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\begin{document}

\title[Lecture 3. RG for the BKT transition]{Lecture 3.\\Renormalization group for the Berezinskii\--Kosterlitz\--Thouless transition}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Renormalization group approach}
\unpause

Suppose we consider a field theory system with the correlation length $r_c$. It is described by a {\red bare action} defined the at the UV cutoff $r_0$, which depends on the set of parameters $\lambda_0$. We are interested in correlations functions on a scale $r$, $r_0\ll r\ll r_c$. Let $G_\text{exact}(\lambda_0,r_0;\cdots)$ be exact correlation functions calculated in all orders of the perturbation theory.
\unpause
There exists a \emph{dressed action}, which depend on the set of parameters $\lambda$. Let $G_\text{tree}(\lambda,r;\cdots)$ the \emph{tree} (first nonvanishing order) correlation functions for the dressed action.
\unpause
The dressed action is defined in such a way that
$$
G_\text{exact}(\lambda_0,r_0;x_1,\ldots,x_n)=G_\text{tree}\left(\lambda,r;{x_1\over r},\ldots,{x_n\over r}\right).
$$
if $x_i\sim r$.
\unpause
With this definition the parameters $\lambda$ depend on the scale $r$. The group of transformations $(r,\lambda)\to(r',\lambda')$ for the same $\lambda_0,r_0$ is called the \emph{renormalization group} (RG).
\unpause

In combining the RG approach with the perturbation theory make it possible to find the behavior of correlation functions in a wide range of scales in the case of a nearly\-/marginal perturbation.
\unpause
In our case it is
$$
\delta=\beta^2-2\ll1.
$$

\end{frame}

\begin{frame}
\frametitle{Bare and dressed sine\-/Gordon action}
\unpause

The \emph{bare action} of the sine\-/Gordon model on the Euclidean plane:
\eq$$
S_\text{SG}[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over8\pi}-\alpha_0r_0^{\beta_0^2-2}\cos\beta_0\phi\right),
\label<.>{SG-action}
$$
\unpause
Here $r_0$ is the UV cutoff,
\unpause
$\alpha_0$ is the dimensionless bare coupling constant.
\unpause
Instead of the scale $r$ it is convenient to introduce an infrared scale into the action.
\unpause
It can be done by means of a small mass term
\eq$$
S_{SG}[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over8\pi}
+{m_0^2\phi^2\over8\pi}-\alpha_0r_0^{\beta_0^2-2}\cos\beta_0\phi\right).
\label{SG-action-mod}
$$
\unpause
For ultraviolet regularization, we will replace $x^2$ by $x^2+r_0^2$. Then for $m_0^2x^2\ll1$ the free field propagator (with $\alpha_0=0$) is equal to
\eq$$
G_0(x-x')=\log{R_0^2\over(x-x')^2+r_0^2},
\qquad
R_0=(cm_0)^{-1},
\qquad
c=e^{\gamma_E}/2.
\label{G0-def}
$$
\unpause
Now write down the renormalized (dressed) action
\eq$$
S^R_\text{SG}[\phi]=\int d^2x\,\left({(\d_\mu\phi)^2\over8\pi}
+{m^2\phi^2\over8\pi}-{\alpha\over R^2}\cos\beta\phi\right),
\qquad
R=(cm)^{-1},
\label{SG-action-ren}
$$
\unpause
such that $S_\text{SG}[\phi]=S^R_\text{SG}[Z_\phi^{-1/2}\phi]+S^\text{ct}[Z_\phi^{-1/2}\phi]$.
\unpause
Assume that the counterterms
$$
S^\text{ct}[\phi]=\int d^2x\,(\#(\d_\mu\phi)^2+\#\cos\beta\phi).
$$
do not contain a counterterm for the auxiliary mass term.

\end{frame}

\begin{frame}
\frametitle{The renormalization procedure}
\unpause

Hence we have two renormalization constants $Z_\phi$ and $Z_\alpha$:
\eq$$
\Aligned{
\phi
&=Z_\phi^{1/2}\phi_R,
&\quad
\beta_0
&=Z_\phi^{-1/2}\beta,
\\
m_0
&=Z_\phi^{-1/2}m,
&\quad
\alpha_0
&=Z_\alpha\alpha.
}\label{all-ren}
$$
\unpause
Let $G_0(x-x')=\langle\phi(x)\phi(x')\rangle_0$ be the propagator of the unperturbed ($\alpha_0=0$) theory,
\unpause
$G(x-x')=\langle\phi(x)\phi(x')\rangle$ be the exact propagator of the full theory for the field $\phi$,
\unpause
and $G_R(x-x')=\langle\phi_R(x)\phi_R(x')\rangle=Z_\phi^{-1}G(x-x')$ be the renormalized exact propagator.
\unpause
They can be consider as kernels of operators $G_0=4\pi(-\d_\mu^2+m_0^2)^{-1}$, $G$ and $G_R$ correspondingly.
\unpause

The \emph{mass operator} $\Sigma$ is defined by the relation
\eq$$
G^{-1}=G_0^{-1}+{1\over4\pi}\Sigma.
\label{GG0-rel}
$$
\unpause
Suppose that in the momentum space
\eq$$
G_R(p^2)={4\pi\over p^2+M^2}+O(p^4)\quad\text{as $p^2\to0$,}
\label{rencond}
$$
\unpause
with a constant
\eq$$
M^2=m^2+{4\pi\alpha\beta^2\over R^2}=m^2(1+4\pi c^2\alpha\beta^2).
\label{alpha-def}
$$
This defines the renormalized coupling constant $\alpha$ for a given scale $R$.

\end{frame}

\begin{frame}
\frametitle{The renormalization procedure}
\unpause

The renormalization condition can be rewritten as $\Sigma(p^2)=\Sigma_0+\Sigma_1p^2+O(p^4)$.
\unpause
Indeed,
\Multline$$
4\pi G^{-1}(p^2)=p^2+m_0^2+\Sigma(p^2)\uncover<+->{{}=p^2+m_0^2+\Sigma_0+\Sigma_1p^2+O(p^4))}
\\
\uncover<+->{{}={\red\alt<.(2)-.(3)>{\transparent1}{\transparent0}\underbrace{\transparent1\black(1+\Sigma_1)}_{Z_\phi^{-1}}}
\Bigl(p^2+{\red\alt<.(3)>{\transparent1}{\transparent0}\underbrace{\transparent1\black m^2+\Sigma_0(1+\Sigma_1)^{-1}}_{M^2}}\Bigr)+O(p^4)}
\uncover<+->{{}=4\pi(1+\Sigma_1)G_R^{-1}(p^2).}
\label{G-1-expansion}
$$
\unpause
We obtain
\eq$$
Z_\phi={1\over1+\Sigma_1},
\qquad
\uncover<+->{M^2=m^2+{\Sigma_0\over1+\Sigma_1},}
\qquad
\uncover<+->{m^2=
\uncover<.>
{\tikz[overlay,red] {\node[above,xshift=-.7em] (m2eq) {}; \node[above=1cm,draw] (m2r02const) {$m_0^2\phi^2=m^2\phi_R^2$}; \draw[->] (m2r02const) -- (m2eq);}}
{m_0^2\over1+\Sigma_1}.}
\label{phi-m-ren}
$$
\unpause
Instead of calculating $\Sigma$ it is more convenient to calculate the correlation function
\Multline*$$
G(x-x')=\left\langle\phi(x)\phi(x')\right\rangle
={\left\langle\phi(x)\phi(x')e^{-S_1[\phi]}\right\rangle_0\over\left\langle e^{-S_1[\phi]}\right\rangle_0}
\\
=\left\langle\phi(x)\phi(x')\right\rangle_0
-\langle\phi(x)\phi(x')S_1[\phi]\rangle_{0,\text{c}}+{1\over2}\langle\phi(x)\phi(x')S_1^2[\phi]\rangle_{0,\text{c}}
\\
-{1\over6}\langle\phi(x)\phi(x')S_1^3[\phi]\rangle_{0,\text{c}}
+O(\alpha_0^4).
$$
\unpause
The connected averages $\langle\cdots\rangle_{0,\text{c}}$ will be extracted on the fly.
\unpause
Then the mass operator will be extracted by removing `legs' from the diagrams.

\end{frame}

\begin{frame}
\frametitle{RG: first order}
\unpause

Let us calculate
$$
-\langle\phi(x)\phi(x')S_1[\phi]\rangle
=\alpha_0r_0^{\delta_0}\int d^2y\,\langle\phi(x)\phi(x')\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0.
$$
\unpause
We have
\Multline*$$
\langle\phi(x)\phi(x')\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0
=\langle\phi(x)\phi(x')\rangle_0\langle\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0
\\
-\beta_0^2\langle\phi(x)\phi(y)\rangle_0\langle\phi(x')\phi(y)\rangle_0\langle\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0.
$$
\unpause
The first term is disconnected, \uncover<.(1)->{the second one contains two external lines:}
$$
\begin{tikzpicture}[baseline=0pt]
\draw[thick] (-1,-1) node[left] {$x$} -- (1,-1) node[right] {$x'$};
\draw[thick] (-1,0) node[left] {$y$} to[out=80,in=90] (1,0) to[out=-90,in=-80] cycle;
\draw[thick] (-1,0) to[out=60,in=90] (.8,0) to[out=-90,in=-60] cycle;
\draw[thick] (-1,0) to[out=40,in=90] (.6,0) to[out=-90,in=-40] cycle;
\draw[thick] (-1,0) to[out=20,in=90] (.4,0) to[out=-90,in=-20] cycle;
\end{tikzpicture}
\qquad
\uncover<+->{\begin{tikzpicture}[baseline=0pt]
\draw[thick] (-1,-1) node[left] {$x$} -- (1,-1) node[right] {$x'$};
\draw[thick] (0,-1) node[below] {$y$} to[out=170,in=180] (0,1) to[out=0,in=10] cycle;
\draw[thick] (0,-1) to[out=150,in=180] (0,.8) to[out=0,in=30] cycle;
\draw[thick] (0,-1) to[out=130,in=180] (0,.6) to[out=0,in=50] cycle;
\draw[thick] (0,-1) to[out=110,in=180] (0,.4) to[out=0,in=70] cycle;
\end{tikzpicture}}
$$
\unpause
Hence,
$$
-{1\over4\pi}\Sigma^{(1)}(y-y')=-\alpha_0\beta_0^2r_0^{\delta_0
\tikz[overlay,red] {\node[above left] (delta0) {}; \node[draw,above right=.3cm] (deltadef) {$\Gathered{\delta_0=\beta_0^2-2\\\delta=\beta^2-2}$};
  \draw[->] (deltadef) -- (delta0);}
}\langle\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0\delta(y-y')
\uncover<+->{{}=
\tikz[overlay,red] {\node[above,xshift=-.7em] (Sigma1eq2) {};
  \node[draw,above right=.7cm] (cosR) {$\langle\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0=R_0^{-\beta_0^2}$}; \draw[->] (cosR) -- (Sigma1eq2);}
-{\alpha_0\beta_0^2\over R_0^2}\left(r_0\over R_0\right)^{\delta_0}\delta(y-y').}
$$
\unpause
In the momentum space:
\eq$$
\Sigma^{(1)}(p^2)=\Sigma^{(1)}_0={4\pi\alpha_0\beta_0^2\over R_0^2}\left(r_0\over R_0\right)^{\delta_0},
\qquad
\Sigma^{(1)}_1=0.
\label{Sigma(1)-result}
$$

\end{frame}

\begin{frame}
\frametitle{RG: first order}
\unpause

By comparing this with the formulas
\eq$$
M^2=m^2+{4\pi\alpha\beta^2\over R^2}=m^2(1+4\pi c^2\alpha\beta^2).
\tag{\ref{alpha-def}}
$$
\unpause
and
\eq$$
Z_\phi={1\over1+\Sigma_1},
\qquad
M^2=m^2+{\Sigma_0\over1+\Sigma_1},
\qquad
m^2={m_0^2\over1+\Sigma_1},
\tag{\ref{phi-m-ren}}
$$
\unpause
we obtain $Z_\phi=1$ $\Rightarrow$ $m=m_0$, $\beta=\beta_0$.
\unpause
Therefore
\eq$$
Z_\alpha=\left(R\over r_0\right)^\delta.
\label{Z(1)-result}
$$
\unpause
Though the answer is quite clear and applicable to any value of $\delta$, let us formally perform the standard RG procedure.
\unpause
For $\delta\ll1$ we have
$$
Z_\alpha=1+\delta\log{R\over r_0},
$$
\unpause
Take the derivative of $\alpha=Z_\alpha^{-1}\alpha_0$:
$$
{d\alpha\over dt}=-\alpha_0\delta,\qquad t=\log R.
$$
\unpause
Substitute $\alpha_0$ by $\alpha$ in the r.h.s.:
$$
{d\alpha\over dt}=-\alpha\delta,
$$
\unpause
The solution is $\alpha\sim R^{-\delta}$ in consistency with (\ref{Z(1)-result}).

\end{frame}

\begin{frame}
\frametitle{RG: first order}
\unpause

The RG trajectories look like
$$
\begin{tikzpicture}[>=stealth]
\draw[very thick,->] (-3.5,0) -- (3.5,0) node[above right] {$\delta$};
\draw[very thick,->] (0,0) -- (0,3.5) node[above right] {$\alpha$};
\foreach \x in {-3,...,-1}
  {
  \draw[->] (\x,0) -- +(0,1);
  \draw[->] (\x,.9) -- +(0,1.1);
  \draw[->] (\x,1.9) -- +(0,1.1);
  \draw (\x,2.9) -- +(0,.35);
  }
\foreach \x in {-2.5,...,-.5}
  {
  \draw[->] (\x,0) -- +(0,.5);
  \draw[->] (\x,.4) -- +(0,1.1);
  \draw[->] (\x,1.4) -- +(0,1.1);
  \draw (\x,2.4) -- +(0,.85);
  }
\foreach \x in {1,...,3}
  {
  \draw[->] (\x,3.25) -- +(0,-.25);
  \draw[->] (\x,3.1) -- +(0,-1.1);
  \draw[->] (\x,2.1) -- +(0,-1.1);
  \draw (\x,1.1) -- +(0,-1.1);
  }
\foreach \x in {.5,...,2.5}
  {
  \draw[->] (\x,3.25) -- +(0,-.75);
  \draw[->] (\x,2.6) -- +(0,-1.1);
  \draw[->] (\x,1.6) -- +(0,-1.1);
  \draw (\x,.6) -- +(0,-.6);
  }
\end{tikzpicture}
$$
\unpause
The transition point $\delta=0$ here is a line of fixed points for any value of $\alpha$. Is it really the case?

\end{frame}

\begin{frame}
\frametitle{RG: second order}
\unpause

Consider the second order contribution. The connected contribution to the pair correlation function is
\Multline*$$
{1\over2}\langle\phi(x)\phi(x')S_1^2[\phi]\rangle_{\rm 0,c}
={\alpha_0^2r_0^{2\delta_0}\over2}\int d^2y_1\,d^2y_2\,
\langle\phi(x)\phi(x')\lcolon\cos\beta_0\phi(y_1)\rcolon\lcolon\cos\beta_0\phi(y_2)\rcolon\rangle_{\rm 0,c}
\\
\uncover<.(1)->{{}={\alpha_0^2\beta_0^2r_0^{2\delta_0}}\int d^2y_1\,d^2y_2\,
\Bigl(\langle\phi(x)\phi(y_1)\rangle_0\langle\phi(x')\phi(y_2)\rangle_0
\langle\lcolon\sin\beta_0\phi(y_1)\rcolon\lcolon\sin\beta_0\phi(y_2)\rcolon\rangle_0}
\\*
\uncover<.(2)->{{}-\langle\phi(x)\phi(y_1)\rangle_0\langle\phi(x')\phi(y_1)\rangle_0
\Bigl(\langle\lcolon\cos\beta_0\phi(y_1)\rcolon\lcolon\cos\beta_0\phi(y_2)\rcolon\rangle_0
-R_0^{-2\beta_0^2}\Bigr)\Bigr).}
$$
\unpause
These terms correspond to the diagrams
\vspace{-\baselineskip}
$$
\begin{tikzpicture}[baseline=0pt]
\draw[thick] (-2,0) node[left] {$x$} -- (-1,0) node[above left] {$y_1$} -- (1,0) node[above right] {$y_2$} -- (2,0) node[right] {$x'$};
\draw[thick] (-1,0) to[out=80,in=100] (1,0);
\draw[thick] (-1,0) to[out=50,in=130] (1,0);
\draw[thick] (-1,0) to[out=20,in=160] (1,0);
\draw[thick] (-1,0) to[out=-80,in=-100] (1,0);
\draw[thick] (-1,0) to[out=-50,in=-130] (1,0);
\draw[thick] (-1,0) to[out=-20,in=-160] (1,0);
\end{tikzpicture}
\qquad
\uncover<+->{\begin{tikzpicture}[baseline=0pt]
\draw[thick] (-1,-1) node[left] {$x$} -- (0,-1) node[below] {$y_1$} -- (1,-1) node[right] {$x'$};
\draw[thick] (0,-1) to[out=165,in=-165] (0,1) node[above] {$y_2$};
\draw[thick] (0,-1) to[out=130,in=-130] (0,1);
\draw[thick] (0,-1) to[out=105,in=-105] (0,1);
\draw[thick] (0,-1) to[out=75,in=-75] (0,1) node[above] {$y_2$};
\draw[thick] (0,-1) to[out=50,in=-50] (0,1);
\draw[thick] (0,-1) to[out=10,in=-10] (0,1);
\end{tikzpicture}}
$$
\unpause
For calculation of $\Sigma^{(2)}$ we have to remove `legs' and to subtract the contribution of one line in the first diagram:
\Align*$$
-{1\over4\pi}\Sigma^{(2)}(x)
&=\alpha_0^2\beta_0^2r_0^{2\delta_0}
\biggl(\langle\lcolon\sin\beta_0\phi(x)\rcolon\lcolon\sin\beta_0\phi(0)\rcolon\rangle_0
-\beta_0^2R_0^{-2\beta_0^2}\langle\phi(x)\phi(0)\rangle_0
\\*
&\quad
-\delta(x)\int d^2y\,\left(\langle\lcolon\cos\beta_0\phi(0)\rcolon\lcolon\cos\beta_0\phi(y)\rcolon\rangle_0
-R_0^{-2\beta_0^2}\right)\biggr)
$$

\end{frame}

\begin{frame}
\frametitle{RG: second order}
\unpause

Explicitly,
\Align*$$
-{1\over4\pi}\Sigma^{(2)}(x)
&={\alpha_0^2\beta_0^2r_0^{2\delta_0}\over2R_0^{2\beta_0^2}}
\Biggl({\red\left(R_0\over x\right)^{2\beta_0^2}}-{\blue\left(x\over R_0\right)^{2\beta_0^2}}
-2\beta_0^2\log{R_0^2\over x^2}
\\*
&\quad
-\delta(x)\int d^2y\,\left({\red\left(R_0\over y\right)^{2\beta_0^2}}+{\blue\left(y\over R_0\right)^{2\beta_0^2}}
-2\right)\Biggr).
$$
\unpause
In the momentum space we have
\Multline$$
\Sigma^{(2)}(p^2)=-2\pi\alpha_0^2\beta_0^2r_0^{2\delta_0}
\biggl(\int d^2x\,{\red(e^{\i px}-1)x^{-2\beta_0^2}}
\\*
-R_0^{-4\beta_0^2}\int d^2x\,{\blue(e^{\i px}+1)x^{2\beta_0^2}}
-2\beta_0^2R_0^{-2\beta_0^2}G_0(p^2)+2R_0^{2-2\beta_0^2}\biggr).
\label{Sigma(2)-momentum}
$$
\unpause
The second line vanishes as $R_0\to\infty$ for $\delta_0\ll1$.
\unpause
The integral in the first line must be expanded in $p$:
\eq$$
\Sigma^{(2)}(p^2)=\pi\alpha_0^2\beta_0^2r_0^{2\delta_0}\int d^2x\,(px)^2x^{-2\beta_0^2}+O(p^4)
\simeq\pi^2\alpha_0^2\beta_0^2p^2\log{R_0\over r_0}+O(p^4).
\label{Sigma(2)-result}
$$
\unpause
It only contributes to $\Sigma_1$. We have
\eq$$
Z_\phi=1-\pi^2\alpha_0^2\beta_0^2\log{R\over r_0},
\qquad
Z_\alpha=1+\delta_0\log{R\over r_0}.
\label{Z(1)log-result}
$$

\end{frame}

\begin{frame}
\frametitle{RG: second order}
\unpause

Substituting it to $\alpha=Z_\alpha^{-1}\alpha_0$ and $1+\delta/2=Z_\phi(1+\delta_0/2)$, taking the derivation and expressing $\alpha_0,\delta_0$ in terms of $\alpha,\delta$ in the r.h.s., we obtain
\eq$$
{d\alpha\over dt}=-\delta\alpha,
\qquad
{d\delta\over dt}=-4\pi^2\alpha^2,
\qquad
t=\log R.
\label{RG(2)equation}
$$
\unpause
These equations can be rewritten in the form
\eq$$
{d(2\pi\alpha\mp\delta)\over dt}=\pm2\pi\alpha(2\pi\alpha\mp\delta).
\tag{\ref{RG(2)equation}a}
$$
\unpause
This means that the straight lines $2\pi\alpha=\pm\delta$ are RG trajectories. They divide the half\-/plane $\alpha>0$ into three regions:
$$
\begin{tikzpicture}[>=stealth]
\draw[very thick,->] (-3.5,0) -- (3.5,0) node[above right] {$\delta$};
\draw[very thick,->] (0,0) -- (0,3.5) node[above right] {$2\pi\alpha$};
\draw[thick,->] (0,0) -- (-1.5,1.5);
\draw[thick,->] (-1.4,1.4) -- (-3,3);
\draw[thick] (-2.9,2.9) -- (-3.5,3.5) node[above left] {$s_2$};
\draw[thick,->] (3.5,3.5) node[above right] {$s_1$} -- (2,2);
\draw[thick,->] (2.1,2.1) -- (.5,.5);
\draw[thick] (.6,.6) -- (0,0);
\foreach \i/\r in {1/3.35,2/2.9,3/1.8} {
  \draw[domain=-\r:-.9] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=-1:1.1,<-] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=1:\r,<-] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=0:1,->] plot ({-sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=.9:\r] plot ({-sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=0:1.1] plot ({sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=1:\r,<-] plot ({sqrt(\i*\i+\x*\x)},\x);
}
\node at (2.5,.5) {\it I};
\node at (.5,2.5) {\it II};
\node at (-2.5,.5) {\it III};
\end{tikzpicture}
$$

\end{frame}

\begin{frame}
\frametitle{RG: regions}
\unpause

There are three regions:
\begin{itemize}
\item<.(1)->Region~I. $\alpha\to0$ as $R\to\infty$, so that the system looks like a free massless boson at large distances.
\item<.(2)->Region~III. $\alpha\to0$ as $R\to0$, so that the system looks like a free massless boson at small distances. It was conjectured that $\delta\to-1$ as $R\to\infty$ and the system behaves as a massive Dirac fermion. The line $\delta=-1$ was conjectured to be a separatrix.
\item<.(3)->Region~II. $\alpha$ grows for both large and small $R$. The system has no conformal behavior in both IR and UV regions. Since it approaches the line $s_2$ at large $R$, it must be a massive theory.
\end{itemize}

$$
\begin{tikzpicture}[>=stealth]
\draw[very thick,->] (-3.5,0) -- (3.5,0) node[above right] {$\delta$};
\draw[very thick,->] (0,0) -- (0,3.5) node[above right] {$2\pi\alpha$};
\draw[thick,->] (0,0) -- (-1.5,1.5);
\draw[thick,->] (-1.4,1.4) -- (-3,3);
\draw[thick] (-2.9,2.9) -- (-3.5,3.5) node[above left] {$s_2$};
\draw[thick,->] (3.5,3.5) node[above right] {$s_1$} -- (2,2);
\draw[thick,->] (2.1,2.1) -- (.5,.5);
\draw[thick] (.6,.6) -- (0,0);
\foreach \i/\r in {1/3.35,2/2.9,3/1.8} {
  \draw[domain=-\r:-.9] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=-1:1.1,<-] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=1:\r,<-] plot (\x,{sqrt(\i*\i+\x*\x)});
  \draw[domain=0:1,->] plot ({-sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=.9:\r] plot ({-sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=0:1.1] plot ({sqrt(\i*\i+\x*\x)},\x);
  \draw[domain=1:\r,<-] plot ({sqrt(\i*\i+\x*\x)},\x);
}
\node at (2.5,.5) {\it I};
\node at (.5,2.5) {\it II};
\node at (-2.5,.5) {\it III};
\end{tikzpicture}
$$



\end{frame}

\begin{frame}
\frametitle{Seminar}
\unpause






\end{frame}



\end{document}
