# Solving non-separable polynomials over the field of Puiseux series via
golden lifting

RaGon Ebker
email<EMAIL_ADDRESS>
###### Abstract

We develop an iterative method to calculate the roots of arbitrary polynomials
over the field of Puiseux series including non-separable ones. The method
works by transforming the polynomial and its roots into a special form and
then extracting a new, univariate polynomial that contains information about
our roots. We also provide a working implementation of the algorithm in
Python.

Keywords— algebraic geometry, puiseux series, newton puiseux, power series

##  1 Notation

Let $\mathbb{N}$ be the set of natural numbers including $0$. Let $K$ be a
field and $K((x^{\frac{1}{n}}))$ be the field of Puiseux series over $K$. Let
the elements of $K((x^{\frac{1}{n}}))$ have the form
$y=\sum_{k=k_{0}}^{\infty}b_{k}x^{\frac{k}{n}},n\in\mathbb{Z}$. When $y$ has
just finitely many terms, we call $d_{x}$ the degree of $y$. Let
$Q:K((x))\mapsto K((x))$ be a polynomial over the field of Puiseux series,
$Q(y)=a_{d_{y}}y^{d_{y}}+...+a_{1}y+a_{0},d_{y}\in\mathbb{N^{+}}$, $d_{y}$
being the degree of $Q$. Let
$\alpha=\sum_{k=0}^{\infty}b_{k}x^{\frac{k}{n}},n\in\mathbb{Z}$ be a root of
$Q$.

######  Definition 1 (s-multiplicity).

A polynomial has $s-$multiplicty, when there exist $s$ roots
$\alpha_{1},...,\alpha_{s}$ of $Q$ with the coefficients $b_{0}$ all being
$0$.

######  Definition 2 (s-plus-multiplicity).

A polynomial $Q$ has $s^{+}-$Multiplicty, when there exist $s^{+}$ roots
$\alpha_{1},...,\alpha_{s^{+}}$ of $Q$, $s^{+}\in\\{1,...,s\\},s$ being the
$s-$ Multiplicity of Q with the coefficients $b_{0}$ all being $0$ and the
term $b_{1}$ has the same valuation for all
$\alpha_{j},j\in\\{1,...,s^{+}\\}$.

######  Example 1.

The polynomial $Q(y)=(y-(1+x+x^{2}))(y-x^{0.5})(y-x^{0.6})(y-x^{0.5}+x^{2})$
has $s$-Multiplicty 3 and $s^{+}$-Multiplicty 2.

Let $v:K((x))\mapsto\mathbb{Q}$ be the valuation map defined by $v(y)=k_{0}$.

##  2 Introduction

Let’s imagine for a moment, that we have an algorithm to calculate the
smallest root of a polynomial $Q$, for example,
$Q(y)=(y-(1+x+x^{2}))(y-(2+x+x^{2}))$ over the field of Puiseux Series. In
this case, the smallest root is $(1+x+x^{2})$. Small means the root with the
leading coefficient that has the smallest valuation, in this case. But we can
only calculate its term with the highest evaluation, i.e. $1$.
How can we proceed to calculate the next terms of the root? The answer may
seem obvious: By shifting the roots of the polynomial $Q$. Now we have the
shifted Polynomial

$Q_{shift}=(y-(x+x^{2}))(y-(1x+x^{2}))$

and we can easily calculate the next term with our algorithm: $x$. In this
paper we are going to explore this idea, mainly:
1\. How to develop an algorithm to calculate the smallest root under certain
conditions
2\. How to transform and shift our polynomial to fulfill this condition
A good overview of this process can also be seen in section 5. Comparable
methods to solve this problem over the Puiseux series and power series exist
for example in the form of Hensels Lemma and its developed versions [1] or in
the Newton Puiseux method [3].

##  3 Main Result

In this section, we show the main result. It works by reducing our polynomial
to a smaller one, under certain requirements. In section 4 we are going to see
how we can transform every polynomial into one that satisfies these exact
requirements. We calculate the root of the polynomial coefficient by
coefficient just as in the original Newton-Puiseux algorithm.

######  Theorem 1 (Golden Lifting).

Let $Q:K((x))\mapsto K((x))$ be a polynomial over the field of Puiseux series,
$Q(y)=a_{d_{y}}y^{d_{y}}+...+a_{1}y+a_{0},d_{y}\in\mathbb{N^{+}}$. Let
$\alpha_{1},...,\alpha_{n}$ be the roots of our polynomial and let
$v(a_{i})\geq 0,i\in\\{1,...,d_{y}\\}$. Now we assume $v(\alpha_{j})\geq e>0\
,j\in\\{1,...,s\\}$, $s\in\\{1,...,d_{y}-1\\},e\in\mathbb{Q}$, with
$v(\alpha_{j})=e$ for $j\in\\{1,...,s^{+}\\},s^{+}>1$. $s$ is exactly the
$s-$multiplicity of $\alpha_{j}$, $s^{+}$ the $s^{+}$-Multiplicity. $e$ is the
smallest valuation of all $\alpha_{j}$. We can further represent those roots
as
$\alpha_{j}=c_{1_{j}}x^{e}+c_{2_{j}}x^{e+\gamma_{2_{j}}}+c_{3_{j}}x^{e+\gamma_{2_{j}}+\gamma_{3_{j}}}...,\gamma_{j}\in\mathbb{Q},j\in\\{1,...,s^{+}\\}.$
Now the aim of this theorem is to calculate the
$c_{1_{j}},j\in\\{1,...,s^{+}\\}$.
We further assume $v(\alpha_{j})=0,\ \forall j\in\\{s+1,...,d_{y}\\}.$We now
have a look at the coefficients of $Q$. We remember

$a_{i}=b_{i_{0}}x^{\delta_{1}}+b_{i_{1}}x^{\delta 1+\delta_{2}}+...,\
\delta_{k}\in\mathbb{Q},k\in\mathbb{N^{+}}.$

Let now $Q_{R}(x)=b_{s_{min}}x^{d_{y}}+...+b_{1_{min}}x$ be such that
$b_{i_{min}}:=\min_{l\in\\{1,...,d_{x}\\}}\\{b_{i_{l}}\neq
0\\},i\in\\{1,...,s\\}$.
Then the $s$ roots of $Q_{R}$ are exactly the $c_{1_{j}}$. Then $Q_{R}$ is of
degree $s$, and $c^{+}$ roots are exactly the $c_{1_{j}}j\\{1,...,s^{+}\\}$
belonging to the the roots with the lowest valuation. The other $s-s^{+}$
roots are zero.

###### Proof.

We know that $c_{1_{j}},j\in\\{1,...,s^{+}\\}$ being the correct coefficient
of a root of $Q$ is equivalent to $Q(c_{1_{j}}x^{e})\mod x^{e^{\prime}}\equiv
0$, i.e. $c_{1_{j}}x$ is a root of $Q(y)\mod x^{e^{\prime}}$.
$e^{\prime}\in\mathbb{Q}$ is in this case any number bigger than $s\cdot e$
and smaller than the exponent of any term of $Q$, that has a valuation bigger
than $e$. This is can also be seen by looking at the polynomial in its linear
factorization. We show that $Q(\beta x^{e})=Q_{R}(\beta x^{e})\mod
x^{e^{\prime}}$:

$\displaystyle Q(\beta x^{e})$ $\displaystyle\equiv a_{d_{y}}(\beta
x^{e})^{d_{y}}+...+a_{1}\beta x^{e}+a_{0}\mod x^{e^{\prime}}$
$\displaystyle\equiv a_{d_{y}}^{*}(\beta x^{e})^{d_{y}}+...+a_{1}^{*}(\beta
x^{e})+a_{0}^{*}+\left(b_{d_{y_{min}}}(\beta
x^{e})^{d_{y}}+...+b_{1_{min}}(\beta x^{e})+b_{0_{min}}\right)\mod
x^{e^{\prime}}$ $\displaystyle\equiv b_{d_{y_{min}}}(\beta
x^{e})^{d_{y}}+...+b_{1_{min}}(\beta x^{e})+b_{0_{min}}\mod x^{e^{\prime}}$
$\displaystyle\equiv b_{s_{min}}(\beta x^{e})^{s}+...+b_{1_{min}}(\beta
x^{e})+b_{0_{min}}\mod x^{e^{\prime}}$

For $a_{i}^{*}=a_{i}-b_{i_{min}},i\in\\{1,...,d_{y}\\}$. ∎

So to conclude: When we solve $Q_{R}$ we obtain the solutions
$c_{1_{j}}x^{e}$. We will explain how to solve $Q_{R}$ with the help of shifts
in section 4. To apply this theorem, we need to fulfill the condition
$v(\alpha_{j})\geq e>0,j\in\\{1,...,s\\}$ with $v(\alpha_{j})=e$ for at least
one $j\in\\{1,...,s\\}$ and $v(\alpha_{k})=0,\ \forall
k\in\\{s+1,...,d_{y}\\}$ (1). We also need to obtain $e$ (2) and $s$ (3). Once
we have applied this step, we can extract (1), (2), and (3), for the next
step, from the set of roots of $Q_{R}$. We can proceed equally to encounter
the next coefficients $c_{2_{k}},...,c_{d_{y_{l}}}$, after transforming our
polynomial to fulfill our assumption again. $k$ and $l$ are the corresponding
indices.

##  4 Initial shift of the Polynomial

In this section, we are going to prove how to transform a polynomial
$Q:K((x^{\frac{1}{n}}))\mapsto K((x^{\frac{1}{n}}))$ into one that fulfills
the condition (1), if it does not already, and how to extract (2) and (3). We
first start with a commonly know Lemma:

######  Lemma 1.

Suppose we have a polynomial $P$ over the $K$, with
$\alpha_{1},...,\alpha_{n},n\in N^{+}$ being its roots. Then for any constant
$c\in K$ the polynomial with roots $\alpha_{1}+c,...,\alpha_{n}+c$ has the
form

$Q(y)=P(y-c)=a_{n}(y-c)^{n}+a_{n-1}(y-c)^{{n-1}}+\cdots+a_{{0}}.$

and the polynomial with the roots $c\alpha_{1},...,c\alpha_{n}$ has the form

$Q(y)=a_{n}y^{n}+a_{n-1}cy^{{n-1}}+\cdots+a_{{0}}c^{n}.$

We now start with (1). We can check (1) by calculating the constant parts of
the roots via $Q_{|x=0}$. If we have at least one root of $Q_{|x=0}$ which is
unequal to zero and one that is equal to zero we fulfill the conditions.
Checking $Q_{|x=0}$, mainly to get the constant part of a root of $Q$, is a
common technique and can be seen by looking at $Q$ in linear factor
representation.
CASE 1
First we want to ensure that $v(\alpha_{j})=0$ holds for at least one
$i\in\\{1,...,d_{y}\\}$. What if all roots of $Q_{|x=0}$ are zero for all
$j\in\\{1,...,d_{y}\\}$? In this case, we shift our polynomials via
multiplication as in Lemma 1. For this end, we actually need to know $e$, the
valuation of $\alpha_{j}$, which we can obtain via the Newton Polygon of our
polynomial. We thus multiply our roots with $a^{e}$.
Going back to our now eventually shifted polynomial, which has
$v(\alpha_{j})=0$ holds for at least one $j\in\\{1,...,d_{y}\\}$.

CASE 2
Now we check if $v(\alpha_{j})>0$ for at least one $j\in\\{1,...,d_{y}\\}$
Suppose we have $v(\alpha_{j})=0,\forall i\in\\{1,...,n\\}$, then we can get
the constant term of the $\alpha_{j}$ by evaluating $Q_{|x=0}$ and taking its
roots, as we have already discussed. Once we have obtained all the constant
parts $c_{j}$ of $\alpha_{j}$ we are going to use them to shift our polynomial
with help of Lemma 1. If we end up with a polynomial with $v(\alpha_{j})>0$
for all $j\in\\{1,...,d_{y}\\}$, we go back to case 1. If not we have one with
the desired condition.
Now we can finally talk about the case when (1) is fulfilled: We calculate the
roots of $Q_{|x=0}$ or take the already calculated roots, and one of them is
now unequal to zero. Let’s call it $\alpha_{k},k\in\\{1,...,d_{y}\\}.$ In this
step we already obtain the $s$ -multiplicity of that root, which is exactly
the multiplicity we need for our next step. After this calculation, we shift
again, and so on, reaching our root iteratively.

##  5 Algorithm

In this section, we are going to explore the algorithm in detail. An
implementation can be found at [4]. First, we have a look at the flow diagram
in figure 1.

Figure 1: Flowchart of the algorithm

1 def info_constructor(self,d,root_dict_list,x_lift):

2 self.d = d

3 self.root_dict_list = root_dict_list

4 self.x_lift = x_lift

Algorithm 1 info class

For the implementation, we first define an info class that saves all of our
important information. When initializing the algorithm, we create an object
from the info class, with specified precision $d$, an empty root_dict_list,
and an x_lift of $0$. The current alpha can always be calculated by combining
the parts of our root from root_dict_list. We then give the info object to the
method calculate_smallest_root, our main method:

1

Input : Polynomial $p$, info object

Output : The info object that contains, among other information, the root
alpha of $p$ with the lowest valuation

2 p_shift $\leftarrow$ calculate_initial_shift(p,info) ;

3 if  _p_shift(info.alpha) == 0_ then

4 return

5 for  _$i\leftarrow 1$ to info.d_ do

6 info.d $\leftarrow$ info.d -1 ;

7 p_shift = shift_horizontally(p_shift,calculate_h_shift(info)) ;

8 multiplicity = info.last_root.multiplicity ;

9 golden_lifting(p_shift,multiplicity,info) ;

10 if  _p_shift(info.alpha) == 0_ then

11 return

12

return

Algorithm 2 calculate_smallest_root

As we can see in 2, we follow the process described in sections 3 and 4. We
first calculate the initial shift, to bring our polynomial into the form
$v(\alpha_{j})\geq e>0,j\in\\{1,...,s\\}$ with $v(\alpha_{j})=e$ for at least
one $j\in\\{1,...,s\\}$ and $v(\alpha_{k})=0,\ \forall
k\in\\{s+1,...,d_{y}\\}$, as described in (1). In this process, we already
obtain the first part of our first root alpha. This also brings us the $s-$
multiplicity of the next part of the root. In the for loop, starting on line
4, we start shifting our polynomial horizontally, so it attains the form (1)
again. Shifting a polynomial horizontally means adding a constant to its root.

Input : Polynomial $p$, info object

Output : Shifted polynomial $p$ or p_shift

1 root_dict, shift_number = get_sub_x_root(p,info)

2 if  _shift_number == 0_ then

3 info.root_dict_list.append(root_dict)

4 return $p$

5 else

6 slopes $\leftarrow$ get_newton_slopes(p)

7 min_slope $\leftarrow$ min_not_zero(slopes)

8 p_shift $\leftarrow$ shift_vertically(p,x**(-min_slope))

9 info.x_lift $\leftarrow$ min_slope

10 info.d $\leftarrow$ info.d-1

11 info $\leftarrow$ calculate_smallest_root_q_x(p_shift,info)

12 return p_shift

Algorithm 3 calculate_initial_shift

The initial shift method calculates the roots of $p_{|x=0}$ in line 1 with
get_sub_x_root. Then we shift our polynomial vertically if all of the roots,
in this case described by shift_number, are unequal to zero. Vertical shifting
means the multiplication of the roots with a constant.

Input : Polynomial $p$, info object

Output : Shifted polynomial $p$ or p_shift

1 root_dict, shift_number = get_sub_x_root(p,info)

2 if  _shift_number == 0_ then

3 info.root_dict_list.append(root_dict)

4 return

5 else

6 slopes $\leftarrow$ get_newton_slopes(p)

7 min_slope $\leftarrow$ min_not_zero(slopes)

8 p_shift $\leftarrow$ shift_vertically(p,x**(-min_slope))

9 info.x_lift $\leftarrow$ min_slope

10 info.d $\leftarrow$ info.d-1

11 info $\leftarrow$ calculate_smallest_root_q_x(p_shift,info)

12 return

Algorithm 4 calculate_smallest_root_q_x

Algorithm 4 is the same as 3 but it does not return anything.

Input : p_shift,multiplicity,info

1 shifted_coeffs $\leftarrow$ reversed(p_shift.coeffs())

2 cutoff_coeffs $\leftarrow$ []

3 for  _$i\leftarrow 1$ to info.d_ do

4 coeff $\leftarrow$ shifted_coeffs[i]

5 cutoff_coeffs.append(term_with_lowest_valuation(coeff))

6new_poly $\leftarrow$ Monic(Poly(reversed(cutoff_coeffs)))

7 if  _multiplicity $>$ 1_ then

8 calculate_smallest_root_q_x(new_poly,info)

9 return

10 else

11 r $\leftarrow$ roots(new_poly)

12 info.root_dict_list.append(order_roots(r)) return

Algorithm 5 golden_lifting

The algorithm 5 describes exactly the process of extracting $p_{R}$ from our
polynomial $p$, just as we extracted $Q_{R}$ from $Q$ in theorem 1. Here it is
also possible to check if the polynomial $p_{R}$, which is called new_poly in
the code and pseudocode, has a very simple form for example when it consists
only of a single root with multiplicity $s$. This would mean that its
$s$-multiplicity and $s^{+}$ multiplicity are the same. We then start the same
process as in the initial_shift method to calculate the roots of $p_{R}$ i.e.
new_poly.

## References

  * [1] Neiger, V., Rosenkilde, J. & Schost, E. Fast Computation of the Roots of Polynomials Over the Ring of Power Series. (2017,5)
  * [2] Willis, N., Didier, A. & Sonnanburg, K. How to Compute a Puiseux Expansion. (2008)
  * [3] Brieskorn, E. & Knörrer, H. Plane Algebraic Curves: Translated by John Stillwell. (Springer Science & Business Media,2012)
  * [4] Ebker, R. puiseux_solver. (GitHub,2023), https://github.com/RagonEbker/puiseux_solver


# On a non trivial self-consistent backreaction of quantum fields in 2D
dilaton gravity

E. T. Akhmedov  Moscow Institute of Physics and Technology, Institutskii per.
9, 141700, Dolgoprudny, Russia  NRC “Kurchatov Institute”, 123182, Moscow,
Russia  P. A<EMAIL_ADDRESS>Moscow Institute of
Physics and Technology, Institutskii per. 9, 141700, Dolgoprudny, Russia  NRC
“Kurchatov Institute”, 123182, Moscow, Russia  K. V.
<EMAIL_ADDRESS>Moscow Institute of Physics and Technology,
Institutskii per. 9, 141700, Dolgoprudny, Russia  NRC “Kurchatov Institute”,
123182, Moscow, Russia

###### Abstract

We consider $(1+1)$-dimensional dilatonic black hole with two horizons,
canonical temperatures of which do not coincide. We show that the presence of
quantum fields in such a background leads to a substantial backreaction on the
metric: 2D dilatonic analog of the semiclassical Einstein equations are solved
self-consistently, and we demonstrate that taking into account of backreaction
leads to a geometry with two horizons with coinciding temperatures.

##  1 Introduction

Spacetimes with multiple horizons are interesting backgrounds to consider from
the perspective of consistency and/or stability when quantum fields are added.
The notable example is the Schwarzschild-de Sitter spacetime. It is known that
quantum fields in such a background in Euclidean signature inevitably lead to
the conical singularity [1, 2, 3, 4, 5], whose presence poses a question of
the consistency of quantum field theory in the backgrounds with multiple
horizons.

Here, however, we adopt a different logic. Namely, one can consider quantum
fields in any state, which has a sensible stress energy tensor (e.g. conserved
etc.) in backgrounds with Lorentzian signature. But if a quantum field is
taken in a spacetime with a horizon e.g. in a thermal state which does not
coincide with the canonical temperature (Unruh, Hawking or Gibbons-Hawking,
correspondingly) then its regularized stress-energy tensor diverges at the
horizon. We interpret such a situation as leading just to a strong
backreaction on the geometry by the quantum fields (see e.g. [6, 7] and [8, 9,
10, 11, 12, 13, 14, 15]).

Furthermore, if the geometry in question has multiple horizons, and the
canonical temperatures of these horizons do not coincide, any thermal state of
quantum fields leads to divergence of the regularized stress-energy tensor on
at least one of the horizons. Then in the situation with multiple horizons the
regularized stress-energy tensor inevitably has a divergence, which causes
strong backreaction on the background geometry. In this paper we consider a
solvable example of 2D quantum field theory where such a conclusion can be
checked. In this note we restrict our attention to the case when quantum
fields are considered on a classical gravitational background, i.e.
gravitational field is considered to be classical.

In [16] the authors have considered the two-dimensional analogue of
Schwarzschild-de Sitter spacetime, and argued that for the thermal states the
stress-energy tensor diverges at least on one of the horizons. They also
conclude that the divergence of stress-energy tensor signals that the state of
quantum fields is unphysical. Then, they attempt to construct a state which is
regular on both horizons. We, however, adopt a different logic, that we have
mentioned above and which is in accordance with [17]. The point is that
thermal states in stationary backgrounds have a notable advantage in
comparison with other states: according to the fluctuation-dissipation theorem
they are stable under quantum corrections for self-interacting fields333In
[16] they consider a gaussian theory in the background in question. In this
note we also restrict our attention to the gaussian theory, but keep in mind
the effect of self-interactions. Also in two dimensions the situation with the
fluctuation-dissipation theorem is different from higher dimensions due to the
kinematic reasons, but we consider 2D case as the model example..

The problem of backreaction of matter fields on background geometries has been
addressed by many authors (see e.g. [6, 7, 18] for the related works). In
these papers the Einstein equations

$\displaystyle G^{\,\,\,\mu}_{\nu}+\Lambda\delta^{\,\,\,\mu}_{\nu}=8\pi
G\langle:\hat{T}^{\,\,\,\mu}_{\nu}:\rangle,$ (1.1)

are solved perturbatively:

  * •

First, solution $g^{(0)}$ of the Einstein equations in the vacuum (i.e. with
the zero on the r.h.s.) is considered (usually a black hole with $\Lambda=0$);

  * •

Second, the stress-energy tensor for quantum fields in the background
$g^{(0)}$ is calculated;

  * •

Third, the Einstein equations are solved perturbatively for a new metric $g$
on the l.h.s., while on the r.h.s. one plugs the stress-energy tensor
calculated with the metric $g^{(0)}$.

In this approach, the new metric $g$ is usually divergent at the position of
the horizon.

In our paper we take a different approach. Namely, we solve the 2D analog of
Einstein equations (in dilatonic gravity theory) self-consistently, i.e. the
stress-energy tensor is calculated for the same metric which is plugged on the
l.h.s. of these equations. Of course, for this method to work, one has to know
the form of the expectation value of the stress-energy tensor in a generic
background. That is the reason why at this stage we have to confine our
analysis to the 2D situation, where the exact form of the regularized stress-
energy tensor is known at any point in spacetime for arbitrary metric.

Note that there is another method to tackle the backreaction problem using the
effective action approach (see, e.g., [19, 20, 21, 22, 23, 24, 25, 26, 27,
28]). The advantage of our method is that we can explicitly calculate the
quantum average of stress-tensor for any level population (see (3.10)).

Usually to make the gravity in 2D dynamical one has to add the dilaton field:

$\displaystyle S^{\text{grav}}=\frac{1}{16\pi G}\int
d^{2}x\sqrt{-g}e^{-2\phi}\Big{[}R-4\omega(\partial_{\mu}\phi)^{2}+4\lambda^{2}\Big{]},$
(1.2)

where $\omega$ is an arbitrary constant. For particular values of this
$\omega$ parameter this action reduces to the known theories:

  * •

$\omega=0$ case is the Jackiw-Teitelboim theory;

  * •

if $\omega=-\frac{1}{2}$ one obtains planar general relativity;

  * •

$\omega=-1$ one has the first-order string theory.

To make the kinetic term canonical one can always redefine the dilaton field,
but then the dilaton coupling is modified.

The equations of motion following from the variation of this action with
respect to the metric and dilaton field, correspondingly, are as follows:

$\displaystyle e^{2\phi}T_{\mu\nu}^{\text{grav}}\equiv-2(\omega+1)D_{\mu}\phi
D_{\nu}\phi+D_{\mu}D_{\nu}\phi-
g_{\mu\nu}D_{\mu}D^{\nu}\phi+(\omega+2)g_{\mu\nu}D_{\mu}\phi D^{\mu}\phi-
g_{\mu\nu}\lambda^{2}=0,$ (1.3) $\displaystyle R-4\omega
D_{\mu}D^{\mu}\phi+4\omega D_{\mu}\phi D^{\mu}\phi+4\lambda^{2}=0.$ (1.4)

The comprehensive analysis of solutions of these equations of motion for
arbitrary $\omega$ can be found in [29]. For a particular choice of the
integration constant for the dilaton, the solution of these equations of
motion can be represented in the following form:

$\displaystyle\phi=-\log(ar)^{\frac{1}{2(\omega+1)}},$ $\displaystyle
ds^{2}=-[a^{2}r^{2}-(ar)^{\frac{\omega}{\omega+1}}]dt^{2}+\frac{dr^{2}}{a^{2}r^{2}-(ar)^{\frac{\omega}{\omega+1}}},$
(1.5)

for some $a$ related to $\lambda$ and to be defined below.

We begin in Sec. 2 by discussing the properties of this solution for the case
when $\omega=-\frac{4}{3}$. The most important property of such a solution is
the presence of two horizons with different temperatures. Then, in Sec. 3 we
add gaussian quantum fields that live in the background of this geometry, and
demonstrate that the regularized stress-energy tensor diverges at least on one
of the horizons. In Sec. 4 we solve semiclassical gravitational equations and
show that the resulting backreacted geometry has two horizons with coinciding
temperatures.

##  2 Starting geometry

Solutions (1) with different $\omega$ correspond to different physical
situations, and we want to consider the most similar one to the black hole in
de Sitter space-time. We are looking for multiple horizon scenario, where
horizons have different temperatures (surface gravity). To model this
situation, we propose to consider the case $\omega=-\frac{4}{3}$. Then the
metric takes the form:

$ds^{2}=-[a^{2}r^{2}-a^{4}r^{4}]dt^{2}+\frac{dr^{2}}{a^{2}r^{2}-a^{4}r^{4}},$
(2.1)

while the dilaton field acquires the form:

$\displaystyle\phi=\frac{3}{2}\log(ar),$ (2.2)

where $a=-\sqrt{\frac{2\lambda^{2}}{3}}$. The Ricci scalar for the metric in
question is $R=-\frac{4}{3}\lambda^{2}\left[1-6(ar)^{2}\right]$. This metric
is defined for $r>0$ and possesses two horizons: at $r=0$ and at $r=1/a$. The
canonical temperatures at the horizons are:

$T=\begin{cases}\frac{a}{2\pi},\qquad r\to 1/a,\\\ 0,\,\,\,\,\qquad r\to
0.\end{cases}$ (2.3)

Note that the zero of the function $[a^{2}r^{2}-a^{4}r^{4}]$ at $r=0$ is
degenerate. As a result, the corresponding temperature is vanishing. The
Penrose-Carter diagram for this spacetime is depicted on the Fig.1.

Figure 1: Maximal analytical extension of the metric (2.1). There are infinite
copies of this diagram extending both in vertical and horizontal direction.
Coordinates (2.1) cover only chart I (highlighted by green on the diagram).
The boundary of the chart are lightlike surfaces corresponding to $ar=0$ and
$ar=1$.

For such space-times as we consider it is useful to use other (tortoise)
coordinates:

$ds^{2}=-\left[a^{2}r^{2}-a^{4}r^{4}\right]\,\left(dt^{2}-dr_{*}^{2}\right),$
(2.4)

where

$r_{*}=-\frac{1}{a^{2}r}+\frac{1}{a}\text{arctanh}(ar).$ (2.5)

Or using Eddington-Finkelstein coordinates

$u=t-r_{*},\qquad v=t+r_{*},$ (2.6)

we can rewrite the metric in the form:

$ds^{2}=C(u,v)dudv,\qquad C(u,v)=-a^{2}r^{2}(u,v)+a^{4}r^{4}(u,v),$ (2.7)

where $r(u,v)$ is determined from

$\frac{v-u}{2}=-\frac{1}{a^{2}r}+\frac{1}{a}\text{arctanh}(ar).$ (2.8)

Using substitution

$\displaystyle r=\frac{1}{a\cosh(aX)},$ (2.9)

we can write the metric in the form

$\displaystyle ds^{2}=-e^{2\nu(X)}dt^{2}+dX^{2},$ (2.10)

where

$e^{2\nu(X)}=\frac{\sinh^{2}(aX)}{\cosh^{4}(aX)}.$ (2.11)

In these coordinates, the dilaton field is equal to

$\phi(X)=-\frac{3}{2}\log\cosh(aX).$ (2.12)

In this note we use all these coordinate systems in different situations,
depending on where it is convenient to use one of them.

##  3 Adding quantum fields

The next step in our reasoning is to add quantum fields. We consider real
conformal field $\varphi$ with the action:

$\displaystyle
S=S^{\text{grav}}+S^{\text{matter}}=S^{\text{grav}}-\frac{1}{2}\int
d^{2}x\partial_{\mu}\varphi\partial^{\mu}\varphi,$ (3.1)

just as in e.g. [19, 20, 21, 22, 23], but rather than integrating the field
$\varphi$ out we explicitly quantize it and calculate the regularized stress-
energy tensor. We will adopt the canonical quantization.

It is useful to use the Eddington-Finkelstein coordinates (2.7), in which the
equations of motion for the scalar field take the form:

$\displaystyle\partial_{U}\partial_{V}\varphi(U,V)=0.$ (3.2)

Thus, we can write down the mode decomposition of the field operator as:

$\displaystyle\hat{\varphi}(U,V)=\int_{0}^{\infty}\frac{d\omega}{\sqrt{4\pi\omega}}\Big{[}\hat{a}^{\dagger}_{\omega}e^{i\omega
U}+\hat{b}^{\dagger}_{\omega}e^{i\omega V}+\text{h.c.}\Big{]}.$ (3.3)

Commutation relations for the creation and annihilation operators for the left
and right moving modes are:

$\displaystyle[\hat{a}^{\dagger}_{\omega},\hat{a}_{\omega^{\prime}}]=\delta(\omega-\omega^{\prime}),\quad[\hat{b}^{\dagger}_{\omega},\hat{b}_{\omega^{\prime}}]=\delta(\omega-\omega^{\prime}).$
(3.4)

We take the quantum field in the thermal state, for which:

$\displaystyle\langle\hat{a}_{\omega}^{\dagger}\hat{a}_{\omega^{\prime}}\rangle=\langle\hat{b}_{\omega}^{\dagger}\hat{b}_{\omega^{\prime}}\rangle=\frac{1}{e^{\beta\omega}-1}\delta(\omega-\omega^{\prime}),$
(3.5)

where $\beta=1/T$ is the inverse temperature.

Then, the regularized stress-energy tensor is given by [30]:

$-\langle:T_{\mu\nu}^{\text{matter}}:\rangle=\Theta_{\mu\nu}+\frac{R}{48\pi}g_{\mu\nu},$
(3.6)

where

$\displaystyle\Theta_{uu}$
$\displaystyle=\frac{1}{48\pi}\bigg{(}\frac{2\pi}{\beta}\bigg{)}^{2}-\frac{1}{12\pi}C^{1/2}\partial^{2}_{u}C^{-1/2},$
$\displaystyle\Theta_{vv}$
$\displaystyle=\frac{1}{48\pi}\bigg{(}\frac{2\pi}{\beta}\bigg{)}^{2}-\frac{1}{12\pi}C^{1/2}\partial^{2}_{v}C^{-1/2},$
(3.7) $\displaystyle\Theta_{uv}$ $\displaystyle=\Theta_{vu}=0,$

and $C(u,v)$ here is equal to (2.7). Then, plugging the expression for the
conformal factor, $C(u,v)$, we find that

$\Theta_{uu}=\begin{cases}\frac{1}{48\pi}\bigg{[}\big{(}\frac{2\pi}{\beta}\big{)}^{2}-a^{2}\bigg{]},\qquad\text{as}\quad
r\to\frac{1}{a},\\\
\frac{1}{48\pi}\big{(}\frac{2\pi}{\beta}\big{)}^{2},\qquad\qquad\quad\text{as}\quad
r\to 0,\end{cases}$ (3.8)

and the same expression for $\Theta_{vv}$. We see that one can choose the
inverse temperature $\beta$ such that $\Theta_{\mu\nu}$ is vanishing only on
one of the horizons. If $\Theta_{\mu\nu}$ has a non-zero value on a horizon,
it implies that $T^{\mu}_{\nu}$ and $T^{\mu\nu}$ blow up on this horizon,
which leads to the substantial backreaction on the geometry.

In the coordinates (2.10) the stress energy tensor has the following
form444Note, that the stress energy tensor is conserved
$D_{\mu}\langle:T^{\mu\nu}_{\text{matter}}:\rangle=0$.:

$\displaystyle-\langle:$ $\displaystyle
T_{tt}^{\text{matter}}:\rangle=\frac{\pi}{6\beta^{2}}+\frac{e^{2\nu(X)}}{24\pi}\Big{[}\nu^{\prime}(X)^{2}+2\nu^{\prime\prime}(X)\Big{]},$
$\displaystyle-\langle:$ $\displaystyle
T_{XX}^{\text{matter}}:\rangle=\frac{e^{-2\nu(X)}\pi}{6\beta^{2}}-\frac{\nu^{\prime}(X)^{2}}{24\pi}.$
(3.9)

Below we will use these expressions to find a self-consistent solution of the
modified equations of motion.

We consider only thermal states in our calculations. Such a state has level
population given by the planckian distribution. However, our method allows one
to extend the calculation for any level population. For example, for generic
level population $n(\omega)$ energy density is given by the following
expression:

$\displaystyle-\langle:$ $\displaystyle
T_{tt}^{\text{matter}}:\rangle=\int_{-\infty}^{\infty}\frac{\omega
d\omega}{2\pi}\Big{[}n(\omega)+\Theta(-\omega)\Big{]}+\frac{e^{2\nu(X)}}{24\pi}\Big{[}\nu^{\prime}(X)^{2}+2\nu^{\prime\prime}(X)\Big{]},$
(3.10)

where

$\displaystyle\langle\hat{a}_{\omega}^{\dagger}\hat{a}_{\omega^{\prime}}\rangle=\langle\hat{b}_{\omega}^{\dagger}\hat{b}_{\omega^{\prime}}\rangle=n(\omega)\delta(\omega-\omega^{\prime}),$
(3.11)

and $\Theta(-\omega)$ is the Heaviside step function. If
$n(\omega)=[e^{\beta\omega}-1]^{-1}$, then:

$\displaystyle\int_{-\infty}^{\infty}\frac{\omega
d\omega}{2\pi}\Big{[}\frac{1}{e^{\beta\omega}-1}+\Theta(-\omega)\Big{]}=\frac{\pi}{6\beta^{2}}$
(3.12)

and we obtain (3).

##  4 Backreaction of quantum fields on the background geometry

Now, let us continue with the solution of the 2D dilatonic analog of the
semiclassical Einstein equations (1.1) self-consistently, i.e. by plugging in
it an arbitrary metric and regularized stress-energy tensor of the matter
fields calculated in the background of the same metric. In the 2D dilaton
gravity that we consider here these equations of course look differently from
(1.1).

Let us start with writing down non-zero components of the gravitational stress
energy tensor in the unitary gauge (2.10)555Note, that the third equation
$\frac{\delta S^{\text{grav}}}{\delta\phi(X)}=0$ is satisfied if
$T^{\text{grav}}_{\mu\nu}=8\pi G\langle:T^{\text{matter}}_{\mu\nu}:\rangle$.:

$\displaystyle
e^{2\phi(X)}T^{\text{grav}}_{tt}=e^{2\nu(X)}\big{[}\lambda^{2}-\frac{2}{3}\phi^{\prime}(X)^{2}+\phi^{\prime\prime}(X)\big{]},$
$\displaystyle
e^{2\phi(X)}T^{\text{grav}}_{XX}=-\lambda^{2}-\phi^{\prime}(X)\nu^{\prime}(X)+\frac{4}{3}\phi^{\prime}(X)^{2}.$
(4.1)

Then, the 2D dilatonic analog of Einstein equations (1.1) reduces to:

$\displaystyle\lambda^{2}-\frac{2}{3}\phi^{\prime}(X)^{2}+\phi^{\prime\prime}(X)=8\pi
Ge^{2\phi(X)}\bigg{(}\frac{e^{-2\nu(X)}\pi}{6\beta^{2}}+\frac{1}{24\pi}\Big{[}\nu^{\prime}(X)^{2}+2\nu^{\prime\prime}(X)\Big{]}\bigg{)},$
$\displaystyle-\lambda^{2}-\phi^{\prime}(X)\nu^{\prime}(X)+\frac{4}{3}\phi^{\prime}(X)^{2}=8\pi
Ge^{2\phi(X)}\bigg{(}\frac{e^{-2\nu(X)}\pi}{6\beta^{2}}-\frac{\nu^{\prime}(X)^{2}}{24\pi}\bigg{)},$
(4.2)

where now $\phi(X)$ and $\nu(X)$ are unknown functions to be determined from
these equations.

When $\beta=0$ (the case of interest for us, as is explained below), these
equations have an obvious solution with $\nu(X)$ being a linear function of
$X$ and $\phi$ being a constant. The constants in the solution should be
suitably adjusted according to the values of $\lambda$ and $G$. However, that
is not a solution that we are looking for. Because our solution should have an
appropriate asymptotic behavior at one of the horizons: we want to find what
happens under backreaction (due to the quantum fields) with the solution with
two horizons.

To find the solution of (4.2) with appropriate asymptotic behavior we have to
apply numerical methods. To do that, we need to specify boundary conditions.
We will assume, that the horizon at $r=0$ ($X=\infty$) remains unchanged.
Comparing with (3.8), one can see that this conditions implies that
$1/\beta=0$ (in other words, the state of the field is the Fock space ground
state). Hence, Eq. (4.2) for the dimensionless variable $x=|a|X$ (with
$a=-\sqrt{\frac{2\lambda^{2}}{3}}$) reads666Here we have set $G=1$.:

$\displaystyle\frac{3}{2}-\frac{2}{3}\phi^{\prime}(x)^{2}+\phi^{\prime\prime}(x)=\frac{e^{2\phi(x)}}{3}\Big{[}\nu^{\prime}(x)^{2}+2\nu^{\prime\prime}(x)\Big{]},$
$\displaystyle-\frac{3}{2}-\phi^{\prime}(x)\nu^{\prime}(x)+\frac{4}{3}\phi^{\prime}(x)^{2}=-\frac{e^{2\phi(x)}}{3}\nu^{\prime}(x)^{2}.$
(4.3)

The solution of these equations for the dilaton field and for the
$e^{2\nu(x)}$ have the form shown on the Fig. 2.

Figure 2: Numerical plot of the dilaton field $\phi(x)$ and for the
$e^{2\nu(x)}$ with and without account of backreaction. Note that the solution
without backreaction is defined only for $X>0$, while solution with the
backreaction is defined on the whole $X$ axis.

These forms of the solutions could have been guessed from the following
perspective: the solutions of (4.3) should interpolate from the form of (2.12)
and (2.11) at $x\gg x_{0}$ (where $x_{0}$ is a reference gluing point) to the
solution

$\phi(x)\approx\phi_{0}=\text{const},\qquad\nu(x)\approx\frac{3}{\sqrt{2}}e^{-\phi_{0}}x+\text{const},$
(4.4)

at $x\ll x_{0}$, as follows from (4.3). As can be seen from the Fig. 2 these
are precisely solutions obeying such boundary conditions.

There is one important point to note here. The solutions (2.12) and (2.11) are
defined in the region $X>0$ (from single-valuedness of (2.9)), but the
backreacted geometry of the Fig. 2 is defined on the entire axis
$X\in(-\infty;+\infty)$. The new geometry has two horizons at $X=\pm\infty$,
and canonical temperatures at these two horizons are vanishing777This can be
seen from the examination of the contribution
$C^{1/2}\partial^{2}_{u}C^{-1/2}$ to the stress energy tensor at
$X=\pm\infty$.. Thus, the backreacted geometry with the quantum state under
consideration is in fact stable.

##  5 Conclusions

We have started with the geometry that possessed two horizons with different
canonical temperatures. The fact that the two horizons have different
temperatures leads to substantial backreaction of quantum fields on the
background metric – the divergence of the stress-energy tensor on one of the
horizons does not allow us to neglect the influence of quantum fields on the
metric of the spacetime at this horizon. This happens if one considers fields
to be in a thermal state, and one could, in principle, define a very specific
state for which the regularized stress-energy tensor is regular on both
horizons (see e.g. [16]). But unlike the thermal one any another state will
not necessarily survive quantum corrections in self-interacting theory (at
least in dimensions higher than two).

Then, we have written down 2D dilatonic analog of the semiclassical Einstein
equations in a self-consistent manner, i.e. the stress-energy tensor on the
r.h.s. of these equations was calculated for arbitrary metric, which is to be
determined from the resulting equations. Then, we have solved these equations
numerically with the boundary condition that one of the horizon is unchanged.

We have found quite a remarkable result that the account of the backreaction
changes the geometry in a way that the two horizons have the same (vanishing)
temperature. Hence, after the introduction of quantum fields and accounting of
the backreaction we obtain a geometry that is stable for the vacuum state (the
thermal state with vanishing temperature) of the quantum field.

##  6 Acknowledgements

We would like to thank Artem Alexandrov for valuable discussions regarding
numerical solution of equations. This work was supported by the Foundation for
the Advancement of Theoretical Physics and Mathematics “BASIS” and by Russian
Ministry of education and science.

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