strings.tex

\documentclass{article}
\usepackage{yehyun}

\author{Julian Choi}
\title{Notes on Polchinski's String Theory}

\begin{document}

\maketitle
\tableofcontents

\section{Bosonic String Theory}
\subsection{Introduction}
\textcolor{red}{Lec. 1, 2, 3; Polchinski 1, BBS 2}

We start by reviewing the action of a point-like particle. If we let $X^\mu(\tau)$ be a parametrization of the particle's worldline, the simplest Poincare-invariant action is 
\begin{equation}
    S_\text{pp}=-m\int\dd\tau(-\dot X^\mu\dot X_\mu)^{1/2}\implies\delta S_\text{pp}=-m\int\dd\tau\dot u_\mu\delta X^\mu.
\end{equation}
The equation of motion from varying $X^\mu$ is $\dot u^\mu=0,\quad u^\mu=\dot X^\mu(-\dot X^\nu\dot X_\nu)^{-1/2}.$

Consider now an action with an additional parameter. If we let an independent field on the world-line be $\gamma_{\tau\tau}$ and its tetrad $\eta\equiv\sqrt{-\gamma_{\tau\tau}}$, 
\begin{equation}
    S'_{pp}=\frac 12\int\dd\tau\left(\eta^{-1}\dot X^\mu\dot X_\mu-\eta m^2\right).\label{ppPolyakov}
\end{equation}
This action is known as the sigma model. Its equation of motion from varying $\eta$ is $\eta^2=-\frac 1{m^2}\dot X^\mu\dot X_\mu$ and from varying $X^\mu$ is $\ddot X^\mu+\Gamma^\mu_{\alpha\beta}\dot X^\alpha\dot X^\beta=0$, where $\Gamma$ is the Christoffel symbol. Turns out, eliminating $\eta$ in $S'_{pp}$ with the equation of motion yields $S_{pp}$ - in other words, $S'_{pp}$ is the classical theory of $S_{pp}$.  

Now, we consider the action of a one-dimensional string. We let the coordinates on the worldsheet by $y^1,y^2$, with embedding $X^\mu(y^1,y^2)$. The induced metric is written $h_{ab}=\pde{X^\mu}{y^a}\pde{X^\nu}{y^b}g_{\mu\nu}$.

The simplest action (Nambu-Goto action) is then
\begin{equation}
    S_{NG}=-\frac 1{2\pi\alpha'}\int\dd^2y\sqrt{-h},\label{nambugoto}
\end{equation}
where $T=\frac 1{2\pi\alpha'}$ is the tension of the string. $\alpha'$, then, can be interpreted as being on the order $l_\text{string}=\sqrt{\alpha'}$. 

One can check that under the static gauge (i.e. $y^1=X^0,y^2=\sigma$ where $\sigma$ is the transverse direction of the string), in the nonrelativistic limit, we obtain $S_{NG}=-LT\int\dd\tau$, which is what we expect for the classical string.  

In lieu of \eqref{ppPolyakov}, we can write the Nambu-Goto action as (Polyakov action)
\begin{equation}
    S_P=-\frac 1{4\pi\alpha'}\int\dd y^2(-\gamma)^{1/2}\gamma^{ab}h_{ab},\label{polyakov}
\end{equation}
where $\gamma$ is again, an independent field on $M$. To derive the equivalence, we recall that (from Jacobi's identity $\dd(\det A)=\det A\Tr(A^{-1}\dd A)$; also, recall $0=\delta(\delta^a_b)=g^{ac}\delta g_{cb}+g_{cb}\delta g^{ac}$)
\begin{equation}
    \delta\gamma=\gamma\gamma^{ab}\delta\gamma_{ab}=-\gamma\gamma_{ab}\delta\gamma^{ab}.
\end{equation}
The equation of motion ($\delta_\gamma S_P=0$) yields, after dividing each side of the equation by own square of its determinant (for RHS, recall $\det cA=c^2A$)
\begin{equation}
    h_{ab}=\frac 12\gamma_{ab}\gamma^{cd}h_{cd}\implies h_{ab}(-h)^{-1/2}=\gamma_{ab}(-\gamma)^{-1/2}.
\end{equation}
In conjunction with \eqref{polyakov}, we retrieve \eqref{nambugoto}.

Lastly, we note that, given a $p$-brane non-linear sigma model action with a cosmological constant:
\begin{equation}
    S_\sigma=-\frac{T_p}2\int\dd^{p+1}\sigma\sqrt{-\gamma}\gamma^{\alpha\beta}h_{\alpha\beta}+\Lambda_p\int\dd^{p+1}\sigma\sqrt{-\gamma},
\end{equation}
the equation of motion requires, for $\gamma\ne 0$, $\Lambda_p=0$ for $p=1$, i.e. strings, and $\Lambda_p\ne 0$ for $p>1$. 

\subsubsection{Symmetries}

From \eqref{polyakov}, three symmetries are evident:
\begin{itemize}
    \item (Global) Poincare: $X'^\mu(\tau,\sigma)=\Lambda^\mu_\nu X^\nu(\tau,\sigma)+a^\mu,\gamma'_{ab}(\tau,\sigma)=\gamma_{ab}(\tau,\sigma)$
    \item (Local) Reparametrization: $X'^\mu(\tau',\sigma')=X^\mu(\tau,\sigma),\pde{\sigma'^c}{\sigma^a}\pde{\sigma'^d}{\sigma^b}\gamma'_{cd}(\tau',\sigma')=\gamma_{ab}(\tau,\sigma)$
    \item (Local) Weyl $X'^\mu(\tau,\sigma)=X^\mu(\tau,\sigma),\gamma'_{ab}(\tau,\sigma)=\exp\left(2\omega(\tau,\sigma)\right)\gamma_{ab}(\tau,\sigma)$.
\end{itemize}
Note that a diffeomorphism is a Weyl transformation (and vice versa) if $\partial^\alpha\epsilon^\beta+\partial^\beta\epsilon^\alpha=\Lambda g^{\alpha\beta}.$
As an example, for $g^{\alpha\beta}=\eta^{\alpha\beta}$, for $y^\pm = y^1\pm y^2$ and $\epsilon^\pm=\epsilon^1\pm\epsilon^2$, $y^\pm\to y^\pm+\epsilon^\pm(y^\pm)$ is a diffeomorphism that is also a Weyl transformation. In general, in $D$ flat spacetime dimensions, the condition for a diffeomorphism to be a Weyl rescaling is $\partial^2\Lambda\eta_{\mu\nu}=(2-D)\partial_\mu\partial_\nu\Lambda$. Note that, for $D=2$, we have $\Lambda=\Lambda^+(y^+)+\Lambda^-(y^-)$ but for $D>2$, $\Lambda=\Lambda_0+\Lambda_\mu x^\mu$.

An important feature of these symmetries is that they allow gauge fixing of $\gamma$. It turns out, as long as $M$ has a vanishing Euler characteristic, one can choose the unit gauge, $\gamma_{\alpha\beta}=\eta_{\alpha\beta}$. 

\subsubsection{Equations of Motion}

Varying \eqref{polyakov} with respect to $\gamma$, we obtain the condition
\begin{equation}
    T^{ab}\equiv -\frac 1\alpha\left(h^{ab}-\frac 12\gamma^{ab}\gamma_{cd}h^{cd}\right)=0.
\end{equation}
This quantity, the stress-energy tensor, is invariant under diffeomorphism. Furthermore, Weyl invariance implies that the energy-momentum tensor is traceless as well, because $\delta S=\int\delta g_{\mu\nu}\frac{\delta S}{\delta\gamma_{\mu\nu}}\propto \int\gamma_{\mu\nu}\frac{\delta S}{\delta g_{\mu\nu}}=0\implies \gamma_{ab}T^{ab}=0.$

A common choice of $\gamma$ is the unit gauge, $\gamma_{ab}=\eta_{ab}$, which reduces the $T^{\alpha\beta}=0$ condition to the Virasoro constraints:
\begin{equation}
    \dot X^2+X'^2=0,\quad \dot X\cdot X'=0.
\end{equation}

Varying $X^\mu$, we obtain 
\begin{equation}
    \delta S=\frac 1{2\pi\alpha'}\int^\infty_{-\infty}\dd\tau\int^l_0\dd\sigma(-\gamma)^{1/2}\delta X^\mu\nabla^2X_\mu-\left.\frac 1{2\pi\alpha'}\int^\infty_{-\infty}\dd\tau(-\gamma)^{1/2}\delta X^\mu\partial^\sigma X_\mu\right|^{\sigma=l}_{\sigma=0},
\end{equation}
giving the equation of motion
\begin{equation}
    \nabla^2X^\mu=\gamma^{ab}\nabla_a\nabla_b x^\mu=0,
\end{equation}
and three boundary conditions: 
\begin{itemize}
    \item Neumann: $\left.n^a\partial_aX_\mu\right|_{\partial M}=0$;
    \item Dirichlet: $\left.X^\mu\right|_{\partial M}=$ constant;
    \item Closed: $X^\mu(\tau,l)=X^\mu(\tau,0),\partial^\sigma X^\mu(\tau,l)=\partial^\sigma X^\mu(\tau,0),\gamma_{ab}(\tau,l)=\gamma_{ab}(\tau,0)$.
\end{itemize}

\subsubsection{Open Strings, Neumann BC}
Let's now find a general solution. We want to solve $\nabla^2 X^\mu=0$ given the Neumann boundary condition $\left.\partial^\sigma X^\mu\right|_{\sigma=0,\pi}=0$. The general solution for this is 
\begin{equation}
    X^\mu=X^\mu_L(\tau+\sigma)+X^\mu_R(\tau-\sigma).
\end{equation}
Note that these are independent solutions because $\partial_\sigma=\frac 12(-\partial_-+\partial_+)$ where $\partial_\pm=\partial_\tau\pm\partial_\sigma$. So, the $\partial_\sigma$ boundary condition is 
\begin{equation}
    0=X'^\mu_L(\tau,0)-X'^\mu_R(\tau,0)\implies X^\mu(\tau,\sigma)=\frac 12\left[f^\mu(\tau+\sigma)+f^\mu(\tau-\sigma)\right].
\end{equation}
With the $\sigma=\pi$ boundary condition, we find $f'^\mu$ is periodic with period $2\pi$. Hence, for general argument $u$,
\begin{equation}
    f^\mu(u)=f^\mu_0+f^\mu_1u+\sum^\infty_{n=1}\left(A^\mu_n\cos(nu)+B^\mu_n\sin(nu)\right).
\end{equation}
With trig identities, we finally obtain 
\begin{align}
    X^\mu&=f^\mu_0+f^\mu_1\tau+2\sum^\infty_{n=1}\left[A^\mu_n\cos(n\tau)\cos(n\sigma)+B^\mu_n\sin(n\tau)\cos(n\sigma)\right]\\
    &=X^\mu_0+\sqrt{2\alpha'}\alpha^\mu_0\tau-i\sqrt{2\alpha'}\sum^\infty_{n=1}\left(\alpha^\mu_{-n}e^{in\tau}-\alpha^\mu_ne^{-in\tau}\right)\frac{\cos(n\sigma)}n\\
    &=X^\mu_0+\sqrt{2\alpha'}\alpha^\mu_0\tau+i\sqrt{2\alpha'}\sum_{n\in\mathbb R\setminus0}\frac 1n\alpha^\mu_ne^{-in\tau}\cos(n\sigma),
\end{align}
where we defined $\alpha^\mu_{-m}=\left(\alpha^\mu_m\right)^\dag$ for $m>0$. The canonical commutation relations are:
\begin{equation}
    \comm{X^\mu(\tau,\sigma)}{\Pi^\nu(\tau',\sigma')}=i\eta^{\mu\nu}\delta^2\left(\tau-\tau',\sigma-\sigma'\right),\quad\comm{\alpha^\mu_m}{\alpha^\nu_n}=m\eta^{\mu\nu}\delta_{m+n,0}.
\end{equation}
Note that, however, that this yields negative norm states. At the cost of making a gauge choice that is not Lorentz covariant, we can solve the Virasoro constraints and have positive possibilities. This is done through light cone quantization.

We now impose the light cone gauge. The light cone coordinates are defined as, 
\begin{equation}
    X^+=\frac 1{\sqrt 2}(X^0+X^1),\quad X^-=\frac 1{\sqrt 2}(X^0-X^1),\quad X^i=X^i\implies\eta_{\mu\nu}=\begin{pmatrix}0&-1&&\\-1&0&&\\ &&1&\\ &&&1\end{pmatrix}.
\end{equation}
The light cone gauge, similarly defined, is defined to take $X^+(\tau)=\tau$. For a point particle, 
\begin{align*}
    \mathcal L&=\frac 12\left(e^{-1}\dot X^\mu\dot X_\mu-em^2\right)\\
    &=\frac 12e^{-1}\left(-2\dot X^-+\dot X^i\dot X^i\right)-\frac 12em^2\\
    \implies H&=\frac 1{p^+}\left(\frac 12p^ip^i+\frac 12m^2\right).
\end{align*}
Given $H=p^-$, we retrieve $2p^+p^-=p^ip^i+m^2$, the mass shell condition. 

Similarly, for a string, we can take $C\tau=X^+(\tau)$ for some constant $C$. This is because we can use diffeomorphism $\bigcap$ Weyl invariance (this is a redundancy among the intermediate steps in $\gamma_{\alpha\beta}\to e^{2\Phi}\delta_{\alpha\beta}\to\delta_{\alpha\beta}$. I.e. in another process $\gamma_{\alpha\beta}\to e^{2\Phi'}\delta_{\alpha\beta}\to\delta_{\alpha\beta}$, $e^{2\Phi}\delta_{\alpha\beta}$ is related to $e^{2\Phi}\delta_{\alpha\beta}$ through a diffeomorphism $\cap$ Weyl transformation. Note that this isn't a simple rescaling/Weyl since $\Phi$ are not constants.)

Recall that the condition for a diffeomorphism to be a Weyl transformation is $\eta_{\mu\nu}\partial^2\Lambda=(2-D)\partial_\mu\partial_\nu\Lambda$ where $\Lambda$ is the Weyl factor. Since $\nabla^2X^+=0$, we can set $\tau$ to be proportional to $X^+$. We set $X^+=\tau[2\alpha'p^+]$ for convenience. 

The Virasoro constraints in the light cone gauge are:
\begin{equation}
    (\dot X\pm X')^2=-2(\dot X^+\pm X'^+)(\dot X^-\pm X'^-)+(\dot X^i\pm X'^i)^2=0\implies \dot X^-\pm X'^-=\frac{(\dot X^i\pm X'^i)^2}{4\alpha'p^+}.
\end{equation}
This becomes, with explicit computations, 
\begin{equation}
    \sqrt{2\alpha'}\sum_n\alpha^-_ne^{-in(\tau\pm\sigma)}=\frac{2\alpha'}{4\alpha'p^+}\sum_{n',m''}\alpha^i_{n'}\alpha^i_{n''}e^{-i(n'+n'')(\tau\pm\sigma)}=\frac 1{2p^+}\sum_{n\in\mathbb Z}\left(2L^\perp_n\right)e^{-in(\tau\pm\sigma)},
\end{equation}
where, 
\begin{equation}
    L^\perp_n=\frac 12\sum_{p\in\mathbb Z}\alpha^i_p\alpha^i_{n-p},
\end{equation}
are called the transverse Virasoro modes. 

Hence, the Virasoro constraints imply 
\begin{equation}
    \sqrt{2\alpha'}\alpha^-_n=\frac 1{p^+}L^\perp_n,\quad \alpha^-_0=\sqrt{2\alpha'}p^-\implies 2\alpha'p^+p^-=\frac 12\sum_p\alpha^i_{-p}\alpha^i_p,
\end{equation}
and the second equation comes from setting $X^+=\sqrt{2\alpha'}\alpha_0^+\tau$ to $2\alpha'p^+\tau$. This yields the mass spectrum:
\begin{equation}
    M^2\equiv -p^\mu p_\mu=2p^+p^--p^ip^i=\frac 1{\alpha'}\sum^\infty_{n=1}\alpha^i_n\alpha^i_{-n}.
\end{equation}
To summarize, 
\begin{align}
    X^+&=2\alpha'p^+\tau\\
    X^i&=X^i_0+\sqrt{2\alpha'}\alpha^i_0\tau+i\sqrt{2\alpha'}\sum_{n\ne 0}\frac 1n\alpha^i_ne^{-in\tau}\cos(n\sigma)\\
    X^-&=X^-_0+\sqrt{\alpha'}\alpha^-_0\tau+i\sqrt{2\alpha'}\sum_{n\ne 0}\frac 1n\alpha^-_ne^{-in\tau}\cos(n\sigma),
\end{align}
with Virasoro constraint 
\begin{equation}
    \alpha^-_n=\frac 1{2p^+}\frac 1{\sqrt{2\alpha'}}\sum_{p\in\mathbb Z}\alpha^i_p\alpha^i_{n-p}=\frac 1{\sqrt{2\alpha'}}\frac 1{p^+}L^\perp_n.
\end{equation}
To quantize, we impose
\begin{equation}
    \comm{X^-}{p^+}=i\eta^{+-}=-i,\quad\comm{X^i(\sigma)}{\Pi^j(\sigma')}=i\delta^{ij}(\sigma-\sigma'),\quad\comm{\alpha^i_m}{\alpha^j_n}=m\delta^{ij}\delta_{m+n,0}.
\end{equation}
We now normalize $\alpha$:
\begin{equation}
    a^i_m=\frac{\alpha^i_m}{\sqrt m},\quad(a^i_m)^\dag=\frac{\alpha^i_{-m}}{\sqrt m}\implies\comm{a^i_m}{(a^i_m)^\dag}=1.
\end{equation}
The general state and its corresponding mass is 

\begin{equation}
    \ket{\{N_{i,n}\};k}=\prod^{D-1}_{i=2}\prod^\infty_{n=1}\frac{(\alpha^i_{-n})^{N_{i,n}}}{\sqrt{n^{N_{i,n}}N_{i,n}!}}\ket{0;k}\implies M^2=\frac 1{\alpha'}\left(\sum^\infty_{n=1}\alpha^i_{-n}\alpha^i_n+A\right)\equiv\frac 1{\alpha'}(N+A),
\end{equation}
where $A$ is some normal ordering ambiguity. A QFT-style approach asserts that $A$ comes from summing zero-point energies of each mode, i.e. $\frac 12\omega$. Multiplying by $D-2$, the number of transverse directions, and regularizing it, we find 
\begin{equation}
    A=\frac{D-2}2\sum^\infty_{n=1}n=\frac{2-D}{24}.
\end{equation}


The first couple states are:
\begin{itemize}
    \item At $N=0$, have $\ket{0;k}\implies M^2=\frac 1{\alpha'}A$
    \item At $N=1$, have $\alpha^i_{-1}\ket{0;k}\implies M^2=\frac 1{\alpha'}(1+A)$. Note, however, that there are $D-2$ such states, which means only a massless particle is possible, giving $A=-1\implies D=26$. This is necessary but not sufficient for the absence of a Lorentz anomaly. Thus, the bosonic string theory is Lorentz invariant  only at $D=26$, which is known as the critical dimension. 
\end{itemize}

\subsubsection{Closed Strings}

We now look at closed strings, which correspond to solving the equation of motion $(-\partial^2_\tau+\partial^2_\sigma)X^\mu=0$ with boundary condition $X^\mu(\tau,\sigma)=X^\mu(\tau,\sigma+2\pi)$. 

Since the general solution is given $X^\mu(\tau,\sigma)=X^\mu_L(\tau+\sigma)+X^\mu_R(\tau-\sigma)$, we obtain 
\begin{equation}
    X^\mu_L(\tau+\sigma)-X^\mu_L(\tau+\sigma+2\pi)=X^\mu_R(\tau-\sigma-2\pi)-X^\mu_R(\tau-\sigma),
\end{equation}
but since $\tau+\sigma$ and $\tau-\sigma$ are independent, we obtain that $X_L$ and $X_R$ have periodic derivatives. Hence, we get 
\begin{align*}
    X^\mu_{L,R}&=\frac 12X^{L,R\mu}_0+\sqrt{\frac{\alpha'}2}\alpha_0^{L,R\mu}(\tau\pm\sigma)+i\sqrt{\frac{\alpha'}2}\sum_{n\ne 0}\frac{\alpha^{L,R\mu}_n}ne^{-in(\tau\pm\sigma)}\\
    \implies X^\mu&=X^\alpha_0+\sqrt{2\alpha'}\alpha^\mu_0\tau+i\sqrt{\frac{\alpha'}2}\sum_{n\ne 0}\frac 1n\left(\alpha^\mu_ne^{-in(\tau-\sigma)}+\tilde\alpha^\mu_ne^{-in(\tau+\sigma)}\right)
\end{align*}
where, we let $\alpha^L=\tilde\alpha$ and $\alpha^R=\alpha$ and used the periodict boundary condition to conclude that $\alpha^\mu_0=\tilde\alpha^\mu_0$.

Now, we impose the light cone gauge, setting $X^+=\alpha' p^+\tau$, with the Virasoro constraint $\dot X^-\pm X'^-=\frac 1{2p^+\alpha'}(\dot X^i\pm X'^i)^2$. Noting that, 
\begin{equation}
    \dot X^i\pm X'^i=\sqrt{2\alpha'}\sum_{n\in\mathbb Z}\begin{cases}\tilde\alpha^i_ne^{-in(\tau+\sigma)}&\text{for +}\\\alpha^i_ne^{-in(\tau-\sigma)}&\text{for -}\end{cases},
\end{equation}
we find 
\begin{align*}
    \sqrt{2\alpha'}\tilde\alpha_n^-&=\frac 2{p^+}\tilde L_n^\perp\\
    \sqrt{2\alpha'}\alpha_n^-&=\frac 2{p^+} L_n^\perp,
\end{align*}
thus, $L^\perp_0=\tilde L_0^\perp$. This gives, by definition, the level-matching condition $N=\tilde N$. The mass spectrum is therefore 
\begin{equation}
    M^2=\frac 2{\alpha'}\left(N+\tilde N+A+\tilde A\right)=\frac 4{\alpha'}\left(N-1\right),
\end{equation}
where we imposed the same commutation relations as before. 

The first couple states are:
\begin{itemize}
    \item At $N=0$, have $\ket{0;k}\implies M^2=-\frac 4{\alpha'}$, 
    \item At $N=\tilde N=1$, have $\alpha^i_{-1}\tilde\alpha^i_{-1}\ket{0;k}$, the tensor product of left-movers and right-movers of the open string spectrum. 
\end{itemize}

\subsection{Conformal Field Formalism}

Recall that the diffeomorphisms that leave $g_{\alpha\beta}$ invariant up to scale (i.e. intersection of diff and Weyl) are called conformal transformations. The world-sheet coordinates in the Polyakov theory is thus 2D conformal symmetric. From $\partial_\alpha\epsilon_\beta+\partial_\beta\epsilon_\alpha=\lambda\delta_{\alpha\beta}$, we find that $\epsilon$ and $y$ satisfy the Cauchy-Riemann equations. It is therefore natural to consider $\epsilon$ and $y$ in the complex coordinates. 

\subsubsection{Complex Formalism}

We define $z=y_1+iy_2$ and $\partial_z=\frac 12(\partial_1-i\partial_2)$ and let $\overline z$ and $\overline\partial_z$ be naturally defined. This gives 
\begin{equation}
    g_{ab}=\frac 12\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad g^{ab}=2\begin{pmatrix}0&1\\1&0\end{pmatrix}.
\end{equation}
The conformal transformation in this formalism is 
\begin{equation}
    z\to w(z),\quad \overline z\to\overline w,\quad g_{ab}\to e^{2\omega}g_{ab},\quad \omega=\ln\abs{\partial_zw},
\end{equation}
where the last bit comes from preserving the unit gauge. In other words, conformal transformations are holomorphic. 

In the complex coordinates, the global conformal group is 
\begin{equation}
    z\to\frac{az+b}{cz+d},\quad\overline z\to\frac{\overline a\overline z+\overline b}{\overline c\overline z+\overline d},\quad ad-bc=1.
\end{equation}
Then, $G=\text{SL}(2,\mathbb C)/\mathbb Z_2\simeq\text{SO}(3,1)$. In general, given $D=p+q$, we have the conformal group $G=\text{SO}(p+1,q+1)$. For example, 
\begin{itemize}
    \item $D=2$ Euclidean space has $G=\text{SO}(1,3)$
    \item $D=2$ Lorentzian space has $G=\text{SO}(2,2)\simeq\text{SL}(2,\mathbb R)\times\text{SL}(2,\mathbb R)$
    \item $D=4$ Lorentzian space has $G=\text{SO}(2,4)$.
\end{itemize}

The Polyakov action and the equation of motion are:
\begin{equation}
    S=\frac 1{2\pi\alpha'}\int\dd z^2\partial X^\mu\overline\partial X_\mu\implies \partial\overline\partial X^\mu(z,\overline z)=0.
\end{equation}
In other words, the classical solutions are harmonic and derivatives are holomorphic ($\partial X^\mu$)/antiholomorphic ($\overline\partial X^\mu$). Note that the classical equation's operator statement holds true as well (Ehrenfest theorem), as we will show once we define the expectation value. 

The other conservation law, the conservation of the stress tensor, is given 
\begin{equation}
    \partial T_{\overline{zz}}=\overline\partial T_{zz}=0,\quad T^{\mu\nu}=-\frac 1{\alpha'}\begin{pmatrix}\partial X^\mu\partial X_\mu&0\\0&\overline\partial X^\mu\overline\partial X_\mu\end{pmatrix}.
\end{equation}
This gives a bigger family of conserved currents, $j(z)=v(z)T(z)$ and $\tilde j(\overline z)=v(z)^*\tilde T(\overline z)$. These generate conformal transformations, $z\to z+v(z)$ and $\overline z\to\overline z+v(z)^*$, respectively. \textcolor{red}{why? Lec 5, pg 10}

\subsubsection{Expectation Value, Operator Product Expansion}

We start by defining the expectation value of a functional:
\begin{equation}
    \ev{\mathcal F[X]}=\int[\dd X]\exp(-S)\mathcal F[X].
\end{equation}
Since the expectation value of a total derivative is zero, we find 
\begin{equation}
    0=\int[\dd X]\frac\delta{\delta X_\mu(z,\overline z)}\exp(-S)=-\int[\dd X]\exp(-S)\frac{\delta S}{\delta X_\mu(z,\overline z)}=-\ev{\frac{\delta S}{\delta X_\mu(z,\overline z)}}=\frac 1{\pi\alpha'}\ev{\partial\overline\partial X^\mu(z,\overline z)}.
\end{equation}
The same calculation goes if we include insertions, as long as they aren't at $z$. Thus, $\partial\overline\partial X^\mu=0$ as an operator equation.

If there is an insertion at $z$, we obtain 
\begin{align*}
    0&=\int[\dd X]\frac\delta{\delta X_\mu(z,\overline z)}\left[\exp(-S)X^\nu(w,\overline w)\right]\\
    &=\int[\dd X]\exp(-S)\left[\eta^{\mu\nu}\delta^2(z-w,z-\overline w)+\frac 1{\pi\alpha'}\partial_z\partial_{\overline z}X^\mu(z,\overline z)X^\nu(w,\overline w)\right]\\
    \implies&\frac 1{\pi\alpha'}\partial_z\partial_{\overline z}X^\mu(z,\overline z)X^\nu(w,\overline w)=-\eta^{\mu\nu}\delta^2(z-w,\overline z-\overline w).
\end{align*}
Now, we define the normal ordered operators to not have this singular behavior (i.e. $\partial\overline\partial X$ be harmonic)
\begin{equation}
    \normord{X^\mu(z_1,\overline z_1)X^\nu(z_2,\overline z_2)}=X^\mu(z_1,\overline z_1)X^\nu(z_2,\overline z_2)+\frac{\alpha'}2\eta^{\mu\nu}\left[\ln(z_1-z_2)+\ln(\overline z_1-\overline z_2)\right].
\end{equation}
Note that $\partial\overline\partial(\ln z+\ln\overline z)=2\pi\delta^2(z,\overline z)$, which gives $\partial_1\overline\partial_1:X^\mu(z_1,\overline z_1)X^\nu(z_2,\overline z_2):=0$. Since the normal ordered product is harmonic, we can Taylor expand within the normal ordering brackets. 

The definition can be summarized as 
\begin{equation}
    \normord{\mathcal F}=\exp\left(-\frac{\alpha'}4\int\dd^2z_1\dd^2z_2\ln\abs{z_{12}}^2\frac\delta{\delta X^\mu(z_1,\overline z_1)}\frac\delta{\delta X_\mu(z_2,\overline z_2)}\right)\mathcal F,
\end{equation}
and for any pair of operators 
\begin{equation}
    \normord{\mathcal F}\normord{\mathcal G}=\exp\left(-\frac{\alpha'}2\int\dd^2z_1\dd^2z_2\ln\abs{z_{12}}^2\frac\delta{\delta X^\mu_F(z_1,\overline z_1)}\frac\delta{\delta X_{G\mu}(z_2,\overline z_2)}\right)\normord{\mathcal{FG}}
\end{equation}
Note that the normal ordering operator commutes with the derivative operator. With this formalism, we can take contractions similar to the Wick formalism in ordinary QFT.

This is an example of an operator product expansion. In general, $\mathcal A_i(\sigma_1)\mathcal A_j(\sigma_2)=\sum_kc^k_{ij}(\sigma_1-\sigma_2)\mathcal A_k(\sigma_2)$. OPEs are usually used as asymptotic expressions. We denote equality up to nonsingular terms by the tilde $\sim$. 

\paragraph{Example Expansion}

Since normal ordered operators are harmonic, we can expand them within the brackets:
\begin{align*}
    X^\mu(z_1,\overline z_1)X^\nu(z_2,\overline z_2)&=-\frac{\alpha'}2\eta^{\mu\nu}\ln\abs{z_1-z_2}^2+\normord{X^\mu(z_1,\overline z_1)X^\nu(z_2,\overline z_2)}\\
    &=-\frac{\alpha'}2\eta^{\mu\nu}\ln\abs{z_1-z_2}^2+\normord{X^\mu X^\nu(z_2,\overline z_2)}+\sum^\infty_{k=1}\frac 1{k!}\left[(z_{12})^k\normord{\partial^k X^\mu X^\nu(z_2,\overline z_2)}+(\overline z_{12})^k\normord{\overline\partial^k X^\mu X^\nu(z_2,\overline z_2)}\right]
\end{align*}

\paragraph{Example OPE}

Be careful with the variables, to not mix up $z_{1,2}$ and $z$.
\begin{align*}
    &\normord{\partial X^\mu(z)\partial X_\mu(z)}\normord{\partial' X^\nu(z')\partial' X_\nu(z')}\\
    &=\normord{\partial X^\mu(z)\partial X_\mu(z)\partial' X^\nu(z')\partial' X_\nu(z')} +4\cdot-\frac{\alpha'}2\int\dd^2z_1\dd^2z_2\ln\abs{z_{12}}^2\partial_z\delta(z-z_1)\partial_{z'}\delta(z'-z_2)\normord{\partial X_\mu(z)\partial'X^\mu(z')}\\
    &\quad +2\cdot\frac{\alpha'^2}{4}\int\dd^2z_1\dd^2z_2\dd^2z_3\dd^2z_4\ln\abs{z_{12}}^2\ln\abs{z_{34}}^2\partial\delta(z-z_1)\partial\delta(z-z_3)\partial'\delta(z'-z_2)\partial'\delta(z'-z_4)\normord{\eta^\mu_\mu}+\\
    &=\normord{\partial X^\mu(z)\partial X_\mu(z)\partial' X^\nu(z')\partial' X_\nu(z')}-2\alpha'(\partial\partial'\ln\abs{z-z'}^2)\normord{\partial X^\mu(z)\partial'X_\mu(z')}+\frac D2\alpha'^2(\partial\partial'\ln\abs{z-z'}^2)^2\\
    &=\normord{\partial X^\mu(z)\partial X_\mu(z)\partial' X^\nu(z')\partial' X_\nu(z')}-\frac{2\alpha'}{(z-z')^2}\normord{\partial X^\mu(z)\partial'X_\mu(z')}+\frac{D\alpha'^2}{2(z-z')^4},
\end{align*}
where we used 
\begin{equation}
    \fde{(\partial_zX^\nu(z))}{(X^\mu(z_1))}=\partial_z\eta^\nu_\mu\delta(z-z_1),\quad \partial_z\delta(z-z_1)=-\partial_{z_1}\delta(z-z_1).
\end{equation}
Taylor expanding the $\partial X^\mu(z)$ around $z'$, we finally find 
\begin{equation}
    \normord{\partial X^\mu(z)\partial X_\mu(z)}\normord{\partial' X^\nu(z')\partial' X_\nu(z')}\sim\frac{D\alpha'^2}{2(z-z')^4}-\frac{2\alpha'}{(z-z')^2}\normord{\partial'X^\mu(z')\partial'X_\mu(z')}-\frac{2\alpha'}{z-z'}\normord{{\partial'}^2X^\mu(z')\partial'X_\mu(z')}
\end{equation}

\subsubsection{Noether's theorem}

Noether's theorem relates a continuous symmetry to the conserved current. Suppose there is a global symmetry $\phi\to\phi+\delta\phi$. In general, $\phi\to \phi+\rho(x)\delta\phi$ is not a symmetry unless $\rho(x)$ is a constant. Thus, the variation must be proportional to $\delta\rho$. Looking at the invariance of $\ev{1}$,
\begin{align*}
    0&=\int\mathcal D\phi'e^{-S[\phi']}-\int\mathcal D\phi e^{-S[\phi]}\\
    &=\int\mathcal D\phi e^{-S[\phi]}\left(\frac{i\epsilon}{2\pi}\int\dd^d\sigma g^{1/2}j^a(\sigma)\partial_a\rho(\sigma)+\mathcal O(\epsilon^2)\right)\\
    &=\frac\epsilon{2\pi i}\int\dd^d\sigma g^{1/2}\rho(\sigma)\ev{\nabla_aj^a(\sigma)}.
\end{align*}
And thus $\nabla_aj^a=0$. This is known as Noether's theorem.

For $\delta\mathcal L=\epsilon\partial_aK^a$, we find the classical result:
\begin{equation}
    j^a=\frac{2\pi i}\epsilon\left(\pde{\mathcal L}{(\partial_a\phi_\mu)}\delta\phi_a-\epsilon K^a\right).
\end{equation}

\subsubsection{Ward Identites}

Ward identities relate a continuous symmetry to the operator products of the current. To derive the Ward identity, we let $\rho(x)=1$ be $1$ on some compact region $M$ and $0$ otherwise. This is a change of variable with unit Jacobian, meaning that the functional derivative of an expectation value,
\begin{equation}
    \ev{\mathcal O}=\frac 1Z\int\mathcal D\phi e^{-S}\mathcal O,\quad Z=\int\mathcal D\phi e^{-S},
\end{equation}
is zero, since it is a symmetry (i.e., $\delta\ev{\mathcal O}=0$). So, 
\begin{align*}
    \ev{\mathcal O}&=\frac 1Z\int\mathcal D\phi\left[\mathcal O+\delta\mathcal O\right]e^{-(S+\delta S)}=\ev{\mathcal O}+\ev{\delta\mathcal O}-\ev{\mathcal O\delta S},
\end{align*}
to the first order \textcolor{red}{why?}. Thus, 
$$\ev{\delta\mathcal O(x^0)}=\frac{i\epsilon}{2\pi}\ev{\int\dd^Dxj^\mu(x)\partial_\mu\rho(x)\mathcal O(x_0)}=-\frac{i\epsilon}{2\pi}\ev{\int\dd^Dx\partial_\mu j^\mu(x)\mathcal O(x_0)\rho(x)}.$$
\textcolor{red}{Note that the factor of $i\epsilon/2\pi$ is Polchinski's convention which McAllister doesn't.} So, 
\begin{equation}
    \frac{i\epsilon}{2\pi}\partial_\mu j^\mu(x)\mathcal O(x_0)=-\delta^D(x-x_0)\delta\mathcal O(x_0),
\end{equation}
plus some total derivative. This is the Ward identity.

Note that this is an active transformation (i.e. the physical field configuration $\phi$ changes, $\phi(x)\to\phi'(x')$. For a passive transformation, it would be $\phi(x)\to\phi(x')$.)

We can take the volume integral and use the divergence theorem. In two flat dimensions, (2.3.11),
\begin{equation}
    \text{Res}_{z\to z_0}j(z)\mathcal A(z_0,\overline z_0)+\overline{\text{Res}}_{\overline z\to\overline z_0}\tilde j(\overline z)\mathcal A(z_0,\overline z_0)=\frac 1{i\epsilon}\delta\mathcal A(z_0,\overline z_0).
\end{equation}
For example,
\begin{enumerate}
    \item Under spacetime translation, $\delta X^\mu=\epsilon a^\mu\implies j^\mu_a=\frac i{\alpha'}\partial_aX^\mu$
    \item Under worldsheet translation, $\delta\sigma^a=-\epsilon v^a\partial_aX^\mu$, which gives 
    \begin{equation}
        j_a=iv^bT_{ab},\quad T_{ab}=-\frac 1{\alpha'}\normord{\left(\partial_aX^\mu\partial_bX_\mu-\frac 12\delta_{ab}\partial_cX^\mu\partial^cX_\mu\right)}
    \end{equation}
    \item Under conformal transformation, $\delta X^\mu=-\epsilon v(z)\partial X^\mu-\epsilon v(z)^*\overline\partial X^\mu$, where $v(z)$ is holomorphic.
\end{enumerate}

\subsubsection{Conformal Field Theory, OPEs}
Since the energy-momentum tensor is traceless, there exist a larger symmetry, known as the conformal symmetry, corresponding to an infinitesimal coordinate transformation $z'=z+\epsilon v(z)$ with $v$ holomorphic. These strongly constrains the OPEs, especially with the energy-momentum tensor per the Ward identities. \textcolor{red}{Look at problem set 3/4.} Some useful OPEs derived from these constraints are as follows. For general and primary operator with known weights (defined $\mathcal A'(\zeta z,\overline{\zeta z})=\zeta^{-h}\overline\zeta^{-\tilde h}\mathcal A(z,\overline z)$,)
\begin{align}
    T(z)\mathcal A(0)&\sim\cdots+\frac h{z^2}\mathcal A(0)+\frac 1z\partial\mathcal A(0)\\
    T(z)\mathcal O(0)&\sim\frac h{z^2}\mathcal O(0)+\frac 1z\partial\mathcal O(0).
\end{align}
As for definitive operators, we have 
\begin{align}
    T(z)X^\mu(0)&\sim\frac 1z\partial X^\mu(0)\\
    T(z)\partial X^\mu&\sim\frac 1{z^2}\partial X^\mu(0)+\frac 1z\partial^2X^\mu(0)\\
    X^\mu(z)X^\nu(0)&\sim-\frac{\alpha'}2\eta^{\mu\nu}\ln\abs{z}^2\\
    \partial X^\mu(z)X^\nu(0)&\sim-\frac{\alpha'}2\eta^{\mu\nu}\frac 1z\\
    \partial X^\mu(z)\partial X^\nu(0)&\sim-\frac{\alpha'}2\eta^{\mu\nu}\frac 1{z^2}\\
    T(z)T(0)&\sim\frac c{2z^4}+\frac 2{z^2}T(0)+\frac 1z\partial T(0),
\end{align}
where $c$ is the central charge. In the $X^\mu$ field theory, $c=D$. 

This yields the transformation law 
\begin{align*}
    \epsilon^{-1}\delta T(z)&=-\frac c{12}\partial^3_zv(z)-2\partial_zv(z)T(z)-v(z)\partial_zT(z)\\
    (\partial_zz')^2T'(z')&=T(z)-\frac c{12}\{z',z\},
\end{align*}
where $\{f,z\}$ is the Schwarzian derivative $\{f,z\}=\frac{2\partial^3_zf\partial_zf-3\partial^2_zf\partial^2_zf}{2\partial_zf\partial_zf}$. Note the nontensoral behavior (under conformal transformatinos) of the energy-momentum tensor here. 

\subsubsection{Virasoro Algebra}

We define the Virasoro generators as the coefficients of the Laurent expansion of $T$. They have an algebra due to the fact that the gauge choice doesn't fully fix the reparametrization symmetry:
\begin{equation}
    T_{zz}(z)=\sum^\infty_{m=-\infty}\frac{L_m}{z^{m+2}},\quad\tilde T_{\overline z\overline z}(\overline z)=\sum^\infty_{m=-\infty}\frac{\tilde L_m}{\overline z^{m+2}}\implies L_m=\oint_C\frac{\dd z}{2\pi iz}z^{m+2}T_{zz}.
\end{equation}
From the holomorphicity of the $L_m$ integrand, $L_m$ are independent of $C$, and we conclude that $L_m$ is conserved charges associated with the conformal transformations. Indeed, the Virasoro generators are related to the Hamiltonian:
\begin{equation}
    H=L_0+\tilde L_0-\frac{c+\tilde c}{24}.
\end{equation}
Furthermore, for the ground state, $L_0=\tilde L_0=0$, and thus the Casimir energy is given $E=-\frac{c+\tilde c}{24}$. 

Recall that, for a charge defined by $Q_i\{C\}\equiv\oint_C\frac{\dd z}{2\pi i}j_i$, we have 
\begin{align*}
    \comm{Q_1}{Q_2}\{C_2\}&=\oint_{C_2}\frac{\dd z_2}{2\pi i}\text{Res}_{z_1\to z_2}j_1(z_1)j_2(z_2)\\
    \comm{Q}{\mathcal A(z_2,\overline z_2)}&=\text{Res}_{z_1\to z_2}j(z_1)\mathcal A(z_2,\overline z_2)=\frac 1{i\epsilon}\delta\mathcal A(z_2,\overline z_2),
\end{align*}
and similarly for antiholomorphic charge and currents. This indeed is the familiar statement that charges generate transformations.

Applying this to the Virasoro generators, we obtain the Virasoro algebra:
\begin{equation}
    \comm{L_m}{L_n}=(m-n)L_{m+n}+\frac c{12}(m^3-m)\delta_{m,-n},
\end{equation}
and the same algebra holds for $\tilde L_m$ with central charge $\tilde c$. 
Note that:
\begin{itemize}
    \item This is an infinite-dimensional algebra, but there exists a finite subgroup, the three generators $L_0,L_{\pm 1}$ which form the algebra SL$(2,\mathbb R)$. 
    \item This is a central extension of Witt (also thought of as classical) algebra, which is the algebra of the generators of the holomorphic transformation $z\to z-\epsilon_nz^{n+1}$
    \item Its unitary representations can be obtained from a Kac-Moody algebra using the coset construction (\textcolor{red}{BBS 3.1})
    \item While this derivation is valid only for closed strings due to the fact that $z$ coordinate is naturally defined, the same construction holds for open strings (\textcolor{red}{Polchinski 2.6})
\end{itemize}

\subsubsection{Modes}
To expand the fields into modes, we define 
\begin{align}
    \partial X^\mu(z)&=-i\left(\frac{\alpha'}2\right)^{1/2}\sum^\infty_{m=-\infty}\frac{\alpha^\mu_m}{z^{m+1}}\implies\alpha^\mu_m=\left(\frac 2{\alpha'}\right)^{1/2}\oint\frac{\dd z}{2\pi}z^m\partial X^\mu(z)\\
    \overline\partial X^\mu(\overline z)&=-i\left(\frac{\alpha'}2\right)^{1/2}\sum^\infty_{m=-\infty}\frac{\tilde\alpha^\mu_m}{\overline z^{m+1}}\implies\tilde\alpha^\mu_m=-\left(\frac 2{\alpha'}\right)^{1/2}\oint\frac{\dd\overline z}{2\pi}z^m\overline\partial X^\mu(\overline z).
\end{align}
This yields
\begin{align}
    p^\mu&=\left(\frac 2{\alpha'}\right)^{1/2}\alpha^\mu_0\\
    X^\mu(z,\overline z)&=x^\mu-i\frac{\alpha'}2p^\mu\ln\abs{z}^2+i\left(\frac{\alpha'}2\right)^{1/2}\sum^\infty_{m=-\infty,m\ne 0}\frac 1m\left(\frac{\alpha^\mu_m}{z^m}+\frac{\tilde\alpha^\mu_m}{\overline z^m}\right).
\end{align}
This yields the typical canonical commutators, using contour integrals (note that $\partial X^\mu$, $\overline\partial X^\mu$ are (anti)holomorphic)
\begin{equation}
    \comm{\alpha^\mu_m}{\alpha^\nu_n}=\comm{\tilde\alpha^\mu_m}{\tilde\alpha^\nu_n}=m\delta_{m+n,0}\eta^{\mu\nu},\quad\comm{x^\mu}{p^\nu}=i\eta^{\mu\nu}.
\end{equation}
Expanding the Virasoro generators, we obtain 
\begin{equation}
    L_m\sim\frac 12\sum^\infty_{m=-\infty}\alpha^\mu_{m-n}\alpha_{\mu n},\quad L_0=\frac{\alpha'p^2}4+\sum^\infty_{n=1}\left(\alpha^\mu_{-n}\alpha_{\mu n}\right),
\end{equation}
with the first expression is equal up to ordering of the annihilation operators. Note that, for $L_m,m\ne 0$, the definition is well-defined, but for $m=0$, we can use $2L_0=L_1L_{-1}-L_{-1}L_1$ to determine that the normal-ordering constant is zero. It turns out that in free field theory, the creation-annihilation normal ordering operator is equivalent to the CFT normal ordering operator.

\subsubsection{State-Operator Correspondence}
We now have the states, defined from the ground state $\ket{0;k}$s, defined such that they have momentum $k^\mu$ and are annihilated by all lowering modes $\alpha^\mu_m$ for $m>0$. The higher modes are defined raising the ground state repeatedly. 

One central result in string theory is the state-operator correspondence, which says one can map the states to operators at the origin (these are known as vertex operators). Starting with the identity operator, we have
\begin{equation}
    \ket 1\simeq\ket{0;0},
\end{equation}
 since when we act on $\ket 1$ with the raising/lowering operators, there are no poles, giving zero for $\alpha^\mu_m\ket 1$ for $m\ge 0$. Generalizing, we obtain 
\begin{equation}
    \ket{\normord{e^{ik\cdot X(0,0)}}}=\ket{0;k}.
\end{equation}

\subsection{Free Theories}
\subsubsection{$bc$ Theory}
The $bc$ theory is a free CFT with anticommuting fields $b$ of weight $(\lambda,0)$ and $c$ of weight $(1-\lambda,0)$ with the action 
\begin{equation}
    S=\frac 1{2\pi}\int\dd^2zb\overline\partial c.
\end{equation}
The equations of motion are 
\begin{equation}
    \overline\partial c(z)=\overline\partial b(z)=0,\quad \overline\partial b(z)c(0)=2\pi\delta^2(z,\overline z).
\end{equation}
The normal ordered $bc$ product is given 
\begin{equation}
    \normord{b(z_1)c(z_2)}=b(z_1)c(z_2)-\frac 1{z_{12}}.
\end{equation}
From this definition, $b(z_1)c(z_2)\sim\frac 1{z_{12}}$ and $c(z_1)b(z_2)\sim\frac 1{z_{12}}$. Furthermore, the other OPEs are nonsingular. 

The energy-momentum tensor are given 
\begin{equation}
    T(z)=\normord{(\partial b)c}-\lambda\partial(\normord{bc}),\quad \tilde T(z)=0.
\end{equation}
The $TT$ OPEs are the standard form with the central charge $c=-3(2\lambda-1)^2+1$ and $\tilde c=0$.

The $b$ and $c$ fields have the following expansions:
\begin{equation}
    b(z)=\sum^\infty_{m=-\infty}\frac{b_m}{z^{m+\lambda}},\quad c(z)=\sum^\infty_{m=-\infty}\frac{c_m}{z^{m+1-\lambda}},
\end{equation}
for which they have the algebra 
\begin{equation}
    \acomm{b_m}{c_n}=\delta_{m,-n}.
\end{equation} 
The ground states for each oscillator is given $\ket{\downarrow}$ and $\ket{\uparrow}$, with 
\begin{align*}
    b_0\ket{\downarrow}&=0,\quad c_0\ket{\downarrow}=\ket{\uparrow},\quad b_n\ket{\downarrow}=c_n\ket{\downarrow}=0\\
    b_0\ket{\uparrow}&=\ket{\downarrow},\quad c_0\ket{\uparrow}=0,\quad b_n\ket{\uparrow}=c_n\ket{\uparrow}=0.
\end{align*}
It is typical to group $b_0$ with lowering operators and $c_0$ with raising operators, and hence $\ket{\downarrow}$ is the vacuum.

The Virasoro generators are 
\begin{equation}
    L_m=\sum^\infty_{n=-\infty}(m\lambda-n)\textbf{:}b_nc_{m-n}\textbf{:}+\frac{\lambda(1-\lambda)}2\delta_{n,0},
\end{equation}
where $\textbf{::}$ is the creation-annihilation operator normal ordering. 

\subsection{Gauge Fixing}

Reminder that the Polyakov path integral is given 
\begin{equation}
    \int[\dd X\dd g]\exp\left(-S_X-\lambda\chi\right),
\end{equation}
with 
\begin{equation}
    S_X=\frac 1{4\pi\alpha'}\int_M\dd^2\sigma g^{1/2}g^{ab}\partial_aX^\mu\partial_bX_\mu,\quad \chi=\frac 1{4\pi}\int_M\dd^2\sigma g^{1/2}R+\frac 1{2\pi}\int_{\partial M}\dd sk,
\end{equation}
with $k$ the geodesic curvature and $\chi$ is the Euler characteristic per the Gauss-Bonnet equation. The $\chi$ term is topological and its coefficient $\lambda$ controls the coupling strength. 

Before we compute scattering amplitudes, we need to rid of the extra gauge freedom. We use the Fadeev-Popov method to gauge fix the extra diff $\times$ Weyl symmetry degrees of freedom; i.e. divide by the volume of the local symmetry group $V_{\text{diff}\times\text{weyl}}$. We define $\zeta$ to be a combined coordinate and Weyl transformation:
\begin{equation}
    g^\zeta_{ab}(\sigma')=e^{2\omega(\sigma)}\pde{\sigma^c}{\sigma'^a}\pde{\sigma^d}{\sigma'^b}g_{cd}(\sigma).
\end{equation}
We define the Fadeev-Popov measure to be 
\begin{equation}
    1=\Delta_{FP}(g)\int D\zeta \delta(g-\hat g^\zeta),
\end{equation}
where $\hat g_{ab}$ is the fiducial metric, the specific metric fixed through gauge freedom.  Following the standard procedure, 
\begin{align*}
    Z[\hat g]&=\int\frac{[\dd\zeta\dd X\dd g]}{V_{\text{diff}\times\text{weyl}}}\Delta_\text{FP}(g)\delta(g-\hat g^\zeta)\exp(-S[X,g])\\
    &=\int\frac{[\dd\zeta\dd X^\zeta]}{V_{\text{diff}\times\text{weyl}}}\Delta_\text{FP}(\hat g^\zeta)\exp(-S[X^\zeta,\hat g^\zeta])\\
    &=\int\frac{[\dd\zeta\dd X]}{V_{\text{diff}\times\text{weyl}}}\Delta_\text{FP}(\hat g)\exp(-S[X,\hat g])\\
    &=\int[\dd X]\Delta_\text{FP}(\hat g)\exp(-S[X,\hat g])
\end{align*}
where we relabeled $X$ to $X^\zeta$ in the second equation and in the third equation we used the gauge invariance of $\dd X$, $\Delta_\text{FP}(g)$ (this follows from the gague invariance of the delta function), and the action. 

To find the measure $\Delta_{FP}$, we first expand $\zeta$ near the identity:
\begin{equation}
    \delta g_{ab}=(2\delta\omega-\nabla_c\delta\sigma^c)g_{ab}-2(P_1\delta\sigma)_{ab},\quad(P_1\delta\sigma)_{ab}=\frac 12(\nabla_a\delta\sigma_b+\nabla_b\delta\sigma_a-g_{ab}\nabla_c\delta\sigma^c),
\end{equation}
with this substitution, 
\begin{align*}
    \Delta_{FP}(\hat g)^{-1}&=\int[\dd\delta\omega\dd\delta\sigma]\delta[-(2\delta\omega-\hat\nabla\cdot\delta\sigma)\hat g+2\hat P_1\delta\sigma]\\
    &=\int[\dd\delta\omega\dd\beta\dd\delta\sigma]\exp\left(2\pi i\int\dd^2\sigma\hat g^{1/2}\beta^{ab}[-(2\delta\omega-\hat\nabla\cdot\delta\sigma)\hat g+2\hat P_1\delta\sigma]_{ab}\right)\\
    &=\int[\dd\beta'\dd\delta\sigma]\exp\left(4\pi i\int\dd^2\sigma\hat g^{1/2}\beta'^{ab}(\hat P_1\delta\sigma)_{ab}\right).
\end{align*}
From Berezin integrals, we can invert the integral by replacing the bosons with fermions, $\delta\sigma^a\to c^a$, $\beta'_{ab}\to b_{ab}$. This gives 
\begin{equation}
    \Delta_{FP}(\hat g)=\int[\dd b\dd c]\exp\left(-\frac 1{2\pi}\int\dd^2\sigma\hat g^{1/2}b_{ab}\hat\nabla^ac^b\right).
\end{equation}
Thus, the path integral can be written
\begin{equation}
    Z[\hat g]=\int[\dd X\dd b\dd c]\exp(-S_X-S_g)=(\det\hat\nabla^2)^{-D/2}\det\hat P_1.
\end{equation}

\subsubsection{Weyl Anomaly}

In the previous derivation, we had assumed that the gauge-fixed path integral is independent of the fiducial metric, $\hat g$, which requires $\ev{\cdots}_{g^\zeta}=\ev{\cdots}_g$. While a global Weyl transformation is a classical symmetry of the Polyakov action, there is a quantum anomaly, due to the choice of the regulator. Under the Weyl transformation, the expectation value changes as 
\begin{equation}
    \delta_\text{Weyl}\ev{\cdots}_g=-\frac 1{2\pi}\int\dd^2\sigma g(\sigma)^{1/2}\delta\omega(\sigma)\ev{T^a_a(\sigma)\cdots}_g,\quad T^a_a=-\frac c{12}R.
\end{equation}
This is known as the Weyl anomaly. Note that in the bosonic string theory, $c=c^X+c^g=D-26$, and hence the theory is Weyl-invariant/anomaly-free for $D=26$. Wess-Zumino consistency condition states that Weyl invariance holds in the quantum theory if and only if $a_1=-c/12$ vanishes.

\subsubsection{NLSM, Curved Spacetime}

The nonlinear sigma model generalizes the Polyakov by replacing the flat metric with a general metric. The metric is written 
\begin{equation}
    S_\sigma=\frac 1{4\pi\alpha'}\int_M\dd^2\sigma g^{1/2}\left[\left(g^{ab}G_{\mu\nu}(X)+i\epsilon^{ab}B_{\mu\nu}(X)\right)\partial_aX^\mu\partial_bX^\nu+\alpha'R\Phi(X)\right),
\end{equation}
where $G_{\mu\nu}$ is the (symmetric) metric, $B_{\mu\nu}$ is an antisymmetric tensor, and $\Phi$ is the dilaton. The Weyl anomaly is 
\begin{equation}
    T^a_a=-\frac 1{2\alpha'}\beta^G_{\mu\nu}g^{ab}\partial_aX^\mu\partial_bX^\nu-\frac i{2\alpha'}\beta^B_{\mu\nu}\epsilon^{ab}\partial_aX^\mu\partial_bX^\nu-\frac 12\beta^\Phi R,
\end{equation}
where
\begin{align*}
    \beta^G_{\mu\nu}&=\alpha'R_{\mu\nu}+2\alpha'\nabla_\mu\nabla_\nu\Phi-\frac{\alpha'}4H_{\mu\lambda\omega}H\indices{_\nu^{\lambda\omega}}+\mathcal O(\alpha'^2)\\
    \beta^B_{\mu\nu}&=-\frac{\alpha'}2\nabla^\omega H_{\omega\mu\nu}+\alpha'\nabla^\omega\Phi H_{\omega\mu\nu}+\mathcal O(\alpha'^2)\\
    \beta^\Phi&=\frac{D-26}6-\frac{\alpha'}2\nabla^2\Phi+\alpha'\nabla_\omega\Phi\nabla^\omega\Phi-\frac{\alpha'}{24}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+\mathcal O(\alpha'^2),
\end{align*}
where 
\begin{equation}
    H_{\omega\mu\nu}=\partial_\omega B_{\mu\nu}+\partial_\mu B_{\nu\omega}+\partial_\nu B_{\omega\mu}
\end{equation}
is the inariant three-index field strength.

The condition that this theory be Weyl-invariant is thus $\beta^G_{\mu\nu}=\beta^B_{\mu\nu}=\beta^\Phi=0$. The equations of motion look like Einstein's equation, Maxwell's equation, for $\beta^G_{\mu\nu}=0$ and $\beta^B_{\mu\nu}$ respectively. One nontrivial solution to these equations is 
\begin{equation}
    G_{\mu\nu}(X)=\eta_{\mu\nu},\quad B_{\mu\nu}=0,\quad \Phi(X)=V_\mu X^\mu,
\end{equation}
with $V_\mu V^\mu={26-D}{6\alpha'}$. This is the linear dilaton CFT. 


\subsubsection{Scattering Amplitudes, Vertex Operators}

The $n$-particle $S$-matrix is given 
\begin{equation}
    S_{j_1\cdots j_n}(k_1,\cdots,k_n)=\sum_\text{compact}\int\frac{[\dd X\dd g]}{V_{\text{diff}\times\text{weyl}}}\exp(-S_X-\lambda\chi)\prod^n_{i=1}\int\dd^2\sigma_ig(\sigma_i)^{1/2}\mathcal V_{j_i}(k_i,\sigma_i),
\end{equation}
where $\mathcal V_i$ are the diff x Weyl invariant local vertex operators, and the Euler number is given by $\chi=2-2g-b-c$, where $g$ is the number of handles (genus), $b$ is the number of holes, and $c$ is the number of cross-caps. 

The vertex operators can be found from the state-operator mapping. For example, the closed string takyon, related to $\ket{0,0;k}$, is given by 
\begin{equation}
    V_{\ket{0,0;k}}=2g_c\int\dd^2\sigma g^{1/2}e^{ik\cdot X}=g_c\int\dd^2z\normord{e^{ik\cdot X}}.
\end{equation}
Since the vertex operator has to be diff x Weyl invariant (total weight 0), we need $e^{ik\cdot X}$ to be weight of $(1,1)$, which gives 
\begin{equation}
    m^2=-k^2=-\frac 4{\alpha'}.
\end{equation}
Note that, in a curved worldsheet, the condition for the Weyl anomaly to vanish is then $k^2=4/\alpha'$ as well for this vertex operator.

Similarly, the first excited state is given 
\begin{equation}
    V_{\ket{1,1;k}}=\frac{2g_c}{\alpha'}\int\dd^2z\normord{\partial X^\mu\overline\partial X^\nu e^{ik\cdot X}},
\end{equation}
with the weights requiring these states to be massless. 

The open string takyon has the vertex operator 
\begin{equation}
    V_{\ket{0;k}}=g_0\int_{\partial M}\dd s[e^{ik\cdot X}]_r,
\end{equation}
where we defined a renormalized operator by 
\begin{equation}
    [\mathcal F]_r=\exp\left(\frac 12\int\dd^2\sigma\dd^2\sigma'\Delta(\sigma,\sigma')\fde{}{X^\mu(\sigma)}\fde{}{X_\mu(\sigma')}\right)\mathcal F,
\end{equation}
where $\Delta(\sigma,\sigma')=\frac{\alpha'}2\ln d^2(\sigma,\sigma')$ where $d(\sigma,\sigma')$ is the geodesic distance. This evidently reduces to $d^2(\sigma,\sigma')=\abs{z-z'}^2$ in flat worldsheets.

\subsection{Quantization}

It is time to quantize the gauge-fixed theory again, keeping in mind the unphysical field configurations (i.e. ghost fields and negative norm states).

\subsubsection{OCQ}

We want the scattering amplitude to be invariant under a gauge change in the metric. From the definition of the stress tensor, this can be written as $\bra{f}T^{ab}\ket{i}=0$. While there are ghost contributions to this tensor, in old covariant quantization, we ignore them and let the matter stress-tensor be zero - $T^{ab}_m=0$ - for physical fields. This, in turn sets $L^m_n=0$. We define the physical states to be 
\begin{equation}
    (L^m_n+A\delta_{n,0})\ket{\psi_\text{physical}}=0,
\end{equation}
similar to the Gupta-Bleuler quantization. 

We then define the spurious states to be 
\begin{equation}
    \ket\psi_\text{spurious}=\sum^\infty_{n=1}L^m_{-n}\ket{\psi_n}.
\end{equation}
A state that is both spurious and physical is called null. Then, for $\ket\psi$ physical and $\ket\chi$ null, $\ket\psi+\ket\chi$ is physical with the same inner product with any physical state, i.e. indistinguishable. We define the real physical Hilbert space as a quotient space:
\begin{equation}
    \mathcal H_\text{OCQ}=\mathcal H_\text{phys}/\mathcal H_\text{null}.
\end{equation}

\paragraph{Example OCQ}
At the zeroth level, the only state is $\ket{0;k}$, and thus the physical states are defined $(L^m_0+A)\ket\psi=0\implies m^2=A/\alpha'$. As there are no null levels, there is one equivalence class. 

At the next level, there are $D$ states $e\cdot\alpha_{-1}\ket{0;k}$. The $L^m_0$ condition gives $m^2=(1+A)/\alpha'$ and the $L^m_n$ conditions give $k\cdot e=0$. There is one spurious state, being $L^m_{-1}\ket{0;k}=\sqrt{2\alpha'}k\cdot\alpha_{-1}\ket{0;k}$. We consider different values of $A$ to consider the nullity: 
\begin{itemize}
    \item If $A<-1$, the spurious states are not physical and the spectrum consists of $D-1$ massive vector particles. Turns out, there is no known way to give this consistent interactions.
    \item If $A=-1$, the spurious states are null and the spectrum consists of $D-2$ massless vector particles ($e_\mu\simeq e_\mu+\gamma k_\mu$)
    \item If $A>-1$, there are tachyons.
\end{itemize}
Turns out, if $A=-1,D=26$, the OCQ is the same as light-cone quantization. In general, to find $A$ in a consistent theory, we set it as the sum of the zero-modes in the transverse directions. 

\subsubsection{BRST}

BRST quantization is a more systematic way of quantizing string states. Consider, in general, a theory with path integral fields $\phi_i$, with gauge invariance $\epsilon_\alpha\delta^\alpha$ that satisfies algebra $\comm{\delta_\alpha}{\delta_\beta}=f^\gamma_{\alpha\beta}\delta_\gamma$. With the gauge fixing condition $F^A(\phi_i)=0$, our path integral becomes, after Fadeev-Popov procedure,
\begin{equation}
    Z=\int[\dd\phi_i\dd B_A\dd B_A\dd c^\alpha]\exp\left(-S_X+iB_AF^A-b_Ac^\alpha\delta_\alpha F^A\right).
\end{equation}
This path integral is (classically) invariant under the BRST transformation 
\begin{equation}
    \delta_B\phi=-i\epsilon c^\alpha\delta_\alpha\phi_i,\quad \delta_BB_A=0,\quad \delta_Bb_A=\epsilon B_A,\quad \delta_Bc^\alpha=\frac i2\epsilon f^\alpha_{\beta\gamma}c^\beta c^\gamma.
\end{equation}
Consider now a small change $\delta F$ in the gauge fixing condition. The change in the actions give 
\begin{equation}
    \epsilon\delta\braket{f}{i}=-\epsilon\bra f\acomm{Q_B}{b_A\delta F^A}\ket i.
\end{equation}
For this to hold for arbitrary $\delta F$, it must be that, for physical states, $Q_B\ket\psi=0$, i.e. physical states must be BRST-invariant. 

There exist a constraint on BRST charges as well, that it must commute with the Hamiltonian. This gives $Q_B^2=0$, i.e. BRST charges are nilpotent. This lets us define null states, of the form $Q_B\ket\xi$. These states are annihilated by $Q_B$ and thus are physical, but it is orthogonal to all physical states. This vanishes all physical amplitudes and give rise to equivalence classes. 

Our prescription is therefore 
\begin{equation}
    \mathcal H_\text{BRST}=\mathcal H_\text{closed}/\mathcal H_\text{exact}\equiv H(Q_B),
\end{equation}
where closed states correspond to states annihilated by $Q_B$ and exact states correspond to states that can be written in the form $Q_B\ket\xi$. This is known as the cohomology of $Q_B$. 

\paragraph{Example - Point Particle}

For the point particle ($S=\int\dd\tau\left(\frac 12e^{-1}\dot X^\mu\dot X_\mu+\frac 12em^2\right)$), the local symmetry is coordinate reparametrization with $\tau\to\tau'(\tau)$. The basis of transformations are
\begin{equation}
    \delta_{\tau_1}X^\mu(\tau)=-\delta(\tau-\tau_1)\partial_\tau X^\mu(\tau),\quad\delta_{\tau_1}e(\tau)=-\partial_\tau\left[\delta(\tau-\tau_1)e(\tau)\right],
\end{equation}
where these are related to the whole transformation by $\delta X^\mu(\tau)=\int\dd\tau_1\epsilon(\tau_1)\delta_{\tau_1}X^\mu(\tau)$. The structure function is calculated to be $f^{\tau_3}_{\tau_1\tau_2}=\delta(\tau_3-\tau_1)\partial_{\tau_3}\delta(\tau_3-\tau_2)-\delta(\tau_3-\tau_2)\partial_{\tau_3}\delta(\tau_3-\tau_1)$, giving the BRST transformation
\begin{equation}
    \delta_BX^\mu=i\epsilon c\dot X^\mu,\quad \delta_Be=ie(\dot ce),\quad \delta_BB=0,\quad\delta_Bb=\epsilon B,\quad \delta_Bc=i\epsilon c\dot c.
\end{equation}
With $F(\tau)=1-e(\tau)$, the guage-fixed action is 
\begin{equation}
    S=\int\dd\tau\left(\frac 12e^{-1}\dot X^\mu\dot X_\mu+\frac 12em^2+iB(e-1)-e\dot bc\right).
\end{equation}
Classical solution (i.e. integrating out $B$) gives $e=1$:
\begin{equation}
    S=\int\dd\tau\left(\frac 12\dot X^\mu\dot X_\mu+\frac 12m^2-\dot bc\right).
\end{equation}
The BRST transformation becomes (with the equation of motion from $e$)
\begin{equation}
    \delta_BX^\mu=iec\dot X^\mu,\quad\delta_Bb=i\epsilon\left(-\frac 12\dot X^\mu\dot X_\mu+\frac 12m^2-\dot bc\right),\quad\delta_Bc=ie\dot cc.
\end{equation}
The BRST charge can then be found from the commutator ($\delta\phi=i\epsilon\comm{Q_B}{\phi}$) as $Q_B=cH=\frac 12c\left(-\dot X^\mu\dot X_\mu+m^2\right)$. 

There exist two states in this system, defined 
\begin{equation}
    b\ket{k,\downarrow}=c\ket{k,\uparrow}=0,\quad b\ket{k,\uparrow}=\ket{k,\downarrow},\quad c\ket{k,\downarrow}=\ket{k,\uparrow},
\end{equation}
giving us 
\begin{equation}
    Q_B\ket{k,\downarrow}=\frac 12(k^2+m^2)\ket{k,\uparrow},\quad Q_B\ket{k,\uparrow}=0.
\end{equation}
The closed states are $\ket{k,\downarrow}$ with $k^2+m^2=0$ and $\ket{k,\uparrow}$ for all $k$ and the exact states are $\ket{k,\uparrow}$ with $k^2+m^2\ne 0$. Thus, the physical states are $\ket{k,\uparrow},\ket{k,\downarrow}$ with mass-shell condition $k^2+m^2=0$. 

However, there is one extra condition, that $\ket{k,\uparrow}$ don't appear in physical amplitudes. This is due to the fact that amplitudes cannot have delta functions, but $\ket{\uparrow}$ states have delta function filtering out the off-shell states. 

\pagebreak

\section{Appendix}

\subsection{Conformal Group}

Conformal transformations satisfy $\partial^\mu\epsilon^\nu+\partial^\nu\epsilon^\mu=\Lambda(x)\eta^{\mu\nu}$. As we have seen in section 1, the general solution of this requirement is that $\Lambda$ be linear and $\epsilon$ quadratic. Hence, 
\begin{align*}
    \Lambda&=\Lambda_0+\Lambda_\alpha x^\alpha\\
    \epsilon_\mu&=a_\mu+\omega_{\mu\nu}x^\nu+\lambda x_\mu+(b_\mu x^2-2(b\cdot x)x_\mu).
\end{align*}
The four terms in $\epsilon_\mu$ correspond to infinitesimal forms of translation, rotation, dilatation, and special conformal transformation (the finite form is $x^\mu\to\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}$), respectively. To summarize, 

\begin{center}
    \begin{tabular}{c | c | c | c | c}
         & Count & Infinitesimal & Finite & Generator\\
        \hline 
        Translations & D & $x^\mu\to x^\mu+a^\mu$ & $x^\mu\to x^\mu+a^\mu$ & $-i\partial_\mu$\\
        Rotation & $\frac 12D(D-1)$ & $x^\mu\to x^\mu+\omega^\mu_\nu x^\nu,\,\omega^\mu_\nu\in\text{Alt}_D$ & $x^\mu\to\Lambda^\mu_\nu x^\nu,\,\Lambda^\mu_\nu\in\text{SO}(D)$ & $-ix^\mu\partial_\mu$\\
        Dilatation & 1 & $x^\mu\to x^\mu+\lambda x^\mu$ & $x^\mu\to\lambda x^\mu$ & $-ix^\mu\partial_\mu$\\
        SCT & $D$ & $x^\mu\to x^\mu-2(b\cdot x)x_\mu+b_\mu x^2$ & $x^\mu\to\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}$ & $-i(2x_\mu x^\nu\partial_\nu-x^2\partial_\mu)$
    \end{tabular}
\end{center}


\end{document}