For the remaining $8$ kinds, we will show that

(83) $|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$

and $J_{j}$ must contain at least one of those $8$ kinds.

######  Lemma 46.

The segment $J_{j}$ for Type-1 must contain at least one segment of types 1.1,
1.2, 1.5, 1.6, 1.7, 1.8, 1.13 and 1.14 as shown in Figure 9.

###### Proof.

If $X=\mathcal{P}\cong S_{0,3}$, then $\eta$ is a single $J$ and does not
intersect with $2a$ and $2c$. So it is a combination of types $1.1,1.2,1.3$,
and $1.4$. Since $\eta$ is a filling closed curve, it must intersect with
$\alpha$ and $\gamma$. So $\eta$ contains at least one segment from $\alpha$
to $\gamma$, that is of type 1.1 or 1.2.

If $X$ is not $S_{0,3}$, then $\eta$ must intersect with at least one of $2a$
and $2c$, and thus $J_{j}$ must have an endpoint on $a$ or $c$. Suppose
$J_{j}$ do not contain any of types $1.1,1.2,1.5,1.6,1.7,1.8,1.13$ and $1.14$.
Then $J_{j}$ only consists of segments which may be from $a$ to $\gamma$, or
from $\gamma$ to $\gamma$, or from $c$ to $\alpha$ and or from $\alpha$ to
$\alpha$. So if starting at a point on $a$, the geodesic $J_{j}$ can only
travel to $\gamma$, then go to next $\gamma$ several times, and finally return
to $a$. But in this way $J_{j}$ would be homotopic to a piece of geodesic line
$a$, which contradicts to the assumption that $J_{j}$ is part of a shortest
closed geodesic representative for $\eta$ as shown in (65). Similarly the
geodesic $J_{j}$ can not start at a point in $c$. So $J_{j}$ must contain at
least one segment of those 8 kinds. ∎

Next we prove (83) for those $8$ types in Lemma 46 case by case.

######  Lemma 47 (type 1.1).

If $b\geq Am$ and (73) holds, then for type 1.1 in Figure 9 we have

$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

###### Proof.

Recall that $\dot{x}=\dot{y}=0$ and $\dot{b}=-1$. Since $b\geq Am$ and (73)
holds, it follows by Lemma 44 that
$|\dot{\alpha}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ and
$|\dot{\gamma}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. It is clear that
$e^{-b}\leq\frac{2}{b^{2}}$. Next we apply Lemma 43 to the rectangle
$(\gamma-y,b,\alpha-x,d)$ to get

$|\dot{d}+1|=|\dot{d}-\dot{b}|\leq
A^{\prime}(e^{-b}+|\dot{\alpha}|+|\dot{\gamma}|)=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$

for some constant $A^{\prime}>0$ only depending on $A$. ∎

######  Lemma 48 (type 1.2).

If $b\geq Am$ and (73) holds, then for type 1.2 in Figure 9 we have

$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

###### Proof.

Let $h$ be the perpendicular between $\alpha$ and $\gamma$ in type 1.2. Let
$e$ to be part of $\alpha$ between $h$ and $b$, and let $f$ to be part of
$\gamma$ between $h$ and $a$ (see Figure 10). Then we have a rectangle
$(\alpha-e-x,h,y-f,d)$ with $\angle(h,\alpha-
e-x)=\angle(h,y-f)=\frac{\pi}{2}$. We remark here that the length $\alpha-e-x$
and $y-f$ may be negative. Now we compute the lengths of these edges and their
derivatives.

Figure 10.

By Lemma 42 and (77) we have

$\displaystyle\cosh h$ $\displaystyle=$ $\displaystyle\sinh 2a\sinh
b\cosh\gamma-\cosh 2a\cosh b$ $\displaystyle=$ $\displaystyle 2\sinh a\cosh
a\sinh b\frac{\cosh c+\cosh a\cosh b}{\sinh a\sinh b}$
$\displaystyle-(2\cosh^{2}a-1)\cosh b$ $\displaystyle=$ $\displaystyle 2\cosh
a\cosh c+\cosh b$

and

$\displaystyle\dot{h}$ $\displaystyle=$ $\displaystyle\dot{b}\frac{\sinh
b}{\sinh h}$ $\displaystyle=$ $\displaystyle\dot{b}\frac{\cosh h}{\sinh
h}\frac{\sinh b}{2\cosh a\cosh c+\cosh b}.$

So by (73), (8.2), $h\geq b\geq A$ and $\dot{b}=-1$, we have

$\displaystyle|\dot{h}+1|$ $\displaystyle=$ $\displaystyle\left|1-\frac{\cosh
h}{\sinh h}\frac{\sinh b}{2\cosh a\cosh c+\cosh b}\right|$ $\displaystyle\leq$
$\displaystyle\left|1-\frac{\sinh b}{2\cosh a\cosh c+\cosh
b}\right|+\left|1-\frac{\cosh h}{\sinh h}\right|\frac{\sinh b}{2\cosh a\cosh
c+\cosh b}$ $\displaystyle\leq$ $\displaystyle\frac{2\frac{\cosh a\cosh
c}{\cosh b}+\frac{e^{-b}}{\cosh b}}{2\frac{\cosh a\cosh c}{\cosh
b}+1}+\left(\frac{\cosh b}{\sinh b}-1\right)$ $\displaystyle\leq$
$\displaystyle 4\pi^{2}\frac{m^{2}}{b^{2}}+e^{-b}+\frac{e^{-b}}{\sinh A}$
$\displaystyle=$ $\displaystyle O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

By Lemma 42, (77), (79), (8.2) and (8.2) we have

$\displaystyle\sinh e$ $\displaystyle=$ $\displaystyle\sinh
2a\frac{\sinh\gamma}{\sinh h}$ $\displaystyle=$ $\displaystyle\frac{2\cosh
w\cosh a}{\sinh h}$

and

$\displaystyle\quad\ \dot{e}$ $\displaystyle=$ $\displaystyle\frac{1}{\cosh
e}\sinh 2a\bigg{(}\dot{\gamma}\frac{\cosh\gamma}{\sinh
h}-\dot{h}\frac{\sinh\gamma\cosh h}{\sinh^{2}h}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{\sinh 2a}{\cosh e}\bigg{[}\frac{1}{\cosh w}\frac{(\cosh
a+\cosh b\cosh c)}{\sinh^{2}b}\frac{(\cosh c+\cosh a\cosh b)}{\sinh a\sinh
b\sinh h}$ $\displaystyle+\frac{\sinh b}{\sinh h}\frac{\cosh w}{\sinh
a}\frac{\cosh h}{\sinh^{2}h}\bigg{]}$ $\displaystyle=$
$\displaystyle\frac{1}{\cosh e}\bigg{[}\frac{2\cosh a}{\cosh w}\frac{(\cosh
a+\cosh b\cosh c)}{\sinh^{2}b}\frac{(\cosh c+\cosh a\cosh b)}{\sinh b\sinh h}$
$\displaystyle+2\frac{\cosh a}{\sinh h}\frac{\sinh b}{\sinh h}\cosh
w\frac{\cosh h}{\sinh h}\bigg{]}.$

Since $h>b\geq Am$, by (73) and (8.2) it is not hard to see that $\frac{\cosh
a}{\sinh h}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ and $\frac{2\cosh a}{\cosh
w}\frac{(\cosh a+\cosh b\cosh c)}{\sinh^{2}b}\frac{(\cosh c+\cosh a\cosh
b)}{\sinh b\sinh h}=O_{A}\left(\frac{m^{4}}{b^{4}}\right)$. Recall that by
(73) $\cosh
w=\sqrt{\sinh^{2}w+1}\leq\sqrt{4\pi^{2}\frac{m^{2}}{b^{2}}+1}\leq\sqrt{\frac{4\pi^{2}}{A^{2}}+1}$.
So we have that $e\leq\sinh e=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ and
$0\leq\dot{e}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.

By Lemma 42, (8.2) and (8.2) we have

(88) $\sinh f=\sinh b\frac{\sinh\gamma}{\sinh h}$

and

$\displaystyle\quad\quad\dot{f}$ $\displaystyle=$ $\displaystyle\frac{1}{\cosh
f}\bigg{(}\dot{\gamma}\cosh\gamma\frac{\sinh b}{\sinh
h}+\dot{b}\sinh\gamma\frac{\cosh b}{\sinh h}-\dot{h}\sinh\gamma\frac{\sinh
b\cosh h}{\sinh^{2}h}\bigg{)}$ $\displaystyle=$
$\displaystyle\dot{\gamma}\frac{\cosh\gamma}{\cosh f}\frac{\sinh b}{\sinh
h}+\dot{b}\frac{\sinh\gamma}{\cosh f}\bigg{(}\frac{\cosh b}{\sinh
h}-\frac{\sinh^{2}b\cosh h}{\sinh^{3}h}\bigg{)}$ $\displaystyle=$
$\displaystyle\dot{\gamma}\frac{\cosh\gamma}{\cosh f}\frac{\sinh b}{\sinh
h}+\dot{b}\frac{\sinh\gamma}{\cosh f}\frac{(\cosh h-\cosh b)(1+\cosh b\cosh
h)}{\sinh^{3}h}$ $\displaystyle=$
$\displaystyle\dot{\gamma}\frac{\cosh\gamma}{\cosh f}\frac{\sinh b}{\sinh
h}-\frac{\sinh\gamma}{\cosh f}\frac{2\cosh a\cosh c(1+\cosh b\cosh
h)}{\sinh^{3}h}.$

Since $\frac{\sinh\gamma}{\sinh f}=\frac{\sinh h}{\sinh b}\leq\frac{2\cosh
a\cosh c+\cosh b}{\sinh b}=O_{A}(1)$ is bounded (by (8.2) and (8.2)), we know
that $\frac{\cosh\gamma}{\cosh f}=O_{A}(1)$ is also bounded. Apply (73) and
(8.2) and the fact that $h>b\geq Am$, we have that $\frac{2\cosh a\cosh
c(1+\cosh b\cosh h)}{\sinh^{3}h}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. Thus
together with Lemma 44, we have
$|\dot{f}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.

Now consider the rectangle $(\alpha-e-x,h,y-f,d)$. By Lemma 44 and the
discussion above we have
$|\dot{\alpha}-\dot{e}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$,
$|\dot{f}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ and
$|\dot{h}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. Recall that
$\dot{x}=\dot{y}=0$ and $h>b\geq Am$. Then we apply Lemma 43 to the rectangle
$(\alpha-e-x,h,y-f,d)$ to get
$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ as desired. ∎

######  Lemma 49 (types 1.5–1.8).

If $b\geq Am$ and (73) holds, then for types 1.5, 1.6, 1.7 and 1.8 in Figure 9
we have

$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

###### Proof.

We only prove the lemma for type 1.5. For the other three types, the proofs
are similar as the one for type 1.5.

Let $h$ be the perpendicular between $a$ and $\alpha$ in type 1.5. Let $e$ to
be part of $\alpha$ between $h$ and $b$, and let $f$ to be part of $a$ between
$h$ and $\gamma$ (see Figure 11). Then we have a rectangle
$(f+u-a,h,e+x-\alpha,d)$ with
$\angle(h,f+u-a)=\angle(h,e+x-\alpha)=\frac{\pi}{2}$. We remark here that the
length $f+u-a$ and $e+x-\alpha$ may be negative. Now we compute the lengths of
these edges and their derivatives.

Figure 11.

By Lemma 42, (77), (79) and (8.2), we have

$\displaystyle\cosh h$ $\displaystyle=$ $\displaystyle\sinh b\sinh\gamma$
$\displaystyle=$ $\displaystyle\frac{\cosh w}{\sinh a}\sinh b$

and

$\displaystyle\quad\quad\dot{h}$ $\displaystyle=$ $\displaystyle\frac{1}{\sinh
h}\bigg{(}\dot{b}\cosh b\sinh\gamma+\dot{\gamma}\sinh b\cosh\gamma\bigg{)}$
$\displaystyle=$ $\displaystyle-\frac{\cosh h}{\sinh h}\frac{\cosh b}{\sinh
b}$ $\displaystyle+\frac{1}{\cosh w}\frac{(\cosh a+\cosh b\cosh
c)}{\sinh^{2}b}\frac{(\cosh c+\cosh a\cosh b)}{\sinh a\sinh b}\frac{\sinh
b}{\sinh h}$ $\displaystyle=$ $\displaystyle-\frac{\cosh h}{\sinh
h}\frac{\cosh b}{\sinh b}$ $\displaystyle+\frac{1}{\cosh^{2}w}\frac{\cosh
h}{\sinh h}\frac{(\cosh a+\cosh b\cosh c)(\cosh c+\cosh a\cosh
b)}{\sinh^{3}b}.$

Since $b\geq Am$, by (73), (8.2) and (8.2) it is easy to see that
$\frac{1}{\cosh h}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. Which also implies
that $|\frac{\cosh h}{\sinh h}\frac{\cosh b}{\sinh
b}-1|=O_{A}\left(\frac{m^{4}}{b^{4}}\right)$. Then by (73) and (8.2) it is not
hard to see that

$\frac{1}{\cosh^{2}w}\frac{\cosh h}{\sinh h}\frac{(\cosh a+\cosh b\cosh
c)(\cosh c+\cosh a\cosh
b)}{\sinh^{3}b}=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

Thus, we have

$|\dot{h}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

By Lemma 42, we have

$\displaystyle\sinh(a-f)$ $\displaystyle=$ $\displaystyle\frac{\cosh c}{\sinh
h}$ $\displaystyle=$ $\displaystyle\frac{\cosh h}{\sinh h}\frac{1}{\cosh
w}\frac{\sinh a\cosh c}{\sinh b}$

and

$\displaystyle-\dot{f}$ $\displaystyle=$
$\displaystyle-\dot{h}\frac{1}{\cosh(a-f)}\frac{\cosh c\cosh h}{\sinh^{2}h}$
$\displaystyle=$ $\displaystyle-\dot{h}\frac{1}{\cosh(a-f)}\frac{\cosh
h}{\sinh h}\sinh(a-f).$

Then again by (73), (8.2) and the bounds for $h$ and $\dot{h}$ above, it is
easy to see that

$|a-f|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)\quad\text{and}\quad|\dot{f}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

By Lemma 42, we have

(94) $\sinh e=\frac{\cosh f}{\sinh b}\leq\frac{\cosh a}{\sinh b}$

and

$\dot{e}=\frac{1}{\cosh e}\bigg{(}\dot{f}\frac{\sinh f}{\sinh
b}-\dot{b}\frac{\cosh f\cosh b}{\sinh^{2}b}\bigg{)}$

implying that

(95) $|\dot{e}|\leq|\dot{f}|\frac{\sinh a}{\sinh b}+\frac{\cosh a\cosh
b}{\sinh^{2}b}.$

Then by (73) and (8.2) we have $e=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ and
$|\dot{e}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

Recall that $\dot{u}=\dot{a}=\dot{x}=0$. Then we apply Lemma 43, Lemma 44 and
all these bounds above to rectangle $(f+u-a,h,e+x-\alpha,d)$ to obtain
$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ as desired. ∎

######  Lemma 50 (types 1.13 and 1.14).

If $b\geq Am$ and (73) holds, then for types 1.13 and 1.14 in Figure 9 we have

$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

###### Proof.

Recall that $\dot{u}=\dot{v}=0$. For type 1.13, by Lemma 44 and applying Lemma
43 to rectangle $(u,\beta,v,d)$, we obtain
$|\dot{d}+1|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ as desired. For type 1.14,
the proof is similar. ∎

Now we are ready to prove Proposition 40 for Type-1.

######  Proposition 51.

Proposition 40 holds for the case that $\mathcal{P}$ is of Type-1.

###### Proof.

Let $2b$ to be the longest boundary geodesic of $X$, and $\mathcal{P}$ be the
pants with three boundary geodesics $(2a,2b,2c)$. Since $\sum x_{i}\geq
n(2Am+\delta)$, we know $2b\geq 2Am+\delta>\delta$. Hence both the pants
$\mathcal{P}_{\delta}$ and the surface $X_{\delta}$ exist.

As $b$ smoothly decreases to $b-\frac{1}{2}\delta$, we denote $b^{\prime}$ and
$w^{\prime}$ to be the quantities corresponding to $b$ and $w$ respectively.
The lengths $a$ and $c$ are unchanged. It is clear that

$b-\frac{1}{2}\delta\leq b^{\prime}\leq b.$

So $b^{\prime}\geq Am$ always holds.

By Lemma 42, we have $\cosh w=\sinh c\sinh\alpha$ and then

$\displaystyle\sinh^{2}w$ $\displaystyle=$ $\displaystyle\cosh^{2}w-1$
$\displaystyle=$ $\displaystyle\sinh^{2}c\cosh^{2}\alpha-\cosh^{2}c$
$\displaystyle=$ $\displaystyle\sinh^{2}c\frac{(\cosh a+\cosh b\cosh
c)^{2}}{\sinh^{2}b\sinh^{2}c}-\cosh^{2}c$ $\displaystyle=$
$\displaystyle\frac{\cosh^{2}a+\cosh^{2}c+2\cosh a\cosh b\cosh
c}{\sinh^{2}b}.$

Recall that $b-\frac{1}{2}\delta\leq b^{\prime}\leq b.$ Then we have

$\displaystyle\frac{\sinh w^{\prime}}{\sinh w}$ $\displaystyle=$
$\displaystyle\frac{\sinh b}{\sinh
b^{\prime}}\left(\frac{\cosh^{2}a+\cosh^{2}c+2\cosh a\cosh b^{\prime}\cosh
c}{\cosh^{2}a+\cosh^{2}c+2\cosh a\cosh b\cosh c}\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle\frac{\sinh b}{\sinh(b-\frac{1}{2}\delta)}$
$\displaystyle=$
$\displaystyle\cosh\frac{1}{2}\delta+\frac{1}{\tanh(b-\frac{1}{2}\delta)}\sinh\frac{1}{2}\delta$
$\displaystyle\leq$ $\displaystyle\cosh\frac{1}{2}\delta+\frac{1}{\tanh
A}\sinh\frac{1}{2}\delta$ $\displaystyle<$ $\displaystyle 2$

for $\delta\in(0,A_{0}]$ where $A_{0}$ is some constant only depending on $A$.
Recall that (64) says that $\sinh w\leq\frac{\pi m}{b}$. So we have that for
$0<\delta\leq A_{0}$, the inequality that $\sinh w^{\prime}\leq\frac{2\pi
m}{b}$ always holds, i.e, equation(73) always holds. Thus the assumptions in
Lemma 47, 48, 49 and 50 are always satisfied.

As the boundary length $b$ reduces by $\frac{1}{2}\delta$, the length $d$ in
types 1.3, 1,4, 1.9, 1.10, 1.11 and 1.12 are unchanged. For remaining types
1.1, 1.2, 1.5, 1.6, 1.7, 1.8, 1.13 and 1.14, it follows by Lemma 47, 48, 49
and 50 that the length $d$ decreases at least
$\frac{1}{2}\left(1-O_{A}\left(\frac{m^{2}}{b^{2}}\right)\right)\delta$. By
Lemma 46 $J_{j}$ must contain at least one of the decreasing types. So we have

$\ell(J_{j})-\ell(J^{\prime}_{j})\geq\frac{1}{2}\left(1-O_{A}\left(\frac{m^{2}}{b^{2}}\right)\right)\delta$

which together with (66) imply that

$\ell_{\eta}(X)-\ell_{\eta}(X_{\delta})\geq\frac{1}{2}\left(1-O_{A}\left(\frac{m^{2}}{b^{2}}\right)\right)\delta=\frac{1}{2}\left(1-O_{A}\left(\frac{m^{2}n^{2}}{(\sum
x_{i})^{2}}\right)\right)\delta$

where in the last equation we apply $2b\geq\frac{\sum x_{i}}{n}$.

The proof is complete. ∎

######  Remark.

Moreover, if $g=n=1$, then the pair of pants $\mathcal{P}$ must be of Type-1,
and the two boundary curves $2a=2c$ in $\mathcal{P}$. It is not hard to see
that any closed filling curve $\eta$ contains at least two decreasing types.
For this case, actually one can improve Proposition 40 and Theorem 4 to be

######  Proposition 52.

With the same assumptions in Proposition 40, if $g=n=1$ and
$X\in\mathcal{T}_{1,1}(x)$, we have

$\ell_{\eta}(X)-\ell_{\eta}(X_{\delta})\geq\left(1-O_{A}\left(\frac{1}{x^{2}}\right)\right)\delta.$

######  Theorem 53.

For any $\epsilon_{1}>0$, there exists a constant $c(\epsilon_{1})>0$ only
depending on $\epsilon_{1}$ such that for all $T>0$ and any compact hyperbolic
surface $X\cong S_{1,1}$ of geodesic boundary, we have

$\\#_{f}(X,T)\leq c(\epsilon_{1})\cdot e^{T-(1-\epsilon_{1})\ell(\partial
X)}.$

In this paper the bound in Proposition 40 is sufficient for us to prove
Theorem 1 and 2.

###  8.3. Proof of Proposition 40 for Type-2

Now we consider a pair of pants in Figure 4 of Type-2 and the corresponding
right-angled hexagon in Figure 5 of Type-2. Both $2b$ and $2c$ are two
boundary geodesics of the surface.

In this subsection, we always use the notation $\alpha=w$ to denote the
perpendicular between $b$ and $c$ as shown in Figure 5 of Type-2. For the
pants $\mathcal{P}$ of Type-2 we construct, (64) always holds. But when $b$
and $c$ both decrease, (64) may not hold again. Instead, we always assume
$\alpha=w$ satisfy

(96) $\sinh w\leq 2\pi\frac{m}{b}.$

Later in the proof of Proposition 59 we will show (96) holds when $b$ and $c$
do not reduce too much. By Lemma 42 we have

$\displaystyle\cosh a$ $\displaystyle=$ $\displaystyle\sinh b\sinh c\cosh
w-\cosh b\cosh c$ $\displaystyle=$ $\displaystyle 2\sinh^{2}\frac{w}{2}\sinh
b\sinh c-(\cosh b\cosh c-\sinh b\sinh c)$ $\displaystyle\leq$ $\displaystyle
2\sinh^{2}\frac{w}{2}\sinh b\sinh c$

Now we consider to simultaneously reduce the two boundary lengths $2b$ and
$2c$ of $\mathcal{P}$. Let the length $\\{a(t),b(t),c(t)\\}$ be functions in
terms of a common parameter $t$. In our process from $\mathcal{P}$ to
$\mathcal{P}_{\delta}$, we assume $a$ is fixed and both $b$ and $c$ decrease
with constant speed. More precisely, denote $\dot{b}=\frac{db}{dt}$ and
$\dot{c}=\frac{dc}{dt}$, assume

(98) $\dot{a}=0\quad\emph{and}\quad\dot{b}=\dot{c}=-1.$

Let $\alpha=w,\beta$ and $\gamma$ be as shown in type-2 of Figure 4. Denote
$h_{a}$ to be the shortest perpendicular between $a$ and $\alpha$, $h_{b}$ to
be the shortest perpendicular between $b$ and $\beta$, $h_{c}$ to be the
shortest perpendicular between $c$ and $\gamma$. Then by Lemma 42 we have

(99) $\cosh h_{a}=\sinh b\sinh\gamma=\sinh c\sinh\beta,$ (100) $\cosh
h_{b}=\sinh a\sinh\gamma=\sinh c\sinh\alpha,$ (101) $\cosh h_{c}=\sinh
a\sinh\beta=\sinh b\sinh\alpha.$

Recall that by Lemma 42, we have

$\cosh\alpha=\frac{\cosh a+\cosh b\cosh c}{\sinh b\sinh c},$
$\cosh\beta=\frac{\cosh b+\cosh a\cosh c}{\sinh a\sinh c},$
$\cosh\gamma=\frac{\cosh c+\cosh a\cosh b}{\sinh a\sinh b}.$

A direct computation shows that

$\displaystyle\dot{\alpha}$ $\displaystyle=$
$\displaystyle\frac{1}{\sinh\alpha}\bigg{(}\frac{\cosh a(\cosh b\sinh c+\sinh
b\cosh c)}{\sinh^{2}b\sinh^{2}c}$
$\displaystyle+\frac{1}{\sinh^{2}b}\frac{\cosh c}{\sinh
c}+\frac{1}{\sinh^{2}c}\frac{\cosh b}{\sinh b}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{\cosh a}{\sinh w\sinh b\sinh c}\big{(}\frac{\cosh b}{\sinh
b}+\frac{\cosh c}{\sinh c}\big{)}$ $\displaystyle+\frac{1}{\cosh h_{c}\sinh
b}\frac{\cosh c}{\sinh c}+\frac{1}{\cosh h_{b}\sinh c}\frac{\cosh b}{\sinh
b}.$

$\displaystyle\dot{\beta}$ $\displaystyle=$
$\displaystyle\frac{-1}{\sinh\beta}\bigg{(}\frac{(\sinh b+\cosh a\sinh c)\sinh
c}{\sinh a\sinh^{2}c}$ $\displaystyle-\frac{\cosh c(\cosh b+\cosh a\cosh
c)}{\sinh a\sinh^{2}c}\bigg{)}$ $\displaystyle=$ $\displaystyle\frac{\cosh
a+\cosh(b-c)}{\cosh h_{c}\sinh^{2}c}$ $\displaystyle=$
$\displaystyle\frac{\cosh a+\cosh(b-c)}{\sinh\alpha\sinh b\sinh^{2}c}$
$\displaystyle=$ $\displaystyle\frac{\cosh a+\cosh(b-c)}{\cosh h_{b}\sinh
b\sinh c},$ $\displaystyle\dot{\gamma}$ $\displaystyle=$
$\displaystyle\frac{-1}{\sinh\gamma}\bigg{(}\frac{(\sinh c+\cosh a\sinh
b)\sinh b}{\sinh a\sinh^{2}b}$ $\displaystyle-\frac{\cosh b(\cosh c+\cosh
a\cosh b)}{\sinh a\sinh^{2}b}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{\cosh a+\cosh(b-c)}{\cosh h_{b}\sinh^{2}b}$
$\displaystyle=$ $\displaystyle\frac{\cosh a+\cosh(b-c)}{\sinh\alpha\sinh
c\sinh^{2}b}$ $\displaystyle=$ $\displaystyle\frac{\cosh a+\cosh(b-c)}{\cosh
h_{c}\sinh c\sinh b}.$

A direct consequence of all the equations above is

######  Lemma 54.

If $b\geq Am$ and (96) holds, then we have

  1. (1)

$\sinh c\geq\frac{1}{2\pi}\frac{b}{m}$.

  2. (2)

$0<\dot{\alpha}=O_{A}\left(\frac{m}{b}\right),\
0<\dot{\beta}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)\ \emph{and}\
0<\dot{\gamma}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.

###### Proof.

For Part (1), by (96) and (100) we have

$\sinh c\geq\frac{1}{\sinh\alpha}=\frac{1}{\sinh
w}\geq\frac{1}{2\pi}\frac{b}{m}.$

For Part (2), by (8.3), (8.3) and (8.3) we clearly have $0<\dot{\alpha}$,
$0<\dot{\beta}$ and $0<\dot{\gamma}$. By (8.3) we have $\frac{\cosh a}{\sinh
w\sinh b\sinh c}\leq\frac{2\sinh^{2}{\frac{w}{2}}}{\sinh w}\leq 2\sinh w\leq
4\pi\frac{m}{b}$ where we apply (96) in the last inequality. Since $b\geq A$
and $\sinh c\geq\frac{A}{2\pi}$, from (8.3) we have
$\dot{\alpha}=O_{A}\left(\frac{m}{b}\right)$. For $\dot{\beta}$, by (96) and
(8.3) we have $\frac{\cosh a}{\cosh h_{b}\sinh b\sinh c}\leq
2\sinh^{2}{\frac{w}{2}}\leq 8\pi^{2}\frac{m^{2}}{b^{2}}$. Since $b\geq A$ and
$\sinh c\geq\frac{A}{2\pi}$, we know $\frac{\cosh\left(b-c\right)}{\cosh
h_{b}\sinh b\sinh
c}=O_{A}\left(e^{-2b}+e^{-2c}\right)=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.
So by (8.3) we have $\dot{\beta}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. By a
similar argument one can show that
$\dot{\gamma}=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. ∎

Let $J_{j}\subset\mathcal{P}$ to be a geodesic segment in (65). We classify
$J_{j}$ in terms of its possible intersections with $\alpha,\beta,\gamma$ and
$a$ (without orientation). Actually we have

######  Lemma 55.

There are 6 kinds of possible segments as shown in Figure 12. More precisely,
they are: from $\beta$ to $\gamma$ (type 2.1), from $\alpha$ to $\gamma$ (type
2.2), from $\alpha$ to $\beta$ (type 2.3), from $a$ to $\alpha$ (type 2.4),
from $a$ to $\beta$ (type 2.5) and from $a$ to $\gamma$ (type 2.6).

We denote $d$ as the geodesic segment of $J_{j}$ in each kind.

Figure 12. 6 kinds of geodesic segments in Type-2

Similar to what we do for Type-1. Denote $p,q$ and $r$ to be intersection
points of $J_{j}$ and $\alpha,\beta$ and $\gamma$ respectively if they exist.
Denote $x$ to be the part of $\alpha$ that from $p$ to $c$ if it exists, and
$y$ to be the part of $\beta$ that from $q$ to $a$ if it exists, and $z$ to be
the part of $\gamma$ that from $r$ to $a$ if it exists. We fix these points
$p,q$ and $r$ as $b$ and $c$ are simultaneously decreasing, more precisely,
keep the length of each $x,y$ and $z$ to be unchanged. Since $\alpha,\beta$
and $\gamma$ are increasing as $b$ and $c$ are simultaneously decreasing by
Lemma 54,the points $p,q$ and $r$ will still lie in $\alpha,\beta$ and
$\gamma$ respectively during the process. These intersection points separate
$J_{j}$ into several segments. Fixing these points as $b$ and $c$ are
simultaneously decreasing, we get a piecewise geodesic $J^{\prime\prime}_{j}$
homotopic to $J_{j}$ in $\mathcal{P}_{\delta}$.

In type 2.1, the lengths $a,y$ and $z$ are all unchanged during the process,
so the corresponding $d$ is also unchanged. In types 2.5 and 2.6, the
endpoints of $J_{j}$ on boundary $2a$ are fixed, so the length $u$ (part of
$a$ from the endpoint of $J_{j}$ to $\beta$, see Figure 12) is also unchanged.
Since $a,y$ and $z$ are all unchanged, the corresponding $d$ is also
unchanged.

For the remaining $3$ kinds, similar to Type-1 we will show that

(105) $|\dot{d}+1|=O_{A}\left(\frac{m}{b}\right)$

and $J_{j}$ must contain at least two of those $3$ kinds.

######  Lemma 56.

The segment $J_{j}$ for Type-2 must contain at least two segments of types
2.2, 2.3 and 2.4 as shown in Figure 12.

###### Proof.

Since $\eta$ is filling, the segment $J_{j}$ is a geodesic in $\mathcal{P}$
with endpoints on $2a$ (if $X$ is not $S_{0,3}$) or just a closed geodesic in
$\mathcal{P}$ (if $X$ is $S_{0,3}$). Moreover it must have at least one
intersection point with $\alpha$ which is denoted by $o$. On both sides of
$J_{j}$ at point $o$, there is at least one segment $d$ of type 2.2 or 2.3 or
2.4. Then the conclusion follows. ∎

Next we prove (105) for those $3$ types in Lemma 56 case by case.

######  Lemma 57 (types 2.2 and 2.3).

If $b\geq Am$ and (96) holds, then for types 2.2 and 2.3 in Figure 12 we have

$|\dot{d}+1|=O_{A}\left(\frac{m}{b}\right).$

###### Proof.

Recall that $\dot{x}=\dot{y}=\dot{z}=0$ and $\dot{b}=\dot{c}=-1$. For type
2.2, by Lemma 54 and we apply Lemma 43 to rectangle $(\gamma-z,b,\alpha-x,d)$
to obtain that $|\dot{d}+1|=O_{A}\left(\frac{m}{b}\right)$. For type 2.3, the
proof is similar. ∎

######  Lemma 58 (type 2.4).

If $b\geq Am$ and (96) holds, then for type 2.4 in Figure 12 we have

$|\dot{d}+1|=O_{A}\left(\frac{m}{b}\right).$

###### Proof.

Let $h_{a}$ be the perpendicular between $a$ and $\alpha$ in type 2.4. Let $e$
to be part of $\alpha$ between $h_{a}$ and $c$, and let $f$ to be part of $a$
between $h_{a}$ and $\beta$ (see Figure 13). Then we have a rectangle
$(u-f,h_{a},x-e,d)$ with $\angle(h_{a},u-f)=\angle(h_{a},x-e)=\frac{\pi}{2}$.
We remark here that the lengths $u-f$ and $x-e$ may be negative. Now we
compute the lengths of these edges and their derivatives.

Figure 13.

By Lemma 42 we have

(106) $\cosh h_{a}=\sinh c\sinh\beta\ \rm{and}\ \cosh\beta=\sinh e\sinh
h_{a}.$

By (8.3) we have $\dot{\beta}=\frac{\cosh a+\cosh(b-c)}{\sinh\alpha\sinh
b\sinh^{2}c}$. A direct computation shows that

$\displaystyle\dot{h}_{a}$ $\displaystyle=$ $\displaystyle\frac{1}{\sinh
h_{a}}\bigg{(}\dot{c}\cosh c\sinh\beta+\dot{\beta}\sinh c\cosh\beta\bigg{)}$
$\displaystyle=$ $\displaystyle-\frac{\cosh h_{a}}{\sinh h_{a}}\frac{\cosh
c}{\sinh c}+\frac{\cosh\beta}{\sinh h_{a}}\frac{(\cosh
a+\cosh(b-c))}{\sinh\alpha\sinh b\sinh c}$ $\displaystyle=$
$\displaystyle-\frac{\cosh h_{a}}{\sinh h_{a}}\frac{\cosh c}{\sinh
c}+\frac{\sinh e}{\sinh\alpha}\frac{(\cosh a+\cosh(b-c))}{\sinh b\sinh c}.$

Since $\sinh e\leq\sinh\alpha=\sinh w\leq 2\pi\frac{m}{b}$, we have

$\sinh h_{a}=\frac{\cosh\beta}{\sinh e}\geq\frac{1}{2\pi}\frac{b}{m}.$

Since $e\leq\alpha=w$, by (96) and (8.3) we have $\frac{\sinh
e}{\sinh\alpha}\frac{\cosh a}{\sinh b\sinh c}\leq 2\sinh^{2}{\frac{w}{2}}\leq
8\pi^{2}\frac{m^{2}}{b^{2}}$. By Lemma 54 $\frac{\sinh
e}{\sinh\alpha}\frac{\cosh\left(b-c\right)}{\sinh b\sinh
c}=O_{A}\left(e^{-2b}+e^{-2c}\right)=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.
Then we have

$\displaystyle|\dot{h}_{a}+1|$ $\displaystyle\leq$
$\displaystyle\left|\frac{\cosh h_{a}}{\sinh h_{a}}\frac{\cosh c}{\sinh
c}-1\right|+O_{A}\left(\frac{m^{2}}{b^{2}}\right)$ $\displaystyle=$
$\displaystyle O_{A}\left(e^{-2h_{a}}+e^{-2c}+\frac{m^{2}}{b^{2}}\right)$
$\displaystyle=$ $\displaystyle O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

By Lemma 42 and (106) we have

(108) $\sinh e=\frac{\cosh\beta}{\sinh h_{a}}$

and

(109) $\displaystyle\dot{e}$ $\displaystyle=$ $\displaystyle\frac{1}{\cosh
e}\bigg{(}\dot{\beta}\frac{\sinh\beta}{\sinh
h_{a}}-\dot{h}_{a}\frac{\cosh\beta\cosh h_{a}}{\sinh^{2}h_{a}}\bigg{)}$
$\displaystyle=$ $\displaystyle\frac{1}{\cosh e}\bigg{(}\dot{\beta}\frac{\cosh
h_{a}}{\sinh h_{a}}\frac{1}{\sinh c}-\dot{h}_{a}\frac{\cosh h_{a}}{\sinh
h_{a}}\sinh e\bigg{)}.$

By Lemma 54, the fact that $\sinh e\leq\sinh\alpha=\sinh w\leq
2\pi\frac{m}{b}$ and the above bounds, we have
$|\dot{e}|=O_{A}\left(\frac{m}{b}\right)$.

By Lemma 42 and (8.3) we have

(110) $\sinh f=\frac{\cosh e}{\sinh\beta}$

and

$\displaystyle\quad\quad\quad\dot{f}$ $\displaystyle=$
$\displaystyle\frac{1}{\cosh f}\bigg{(}\dot{e}\frac{\sinh
e}{\sinh\beta}-\dot{\beta}\frac{\cosh\beta\cosh e}{\sinh^{2}\beta}\bigg{)}$
$\displaystyle=$ $\displaystyle\dot{e}\frac{\sinh f}{\cosh f}\frac{\sinh
e}{\cosh e}-\frac{(\cosh a+\cosh(b-c))}{\sinh\alpha\sinh
b\sinh^{2}c}\frac{\cosh\beta\cosh e}{\sinh e\sinh c\sinh^{2}\beta}$
$\displaystyle=$ $\displaystyle\dot{e}\frac{\sinh f}{\cosh f}\frac{\sinh
e}{\cosh e}-\frac{(\cosh a+\cosh(b-c))\sinh h_{a}\cosh e}{\sinh\alpha\sinh
b\sinh c\cosh^{2}h_{a}}$ $\displaystyle=$ $\displaystyle\dot{e}\frac{\sinh
f}{\cosh f}\frac{\sinh e}{\cosh e}-\frac{\sinh
e}{\sinh\alpha}\frac{\sinh^{2}h_{a}}{\cosh^{2}h_{a}}\frac{\cosh
e}{\cosh\beta}\frac{(\cosh a+\cosh(b-c))}{\sinh b\sinh c}.$

Since $\sinh e\leq 2\pi\frac{m}{b}$ and
$|\dot{e}|=O_{A}\left(\frac{m}{b}\right)$, we have $\left|\dot{e}\frac{\sinh
f}{\cosh f}\frac{\sinh e}{\cosh
e}\right|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$. By (96) and (8.3) it is not
hard to see that

$\displaystyle\left|\frac{\sinh
e}{\sinh\alpha}\frac{\sinh^{2}h_{a}}{\cosh^{2}h_{a}}\frac{\cosh
e}{\cosh\beta}\frac{(\cosh a+\cosh(b-c))}{\sinh b\sinh c}\right|$
$\displaystyle\leq$ $\displaystyle\cosh e\frac{\cosh a+\cosh(b-c)}{\sinh
b\sinh c}$ $\displaystyle=$ $\displaystyle
O_{A}\left(\frac{m^{2}}{b^{2}}\right).$

Thus, we have $|\dot{f}|=O_{A}\left(\frac{m^{2}}{b^{2}}\right)$.

Recall that $\dot{x}=\dot{u}=0$. By Lemma 54 and all the bounds above, then we
apply Lemma 43 to rectangle $(u-f,h_{a},x-e,d)$ to obtain that
$|\dot{d}+1|=O_{A}\left(\frac{m}{b}\right)$. ∎

Now we are ready to prove Proposition 40 for Type-2.

######  Proposition 59.

Proposition 40 holds for the case that $\mathcal{P}$ is of Type-2.

###### Proof.

Let $2b$ to be the longest boundary geodesic of $X$, and $\mathcal{P}$ be the
pants with boundary geodesics $(2a,2b,2c)$ as in the construction. Since $\sum
x_{i}\geq n(2Am+\delta)$, we know $2b\geq 2Am+\delta>\frac{1}{2}\delta$. By
Lemma 54, we know $\sinh c\geq\frac{b}{2\pi m}\geq\frac{A}{2\pi}$. So if
$0<\delta<4\mathop{\rm arcsinh}(\frac{A}{2\pi})$, we have
$2c>\frac{1}{2}\delta$. Hence both the pants $\mathcal{P}_{\delta}$ and the
surface $X_{\delta}$ exist.

During the process that $b$ and $c$ reduce to $b-\frac{1}{4}\delta$ and
$c-\frac{1}{4}\delta$ respectively, we denote
$b^{\prime},c^{\prime},w^{\prime}$ to be the quantities corresponding to
$b,c,w$ respectively. The length $a$ is unchanged. We have

$b-\frac{1}{4}\delta\leq b^{\prime}\leq b\ \ \rm{and}\ \
c-\frac{1}{4}\delta\leq c^{\prime}\leq c.$

So $b^{\prime}\geq Am$ always holds.

By Lemma 42 $\cosh w=\frac{\cosh a+\cosh b\cosh c}{\sinh b\sinh c}$ and then

$\displaystyle\sinh^{2}w$ $\displaystyle=$ $\displaystyle\cosh^{2}w-1$
$\displaystyle=$ $\displaystyle\frac{\cosh^{2}a+\sinh^{2}b+\sinh^{2}c+1+2\cosh
a\cosh b\cosh c}{\sinh^{2}b\sinh^{2}c}.$

So

$\displaystyle\frac{\sinh w^{\prime}}{\sinh w}$ $\displaystyle=$
$\displaystyle\frac{\sinh b\sinh c}{\sinh b^{\prime}\sinh c^{\prime}}$
$\displaystyle\left(\frac{\cosh^{2}a+\sinh^{2}b^{\prime}+\sinh^{2}c^{\prime}+1+2\cosh
a\cosh b^{\prime}\cosh c^{\prime}}{\cosh^{2}a+\sinh^{2}b+\sinh^{2}c+1+2\cosh
a\cosh b\cosh c}\right)^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle\frac{\sinh b}{\sinh(b-\frac{1}{4}\delta)}\frac{\sinh
c}{\sinh(c-\frac{1}{4}\delta)}$ $\displaystyle=$
$\displaystyle\left(\cosh\frac{\delta}{4}+\frac{1}{\tanh(b-\frac{\delta}{4})}\sinh\frac{\delta}{4}\right)\left(\cosh\frac{\delta}{4}+\frac{1}{\tanh(c-\frac{\delta}{4})}\sinh\frac{\delta}{4}\right)$
$\displaystyle\leq$
$\displaystyle\left(\cosh\frac{\delta}{4}+\frac{\sinh\frac{\delta}{4}}{\tanh
A}\right)\left(\cosh\frac{\delta}{4}+\frac{\sinh\frac{\delta}{4}}{\tanh(\mathop{\rm
arcsinh}(\frac{A}{2\pi})-\frac{\delta}{4})}\right)$ $\displaystyle<$
$\displaystyle 2$

for $\delta\in(0,A_{0}]$ where $A_{0}$ is some constant only depending on $A$.
Recall that (64) says that $\sinh w\leq\frac{\pi m}{b}$. So we have that for
$0<\delta\leq A_{0}$, the inequality that $\sinh w^{\prime}\leq\frac{2\pi
m}{b}$ always holds, i.e, equation(96) always holds. Thus the assumptions in
Lemma 57 and 58 are always satisfied.

As the two boundary lengths $2b$ and $2c$ simultaneously reduce by
$\frac{1}{2}\delta$, the corresponding length $d$ in types 2.1, 2.5 and 2.6
are unchanged. For remaining types 2.2, 2.3 and 2.4, it follows by Lemma 57
and 58 that the length $d$ decreases at least
$\left(1-O_{A}\left(\frac{m}{b}\right)\right)\frac{\delta}{4}$. By Lemma 56
$J_{j}$ must contain at least two of the decreasing types. So we have

$\ell(J_{j})-\ell(J^{\prime}_{j})\geq\frac{1}{2}\left(1-O_{A}\left(\frac{m}{b}\right)\right)\delta$

which together with (66) imply that

$\ell_{\eta}(X)-\ell_{\eta}(X_{\delta})\geq\frac{1}{2}\left(1-O_{A}\left(\frac{m}{b}\right)\right)\delta=\frac{1}{2}\left(1-O_{A}\left(\frac{mn}{\sum
x_{i}}\right)\right)\delta$

where in the last equation we apply $2b\geq\frac{\sum x_{i}}{n}$.

The proof is complete. ∎

###### Proof of Proposition 40.

By assumption, $\sum x_{i}\geq n(2Am+\delta)\geq 2Anm$. So we have

$\left(\frac{mn}{\sum
x_{i}}\right)^{2}\leq\frac{1}{2A}\cdot\left(\frac{mn}{\sum x_{i}}\right).$

Then the conclusion clearly follows by Proposition 51 and 59. ∎

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Renormalizable and unitary Lorentz invariant model of quantum gravity

S.A. Larin

Institute for Nuclear Research of the Russian Academy of Sciences,

60-th October Anniversary Prospect 7a, Moscow 117312, Russia

Keywords: modified theories of gravity, renormalizability, unitarity,
astrophysical and cosmological scales.

###### Abstract

We analyze the $R+R^{2}$ model of quantum gravity where terms quadratic in the
curvature tensor are added to the General Relativity action. This model was
recently proved to be a self-consistent quantum theory of gravitation, being
both renormalizable and unitary. The model can be made practically
indistinguishable from General Relativity at astrophysical and cosmological
scales by the proper choice of parameters.

##  1 Introduction

Creation of quantum gravity still remains a prominent task of modern physics.

The problem is due to well known perturbative non-renormalizability of
Einstein General Relativity. In the work [1] it was shown by direct
calculations that at the one loop level General Relativity is renormalizable
without matter fields but becomes unrenormalizable after inclusion of matter
fields. Then also by explicit calculations it was demonstrated [2] that
General Relativity is non-renormalizable at the two-loop level even without
matter fields.

We mean here that the theory is perturbatively non-renormalizable. There are
of course Quantum Field Theory examples of perturbatively non-renormalizable
theories that lead to clear calculable predictions, such as for example the
non-linear sigma model above two dimensions. We will work within perturbation
theory in the present paper and will not further consider non-perturbative
aspects.

In [3] renormalizability of the $R+R^{2}$-theory was proved. The proof used a
specific covariant gauge for simplicity. For general gauges an assumption was
made that ultraviolet divergences have the so called cohomological structure.
This hypothesis was proved for a class of background gauges in the work [4].
Hence we consider renormalizability of $R+R^{2}$ gravity with four derivatives
of the metric as well established.

We will also call this model briefly as quadratic gravity.

But in the works [3, 5] it was also stated that quadratic gravity is not
physical because it violates unitarity or causality. So this model was
commonly considered as unphysical.

Quite recenyly quadratic quantum gravity was proved to be in fact unitary [6].
Thus the $R+R^{2}$ model is a candidate for the quantum theory of gravitation.

In the present paper we discuss in detail the exact form of the Lagrangian of
quadratic gravity, the questions of unitarity, stability of the vacuum state
and the behavior of the model at astrophysical and cosmological scales.

##  2 Main part

We consider the relativistic $R+R^{2}$ action including all terms quadratic in
the Riemann tensor $R_{\mu\nu\lambda\rho}$ and its simplifications

$S_{sym}=\int d^{D}x\mu^{-2\epsilon}\sqrt{-g}\left(-M_{Pl}^{2}R+\alpha
R_{\mu\nu}R^{\mu\nu}+\beta R^{2}+\delta
R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}+M_{Pl}^{2}\Lambda\right),$ (1)

here the $R$-term is the Einstein-Hilbert Lagrangian. The $\Lambda$-term is
not essential in perturbation theory which we consider.

$M_{Pl}^{2}=1/(16\pi G)$ is the Planck mass squared, $R_{\mu\nu\rho\sigma}$ is
the Riemann tensor, $R_{\mu\nu}$ is the Ricci tensor and $R$ is the Ricci
scalar. $\alpha$, $\beta$ and $\delta$ are coupling constants, $D=4-2\epsilon$
is the dimension of the space-time within dimensional regularization [7].
$\epsilon$ is the regularizarion parameter and $\mu$ is the parameter with the
dimension of a mass in dimensional regularization.

The Riemann tensor reads

$R^{\rho}_{\sigma\mu\nu}=\partial_{\mu}\Gamma_{\nu\sigma}^{\rho}-\partial_{\nu}\Gamma_{\mu\sigma}^{\rho}+\Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\sigma}^{\lambda}-\Gamma_{\nu\lambda}^{\rho}\Gamma_{\mu\sigma}^{\lambda},$
(2)

here are the Christoffell symbols

$\Gamma_{\mu\nu}^{\alpha}=\frac{1}{2}g^{\alpha\beta}\left(\partial_{\nu}g_{\mu\beta}+\partial_{\mu}g_{\nu\beta}-\partial_{\beta}g_{\mu\nu}\right).$
(3)

Let us underline that dimensional regularization [7] is presently the only
known continuous (not discrete like lattice) regularization of ultraviolet
divergences appropriate for perturbative calculations and preserving gauge
invariance of gravity.

The term containing the coupling $\delta$ in the Lagrangian (1) is usually
omitted in the literature, see e.g. [3, 4, 8]. This is because of the Gauss-
Bonnet identity

$\int
d^{4}x\sqrt{-g}\left(R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}-4R_{\mu\nu}R^{\mu\nu}+R^{2}\right)=0.$
(4)

The identity is valid only in four-dimensional space. But the dimension of the
space-time in dimensional regularization is $4-2\epsilon$. Thus it seems that
the term with the coupling $\delta$ must be preserved in the action to have
renormalizability.

From the other side it is in principle possible that one will invent four-
dimensional continuous regularizatiom which preserves gauge invariance of
gravitational Lagrangian. Then the term with the coupling $\delta$ should be
omitted. The number of coupling constants in the Lagrangian most probably
should not depend on the choice of regularization. In this case the term with
the coupling $\delta$ should be omitted in dimensional regularization also.
The point can be checked with direct calculations of counterterms of the
Lagrangian. To establish the full picture it is most probably necessary to
perform two-loop calculations, as it was with the establishing perturbative
non-renormalizability of pure gravity mentioned in the introduction.
Corresponding calculations are rather involved even at the one-loop level.
This is a subject for a seperate publication.

It should be also mentioned that there is the known Regge-Wheeler lattice
regularization which preserves a form of lattice diffeomorphism invariance.

We will work within perturbation theory. Thus a linearized theory is
considered around the flat space metric

$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},$ (5)

here the convention in four dimensions is $\eta_{\mu\nu}=diag(+1,-1,-1,-1)$.
Within dimensional regularization $\eta_{\mu\nu}\eta^{\mu\nu}=D$. Indexes are
raised and lowered by means of the tensor $\eta_{\mu\nu}$.

Gauge transformations of gravity are generated by diffeomorphisms
$x^{\mu}\rightarrow x^{\mu}+\zeta^{\mu}(x)$ and have the form

$h_{\mu\nu}\rightarrow
h_{\mu\nu}+\partial_{\mu}\zeta_{\nu}+\partial_{\nu}\zeta_{\mu}+\left(h_{\lambda\mu}\partial_{\nu}+h_{\lambda\nu}\partial_{\mu}+(\partial_{\lambda}h_{\mu\nu})\right)\zeta^{\lambda},$
(6)

whith arbitrary functions $\zeta_{\mu}(x)$.

Following standard Faddeev-Popov quantization [9], see also [10], one adds to
the Lagrangian a gauge fixing term which can be chosen e.g. in the form

$S_{gf}=-\frac{1}{2\xi}\int
d^{D}x\mu^{-2\epsilon}F_{\mu}\partial_{\nu}\partial^{\nu}F^{\mu},$ (7)

here $F^{\mu}=\partial_{\nu}h^{\nu\mu}$, $\xi$ is the gauge parameter.
Physical results, of course, do not depend on the allowed choice of the form
of the gauge fixing term.

One should also add the ghost term

$S_{ghost}=\int
d^{D}x\mu^{-2\epsilon}d^{D}y\mu^{-2\epsilon}\overline{C}_{\mu}(x)\frac{\delta
F^{\mu}(x)}{\delta\zeta_{\nu}(y)}C_{\nu}(y)=$ (8) $\int
d^{D}x\mu^{-2\epsilon}\partial^{\nu}\overline{C^{\mu}}\left[\partial_{\nu}C_{\mu}+\partial_{\mu}C_{\nu}+h_{\lambda\mu}\partial_{\nu}C^{\lambda}+h_{\lambda\nu}\partial_{\mu}C^{\lambda}+(\partial_{\lambda}h_{\mu\nu})C^{\lambda}\right],$

where $\overline{C}$ and $C$ are ghost fields. Then one gets the generating
functional of graviton Green functions

$Z(J)=N^{-1}\int
dh_{\mu\nu}dC_{\lambda}d\overline{C_{\rho}}\exp{\left[i\left(S_{sym}+S_{gf}+S_{ghost}+\int
d^{D}x\mu^{-2\epsilon}J_{\mu\nu}h^{\mu\nu}\right)\right]},$ (9)

here $N$ is the normalization factor of the functional integral in the usual
notation, $J_{\mu\nu}$ is as usual the source of gravitons.

We work within perturbation theory, hence one makes the shift of the fields

$h_{\mu\nu}\rightarrow M_{Pl}\mu^{-\epsilon}h_{\mu\nu}.$ (10)

Perturbative expansion is in inverse powers of the Plank mass or in other
words in powers of the Newton coupling constant $G\propto 1/M_{Pl}^{2}$.

Let us obtain the graviton propagator. One takes the part of the Lagrangian
quadratic in $h_{\mu\nu}$ and makes the Fourier transform

$Q=\frac{1}{4}\int
d^{D}k~{}h^{\mu\nu}(-k)\left[\left(k^{2}+M_{Pl}^{-2}k^{4}(\alpha+4\delta)\right)P^{(2)}_{\mu\nu\rho\sigma}\right.$
$+k^{2}\left(-2+4M_{Pl}^{-2}k^{2}(\alpha+3\beta+\delta)\right)P^{(0-s)}_{\mu\nu\rho\sigma}$
(11)
$\left.+\frac{1}{\xi}M_{Pl}^{-2}k^{4}\left(P^{(1)}_{\mu\nu\rho\sigma}+2P^{(0-w)}_{\mu\nu\rho\sigma}\right)\right]h^{\rho\sigma}(k),$

$P^{(i)}_{\mu\nu\rho\sigma}$ being projectors to the spin-2, spin-1 and spin-0
components of the field $h_{\mu\nu}$:

$P^{(2)}_{\mu\nu\rho\sigma}=\frac{1}{2}\left(\Theta_{\mu\rho}\Theta_{\nu\sigma}+\Theta_{\mu\sigma}\Theta_{\nu\rho}\right)-\frac{1}{3}\Theta_{\mu\nu}\Theta_{\rho\sigma},$
(12)
$P^{(1)}_{\mu\nu\rho\sigma}=\frac{1}{2}\left(\Theta_{\mu\rho}\omega_{\nu\sigma}+\Theta_{\mu\sigma}\omega_{\nu\rho}+\Theta_{\nu\rho}\omega_{\mu\sigma}+\Theta_{\nu\sigma}\omega_{\mu\rho}\right),$
(13)
$P^{(0-s)}_{\mu\nu\rho\sigma}=\frac{1}{3}\Theta_{\mu\nu}\Theta_{\rho\sigma},$
(14) $P^{(0-w)}_{\mu\nu\rho\sigma}=\omega_{\mu\nu}\omega_{\rho\sigma}.$ (15)

Here $\Theta_{\mu\nu}=\eta_{\mu\nu}-k_{\mu}k_{\nu}/k^{2}$ and
$\omega_{\mu\nu}=k_{\mu}k_{\nu}/k^{2}$ are transverse and longitudinal
projectors correspondingly.

We note that the expression (11) differs from the similar expression presented
in [8] by the absence of $\epsilon$-dependent terms.

To get the graviton propagator $D_{\mu\nu\rho\sigma}$ one inverts the matrix
in square brackets of the expression (11):

$[Q]_{\mu\nu\kappa\lambda}D^{\kappa\lambda\rho\sigma}=\frac{1}{2}(\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma}+\delta_{\mu}^{\sigma}\delta_{\nu}^{\rho}).$
(16)

Then the propagator has the form

$D_{\mu\nu\rho\sigma}=\frac{1}{i(2\pi)^{D}}\left[\frac{4}{k^{2}}\left(\frac{1}{1+M_{Pl}^{-2}k^{2}(\alpha+4\delta)}\right)P^{(2)}_{\mu\nu\rho\sigma}\right.$
(17)
$-\frac{2}{k^{2}}\left(\frac{1+2\epsilon\frac{1-M_{Pl}^{-2}k^{2}(\alpha+4\beta)}{1+M_{Pl}^{-2}k^{2}(\alpha+4\delta)}}{1-\epsilon-
M_{Pl}^{-2}k^{2}\left((2\alpha+6\beta+2\delta)-\epsilon(\alpha+4\beta)\right)}\right)P^{(0-s)}_{\mu\nu\rho\sigma}$
$\left.+4\xi\frac{1}{M_{Pl}^{-2}k^{4}}\left(P^{(1)}_{\mu\nu\rho\sigma}+\frac{1}{2}P^{(0-w)}_{\mu\nu\rho\sigma}\right)\right].$

Then one performs partial fractioning. The propagator takes the forn

$D_{\mu\nu\rho\sigma}=\frac{1}{i(2\pi)^{D}}\left[4P^{(2)}_{\mu\nu\rho\sigma}\left(\frac{1}{k^{2}}-\frac{1}{k^{2}-M_{Pl}^{2}/(-\alpha-4\delta)}\right)\right.$
(18)
$-2\frac{P^{(0-s)}_{\mu\nu\rho\sigma}}{1-\epsilon}\left(1+2\epsilon\frac{1-M_{Pl}^{-2}k^{2}(\alpha+4\beta)}{1+M_{Pl}^{-2}k^{2}(\alpha+4\delta)}\right)$
$\left(\frac{1}{k^{2}}-\frac{1}{k^{2}-M_{Pl}^{2}(1-\epsilon)/(2\alpha+6\beta+2\delta-\epsilon(\alpha+4\beta))}\right)$
$\left.+\frac{4\xi}{M_{Pl}^{-2}k^{4}}\left(P^{(1)}_{\mu\nu\rho\sigma}+\frac{1}{2}P^{(0-w)}_{\mu\nu\rho\sigma}\right)\right].$

In four dimensions one obtains the following graviton propagator

$D_{\mu\nu\rho\sigma}=\frac{4}{i(2\pi)^{D}}\left[\frac{P^{(2)}_{\mu\nu\rho\sigma}-\frac{1}{2}P^{(0-s)}_{\mu\nu\rho\sigma}}{k^{2}}-\frac{P^{(2)}_{\mu\nu\rho\sigma}}{k^{2}-M_{Pl}^{2}/(-\alpha-4\delta)}\right.$
(19)
$\left.+\left(\frac{1}{2}\right)\frac{P^{(0-s)}_{\mu\nu\rho\sigma}}{k^{2}-M_{Pl}^{2}/(2\alpha+6\beta+2\delta)}+\frac{\xi}{M_{Pl}^{-2}k^{4}}\left(P^{(1)}_{\mu\nu\rho\sigma}+\frac{1}{2}P^{(0-w)}_{\mu\nu\rho\sigma}\right)\right],$

We will now consider classical quadratic gravity. In this case for a point
particle having the energy-momentum tensor
$T_{\mu\nu}=\delta^{0}_{\mu}\delta^{0}_{\nu}M\delta^{3}(x)$ one gets the
gravitational field [5]

$V(r)=\frac{M}{2\pi
M_{Pl}^{2}}\left(-\frac{1}{4r}+\frac{e^{-m_{2}r}}{3r}-\frac{e^{-m_{0}r}}{12r}\right).$
(20)

$m_{2}^{2}=M_{Pl}^{2}/(-\alpha-4\delta)$ and
$m_{0}^{2}=M_{Pl}^{2}/(2\alpha+6\beta+2\delta)$ are squared masses
correspondingly of massive spin-2 and spin-0 gravitons. Cupling constants
$\alpha,\beta$ and $\delta$ can be chosen to obtain positive masses. In [3, 5]
it was noted that masses can be chosen large enough to have an agreement with
experiments.

Our propagator (19) reproduces the expression (20). One can see it by means of
the calculation of the tree level Feynman diagram corresponding to an exchange
of two point-like particles by a graviton.

The graviton propagator in the work [3] dos not produce the expression (20).
It contains some technical errors. To see this one puts in the $R+R^{2}$
Lagrangian coupling constants equal to zero except the Newton coupling. The
Lagrangian is reduced then to General Relativity. Hence the graviton
propagator should also be reduced to one of General Relativity:

$D_{\mu\nu\rho\sigma}(k)=\frac{1}{i(2\pi)^{4}}\frac{1/2\eta_{\mu\rho}\eta_{\nu\sigma}+1/2\eta_{\mu\sigma}\eta_{\nu\rho}-1/2\eta_{\mu\nu}\eta_{\rho\sigma}+terms\propto
k}{k^{2}},$ (21)

where the gauge condition with $\xi=0$ is taken for simplicity.

Our propagator (19) reproduces the propagator (21) in this limit. The
propagator of the work [3] has the factor 1 instead of 1/2 in the third term
of the numerator of (21) in the corresponding limit.

The second term in the expression (19) for the graviton propagator has the
non-standard minus sign. Hence one considers it as the massive spin-2 ghost.
For renormalizability of quadratic gravity one must shift poles of all
propagators in Feynman diagrams in the same way $k^{2}\rightarrow k^{2}+i0$.
Hence the spin-2 ghost must be considered as a state with negative metric [3].
That is why violation of either unitarity or causality within the $R+R^{2}$
model was claimed in [3, 5].

But this massive spin-2 ghost is unstable. It unavoidably decays in two or
more physical massless gravitons. The width of the decay is small. However
independently of the value of this decay width spin-2 ghost particles do not
appear as asymptotic states of the $S$-matrix elements. Only physical
gravitons appear as external particles of the $S$-matrix amplitudes. Thus one
concludes that unitarity is preserved in the $R+R^{2}$ model.

There is a statement about instability of theories with ghosts , i.e. their
Hamiltonians are unbounded from below and they do not have stable vacuum
states. This question was raised in [11] within Quantum Mechanics, see also
[12] for the brief review. But this statement is proved only for Quantum
Mechanical systems. Quantum Field Theory is quite a different story and
renomalizability plus unitarity is enough to have a consistent theory.

To see this let us consider the graviton propagator in the opeator formalism:

$D_{\mu\nu\rho\sigma}(x-y)=\frac{\delta^{2}}{\delta J_{\mu\nu}(x)\delta
J_{\rho\sigma}(y)}Z(J)=<0|T\left[h_{\mu\nu}(x)h_{\rho\sigma}(y)\right]|0>.$
(22)

One transforms it to the momentum space and inserts the sum over the complete
set of momentum eigenstates between two graviton fields. The states with
negative norms in the sum have the extra factor $-1$. It gives the negative
residue for the massive spin-2 ghost pole.

There is another way to produce the negative residue for the spin-2 ghost. One
can prescribe negative energy to this ghost. The expansion of the graviton
fields into the creation and annihilation operators produces normalization
factors $1/\sqrt{-2k_{0}}$. This is the reason for the negative residue for
the spin-2 ghost. In this case of negative energy the Hamiltonian would be
indeed unbounded and the vacuum state would be unstable.

But as it was mentioned above, one should choose the variant with negarive
metric in order to have renormalizability in the theory [3]. Thus one has the
consistent theory with the stable ground state. There are no reasons for a
Hamiltonian to be unbounded from below if there are no states with negative
energies.

It should be mentioned that the $S$-matrix by construction automatically
satisfies the unitarity relation

$S^{+}S=1$ (23)

in theories having Hermitian Lagrangians [13].

To see it one considers the $S$-matrix in the operator formalism

$S=T\left(e^{i\int L(x)dx}\right).$ (24)

One introduces a function $g(x)$ having the values in the interval $(0,1)$.
This function describes intensity of interactions. Interactions are switched
off if $g(x)=0$. If $g(x)=1$ then interactions are switched on. Interactions
are switched on partly if $0<g(x)<1$. One substitutes the product $L(x)g(x)$
for the Lagrangian $L(x)$. The $S$-matrix becomes the functional

$S(g)=T\left(exp~{}~{}i\int L(x)g(x)dx\right).$ (25)

One splits the interaction region characterised by the function $g(x)$ into an
infinitely large number of infinitely thin segments $\Delta_{i}$ using the
space-like surfaces $t=const$.

Then one gets

$S(g)=T\left(exp~{}~{}i\int
L(x)g(x)dx\right)=T\left(exp~{}~{}i\sum_{j}\int_{\Delta_{j}}L(x)g(x)dx\right)=$
(26) $T\left(\prod_{j}exp~{}~{}i\int_{\Delta_{j}}L(x)g(x)dx\right).$

$S(g)$ is defined as the limit

$S(g)=\lim_{\Delta_{j}\rightarrow
0}T\left(\prod_{j}\left(1+i\int_{\Delta_{j}}L(x)g(x)dx\right)\right).$ (27)

The r.h.s. of (27) is a product taken in the chronological order of the
segments $\Delta_{j}$. Each factor in this product is unitary up to small
terms of higher orders for sufficiently small $\Delta_{j}$. These higher
orders can be neglected in the considered limit. Hence the whole product is
unitary. Unitarity of $S(g)$ and of the matrix

$S=\lim_{g(x)\rightarrow 1}S(g)$ (28)

is proved.

Sometimes one understands the following thing under unitarity. One derives
from (23) the famous optical theorem stating that imagenary part of an
amplitude of some forward scattering coincides up to a factor with the
corresponding total annihilation crosssection

$Im<i|T|i>=\frac{1}{2}\sum_{n}<i|T^{+}|n><n|T|i>,$ (29)

where $|i>$ is the scattering state, $T$ is the scattering matrix: $S=1+iT$,
and one assumes that all physical states $|n>$ form a complete set in the
theory

$\sum_{n}|n><n|=1.$ (30)

From the other side one can calculate $Im<i|T|i>$ directly from Feynman
diagrams using Cutcosky cuts. Then one assumes that the result should coincide
with (29). But if it does not happen it does not mean violation of unitarity.
It only means that physical states in the theory do not form a complete set
(30) and the complete set is formed by physical plus unphysical states.

Unitarity of theories with negative metric states was previously considered in
[14, 15], see also references therein. Question of causality were also
considered there.

We would like to note that the tree level graviton propagator (19) is modified
by the summation of the chain of one-loop insertions. As it was already
mentioned above the second term of the propagator (19) has the minus sign.
Therefore the summation of the one-loop insertions with the massless graviton
in the loop will shift the pole of the spin-2 ghost from the value
$k^{2}=M_{Pl}^{2}/(-\alpha-4\delta)$ to the complex value
$k^{2}=M_{Pl}^{2}/(-\alpha-4\delta)-i\Gamma$. Here $\Gamma$ is the width of
the spin-2 ghost decay into the pair of massless physical gravitons. This
complex pole is located on the unphysical Riemann sheet. It is analogous to
the known virtual level of the neutron-proton system with opposite spins of
nucleons [16].

It should be noted that one loop corrections in quadratic gravity were studied
in [17].

We would like to underline that we consider not pure $R^{2}$ theory but the
$R+R^{2}$ theory where the $R^{2}$ terms are added to the Einstein-Hilbert
Lagrangian. Gravitational constants $\alpha$, $\beta$ and $\delta$ of these
terms in the Lagrangian can be chosen suffisiently small to ensure that
quadratic gravity will be practically indistinguishable from General
Relativity at astrophysical and cosmological scales. This is independently of
the discussed above in detail question wether the coupling $\delta$ is exactly
zero or not. The $R^{2}$ terms are introduced only to have renormalizability
of quadratic gravity which is valid in particular for arbitrary small
couplings $\alpha$, $\beta$ and $\delta$.

We have analyzed only purely gravitational $R+R^{2}$ action. The inclusion of
the matter fields in the Lagrangian is straightforward and does not change
conclusions.

##  3 Conclusions

We have proved unitarity of quantum gravity with the $R+R^{2}$ action. This
model was previously shown to be renormalizable in the work [3]. The
parameters of quadratic gravity can be adjusted to ensure that the theory will
be practically indistinguishable from General Relativity at astrophysical and
cosmological scales.

One can conclude that the $R+R^{2}$ model is an appropriate candidate for the
fundamental quantum theory of gravity.

##  4 Acknowledgments

The author is grateful to the collaborators of the Theory Division of the
Institute for Nuclear Research for valuable discussions.

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