g^{rr}&g^{r\phi}&g^{rt}\\\ g^{\phi
r}&g^{\phi\phi}&g^{\phi t}\\\ g^{tr}&g^{t\phi}&g^{tt}\end{pmatrix}.$ (24)

where the matrix $K_{\mu\nu}$ and $A_{\mu}$ is given by

$\displaystyle
K_{\mu\nu}=-B_{\mu\nu}-g_{\mu\nu}-\frac{1}{4}A_{\mu}A_{\nu},~{}A_{\mu}=\begin{pmatrix}A_{r}\\\
A_{\phi}\\\ A_{t}\end{pmatrix}.$ (25)

Here, Sen takes a different definition of the matrix; the time component is
placed at the lower right end. However, the new and the old fields obtained in
the (20) equation will satisfy the equation of motion derived from the (12)
action. An explanation of the ideas from Sen is as follows. We know that the
action of low energy string theory in the three dimensions will be reduced to
the Einstein-Hilbert action when all fields
$\\{B^{\prime}_{\mu\nu},~{}A^{\prime}_{\mu},~{}\Phi^{\prime}\\}$ are zero.
That is, the vacuum solution of the Einstein field equation is also a solution
of the equation of motion derived from action (12). So, by using the
transformation (20) we will obtain a new solution with non vanisihing non
gravitational fields
$\\{B^{\prime}_{\mu\nu},~{}A^{\prime}_{\mu},~{}\Phi^{\prime}\\}$. Ansatz
metrics that obey stationary and axial symmetry conditions are as follows

$g_{\mu\nu}dx^{\mu}dx^{\nu}=g_{tt}dt^{2}+2g_{t\phi}dtd\phi+g_{rr}dr^{2}+g_{\phi\phi}d\phi^{2},$
(26)

since the BTZ black hole is a solution of the Einstein vacuum field equation
that meets the above conditions (stationary and axial symmetry), we therefore
will use this solution as an input

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-\left(\frac{r^{2}}{l^{2}}-M\right)dt^{2}-Jdtd\phi+\left(\frac{r^{2}}{l^{2}}-M+\frac{J^{2}}{4r^{2}}\right)^{-1}dr^{2}$
(27) $\displaystyle+r^{2}d\phi^{2}.$

By obtaining the nonzero matrix $\mathcal{M}$ in terms of general metric (26),
combined with the transformation in (20) and substituting into of the BTZ
black hole metric solution, we obtained the BTZ-Sen solution in string frame

$\tilde{g_{tt}}=\frac{M+r^{2}\Lambda}{\left[\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)\right]^{2}},$
(28) $\tilde{g_{rr}}=\frac{4r^{2}}{a^{2}M^{2}-4r^{2}(M+r^{2}\Lambda)},$ (29)
$\tilde{g_{t\phi}}=-\frac{aM\cosh^{2}\left(\frac{\alpha}{2}\right)}{2\left[\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)\right]^{2}},$
(30) $\displaystyle\tilde{g_{\phi\phi}}$ $\displaystyle=$
$\displaystyle\frac{\left(\chi_{1}\cosh^{2}(\alpha)+\chi_{2}\cosh(\alpha)\right)+\chi_{3}}{\left[\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)\right]^{2}},$
(31)

where

$\displaystyle\chi_{1}=4r^{2}\left(M+\Lambda
r^{2}+1\right)^{2}-a^{2}M^{2}\left(M+\Lambda r^{2}+2\right),$
$\displaystyle\chi_{2}=8r^{2}+2\left(M+\Lambda
r^{2}\right)\left[a^{2}M^{2}-4r^{2}\left(M+\Lambda r^{2}\right)\right],$
$\displaystyle\chi_{3}=4r^{2}\left(M+\Lambda
r^{2}-1\right)^{2}-a^{2}M^{2}\left(M+\Lambda r^{2}-2\right).$ (32)

After obtaining the gravitational field, the corresponding non-gravitational
field can be written in the compact form as

$\tilde{A_{\mu}dx^{\mu}}=\frac{\sinh(\alpha)\left[(1+M+\Lambda
r^{2})\tilde{dt}-\frac{aM}{2}\tilde{d\phi}\right]}{\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)},$
(33) $\tilde{B_{t\phi}}=-\tilde{B_{\phi
t}}=\frac{aM\sinh^{2}\left(\frac{\alpha}{2}\right)}{2\left[\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)\right]},$
(34)

and dilaton field

$\tilde{\Phi}=-\ln|\cosh^{2}\left(\frac{\alpha}{2}\right)+\left(M+r^{2}\Lambda\right)\sinh^{2}\left(\frac{\alpha}{2}\right)\bigg{|}.$
(35)

With the transformation in (19), we obtained a black hole solution in Einstein
frame which can be written as

$g_{tt}=\frac{(M-b)(M+r^{2}\Lambda-b)}{M+b(M-b+r^{2}\Lambda)}$ (36)
$g_{rr}=\frac{4r^{2}\left[M+b(M-b+\Lambda
r^{2})\right]}{(M-b)\left[a^{2}(M-b)^{2}+4r^{2}(b-M-r^{2}\Lambda)\right]},$
(37) $g_{t\phi}=-\frac{aM(M-b)}{2\left[M+b(M+\Lambda r^{2})\right]},$ (38)
$\displaystyle g_{\phi\phi}$ $\displaystyle=$
$\displaystyle\frac{a^{2}b(M-b)\left[-2M+b(-M+b+r^{2}\Lambda)\right]}{4\left[M+b(M-b+r^{2}\Lambda)\right]}$
(39) $\displaystyle+\frac{r^{2}}{(M-b)}\left[M+b(M-b+r^{2}\Lambda)\right]$

For non-gravitational field in Einstein frame reads

$\displaystyle
A_{\mu}dx^{\mu}=\frac{Q\left[\left(1+M-b+r^{2}\Lambda\right)dt-\frac{a(M-b)}{2}d\phi\right]}{2\left[M+b(M-b+r^{2}\Lambda)\right]},$
(40) $B_{t\phi}=-B_{\phi
t}=\frac{aMb(M-b)}{2\left[M+b(M-b+r^{2}\Lambda)\right]},$ (41)

and

$\Phi=-\frac{1}{2}\ln|\frac{M+b(M-b+r^{2}\Lambda)}{M-b}\bigg{|}.$ (42)

Where $b=\frac{Q^{2}}{2M}$ and $a=\frac{J}{M}$ is the rotation parameter. This
is a set of solutions from BTZ black holes in heterotic string theory in
string and Einstein frame. From the (36)-(42) solution, if we set $b=0$, then
the metric solution will return to the BTZ black hole. For short hand
purposes, the we will call this solution as BTZ-Sen Black Hole (BTZ-S BH).

As the author mentioned in the previous section that three-dimensional black
holes has similar characteristics to ordinary four-dimensional black holes. A
brief explanation of these statement we start by investigating the event
horizon in detail and what processes occur in BTZ-S BH. A hypersurface has
parametric equation which usually is written as

$f(x^{0},x^{1},x^{2})=0,$ (43)

where $x^{0}=t$, $x^{1}=r$, and $x^{2}=\phi$. The normal vector from the
hypersurface can be written as $n_{\mu}=\frac{\partial f}{\partial x^{\mu}}$.
Hypersurface is said to be null if the conditions below are obeyed

$\displaystyle n^{\mu}n_{\mu}$ $\displaystyle=$ $\displaystyle
g^{\mu\nu}n_{\mu}n_{\nu}=0,$ (44) $\displaystyle g^{rr}=0.$

We can conclude that $g^{rr}=0$ is a way to get the radius of the event
horizon. Applying the method to obtain BTZ black holes in heterotic string
theory, we get

$\displaystyle
r_{\pm}=l\sqrt{\frac{(M-b)}{2}\left[1\pm\sqrt{1-\left(\frac{J}{Ml}\right)^{2}}\right]}.$
(45)

Since the black hole that we are looking at is charged, the event horizon of
this black hole is parametrized by a mass $M$, charge $Q$, cosmological
constant $\Lambda=-\frac{1}{l^{2}}$, and angular momentum $J$. To keep the
event horizon on this BTZ-S BH still exist, the variables $l$, $M$, and $J$
must obey the following conditions

$\displaystyle M>b,~{}|J|\leq Ml.$ (46)

when $b=\frac{Q^{2}}{2M}=0$, this event horizon will be reduced to the radius
in BTZ black holes Banados:1992gq . In extreme conditions $|J|=Ml$, the event
horizon radius $r_{+}$ $r_{-}$ merges into one with a radius
$r_{e}=l\sqrt{\frac{M-b}{2}}$. Again, this extreme radius is reduced to the
extreme radius of the BTZ black hole when $b$ is turned off. For the case of
BTZ-S BH solutions with the condition $g_{tt}=0$

$r_{sl}=l\sqrt{M-b},$ (47)

the value of $r_{sl}$ is often referred to as the static limit radius; the
radius at which the observer will remain stationary even if the black hole is
rotating. The ergosphere of BTZ-S BH can be found in the range of event
horizon radius and static limit of $r_{\pm}<r_{erg}<r_{sl}$. To discuss the
geometry of this black hole in detail, we start from the event horizon graph

Figure 1: The graph $g^{rr}$ with respect to the distance $r$ represents the
geometry of the BTZ black hole in heterotic string theory with $b=-0.2$,
$M=1$, and $\Lambda=-0.03$. The metric solution that satisfies the condition
$g^{rr}(r_{+})=0$ is the radius of the event horizon $r_{+}$. The static limit
radius and ergosphere is presented in table 1.

The Fig. 1 is a plot of the metric tensor $g^{rr}$ with respect to $r$. We
first show the figure with several value of rotation parameters $a$ to see how
much the rotation on BTZ-S BH affects the event horizon(s). From this
information, we get that the greater the rotation of the black hole the
smaller the distance between the horizon(s). When the rotation parameter shows
the value $a=5.8$, the two radii are joined together. After this value, the
black hole has a point of singularity that is not covered by event horizons or
which can be referred as naked singularities.

$a$ |  $r_{-}$ |  $r_{+}$ |  $r_{sl}$ |
---|---|---|---|---
1.0 |  0.44 | 5.14 | 5.16 |
5.8 |  3.65 | 3.65 | 5.16 |
8.0 |  - | - | 5.16 |
Table 1: Table for exact value of $r$ from figure 1. Note that $r_{ls}$ is a
static limit radius. Ergosphere $\left(r_{erg}\right)$ still exist in the
range $r_{+}\leq r_{erg}\leq r_{ls}$.

In addition to the event horizon, the three-dimensional black hole has the
same characteristics as a four-dimensional black hole, namely ergosphere and
static limits. The radius in Eq. (47) depends only on the mass $M$, charge
$Q$, and the cosmological constant of $\Lambda$. We can inferred that, based
on the Fig 1 and the table 1, the rotation parameter $a$ does not affect the
dynamics of the static limit radius. When the naked singularity phenomenon
occurs, the static limit radius $\left(r_{ls}\right)$ will remain.

Figure 2: The graph $g^{rr}$ with respect to $r$ with $a=1$, $M=1$, and
$\Lambda=-0.03$. The metric solution that satisfies the condition
$g^{rr}(r_{+})=0$ is the radius of the event horizon $r_{+}$. The static limit
radius and also the ergosphere is presented in table 2.

From the previous story, the variation of the $a$ parameter does not make the
black hole disappear because there are still static surface limits. Now if we
make a change on the charge, this black hole will produce a different physical
meaning. Therefore, it is very tempting for us to investigate the Fig 2. The
figure is a plot that depict the $g^{rr}$ metric with respect to the distance
with the several value of $b$. It should be reminded again that
$b=\frac{Q^{2}}{2M}$. In this plot, We want to see the dynamics of event
horizons on black holes that begin with $b=0$ to $b=1$. For $b=0$, this black
hole has 2 horizon(s) (explaining the geometry of the BTZ black hole). The
static limit radius of $r_{sl}$ from this black hole with a variation of
$0\leq b\leq 1$ always coincides with the outer horizon of $r_{+}$. This is
the same as the plot in Fig. 1 when the rotation parameter $a=1$. As the $b$
parameter increases, it appears that when the outer horizon touches the static
limit radius $\left(M\approx b\right)$, the BTZ-S BH does not exist.

$b$ |  $r_{-}$ |  $r_{+}$ |  $r_{sl}$ |
---|---|---|---|---
0 |  0.50 |  5.75 |  5.77 |
0.5 |  0.35 |  4.06 |  4.08 |
0.8 |  0.22 |  2.57 |  2.58 |
0.9 |  0.15 |  1.81 |  1.82 |
0.99 |  0.05 |  0.57 |  0.57 |
1.0 |  0 |  0 | 0 |
Table 2: Exact value of $r$ from Fig. 2. Note that $r_{ls}$ is a static limit
radius. The ergosphere $\left(r_{erg}\right)$ exist in the range of $r_{+}\leq
r_{erg}\leq r_{ls}$.

Now look at a particle in the ergosphere. Particles that we can still analyze
must move in the timelike and null geodesics $(ds^{2}\leq 0)$. With $r=\phi=$
constant, we can get

$ds^{2}=\frac{(M-b)(M+r^{2}\Lambda-b)}{M+b(M-b+r^{2}\Lambda)}dt^{2},$ (48)

Since we want to see particles moving in the ergosphere $(r<r_{1})$, we get
the conditions $\frac{r^{2}}{l^{2}}<(M-b)$. The consequences of this
relationship are $ds^{2}>0$; Particles that stand still in the ergosphere tend
to move in a spacelike manner. So other terms in the tensor metric must
contribute and the particle must change position. Fortunately, the case above
corresponds to a black hole BTZ
$ds^{2}=\left(M-\frac{r^{2}}{l^{2}}\right)dt^{2}$, with the condition
$\frac{r^{2}}{l^{2}}<M$.

The limitation above requires us to change the condition of the particles. If
the particle moves in the direction of $\phi$ with $t$ and $r$ constant, then
the particle has an angular velocity around BTZ-S BH that satisfies the
equation

$\Omega^{2}+2\frac{g_{t\phi}}{g_{\phi\phi}}~{}\Omega+\frac{g_{tt}}{g_{\phi\phi}}=0,$
(49)

Thus, we find the angular velocity

$\Omega_{max,~{}min}=-\frac{g_{t\phi}}{g_{\phi\phi}}\pm\sqrt{\left(\frac{g_{t\phi}}{g_{\phi\phi}}\right)^{2}-\frac{g_{tt}}{g_{\phi\phi}}}.$
(50)

The angular velocity obtained above depends only on the distance. In general,
because $g_{tt}<0~{}(\Omega_{min}<0)$ so the observer can see the particle
rotating in the opposite direction to the rotation of the black hole. When the
particle is near the static surface limit of $g_{tt}\rightarrow 0$, the
angular velocity of the particle has two values, namely $\Omega_{min}=0$ and
$\Omega_{max}=-2\frac{g_{t\phi}}{g_{\phi\phi}}$. That is, the particle (at
$\Omega_{min}$), is remain at rest instead of rotating in the opposite
direction of the black hole rotation. Thus, the maximum angular velocity of
$\Omega_{max}$ can be explicitly stated

$\Omega_{max}=\frac{2M\left[M+b(M-b+\Lambda
r^{2})\right]}{ab\left[M+b(M+\Lambda r^{2})\right]\left[b(b-M+\Lambda
r^{2})-2M\right](1+\chi_{4}^{2})},$ (51)

where $\chi_{4}=\frac{2r\left[b(M-b+\Lambda
r^{2})-2M\right]}{a(M-b)\sqrt{b^{2}(b-M+\Lambda r^{2})-2Mb}}$. To ensure the
analysis is still on the right track, we can prove that if the charge term $b$
is turned off, then the maximum angular velocity above belongs to the black
hole BTZ $\left(\Omega_{max}=\frac{a}{2Mr^{2}}\right)$. In addition, Eq. (51)
also satisfies physical conditions. At a very large distance
$(r\rightarrow\infty)$, the angular velocity is zero. This indicates that the
particles that are very far from the black hole do not feel the frame dragging
effects at all. This observation again adds to our belief that the family
solutions of three-dimensional black hole has the same characteristics as the
family solutions of four-dimensional black hole. In the case of Kerr black
hole, zero angular velocity at infinity is caused by an asymptotically flat
space-time background. However, the angular velocity with asymptotically-flat
anti-de Sitter also has the same result. From the previous statement, we can
conclude that the particle will not feel the frame dragging effect no matter
what the asymptotic background is. The angular velocity that gets smaller when
the distance is farther away from the object also obeys the conservation law
of angular momentum. Moreover, particles that enter the ergosphere area will
rotate in the direction of rotation of the BTZSBH around

$L=2\pi r_{+}=2\pi
l\sqrt{\frac{(M-b)}{2}\left[1+\sqrt{1-\left(\frac{J}{Ml}\right)^{2}}\right]}.$
(52)

The formula above is the same as the perimeter of a circle since a
3-dimensional (2+1) black hole has a 2-dimensional surface. Finally, when the
charge contribution is switched off, the perimeter of the BTZ black hole is
again obtained.

With the end of this section, we already know the geometry and characteristics
of BTZSBH. Starting from solutions on the string and Einstein framework, event
horizon, ergosphere, static radius limit, angular velocity, and circumference
of BTZ-S BH. In conclusion, BTZ black holes in the low energy heterotic string
theory still exist in space-time.

##  IV Conclusion

In this paper, we construct the classical exact solution of a three-
dimensional black hole in low-energy heterotic string theory (BTZ-Sen balack
hole). The exact solution we obtain are BTZ black hole Banados:1992gq ;
Banados:1992wn  with Maxwell $F_{\mu\nu}$, nontrivial dilaton $\phi$, and
3-form field $H_{\mu\nu\rho}$. The Hassan-Sen transformation is the path we
choose to obtain the metric solution. This method requires us to have seed
solutions that obey the stationary and axisymmetric conditions. Since this
novelty involves two important concepts, we structured this paper into two
sections; (i) BTZ black hole as a review, and (ii) BTZ Black Hole in the low
energy heterotic string theory (BTZ-Sen). In our discussion on BTZ black hole
in the low energy heterotic string theory, the solution we obtain as a result
of the Hassan-Sen transformation starts from the string frame. The metric
solution is still paramaterized by the mass $M$ and angular momentum $J$. We
employ Sen Sen:1992ua  combined with the method in Magnon:1985sc ;
Brown:1992br ; Kastor:2008xb  to obtain the conserved mass, angular momentum,
and charge since the object obey the lower derivative gravity $(\Lambda\neq
0)$. Expressing the conserved quantity in terms of ADM variable, we obtain the
charge case after conformal transformation to Einstein frame (36)-(39). When
$b$ is switched off, the metric solution automatically reduces to the well
known BTZ black hole.

From our investigation of the event horizons, to keep the black hole exist the
cosmological constant parameters $l$, mass $M$, angular momentum $J$, and
charge $b$ must oeby the conditions $M>b$ and $|J|\leq Ml$. Under these
conditions, we obtain that the number of horizons will decrease as the
rotation parameter $a$ increases. A different story arises when the parameter
$b=\frac{Q^{2}}{2M}$ being varied. For $b=0$, the event horizon
characteristics are the same as BTZ black holes. When the $b$ parameter
increases, the event horizon shrinks until the $b$ parameter equals to the
value of $M$, a black hole does not exist. For another aspect, particles that
stand still in the ergosphere have space-like conditions $(ds^{2}>0)$ so they
must orbit in the same direction of rotation as the black hole. The obtained
angular velocity is reduced to BTZ angular velocity when the $b$ variable is
turned off. In the far region, the particle does not feel the frame dragging
effect of the rotation of the black hole no matter what the asymptotic
background is.

With this background solution, it is tempting to investigate the
phenomenological aspect from the solution. One can obtain the thermodynamical
properties and the first-law of BTZ-S BH novelty1 . Other investigations are
also of interest such as geodesics, deflection, and photon sphere around BTZ-S
BH using the Hamilton-Jacobi equation novelty2 . Since the black holes also
obeys quantum phenomenology, we can also look for massless scalar wave
scattering due to the appearance of BTZ-S BH. This case can be obtained by
using the massless Klein-Gordon equation to obtain the field $\Phi(t,r,\phi)$.
By using JWKB approximation to get the phase shift of the effective potential
that exists from the interaction, and the eikonal approximation to obtain the
scattering amplitude, the scattering cross-section can be obtained from the
absolute square of $f(\theta)$ novelty3 . We leave these three novelties for
our further investigation.

##  V Acknowledgement

We thank Ilham Prasetyo for the insight on Hamiltonian and Lagrangian in
curved space-time. We also thank Haryanto M. Siahaan, Handhika S. Ramadhan,
and A Sayyidina for enlightening discussions. This work is partially funded by
the Q1Q2 grants from Universitas Indonesia under the contract No.
NKB-0270/UN2.R3.1/HKP.05.00/2019.

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# Unramified extensions of quadratic number fields with certain perfect Galois
groups

Joachim König  Korea National University of Education, Department of
Mathematics Education, 28173 Cheongju, South Korea

###### Abstract.

We provide infinite families of quadratic number fields with everywhere
unramified Galois extensions of Galois group $SL_{2}(7)$ and $2.A_{7}$,
respectively. To my knowledge, these are the first instances of infinitely
many such realizations for perfect groups which are not generated by
involutions, a property which makes them difficult to approach for the problem
in question and leads to somewhat delicate local-global problems in inverse
Galois theory.

##  1\. Introduction and main results

The following is a folklore conjecture in inverse Galois theory:

######  Conjecture 1.

For any finite group $G$, there exist infinitely many quadratic number fields
$K$ such that $K$ possesses a Galois extension with Galois group $G$
unramified at all (finite and infinite) primes of $K$.

This is known to be true only in some special cases, including the case of
cyclic groups (from class field theory, cf. [1]), symmetric and alternating
groups (e.g., [22] (which does not consider ramification at infinity) or [6]
(which does)), and a few other almost simple finite groups (e.g., [10]). The
common idea for the nonsolvable cases is to carefully specialize suitable
function field extensions over $\mathbb{Q}$ (or equivalently, multiparameter
polynomials over $\mathbb{Q}$) with prescribed Galois group to obtain
extensions of $\mathbb{Q}$ with a desired ramification behavior. Further
evidence for Conjecture 1 is provided by analogs over global function fields,
which have proved fruitful and led to concrete distribution heuristics, in
[21], for extensions $L/K$ with $[K:\mathbb{Q}]=2$ and $Gal(L/K)=G$. We will
neither present nor use those heuristics in the present paper, although it is
useful to keep in mind that the expected distribution differs depending on the
concrete Galois group $Gal(\tilde{L}/\mathbb{Q})\leq G\wr C_{2}$, where
$\tilde{L}$ denotes the Galois closure of $L/\mathbb{Q}$.

Notably, for groups such as $G=S_{n}$, in order to solve Conjecture 1, it
suffices to obtain Galois extensions $L/\mathbb{Q}$ with group $S_{n}$ in
which all nontrivial inertia groups are generated by transpositions,111Or more
generally, by involutions, although allowing this does not seem to make the
problem much easier. and then base change to suitable quadratic number fields,
killing ramification via Abhyankar’s lemma; in the above terminology, this
corresponds to the case $Gal(\tilde{L}/\mathbb{Q})=G\times C_{2}$. The same
“direct” approach can in principle succeed for all groups $G$ generated by
involutions, but only for those. In particular, all non-abelian simple groups
are generated by involutions (due to the Feit-Thompson theorem). Admittedly,
in practice, realizations with involution inertia are not always easy to
obtain; in particular, their existence still seems unknown for arbitrary
alternating groups, whence those are instead dealt with via the above
$S_{n}$-extensions $L/\mathbb{Q}$, in which $L/L^{A_{n}}$ is automatically
unramified.

In this paper, we wish to progress to nonsolvable groups which are not
generated by involutions, i.e., for which the “$G\times C_{2}$” approach
cannot possibly be employed. Concretely, we show:

######  Theorem 1.1.

Let $G$ be one of $SL_{2}(7)$ and $2.A_{7}$. Then there exist infinitely many
quadratic number fields possessing an everywhere unramified Galois extension
with Galois group $G$.

We remark that for the cases treated in Theorem 1.1, there are explicit
polynomials available, which make a concrete enumeration of infinitely many
quadratic number fields with the above property possible. In Theorem 5.1, we
give such an explicit family in the case $G=SL_{2}(7)$. At the same time, it
should be noted that the proof of Theorem 1.1 itself does not rely on the
concrete shape of these polynomials, instead requiring only general
theoretical results such as the rigidity method, and results on the behavior
of inertia and decomposition groups under specialization of function field
extensions (as well as the subgroup structure of the groups in question).

Note also that the analog of Theorem 1.1 is known for the group $SL_{2}(3)$
(which, however, is a solvable group, namely a double cover of $A_{4}$) due to
[18] (cf. also [23, Proposition 7]).

We also prove the following result applicable to infinitely many groups
$SL_{2}(p)$, although with a weaker conclusion.

######  Theorem 1.2.

Let $p$ be a prime with $p\equiv 3$ mod $4$ and $p\not\equiv-1$ mod $24$. Then
there exist infinitely many quadratic number fields possessing an
$SL_{2}(p)$-extension unramified everywhere outside of the prime
ideal(s)222For what it’s worth, our proof yields quadratic fields in which $p$
is inert, i.e., extended only by a single prime ideal. dividing $p$.

We give some further motivation for the consideration of the groups above, as
well as some insights about the difficulties in dealing with them, in the next
section.

##  2\. Some further motivation

There are different classes of nonsolvable groups not generated by
involutions, all interesting in their own right, and regarding unramified
extensions of quadratic number fields, little seems to be known about any of
them. Just to give one example, even though explicit families of quadratic
number fields are known for each of $G=S_{n}$ ($n\geq 5$) and $H=C_{m}$
($m\geq 3$), considering the same problem for the group $G\times H$ leads to
delicate arithmetic questions (namely, about existence of discriminants
falling into two classes simultaneously; see also Remark 5).

We focus instead on a different class of groups, namely quasisimple groups $G$
not generated by involutions. Here, a group $G$ is called quasisimple if it
fits into an exact sequence $1\to Z\to G\to S\to 1$ with $Z=Z(G)$, $S$ a
simple group, and $G$ is perfect (i.e., equal to its commutator subgroup).
Perfectness is a logical assumption in order to make the problem of
identifying unramified $G$-extensions “disjoint” from the problem of class
number divisibility of quadratic fields. Instead, several other challenges
arise. Notably, in inverse Galois theory, Galois extensions with such groups
are usually obtained via first realizing a suitable quotient group such as
$S=G/Z(G)$ and then solving a central embedding problem. This alone is usually
a nontrivial problem; see, for example, the realization of covering groups of
the alternating and symmetric groups, which was investigated by many authors
([17], [20], [16], [14] are just some of the relevant papers on this topic).

The additional condition that $G$ should not be generated by involutions means
in particular that none of the involutions of $S$ lift to involutions in $G$,
implying that ($|Z|$ is even and) the $2$-Sylow group of $G$ does not contain
a subgroup isomorphic to $C_{2}\times C_{2}$, and hence must be cyclic or
generalized quaternion. Since it is well-known that a nonabelian simple group
cannot have a cyclic $2$-Sylow subgroup, we are left with the quaternion case.
Here, the $2$-Sylow group of $S$ must be (the central quotient of a
generalized quaternion group, and hence) dihedral, and it is then known ([5,
Theorem 2]) that $S$ is either $A_{7}$ or a group $PSL_{2}(q)$ with $q\geq 5$
an odd prime power.

Through this consideration (and via the known Schur multipliers of the simple
groups above), we have arrived naturally at the set of groups to be
considered: they are the double covers $SL_{2}(q)=2.PSL_{2}(q)$ (with $q$ odd)
as well as the even degree covering groups $2.A_{7}$, $6.A_{7}$ and
$6.PSL(2,9)\cong 6.A_{6}$.

The joint idea to attack these groups for our problem is the fact that they
embed as index-$2$ normal subgroups into groups which are generated by
involutions (namely, central extensions of $\Gamma=PGL_{2}(q)$ and of
$\Gamma=S_{7}$ by $C_{2}$ (and in two cases, noncentral extensions by
$C_{6}$), respectively). This is analogous to the embedding of $A_{n}$ into
$S_{n}$ as discussed above, but additional problems arise here. Namely, we
additionally need to begin with a prescribed ramification behavior in the
$\Gamma$-extension, and make sure that the embedding problem with kernel
$C_{2}$ is solvable without introducing extra ramification. We use a function
field approach to control the behavior of inertia and decomposition groups
under specialization. In this setting, it turns out that the embedding problem
usually leads to simultaneous requirements on the ramification index as well
as the residue extension at given branch points of the function field
extension. E.g., recall once more that the problem for the simple group
$A_{n}$ itself is solved via constructing $S_{n}$-extensions of $\mathbb{Q}$
in which all non-trivial inertia groups are generated by transpositions, which
is achieved via finding degree-$n$ extensions with squarefree discriminant,
whose existence is well-known. However in order to ensure that such an
$S_{n}$-extension has no local obstruction to embedding into an extension with
Galois group $2.S_{n}$, one needs to impose conditions on the Frobenius class
of (all!) prime divisors of the squarefree values of such discriminants. This
is usually difficult to guarantee without relying on open conjectures (cf.,
e.g., [6, Section 4] for some informal remarks on the difficulty of such
problems), and in a setting of specialization of function field extensions
amounts to finding values of certain (discriminant) polynomials with all prime
divisors in some prescribed Chebotarev set. In this light, Theorem 1.1 means
that in some special cases, these extra requirements can be obtained
unconditionally.

##  3\. Prerequisites

###  3.1. Local behavior of specialization of function field extensions

We review some known results about the behaviour of inertia and decomposition
groups in specializations of function field extensions. Since these occur in
several predecessor works, we restrict to the most directly relevant results.
For further background on function field extensions and their specialization,
cf., e.g., [9], Sections 2 and 3.

For a finite Galois extension $N/K(t)$ of function fields and any value
$t_{0}\in\mathbb{P}^{1}(K)$, we denote by $N_{t_{0}}/K$ the specialization of
$N/K(t)$ at $t\mapsto t_{0}$, i.e., the residue extension at any place
extending $t\mapsto t_{0}$ in $N$. Recall that $N/K(t)$ is called $K$-regular
if $K$ is algebraically closed in $N$. The following well-known theorem, cf.
[3, Prop. 4.2], ties the ramification in specializations of a function field
extension to the inertia groups at branch points of that function field
extension.

######  Theorem 3.1.

Let $K$ be a number field and $N/K(t)$ a finite $K$-regular Galois extension
with Galois group $G$. Then there exists a finite set $\mathcal{S}_{0}$ of
primes, depending only on $N/K(t)$, such that the following holds for every
prime $\mathfrak{p}$ of $K$ not in $\mathcal{S}_{0}$:
If $t_{0}\in K$ is not a branch point of $N/K(t)$ then the following condition
is necessary for $\mathfrak{p}$ to be ramified in the specialization
$N_{t_{0}}/K$:

$e_{i}:=I_{\mathfrak{p}}(t_{0},t_{i})>0\text{ for some (automatically unique,
up to algebraic conjugates) branch point $t_{i}$.}$

Here $I_{\mathfrak{p}}(t_{0},t_{i})$ is the intersection multiplicity of
$t_{0}$ and $t_{i}$ at the prime $\mathfrak{p}$. Furthermore, the inertia
group of a prime extending $\mathfrak{p}$ in the specialization $N_{t_{0}}/K$
is then conjugate in $G$ to $\langle\tau^{e_{i}}\rangle$, where $\tau$ is a
generator of an inertia subgroup over the branch point $t\mapsto t_{i}$ of
$K(t)$.

Regarding the definition of intersection multiplicity $I_{\mathfrak{p}}$
occurring in Theorem 3.1, note that in the special case $K=\mathbb{Q}$ this
may be defined conveniently in the following way: Let $f(X)\in{\mathbb{Z}}[X]$
be the irreducible polynomial of $t_{i}$ over ${\mathbb{Z}}$ and
$\tilde{f}(X,Y)$ its homogenization (with $\tilde{f}:=Y$ in the special case
$t_{i}=\infty)$. Let $t_{0}=\frac{a}{b}$ with $a,b\in{\mathbb{Z}}$ coprime,
and let $p$ be a prime number. Then $I_{p}(t_{0},t_{i})$ is the multiplicity
of $p$ in $\tilde{f}(a,b)$.

The following result (cf. [9, Thm. 4.1]) clarifies the structure of
decomposition groups in specializations. To state it, we need to introduce a
bit more notation. Given a finite Galois extension $N/K(t)$ with group $G$ and
a $K$-rational place $t\mapsto a\in\mathbb{P}^{1}(K)$, denote by $D_{a}$ and
$I_{a}$ the decomposition and inertia group at the place $t\mapsto a$
respectively. Furthermore, given a prime $p$ of $K$, denote by $D_{a,p}$ the
decomposition group at $p$ in the residue extension $N_{a}/K$. Note that the
Galois group of $N_{a}/K$ is identified with $D_{a}/i_{a}$, and hence
$D_{a,p}$ is identified with a subgroup of this group.

######  Theorem 3.2.

Let $K$ be a number field and $N/K(t)$ a finite $K$-regular Galois extension
with Galois group $G$. Let $t\mapsto t_{i}\in\mathbb{P}^{1}(\overline{K})$ be
a branch point of $N/K(t)$, and let $I_{t_{i}}$ and $D_{t_{i}}$ denote the
inertia and decomposition group at $t\mapsto t_{i}$ in $N(t_{i})/K(t_{i})(t)$.
Then there exists a finite set $\mathcal{S}_{1}$ of primes of $K$, depending
only on $N/K(t)$ and containing the set $\mathcal{S}_{0}$ from Theorem 3.1,
such that for all primes $\mathfrak{p}$ of $K$ not in $\mathcal{S}_{1}$ and
all non-branch points $a\in\mathbb{P}^{1}(K)$ of $N/K(t)$, the following hold:

  * a)

Assume that $\mathfrak{p}$ ramifies in $N_{a}/K$ and
$I_{\mathfrak{p}}(a,t_{i})>0$.333The latter is guaranteed to happen for some
branch point $t_{i}$, unique up to algebraic conjugates, by Theorem 3.1. Let
$D_{t_{i},\mathfrak{p}^{\prime}}$ denote the decomposition group of
$N(t_{i})_{t_{i}}/K(t_{i})$ at the (unique) prime $\mathfrak{p}^{\prime}$ of
$K(t_{i})$ extending $\mathfrak{p}$ such that
$I_{\mathfrak{p}^{\prime}}(a,t_{i})>0$. Then the decomposition group
$D_{a,\mathfrak{p}}$ at $\mathfrak{p}$ in $N_{a}/K$ fulfills
$\varphi(D_{a,\mathfrak{p}})=D_{t_{i},\mathfrak{p}^{\prime}}$, where
$\varphi:D_{i}\to D_{t_{i}}/I_{t_{i}}$ is the canonical epimorphism.

  * b)

In particular, if additionally $I_{\mathfrak{p}}(a,t_{i})=1$, then
$D_{a,\mathfrak{p}}=\varphi^{-1}(D_{t_{i},\mathfrak{p}^{\prime}})$.

######  Remark 1.

  * a)

The finite exceptional set $\mathcal{S}_{1}$ in Theorem 3.2 can be made
explicit, and this is sometimes relevant when doing calculations with concrete
extensions, as we will do in Theorem 5.1. In order to avoid too much
technicality, we only refer to [11, Theorem 2.2] for an effective version of
Theorem 3.2.

  * b)

For many applications, it is enough to obtain some specialization values
$a\in\mathbb{P}^{1}(K)$ yielding a certain prescribed local behavior, rather
than controlling the behavior for all specialization values. In this context,
the notion of a universally ramified prime (see [2] or [10]) becomes useful:
let $\mathcal{U}(N/K(t))$ be the set of all primes of $K$ ramifying in all
specializations $N_{a}/K$ where $a\in\mathbb{P}^{1}(K)$ is a non-branch point
of $N/K(t)$. Let $\mathcal{S}$ be any finite set of primes of $K$ disjoint
from $\mathcal{U}(N/K(t))$. Then there exists a non-empty and
$\mathcal{S}$-adically open set of values $a\in\mathbb{P}^{1}(K)$ such that
$N_{a}/K$ is unramified at all $p\in\mathcal{S}$ (an easy consequence of
Krasner’s lemma). In particular, by choosing
$\mathcal{S}:=\mathcal{S}_{1}\setminus\mathcal{U}(N/K(t))$ with
$\mathcal{S}_{1}$ as in Theorem 3.2, we see that for all such values $a$, the
assertion of Theorem 3.2 actually holds with $\mathcal{S}_{1}$ replaced by
$\mathcal{U}(N/K(t))$, and the latter set is often much more accessible and
can be bounded very effectively by simply comparing some very few
specializations.

We will also require the following lemma, describing the behavior of residue
extensions under certain rational pullbacks.

######  Lemma 3.3.

Let $N/K(t)$ be a $K$-regular Galois extension with group $G$, let $I$ and $D$
denote the inertia and decomposition group at the place $t\mapsto 0$, and
assume that the extension $1\to I\to D\to D/I\to 1$ splits. Then there exists
$\alpha\in K^{\times}$ such the following holds:

If $e:=|I|$ and $u$ is such that $u^{e}=\alpha t$, then the extension
$N(u)/K(u)$ is unramified at $u\mapsto 0$ with decomposition group isomorphic
to $D/I$.

###### Proof.

Let $U$ be a complement to $I$ in $D$, let $\widehat{N}/K((t))$ denote the
completion of $N/K(t)$ at $t\mapsto 0$, and $F/K$ the residue extension of
$N/K(t)$ at $t\mapsto 0$. In particular, $Gal(\widehat{N}/K((t)))\cong D$, and
$\zeta_{e}\in F$. Let $\widehat{N}^{U}$ denote the fixed field of $U$. Then
$\widehat{N}^{U}/K((t))$ is totally ramified of degree $e$, and hence equal to
$K((\sqrt[e]{\alpha t}))=:K((u))$ for a suitable $\alpha\in K$. Consequently,
$\widehat{N}=F((u))$. Now firstly, since $K(u)/K(t)$ is totally ramified of
index $e$ at $t\mapsto 0$, the extension $E(u)/K(u)$ is unramified at
$u\mapsto 0$. Furthermore, the residue extension $E(u)_{0}/K$ is necessarily
contained in the compositum of $K(\zeta_{e})((u))$ (i.e., the Galois closure
of $K((u))/K((t))$) and $\widehat{N}=F((u))$; since $\zeta_{e}\in F$, this
compositum is $F((u))$, and thus its residue extension is $F$. Conversely, the
completion at $u\mapsto 0$ has to be of degree at least
$|D|/[K(u):K(t)]=|D/I|=[F:K]$, yielding the assertion. ∎

######  Remark 2.

The splitting condition in the assumption of Lemma 3.3 is necessary. Indeed,
an analog of Theorem 3.2 (with the base field $K$ replaced by the function
field $K(u)$) shows that the decomposition group at $u\mapsto 0$ in
$N(u)/K(u)$ is a subgroup of $D$ projecting onto $D/I$. This observation is
essentially the reason why we are so far unable to include the (smaller) group
$G=SL_{2}(5)$ in Theorem 1.1.

###  3.2. Embedding problems

We recall some basic terminology around embedding problems; cf., e.g., [15,
Chapter III.5]. A finite embedding problem over a field $K$ is a pair
$(\varphi:G_{K}\to G,\varepsilon:\tilde{G}\to G)$, where $\varphi$ is a
(continuous) epimorphism from the absolute Galois group $G_{K}$ of $K$ onto
$G$, and $\varepsilon$ is an epimorphism between finite groups $\tilde{G}$ and
$G$ fitting in an exact sequence $1\to N\to\tilde{G}\to G\to 1$. The kernel
$N=\ker(\varepsilon)$ is called the kernel of the embedding problem. An
embedding problem is called central if $\ker(\varepsilon)\leq Z(\tilde{G})$. A
(continuous) homomorphism $\psi:G_{K}\to\tilde{G}$ is called a solution to
$(\varphi,\varepsilon)$ if the composition $\varepsilon\circ\psi$ equals
$\varphi$. In this case, the fixed field of $\ker(\psi)$ is called a solution
field to the embedding problem. A solution $\psi$ is called a proper solution
if it is surjective. In this case, the field extension of the solution field
over $K$ has full Galois group $\tilde{G}$.

If $K$ is a number field and $\mathfrak{p}$ is a prime of $K$, every embedding
problem $(\varphi,\varepsilon)$ induces an associated local embedding problem
$(\varphi_{\mathfrak{p}},\varepsilon_{\mathfrak{p}})$ defined as follows:
$\varphi_{\mathfrak{p}}$ is the restriction of $\varphi$ to
$G_{K_{\mathfrak{p}}}$ (well defined up to fixing an embedding of
$\overline{K}$ into $\overline{K_{\mathfrak{p}}}$), and
$\varepsilon_{\mathfrak{p}}$ is the restriction of $\varepsilon$ to
$\varepsilon^{-1}(G({\mathfrak{p}}))$, where
$G(\mathfrak{p}):=\varphi_{\mathfrak{p}}(G_{K_{\mathfrak{p}}})$.

We require two well-known results on central embedding problems. The first
(cf. [13, Chapter IV, Cor. 10.2]) is a local global-principle for the
solvability of such embedding problems, which is essentially a consequence of
the local-global principle for Brauer embedding problems over number fields
([13, Chapter IV, Cor. 7.8]).

######  Proposition 3.4.

Let $\Gamma=C.G$ be a central extension of $G$ by a cyclic group $C$ of prime
order and $\varepsilon:\Gamma\to G$ the canonical projection. Let
$\varphi:G_{\mathbb{Q}}\to G$ be a continuous epimorphism. Then the embedding
problem $(\varphi,\varepsilon)$ is solvable if and only if all associated
local embedding problems $(\varphi_{p},\varepsilon_{p})$ are solvable, where
$p$ runs through all primes of $\mathbb{Q}$ (including the infinite one).

Since the local embedding problem at an unramified prime is always solvable,
the above lemma gives an efficient method to check for global solvability of a
given embedding problem by investigating only the finitely many ramified
primes of $K/\mathbb{Q}$.

The second result (see [17, Prop. 2.1.7]) is a tool to control ramification in
solutions of central embedding problems, assuming solvability.

######  Proposition 3.5.

Let $\Gamma=C.G$ be a central extension of $G$ by a finite abelian group $C$,
let $\varepsilon:\Gamma\to G$ be the canonical projection, and let
$\varphi:G_{\mathbb{Q}}\to G$ be a continuous epimorphism such that the
embedding problem $(\varphi,\varepsilon)$ has a solution. For each finite
prime $p$, let $\tilde{\varphi_{p}}:G_{\mathbb{Q}_{p}}\to\Gamma$ be a solution
of the associated local embedding problem $(\varphi_{p},\epsilon_{p})$, chosen
such that all but finitely many $\tilde{\varphi_{p}}$ are unramified. Then
there exists a (not necessarily proper) solution
$\tilde{\varphi}:G_{\mathbb{Q}}\to\Gamma$ of $(\varphi,\epsilon)$ such that
for all finite primes $p$, the restrictions of $\tilde{\varphi}$ and
$\tilde{\varphi_{p}}$ to the inertia group inside $G_{\mathbb{Q}_{p}}$
coincide. In particular, $\tilde{\varphi}$ is ramified exactly at those finite
primes $p$ for which $\tilde{\varphi_{p}}$ is ramified.

##  4\. Proof of Theorem 1.2

###### Proof.

We first describe the general strategy. We will show the assertion by first
constructing “suitable” $PGL_{2}(p)$-extensions of $\mathbb{Q}$ and then
embedding them into $2.PGL_{2}(p)$-extensions using Propositions 3.4 and 3.5.
Here, by $2.PGL_{2}(p)$, we mean the stem extension of $PGL_{2}(p)$ in which
the “outer” involutions (i.e., the one not contained in $PSL_{2}(p)$) split.
By “suitable” extensions, we mean extensions in which the intermediate
extension of group $PSL_{2}(p)$ is unramified everywhere away from $p$,
meaning that all nontrivial inertia groups other than possibly over $p$ must
be generated by involutions outside of $PSL_{2}(p)$. There is only one class
of such involutions in $PGL_{2}(p)$. The decomposition group $D$ at a prime
with such an inertia subgroup must of course be contained in the centralizer
of such an involution; this centralizer contains (e.g.) a subgroup
$C_{2}\times C_{2}$, which is non-split in $2.PGL_{2}(p)$ and whose preimage
is isomorphic to the dihedral group $D_{4}$. Thus, if in the given
$PGL_{2}(p)$-extension, some prime $q$ had this decomposition and inertia
subgroup, then assuming the existence of a lifting to $2.PGL_{2}(p)$, the
decomposition group at $q$ would be $D_{4}$. Since however the “outer”
involution lifts to a (non-central) element of order $2$ in $D_{4}$, this
would imply that the inertia subgroup in the $2.PGL_{2}(p)$-extension would be
a non-normal of the decomposition group, a contradiction. In other words, a
prime with decomposition group $C_{2}\times C_{2}$ would constitute a local
obstruction to the solvability of the central embedding problem. In our
construction below, we thus need to ensure among others (using Theorem 3.2)
that such primes do not occur. As we will see, for primes with inertia group
generated by an outer involution, the above is in fact the only local
obstruction.

Due to the assumption $p\equiv 3$ mod $4$ and $p\not\equiv-1$ mod $24$, at
least one of $2$ and $3$ is a non-square modulo $p$. Thus, by [13, Chapter I,
Cor. 8.10], there exists a $\mathbb{Q}$-regular Galois extension
$E/\mathbb{Q}(t)$ with Galois group $PGL_{2}(p)$, and with the following
properties: $E$ is ramified at exaxtly three points
$t_{i}\in\mathbb{P}^{1}(\overline{Q})$ ($i=1,2,3$), all $\mathbb{Q}$-rational,
and the corresponding inertia groups are generated by elements of conjugacy
class $pA$ (for $t_{1})$, $4A$ or $6A$ (for $t_{2}$, depending on whether $2$
or $3$ is a non-square modulo $p$), and $2B$ (for $t_{3}$), in Atlas notation.
Here $2B$ is the unique class of involutions outside of $PSL_{2}(p)$. Up to a
fractional linear transformation in $t$, we may and will assume $t_{1}=0$,
$t_{2}=\infty$ and $t_{3}=1$. Now consider the residue extensions at the
branch points $t_{1}$ and $t_{3}$. Since $p\equiv 3$ mod $4$, the elements of
class $2B$ are involutions with exactly two fixed points in the standard
action of $PGL_{2}(p)$ on $p+1$ points. The centralizer of an element of class
$2B$ is $PGL_{2}{p}$ is isomorphic to $C_{2}\times D_{(p-1)/2}$, where
$D_{(p-1)/2}$ denotes the dihedral group of order $p-1$, and thus the residue
extension at $t_{1}$ is of degree dividing $p-1$, with Galois group a subgroup
of $D_{(p-1)/2}$. Assume that it is indeed of even degree. Then it has a
unique quadratic subextension, which we denote by
$\mathbb{Q}(\sqrt{\beta})/\mathbb{Q}$. Consider the extension
$E(s)/\mathbb{Q}(s)$, where $s$ is such that $\beta t=s^{2}$. This is still a
$\mathbb{Q}$-regular $PGL_{2}(p)$-extension (indeed, the quadratic extension
$\mathbb{Q}(s)/\mathbb{Q}(t)$ has only one branch point in common with the
quadratic subextension of $E/\mathbb{Q}(t)$), ramified at $s=0$, $s=\infty$
and $s=\pm\sqrt{\beta}$. In particular, the residue field at the latter two
branch points “swallows” the above quadratic subextension, whence the residue
extension at any of those two branch points in $E(s)/\mathbb{Q}(s)$ is now
cyclic of odd degree.

Now consider the residue extension at $t_{1}$. Since the normalizer of a
subgroup of order $p$ in $PGL_{2}(p)$ is isomorphic to $C_{p}\rtimes C_{p-1}$,
the residue extension at this branch point is of degree dividing $p-1$; and
since furthermore, the residue extension at a branch point of ramification
index $e$ must always contain the $e$-th roots of unity (e.g., [9, Lemma
2.3]), it follows that this residue extension in fact equals
$\mathbb{Q}(\zeta_{p})/\mathbb{Q}$. Since $s\mapsto 0$ is the unique place
over $t\mapsto 0$ in $E(s)$, the same holds for the residue extension of
$E(s)/\mathbb{Q}(s)$ at $s=0$. It then follows from Lemma 3.3 that for
suitable $\alpha\in\mathbb{Q}$, the extension $E(u)/\mathbb{Q}(u)$, where $u$
is such that $s=\alpha u^{p}$, is $\mathbb{Q}$-regular with group
$PGL_{2}(p)$, unramified at $u=0$ and with residue field extension still equal
to $\mathbb{Q}(\zeta_{p})/\mathbb{Q}$ (in particular, unramified outside of
$p$ and $\infty$). Furthermore, at the branch points of $E(u)/\mathbb{Q}(u)$
extending $s=\pm\sqrt{\beta}$, the Galois group of the residue extension is
guaranteed to inject into the one at $s=\pm\sqrt{\beta}$. This is because
$\mathbb{Q}(u)/\mathbb{Q}(s)$ is unramified at those points, whence the
inertia group $I$ at such a branch point in $E(u)/\mathbb{Q}(u)$ is the same
as the corresponding one in $E(s)/\mathbb{Q}(s)$, and the decomposition group
$D$ certainly cannot grow under pullback, implying the quotient of
decomposition over inertia group cannot grow either.

Now from Theorem 3.2, we obtain the following: For all $u_{0}\in\mathbb{Q}$ of
the form $u_{0}=\frac{u_{1}}{u_{2}^{6}}$ (with integers $u_{1},u_{2}$)444The
power in the denominator indeed only needs to be a multiple of the
ramification index at $\infty$ in $E(u)/\mathbb{Q}(u)$, in order to avoid
ramification induced by this branch point; by construction, this ramification
index is either $2$ or $3$, depending on which of the two possible
ramification types we started with. and all but finitely many primes $q$, if
the specialization $E(u)_{u_{0}}/\mathbb{Q}$ is ramified at $q$, then its
inertia group in generated by an element of class $2B$ and its decomposition
group embeds into $C_{2}\times C_{(p-1)/2}\equiv C_{p-1}$. In particular, no
such prime yields a local obstruction to the embedding problem induced by
$2.PGL_{2}(p)\to PGL_{2}(p)$, since the group elements of order $2B$, and a
fortiori the odd order elements of $C_{(p-1)/2}$ are split in the central
extension. To deal with the remaining finitely many primes $q$, it suffices to
additionally choose $u_{0}$ $q$-adically sufficiently close to $u=0$. Indeed,
since $E(u)_{0}/\mathbb{Q}=\mathbb{Q}(\zeta_{p})/\mathbb{Q}$ is unramified
away from $\\{p,\infty\\}$, Remark 1b) yields that for $q\neq p$ finite, the
extension $E(u)_{u_{0}}$ is then unramified at $q$; whereas for $q=p$, it is
cyclic and totally ramified of degree $p-1$. Again, as above, the cyclic
subgroups of order $p-1$ are split in the central extension. Finally, the
infinite prime yields no local obstruction either, since specializing
sufficiently close to $u=0$ (in the archimedean absolute value)555Note that
this is possible within the set of values $u_{0}$ chosen above. yields as
complex conjugation an involution inside $C_{p-1}$, i.e., once again an
element of class $2B$ in $PGL_{2}(p)$.

Via Proposition 3.4, we may thus solve the embedding problem induced by
$G_{\mathbb{Q}}\to Gal(E(u)_{u_{0}}/\mathbb{Q})$ and $2.PGL_{2}(p)\to
PGL_{2}(p)$, and due to Proposition 3.5, this can in fact be achieved without
any ramification inside the kernel (since all non-trivial inertia groups were
split). Thus, if $K$ denotes the fixed field of $SL_{2}(p)$ in the resulting
extension $L/\mathbb{Q}$, then all inertia groups in $Gal(L/\mathbb{Q})$,
except the ones at $p$, are generated by involutions outside of $Gal(L/K)$,
whence $L/K$ is ramified only above $p$. This concludes the proof. ∎

######  Remark 3.

The extensions of $\mathbb{Q}$ with Galois group $PGL_{2}(p)$ constructed in
the above proof are in particular locally cyclic, i.e., all decomposition
groups at primes of $\mathbb{Q}$ are cyclic. This property was previously
studied in [7], and locally cyclic extensions were obtained, e.g., for the
special case $G=PGL_{2}(7)$ ([7, Theorem 5.5]). Our above result (for
infinitely many primes $p$) in particular improves on [7, Theorem 2.1].

##  5\. Proof of Theorem 1.1

We first prove 1.1 in a purely theoretical way, using only ramification data
and group-theoretical properties. Afterwards, we will make the conclusion more
concise using computations with some explicit polynomials.

###  5.1. The case $G=SL_{2}(7)$

###### Proof.

In the following, we adapt the proof of Theorem 1.2 somewhat for the special
case $p=7$, in order to get rid of ramification at $p$. Note that since $3$ is
a non-square modulo $7$, the extensions $E/\mathbb{Q}(t)$ used in the proof of
Theorem 1.2 are of ramification type $(pA,6A,2B)=(7A,6A,2B)$. After pullback
by an extension $\mathbb{Q}(u)/\mathbb{Q}(t)$ of rational function fields as
in the above proof, we obtain a $PGL_{2}(7)$-extension $E(u)/\mathbb{Q}(u)$
with the following ramification data: inertia group of order $3$ at
$u=\infty$, and a total of $14$ branch points of ramification index $2$ and
with residue extension of odd degree; furthermore, no ramification above
$u\mapsto 0$. We specialize $u\mapsto u_{0}$ essentially as above, except that
we will choose $u_{0}$ $7$-adically close to $u=\infty$ rather than to $u=0$.
Note that the normalizer of a subgroup of order $3$ in $PGL_{2}(7)$ is
isomorphic to $S_{3}\times C_{2}\cong C_{3}\rtimes(C_{2}\times C_{2})$, whence
the residue extension $E(u)_{\infty}/\mathbb{Q}$ is of degree dividing $4$,
and on the other hand (due to the ramification index being $3$) contains
$\mathbb{Q}(\zeta_{3})$. Note furthermore that $\mathbb{Q}_{7}$ conains the
third roots of unity, whence the completion of
$E(u)\cdot\mathbb{Q}_{7}/\mathbb{Q}_{7}(u)$ at $u\mapsto\infty$ has Galois
group embedding into the centralizer $C_{PGL_{2}(7)}(C_{3})$,666Here we have
used the general fact that the action of the decomposition group on an inertia
subgroup of order $e$ is via the Galois action of the $e$-th cyclotomic
extension; see, e.g., [11, Lemma 2.1(1)]. and in particular the residue
extension $E(u)_{\infty}\cdot\mathbb{Q}_{7}/\mathbb{Q}_{7}$ is either trivial
or generated by an involution centralizing $C_{3}$. Such an involution
necessarily lies outside of $PSL_{2}(7)$.

As before, Lemma 3.3 yields that there exists $\gamma\in\mathbb{Q}$ such that
for $v$ defined via $u=\gamma v^{3}$, the extension $E(v)/\mathbb{Q}(v)$ is
still $\mathbb{Q}$-regular of group $PGL_{2}(7)$, unramified at $v=\infty$ and
with residue extension $E(v)_{\infty}/\mathbb{Q}$ contained in the above
residue extension $E(u)_{\infty}/\mathbb{Q}$.

Specialize $v\mapsto v_{0}$ $7$-adically close to $v=\infty$. Then by
Krasner’s lemma, the local behavior of $E(v)_{v_{0}}/\mathbb{Q}$ at $p=7$ is
the same as that of $E(v)_{\infty}/\mathbb{Q}$. In particular, by the above,
the completion $E(v)_{0}\cdot\mathbb{Q}_{7}/\mathbb{Q}_{7}$ is either trivial
or quadratic with Galois group generated by an involution of class $2B$. As
before, this constitutes no local obstruction to the central embedding problem
with kernel $C_{2}$, and we thus obtain (for $p=7$) the conclusion of Theorem
1.2 without the exception at $p$. ∎

######  Remark 4.

Theorem 1.2 and its proof explicitly include also the case $p=3$. We leave it
as an exercise to the reader to adapt the argument in the proof of Theorem 1.1
above to regain the analogous (and already known) statement for $p=3$ (i.e.,
infinitely many quadratic number fields with everywhere unramified
$SL_{2}(3)$-extensions).

In the following, we make the assertion of Theorem 1.1 more precise by giving
a fully explicit family of quadratic number fields possessing unramified
$SL_{2}(7)$-extensions. Theorem 5.1 should be compared with previous results
on explicit families of number fields with class number divisible by
prescribed integers $n$ (see, e.g., [4] for an overview as well as new
results).

######  Theorem 5.1.

For all odd integers $t\in 2\mathbb{Z}+1$, the quadratic number field
$\mathbb{Q}(\sqrt{-7\cdot(108+7^{10}t^{6})})$ possesses an everywhere
unramified $SL_{2}(7)$-extension.

###### Proof.

We consider the splitting field $E$ of the degree-$8$ polynomial
$f(t,X)=7^{3}X^{7}(X+7)-t(X^{2}+X+7)$ over $\mathbb{Q}(t)$. This has Galois
group $G=PGL_{2}(7)$ over $\mathbb{Q}(t)$ (e.g., it is, up to multiplication
of the variables by suitable constants, the special case $a=0$ of the
polynomial in [12, Theorem 5.1]). Furthermore, $E/\mathbb{Q}(t)$ is ramified
exactly over $t\mapsto 0$, $t\mapsto\infty$ and one more rational point
$t\mapsto t_{0}:=-7^{10}/(2^{2}\cdot 3^{3})$, with inertia group generator of
cycle structure $(7.1)$, $(6.1^{2})$ and $(2^{3}.1^{2})$ respectively.
Following the above proofs, we manipulate $f$ in such a way that all
ramification in specializations will be “decided” by the branch point of
ramification index $2$. We first find out about the residue extension at this
branch point. Firstly, the decomposition group at this branch point needs to
be contained in the centralizer of an element of cycle structure
$(2^{3}.1^{2})$ in $PGL_{2}(7)$, which is of order $12$. Thus the residue
extension at $t\mapsto t_{0}$ is a Galois extension of group embedding into
$C_{G}(I)/I\cong S_{3}$, and from factorizing $f(t_{0},X)$ (which is the
product of a quadratic polynomial and the square of a cubic polynomial), we
find that it is indeed an $S_{3}$-extension, with quadratic subextension
$\mathbb{Q}(\sqrt{-3})/\mathbb{Q}$. This happens to be the same quadratic
extension as $\mathbb{Q}(\sqrt{t_{0}})/\mathbb{Q}$. Thus, setting
$t(s):=1/s^{2}$, the splitting field $E(s)$ of $f(t(s),X)$ over
$\mathbb{Q}(s)$ is a $PGL_{2}(7)$-extension ramified exactly over
$s\in\\{0,\infty,\pm\sqrt{t_{0}}\\}$ (with ramification index $3$ at $s\mapsto
0$ due to Abhyankar’s lemma, since this place extends $t\mapsto\infty$ of
ramification index $2$), and now, the residue extension at
$s\mapsto\pm\sqrt{1/t_{0}}$ is a $C_{3}$-extension of the base field
$\mathbb{Q}(\sqrt{t_{0}})$. Pulling back further by setting $s(u):=u^{3}$, we
obtain a $PGL_{2}(7)$=extension $E(u)/\mathbb{Q}(u)$ ramified only at the
$6$-th roots of $1/t_{0}$, as well as $u\mapsto\infty$ (and in particular,
unramified at $u\mapsto 0$ due to Abhyankar’s lemma). Since the residue field
$\mathbb{Q}(\sqrt[6]{t_{0}})$ contains
$\mathbb{Q}(\sqrt{t_{0}})=\mathbb{Q}(\sqrt{-3})$, it still remains true that
the residue extension of $E(u)/\mathbb{Q}(u)$ at any of these finite branch
points is a $C_{3}$-extension. Theorem 3.2 then gives the following: There
exists a finite set $\mathcal{S}$ of primes such that, for any integral
specialization $u\mapsto u_{0}\in\mathbb{Z}$ and any prime $p$ not contained
in $\mathcal{S}$, if $p$ ramifies in $E(u)_{u_{0}}/\mathbb{Q}$, then its
inertia group must be generated by an involution of cycle type
$(2^{3}.1^{2})$, and its decomposition group must be either $C_{2}$ or
$C_{2}\times C_{3}$. Since both these groups split in $2.PGL_{2}(7)$, the
corresponding central embedding problem then has no local obstruction coming
from any finite prime $p$ outside $\mathcal{S}$. We now verify computationally
that in fact the primes in the exceptional set $\mathcal{S}$ pose no
obstruction either. This is done by specializing $f(t(s(u)),X)$, which up to
constant factor equals

$F(u,X)=(7u^{2})^{3}X^{7}(X+7)-(X^{2}+X+7),$

at odd integer values $u\mapsto u_{0}\in 2\mathbb{Z}+1$. Calculating the
discriminant of $F(u,X)$ with respect to $X$ and using [11, Theorem 2.2], as
indicated in Remark 1a), one verifies that one may in fact take
$\mathcal{S}=\\{2,3,7,\infty\\}$. Firstly note that $F(u_{0},X)$ mod $2$ is
separable for odd $u_{0}$, whence $2$ is unramified in all such
specializations and thus poses no problem. Neither does the infinite prime,
since our $PGL_{2}(7)$-extensions are such that the quadratic subextension is
imaginary, i.e., complex conjugation is given by an involution of cycle
structure $(2^{3}.1^{2})$. For the remaining primes $p=3$ and $7$, it may be
directly verified, aided by Magma, that all specializations which ramify at
$p$ (in the case of $p=7$, these are actually all specializations $u\mapsto
u_{0}\in 2\mathbb{Z}+1$; in the case of $p=3$, ramification happens only when
$3|u_{0}$) have cyclic decomposition group generated by an involution of cycle
type $(2^{3}.1^{2})$.777Note that, in order to strictly verify this claim, one
of course need not do calculations for infinitely many specializations, since
the behavior of the decomposition group is the same for all values $u_{0}$
which are sufficiently close to each other $p$-adically, by Krasner’s lemma.
Further tricks are available to ease the task, e.g., over
$\mathbb{Q}_{7}(\sqrt{7})$, one may perform a change of variable
$X\mapsto\frac{X}{\sqrt{7}u}$ and consider the polynomial
$7u^{2}F(u,X/\sqrt{7}u)=X^{7}(X+7\sqrt{7}u)-(X^{2}+\sqrt{7}uX+49u^{2})$,
which, as already a low-precision calculation verifies, splits into distinct
linear factors over $\mathbb{Q}_{7}(\sqrt{7})$.

Note furthermore that, since no odd integer is a branch point of the
$PGL_{2}(7)$-extension $E(u)/\mathbb{Q}(u)$, all specializations at such
values are Galois extensions of $\mathbb{Q}$ with group a subgroup of
$PGL_{2}(7)$. Since the Galois group is generated by the set of all inertia
subgroups, we obtain a subgroup of $PGL_{2}(7)$ generated by involutions of
cycle type $(2^{3}.1^{2})$. Furthermore, the polynomial $F(1,X)$ is
irreducible (of degree $8$) modulo $2$, whence the Galois group for all
specializations at odd integers must contain an $8$-cycle (namely, the
Frobenius at $2$)! One may now verify directly that $PGL_{2}(7)$ has no proper
subgroup which is generated by involutions as above while also containing an
$8$-cycle. Hence $Gal(E(u)_{u_{0}}/\mathbb{Q})=PGL_{2}(7)$ without exception.
Propositions 3.4 and 3.5 now yield that $(E(u))_{u_{0}}/\mathbb{Q}$ embeds
into an extension $L/\mathbb{Q}$ with group $SL_{2}(7).2$, and if $F$ denotes
the fixed field of $SL_{2}(7)$ in $L$, then $L/F$ can be chosen everywhere
unramified. It remains to calculate the quadratic field $F$, which is just
given by the square-root of the discriminant of $F(u_{0},X)$ above.
Computation with Magma confirms the shape claimed in the statement of the
theorem. ∎

######  Remark 5.

A logical next step would be to try to realize more general central extensions
of $PSL_{2}(7)$ as Galois groups of unramified extensions of quadratic number
fields, using, e.g., Theorem 5.1. This is easy for the direct product
$SL_{2}(7)\times C_{2}$. Indeed, the values $-7\cdot(108+7^{10}t^{6})$
featuring in the assertion of Theorem 5.1 are all strictly divisible by $7$,
and the factor ($108+7^{10}t^{6}$) is not a square (due to being too close to
another square). Hence, the extension
$\mathbb{Q}(\sqrt{-7\cdot(108+7^{10}t^{6})},\sqrt{-7})/\mathbb{Q}(\sqrt{-7\cdot(108+7^{10}t^{6})})$
is an unramified quadratic extension (necessarily linearly disjoint from the
$SL_{2}(7)$-extension constructed above, which has no quadratic
subextensions). In comparison, the same question for the group
$SL_{2}(7)\times C_{3}$ already seems challenging: quadratic number fields
with class number divisible by $3$ can be parameterized effectively, but
requiring them to simultaneously have the shape of Theorem 5.1 leads to
difficult diophantine conditions. E.g., it is easy to exhibit infinitely many
quadratic number fields of the shape $\mathbb{Q}(\sqrt{-u(4u+27)})$ with class
number divisible by $3$ (for a more general result, see [8, Main Theorem]).
Demanding those to simultaneously fall into the shape of Theorem 5.1 amounts
to finding (suitable) integral points on the conic $U(4U+27)-7DV^{2}=0$, with
$D=108+7^{10}t^{6}$ for some odd integer $t$, as in the Theorem. I do not know
whether such points exist for infinitely many $D$ of the prescribed shape.

###  5.2. The case $G=2.A_{7}$

###### Proof.

We consider the splitting field $E/\mathbb{Q}(t)$ of $f(t,X)=X^{7}-7t(X-1)$.
This is well-known to be $\mathbb{Q}$-regular with Galois group $S_{7}$, and
inertia groups generated by a $7$-cycle (at $t\mapsto 0$), a $6$-cycle (at
$t\mapsto\infty$) and a transposition (at the only further branch point
$t_{3}$); indeed, this is a special case of the family $X^{n}-t(X-1)$
corresponding to a rigid class triple in $S_{n}$ for all $n\geq 3$ (e.g., [13,
Chapter I, Prop. 5.2]); our choice to replace $t$ by $7t$ is merely to make
$t_{3}$ a rational square $t_{3}=\frac{7^{6}}{6^{6}}$ and thus have
$\mathbb{Q}(\sqrt{t_{3}})=\mathbb{Q}$ as in the proof of Theorem 1.2. Note
that the normalizers of the cyclic groups generated by the $7$\- and the
$6$-cycle are the same as in $PGL_{2}(7)$, whence much of the argument from
that case carries over. The difference is that the centralizer of the
transposition is larger than the one of the involution dealt with in the case
$G=SL_{2}(7)$, namely isomorphic to
$\langle(1,2),\textrm{Sym}(\\{3,4,5,6,7\\})\rangle\cong C_{2}\times S_{5}$. We
need to exclude local obstructions to the embedding problem induced by
$2.S_{7}\to S_{7}$ (where $2.S_{7}$ denotes the double cover of $S_{7}$ in
which the transpositions - as well as the triple transpositions - are split),
for primes whose inertia group in an $S_{7}$-extension is generated by a
transposition. Obviously there is no obstruction if the Frobenius element
(inside $S_{5}$) is of odd order; furthermore, note that if the decomposition
group at $p$ is conjugate in $S_{7}$ to $\langle(1,2),(3,4)(5,6)\rangle$ (with
inertia group generated by $(1,2)$), then there is no local obstruction at $p$
either, since that group lifts to $C_{2}\times C_{4}\leq 2.S_{7}$, i.e., the
local embedding problem at $p$ is solvable (and without ramification in the
kernel) via extending the unramified quadratic extension of $\mathbb{Q}_{p}$
to the degree $4$ one. In total, there is no local obstruction as long as the
Frobenius is contained in $\textrm{Alt}(\\{3,4,5,6,7\\})\cong A_{5}$. Now we
are in a situation entirely analogous to the case $G=SL_{2}(7)$; namely, let
$\mathbb{Q}(\sqrt{\beta})/\mathbb{Q}$ be the quadratic subextension of the
residue extension at the branch point $t_{3}$.888One verifies quickly that
this residue extension is indeed an $S_{5}$-extension, and hence, has a unique
quadratic subextension, which in this concrete case equals
$\mathbb{Q}(\sqrt{3})$. Set $\beta t=s^{2}$, and proceed as in the proof of
Theorem 1.2 and the case $G=SL_{2}(7)$ of Theorem 1.1. ∎

######  Remark 6.

Turning the above proof into an explicit family of quadratic number fields
possessing unramified $2.A_{7}$-extensions works just like in the previous
section. We leave this to the interested reader.

Acknowledgement: I would like to thank Kwang-Seob Kim for many helpful
discussions.

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