\begin{flushright}
Peter Battaglino\\ 7. April, 2005
\end{flushright}
\begin{hwproblem}{13.6}
Orientifold O$p$-planes.
\end{hwproblem}
\newline
{\bf{a.}} For an O23-plane the two normal directions $x^{24}$, $x^{25}$ can be represented
by a plane. A closed string at a fixed $\tau$ appears as a parametrized closed curve $X^a(\tau,\sigma)$
in this plane. Draw such an oriented closed string that lies fully in the first quadrant of the
$(x^{24},x^{25})$ plane. Draw also the string $\tilde{X}^a(\tau,2\pi-\sigma)$.
\begin{center}
\includegraphics{O23plane}
\end{center}
{\bf{b.}} For any operator $\mathcal{O}$ the action of $\Omega_p$ is defined by
$\mathcal{O}\rightarrow \Omega_p\mathcal{O}\Omega_p^{-1}$. Use the expansion (13.24) to
calculate the action of $\Omega_p$ on the operators $x_0^a$, $p^a$, $\alpha_n^a$,
$\bar{\alpha}_n^a$, $x_0^i$, $p^i$, $\alpha_n^i$, and $\bar{\alpha}_n^i$.
\newline\newline
The closed string coordinates are represented by
\begin{equation}
X^\mu(\tau,\sigma) = x_0^\mu + \sqrt{2\alpha^\prime}\alpha_0^\mu \tau
+ i\sqrt{\frac{\alpha^\prime}{2}} \sum_{n\neq0} \frac{e^{-in\tau}}{n}
(\alpha_n^\mu e^{in\sigma} + \bar{\alpha}_n^\mu e^{-in\sigma}).
\end{equation}
The action of $\Omega_p$ on each of the tangential coordinates is the same as that of the
twist operator in problem 13.5. Thus,
\begin{eqnarray}
\Omega\, x_0^i \,\Omega^{-1} & = & x_0^i \\
\Omega\, p^i \,\Omega^{-1} & = & p^i \\
\Omega\, \alpha_n^i \,\Omega^{-1} & = & \bar{\alpha}_n^i \\
\Omega\, \bar{\alpha}_n^i \,\Omega^{-1} & = & \alpha_n^i.
\end{eqnarray}
For the transverse dimensions, we have
\begin{eqnarray}
\Omega\, x_0^a \,\Omega^{-1} & = & -x_0^a \\
\Omega\, p^a \,\Omega^{-1} & = & -p^a \\
\Omega\, \alpha_n^a \,\Omega^{-1} & = & -\bar{\alpha}_n^a \\
\Omega\, \bar{\alpha}_n^a \,\Omega^{-1} & = & -\alpha_n^a.
\end{eqnarray}
For the $+$ and $-$ coordinates, $\Omega$ simply reverses the orientation.
This is obviously the case for $X^+$. For $X^-$ we know that the only non-trivial
actions will involve products like $\alpha_n^a\alpha_{q-n}^a$ and their barred equivalents.
The negative signs produced by the $\Omega$ action will cancel, and we will obtain
the same result as in problem 13.5. The orientifold transformations are symmetries of closed
string theory because they leave the closed string Virasoro operators invariant for exactly
the same reason that they leave the $X^-$ coordinate invariant. The closed string Virasoro
operators are the operator-ingredients for the closed string Hamiltonian, so
\begin{equation}
\Omega\,H\,\Omega^{-1} = H \Rightarrow [\Omega,H]=0
\end{equation}
\newline\newline
{\bf{c.}} Denote the ground states by $\ket{p^+,p^i,p^a}$. Assume that the states
$\ket{p^+,p^i,\vec{0}}$ are invariant under $\Omega$: $\Omega\ket{p^+,p^i,\vec{0}}
= \ket{p^+,p^i,\vec{0}}$. Prove that
\[
\Omega\ket{p^+,p^i,p^a} = \ket{p^+,p^i,-p^a}.
\]
\newline
We know that $p^a$ transforms as $\Omega\,p^a\,\Omega^{-1}=-p^a$, so $\Omega$ and
$p^a$ anticommute. Hence,
\begin{eqnarray}
p^a_{\mathrm{op}}\Omega\ket{p^+,p^i,p^a} & = & -\Omega p^a_{\mathrm{op}}\ket{p^+,p^i,p^a} \\
{} & = & -p^a\Omega\ket{p^+,p^i,p^a},
\end{eqnarray}
so we see that
\begin{equation}
\Omega\ket{p^+,p^i,p^a} = \ket{p^+,p^i,-p^a}.
\end{equation}
\newline\newline
{\bf{d.}} The massless states of the oriented closed string theory are given by
\[
\ket{\Phi} = \int \dd p^+\, \dd^{p-1} p^i\, \dd^{d-p} p^a\,\,
\Phi^{\pm}_{IJ}(\tau,p^+,p^i,p^a)(\alpha^I_{-1}\bar{\alpha}^J_{-1}
\pm \alpha^J_{-1}\bar{\alpha}^I_{-1})\ket{p^+,p^i,p^a}.
\]
Find the conditions that must be satisfied by $\Phi^\pm_{ab}$, $\Phi^\pm_{ia}$, and
$\Phi^\pm_{ij}$ to guarantee $\Omega$ invariance. All of the conditions are of the form
\begin{equation*}
\Phi^\pm_{IJ}(\tau,p^+,p^i,p^a)=\cdots \Phi^\pm_{IJ}(\tau,p^+,p^i,-p^a),
\end{equation*}
where the dots represent a sign factor that you must determine for each case.
\newline\newline
We can write the action of $\Omega$ on $\ket{\Phi}$ as
\begin{eqnarray}
\Omega\ket{\Phi} & = & \int \dd^Dp/\dd p^- \Phi_{IJ}^\pm(\dots)\Big(\Omega\alpha_{-1}^I\Omega^{-1}
\Omega\bar{\alpha}_{-1}^J\Omega^{-1} \nonumber \\
{} & \pm & \Omega\alpha_{-1}^J\Omega^{-1}\Omega\bar{\alpha}_{-1}^I
\Omega^{-1}\Big)\Omega\ket{p^+,p^i,p^a}.
\end{eqnarray}
When $(IJ)=(ab)$ the operator sum in the expression for $\ket{\Phi}$ is going to
undergo the transformation $\alpha\leftrightarrow\bar{\alpha}$, while the action of
$\Omega$ on the basis states will invert their transverse momenta. The operator sum
is symmetric under the $\alpha\leftrightarrow\bar{\alpha}$ transformation when the $\pm$
is a $+$, and antisymmetric when the $\pm$ is a $-$. Thus we need a factor of $-$ to cancel
the asymmetry and nothing to cancel the symmetry, so we need a factor of $\pm$:
\begin{equation}
\Phi^\pm_{ab}(p^a) = \pm\Phi^\pm_{ab}(-p^a).
\end{equation}
When $(IJ)=(ia)$, we pick up negative signs in the operator sum under the action of $\Omega$,
and upon reordering terms we obtain, schematically:
\begin{equation}
\mp\Phi_{ia}^\pm(p^a)(\alpha_{-1}^i\bar{\alpha}_{-1}^a\pm \alpha_{-a}^a\bar{\alpha}_{-1}^i)
\ket{p^+,p^i,-p^a}.
\end{equation}
So the action of $\Omega$ contributes a factor of $\mp$ to the operator terms and inverts
the basis states, and we thus require, for $(IJ)=(ia)$, that
\begin{equation}
\Phi_{ia}^\pm(p^a) = \mp\Phi_{ia}^\pm(-p^a).
\end{equation}
The action of $\Omega$ yields the same results when $(IJ)=(ij)$ as for when $(IJ)=(ab)$
so we need a factor of $\pm$ in this case as well:
\begin{equation}
\Phi_{ij}^\pm(p^a) = \pm\Phi_{ij}^\pm(-p^a).
\end{equation}
\clearpage
\begin{hwproblem}{13.7}
Massive level in the open superstring.
\end{hwproblem}
\newline
{\bf{a.}} Consider eight anticommuting variables $b^i$, with $i=1,\dots,8$. Ignoring signs,
how many inequivalent products of the form $b^{i_1}b^{i_2}$ can be built? How many
$b^{i_1}b^{i_2}b^{i_3}$? How many $b^{i_1}b^{i_2}b^{i_3}b^{i_4}$?
\newline\newline
Ignoring signs, the question we are trying to answer is ``How many different ways can we choose $N$
out of 8 objects?''. This is the case because permutations don't matter when we are ignoring signs,
and we can't choose the same object twice, or else we get zero (since the $b$'s are Grassmannian).
Thus, for $b^{i_1}b^{i_2}$ we can choose ${8 \choose 2} = 28$ different products, for
$b^{i_1}b^{i_2}b^{i_3}$ we have ${8 \choose 3} = 56$, and for $b^{i_1}b^{i_2}b^{i_3}b^{i_4}$ we
have ${8 \choose 4} = 70$ choices.
\newline\newline
{\bf{b.}} Consider the first excited level of the open superstring $(\alpha^\prime M^2=1)$. List
the states in the NS sector and the states in the R sector. Confirm that you get the same number
of states (the number is more than 64).
\newline\newline
In the NS sector we have the following allowed states:
\[
\begin{array}{rcr}
b^I_{-3/2}\ket{{\mathrm{NS}}}\otimes\ket{p^+,\vec{p}_T} & \qquad & 8\\
b^I_{-1/2}\,b^J_{-1/2}\,b^{K}_{-1/2}\ket{{\mathrm{NS}}}\otimes\ket{p^+,\vec{p}_T} & \qquad & {8\choose3}=56\\
\alpha^I_{-1}\,b^J_{-1/2}\ket{{\mathrm{NS}}}\otimes\ket{p^+,\vec{p}_T} & \qquad & 8\times8=64
\end{array}
\]
yielding a total of 128 states. In the R sector we have the following allowed states:
\[
\begin{array}{rcr}
d^I_{-1}\ket{{\mathrm{R_1^a}}}\otimes\ket{p^+,\vec{p}_T} & \qquad & 8\times8=64\\
d^I_{-1}\ket{{\mathrm{R_2^a}}}\otimes\ket{p^+,\vec{p}_T} & \qquad & 8\times8=64
\end{array}
\]
also yielding a total of 128 states. Thus we see that the NS and R sectors have the same number
of $\alpha^\prime M^2 = 1$ states.
\clearpage
\begin{hwproblem}{14.1}
A D$p$-brane with orientifolds.
\end{hwproblem}
\newline
{\bf{a.}} The $\Omega$ actions are given by
\begin{eqnarray}
\Omega\,X^a(\tau,\sigma)\,\Omega^{-1} & = & X^a(\tau,\pi-\sigma),\label{adir} \\
\Omega\,X^i(\tau,\sigma)\,\Omega^{-1} & = & X^i(\tau,\pi-\sigma).\label{idir}
\end{eqnarray}
Give the $\Omega$ action on the oscillators $\alpha_n^a$ and $\alpha_n^i$. What is the
expected $\Omega$ action on $\alpha_n^-$? Does it work out?
\newline\newline
We have the expansions
\begin{eqnarray}
X^a(\tau,\sigma) & = & \bar{x}^a + \sqrt{2\alpha^\prime}\sum_{n\neq0}\frac{1}{n}\alpha_n^ae^{-in\tau}
\sin n\sigma \\
X^i(\tau,\sigma) & = & x_0^i + \sqrt{2\alpha^\prime}\alpha_0^i\tau \nonumber \\
{} & {} & + i\sqrt{2\alpha^\prime}
\sum_{n=1}^{\infty}\frac{1}{n}(\alpha_n^ie^{-in\tau}-\alpha_{-n}^ie^{in\tau})\cos n\sigma
\end{eqnarray}
For the transverse oscillators, we see that a factor of $\sin(n\pi-n\sigma)$ contributes an
additional minus sign to the usual $(-1)^n$ for the tangential oscillators, so we get
\begin{eqnarray}
\Omega\,\alpha_n^i\,\Omega^{-1} & = & (-1)^n\alpha_n^i \\
\Omega\,\alpha_n^a\,\Omega^{-1} & = & (-1)^{n+1}\alpha_n^a
\end{eqnarray}
For $\alpha_n^-$ we expect to pick up the same sign change as for the tangential oscillators.
We just need to check the action of $\Omega$ on the Virasoro operators:
\begin{eqnarray}
\sum_{p\in\mathbb{Z}}\alpha_p^I\alpha_{n-p}^I & = & \sum_{p\in\mathbb{Z}}(\alpha_p^i\alpha_{n-p}^i
+ \alpha_p^a\alpha_{n-p}^a)\\
{} & \rightarrow & \sum_{p\in\mathbb{Z}}((-1)^{p+n-p}\alpha_p^i\alpha_{n-p}^i
+ (-1)^{p+1+n-p+1}\alpha_p^a\alpha_{n-p}^a) \\
{} & = & (-1)^n\sum_{p\in\mathbb{Z}}\alpha_p^I\alpha_{n-p}^I,
\end{eqnarray}
which is what we expect.
\newline\newline
{\bf{b.}} Assume that the ground states $\ket{p^+,\vec{p}}$ are $\Omega$ invariant. Find the
states of the theory for $N^\perp \leq 2$. As you will see, some massless sates survive. Interpret
these states along the lines of the discussion below (14.37).
\newline\newline
The possible $N^\perp\leq2$ states that we can build are
\begin{equation*}
\begin{array}{rcr}
\alpha_{-1}^a \ket{p^+,\vec{p}} & \qquad & +1\\
\alpha_{-1}^i \ket{p^+,\vec{p}} & \qquad & -1\\
\alpha_{-1}^a\alpha_{-1}^b \ket{p^+,\vec{p}} & \qquad & +1\\
\alpha_{-1}^a\alpha_{-1}^i \ket{p^+,\vec{p}} & \qquad & -1\\
\alpha_{-1}^i\alpha_{-1}^j \ket{p^+,\vec{p}} & \qquad & +1\\
\alpha_{-2}^a \ket{p^+,\vec{p}} & \qquad & -1\\
\alpha_{-2}^i \ket{p^+,\vec{p}} & \qquad & +1
\end{array}
\end{equation*}
If we only keep states that satisfy $\Omega = +1$ we still end up with massless states,
namely those that we get by acting with $\alpha_{-1}^a$ on the ground states. There
are $(d-p)$ of these massless scalar states. These can be viewed as excitations of the
D$p$-brane in its normal directions.
\newline\newline
{\bf{c.}} Replace the O25-plane by an O$p$-plane coincident with the D$p$-brane at $\bar{x}^a=0$.
Let $\Omega_p$ denote the operator for which this theory keeps only the states with $\Omega_p=+1$.
How should equations (\ref{adir}) and (\ref{idir}) change when $\Omega$ is replaced by $\Omega_p$?
Give the $\Omega_p$ action on the oscillators $\alpha_n^a$ and $\alpha_n^i$.
\newline\newline
Equations (\ref{adir}) and (\ref{idir}) change to
\begin{eqnarray}
\Omega_p\,X^a(\tau,\sigma)\,\Omega_p^{-1} & = & -X^a(\tau,\pi-\sigma) \\
\Omega_p\,X^i(\tau,\sigma)\,\Omega_p^{-1} & = & X^i(\tau,\pi-\sigma).
\end{eqnarray}
Since $\Omega_p$'s action hasn't changed for the tangential coordinates, its action on $\alpha_n^i$
stays the same. However, we pick up a minus sign for the normal oscillators:
\begin{equation}
\Omega_p\,\alpha_n^a\,\Omega_p^{-1} = -(-1)^{n+1}\alpha_n^a = (-1)^n\alpha_n^a,
\end{equation}
so we have
\begin{equation}
\Omega_p\,\alpha_n^I\,\Omega_p^{-1} = (-1)^n\alpha_n^I,
\end{equation}
which means that we have the same allowed spectrum as for the twist-operator invariant
spectrum for open strings, with a D25-brane present. This excludes the massless scalar
states that we did allow for the operator $\Omega$. We are left over with only the states
for which $N^\perp$ is even. This means that we don't get the same normal excitations of
the D$p$-brane as we did for the case with an O25-plane.
\clearpage
\begin{hwproblem}{4}
Yang-Mills theory.
\end{hwproblem}
\newline
$U(N)$ has $N^2$ independent generators. The elements of $SU(N)$ must satisfy the additional
constraint of $\det(M)=1$. Since $M=\exp(i\alpha\cdot J)$ in general, and
$\det(M)=\exp(i{\mathrm{Tr}}(\alpha\cdot J))$, we require that the generators be traceless,
so this additional constraint reduces the dimensionality to $N^2-1$ for $SU(N)$. $SO(N)$ is
just the group of rotations in $N$ dimensions. There are ${N\choose 2} = N(N-1)/2$ dimensions
in this group because there are that many planes of rotation in $N$ dimensions.
\newline\newline
To generalize the QED action $-(1/4)F^{\mu\nu}F_{\mu\nu}$ to non-Abelian groups, we allow the
gauge field to transform as
\begin{equation}
\delta A^a_\mu = gf^{abc}\alpha^bA_\mu^c+\partial_\mu\alpha^a.
\end{equation}
The generators of the gauge group satisfy
\begin{equation}
[T^a,T^b]=if^{abc}T^c.
\end{equation}
When the group is U(1), there is only one generator for the group. Since $f^{abc}$ can be made
completely antisymmetric, the fact that $[T^a,T^b]=0$ means that the structure constants are
zero, and the gauge transformation reduces to the usual transformation in QED.
Define
\begin{equation}
F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A^a_\mu + gf^{abc}A_\mu^bA_\nu^c.
\end{equation}
This then transforms as
\begin{eqnarray}
F_{\mu\nu}^a &\rightarrow& \partial_\mu(A_\nu^a+gf^{abc}\alpha^bA_\nu^c+\partial_\nu\alpha^a)
- \partial_\nu(A_\mu^a+gf^{abc}\alpha^bA_\mu^c+\partial_\mu\alpha^a) \nonumber \\
{}&{}& +gf^{ade}(A_\mu^d+gf^{abc}\alpha^bA_\mu^c+\partial_\mu\alpha^d)
(A_\nu^e+gf^{ebc}\alpha^bA_\nu^c+\partial_\nu\alpha^e) \nonumber \\
{}&=& F^a_{\mu\nu}+gf^{abc}(\partial_\mu(\alpha^bA_\nu^c)-\partial_\nu(\alpha^bA_\mu^c))
+ gf^{ade}A_\mu^d(gf^{ebc}\alpha^bA_\nu^c+\del_\nu\alpha^e) \nonumber \\
{}&{}& + gf^{ade}(gf^{dbc}\alpha^bA_\mu^c+\del_\mu\alpha^d)A^e_\nu+\mathcal{O}(\alpha^2) \nonumber \\
{}&=& F_{\mu\nu}^a+2gf^{abc}((\del_\mu\alpha^b)A_\nu^c-(\del_\nu\alpha^b)A_\mu^c) \nonumber \\
{}&{}&+gf^{abc}\alpha^b(\del_\mu A_\nu^c-\del_\nu A_\mu^c)-g^2f^{dce}f^{abc}\alpha^bA\mu^dA_\nu^e+\mathcal{O}(\alpha^2) \nonumber\\
{}&=&F^a_{\mu\nu}+gf^{abc}\alpha^bF^c_{\mu\nu}+2gf^{abc}((\del_\mu\alpha^b)A_\nu^c-(\del_\nu\alpha^b)A_\mu^c)+\mathcal{O}(\alpha^2)
\end{eqnarray}
to first order in $\alpha^i$.
\end{document}
