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\newenvironment{hwproblem}[1]
{\noindent{\bf{Problem #1:}}\em }
{\vspace{0.2 in}}

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{\noindent{\underline{Solution:}} }
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\begin{document}
\begin{center}
\Large
33-652: String Theory\\
Assignment 7
\end{center}

\begin{flushright}
Peter Battaglino\\ 11. April, 2005
\end{flushright}

\begin{hwproblem}{1}
Differential forms
\end{hwproblem}

\noindent
A $p$-form is a totally antisymmetric tensor, so it has ${D \choose p}$
independent components, or
\begin{equation}
\frac{D!}{p!(D-p)!}
\end{equation}
independent components. Since any $p$-form has the representation
\begin{equation}
\omega = \frac{1}{p!}A_{a_1a_2\dots{}a_p}dx^{a_1}\wedge dx^{a_2}
\wedge \dots \wedge dx^{a_p},
\end{equation}
it can be represented as a linear combination of wedge products of $p$
basis one-forms. Examine the form
\[
dx^{a_1}\wedge dx^{a_2}\wedge \dots \wedge dx^{a_p}.
\]
In order for this form to be different from zero, each of the
$\set{a_i}$ must take on a different value, since
\begin{equation}\label{form_squared_zero}
dx^{a_i}\wedge dx^{a_i} = 0
\end{equation}
for all $a_i$. Since each of the $a_i$'s can run over the index set
$\set{1,\dots,D}$, we see that if we had more than $D$ forms wedging
to create our basis form, then at least two of them would have to
have the same index. We can then push one of them through to the other,
and by (\ref{form_squared_zero}) we see that the form we have constructed
vanishes. Hence the maximum possible value for $p$ is $D$.
Denoting the space of $p$-forms on a manifold $M$ by $\Omega_p(M)$, we
have the exterior derivative $d$ as a mapping $d:\Omega_p(M)\rightarrow
\Omega_{p+1}(M)$, whose action on a $p$-form $\omega_p$ is
\begin{equation}
d\omega_p=\frac{1}{p!}\partial_{a_{p+1}}A_{a_1a_2\dots a_p}dx^{a_{p+1}}
\wedge dx^{a_2}\wedge \dots \wedge dx^{a_p}.
\end{equation}
If we act on this new $p+1$-form with $d$ a second time, we obtain
\begin{equation}
d^2\omega_p = \frac{1}{p!}\partial_{a_{p+2}}\partial_{a_{p+1}}
A_{a_1a_2\dots a_p}dx^{a_{p+2}}\wedge dx^{a_{p+1}}\wedge dx^{a_1}\wedge
dx^{a_2}\wedge \dots \wedge dx^{a_p}.
\end{equation}
Since the partial derivatives commute, we can commute them and relabel
indices to arrive at
\begin{equation}
d^2\omega = -d^2\omega,
\end{equation}
which proves that $d^2=0$ when acting on any $p$-form.
Considering Maxwell's equations, we can write the gauge field $A$ as
\begin{equation}
A = A_\mu dx^\mu.
\end{equation}
We then have 
\begin{eqnarray}
dA & = & \del_\nu A_\mu dx^{\nu}\wedge dx^{\mu} \nonumber \\
{} & = & \del_\nu A_\mu (dx^\nu\otimes dx^\mu - dx^\mu\otimes dx^\nu)\nonumber \\
{} & = & (\del_\mu A_\nu - \del_\nu A_\mu)dx^\mu\otimes dx^\nu \nonumber \\
{} & = & F_{\mu\nu} dx^\mu\otimes dx^\nu \nonumber \\
{} & = & F,
\end{eqnarray}
where $F$ is the field strength two-form.
Now let's examine the components of $dA$:
\begin{eqnarray}
dA &=& \del_\mu A_\nu dx^\mu\wedge dx^\nu \nonumber \\
{} &=& (\del_t A_x - \del_x\Phi)dt\wedge dx \nonumber \\
{}&+&(\del_t A_y - \del_y\Phi)dt\wedge dy \nonumber \\
{}&+&(\del_t A_z - \del_z\Phi)dt\wedge dz \nonumber \\
{}&+&(\del_x A_y - \del_y A_x)dx\wedge dy \nonumber \\
{}&+&(\del_z A_x - \del_x A_z)dz\wedge dx \nonumber \\
{}&+&(\del_y A_z - \del_z A_y)dy\wedge dz \nonumber \\
{}&=& E_x dt\wedge dx + E_y dt\wedge dy + E_z dt\wedge dz \nonumber \\
{}&+& B_z dx\wedge dy + B_y dz\wedge dx + B_x dy\wedge dz.
\end{eqnarray}
Taking the exterior derivative of the above expression, we have
\begin{eqnarray}
d^2A &=& \del_y E_x dy\wedge dt\wedge dx + \del_z E_x dz\wedge dt\wedge dx \nonumber \\
{}&+& \del_x E_y dx\wedge dt\wedge dy + \del_z E_y dz\wedge dt\wedge dy \nonumber \\
{}&+& \del_x E_z dx\wedge dt\wedge dz + \del_y E_z dy\wedge dt\wedge dz \nonumber \\
{}&+& \del_t B_x dt\wedge dy\wedge dz + \del_x B_x dx\wedge dy\wedge dz \nonumber \\
{}&+& \del_t B_y dt\wedge dz\wedge dx + \del_y B_y dy\wedge dz\wedge dx \nonumber \\
{}&+& \del_t B_z dt\wedge dx\wedge dy + \del_z B_z dz\wedge dx\wedge dy \nonumber \\
{}&=& (\del_z E_y - \del_y E_z + \del_t B_x)dt\wedge dy\wedge dz \nonumber \\
{}&+& (\del_x E_z - \del_z E_x + \del_t B_y)dt\wedge dz\wedge dx \nonumber \\
{}&+& (\del_y E_x - \del_x E_y + \del_t B_z)dt\wedge dx\wedge dy \nonumber \\
{}&+& (\del_x B_x + \del_y B_y + \del_z B_z)dx\wedge dy\wedge dz \nonumber \\
{}&=& 0.
\end{eqnarray}
This is a fancy way of writing
\begin{eqnarray}
\nabla\times\mathbf{E}+\frac{\del\mathbf{B}}{\del t} & = & 0, \\
\nabla\cdot\mathbf{B} & = & 0.
\end{eqnarray}
Now consider the Hodge dual of a form, defined as
\begin{equation}
\star dx^{a_1}\wedge\cdots\wedge dx^{a_p}
\equiv \frac{1}{(D-p)!}
\epsilon^{a_1\cdots a_p}_{\phantom{a_1}\phantom{\cdots}\phantom{a_p}a_{p+1}\cdots a_D}
dx^{a_{p+1}}\wedge\cdots\wedge dx^{a_D}.
\end{equation}
To construct the vector curl in three dimensional Euclidean space, we examine the
Hodge dual of the exterior derivative of a one-form. Let a one-form $\omega$ defined as
\begin{equation}
\omega = \omega_a dx^a
\end{equation}
be given. Its exterior derivative is given by
\begin{equation}\label{cross}
d\omega = \left(\frac{\del\omega_z}{\del y}-\frac{\del\omega_y}{\del z}\right) dy\wedge dz
+\left(\frac{\del\omega_x}{\del z}-\frac{\del\omega_z}{\del x}\right) dz\wedge dx
+\left(\frac{\del\omega_y}{\del x}-\frac{\del\omega_x}{\del y}\right) dx\wedge dy.
\end{equation}
Now consider the Hodge dual of the form $dx^a\wedge dx^b$, given by
\begin{equation}
\star dx^a\wedge dx^b = \epsilon^{abc}dx^c.
\end{equation}
Applying this to equation (\ref{cross}), we see that
\begin{equation}
\star d\omega = \left(\frac{\del\omega_z}{\del y}-\frac{\del\omega_y}{\del z}\right)dx
+ \left(\frac{\del\omega_x}{\del z}-\frac{\del\omega_z}{\del x}\right)dy
+ \left(\frac{\del\omega_x}{\del x}-\frac{\del\omega_x}{\del y}\right)dz.
\end{equation}
Now, for each point $p$ on a manifold $M$, we can consider the metric to be a map
between the tangent space $T_pM$ and the dual space $T^*_pM$ of that manifold.
Denote this operation by $\sharp$, so that $\sharp\,dx^\alpha = \del/\del x^\alpha$
and $\sharp\,\del/\del x^\alpha=dx^\alpha$. From the above
equations we see that the curl of a vector field $V$ is given (in a really retarded way) by
\begin{equation}
\nabla\times V = \sharp\star d\,\sharp\,V.
\end{equation}
\clearpage
\begin{hwproblem}{14.4}
Separated D$p$-branes and an O$p$-plane
\end{hwproblem}

\noindent
{\bf{a.}} Show the two strings obtained by the orientifold symmetry. Since the arguments
$p^+,\vec{p}$ of the ground states are always present, let us omit them for brevity. The ground
states are of four types:
\begin{equation}\label{groundstates}
\ket{[ij]},\quad \ket{[i\bar{j}]},\quad \ket{[\bar{i}j]},\quad \ket{[\bar{i}\bar{j}]}.
\end{equation}
Each class contains $N^2$ ground states since $i$ and $j$ run from $1$ to $N$ and $\bar{i}$
and $\bar{j}$ run from $\bar{1}$ to $\bar{N}$. Define an expected action of $\Omega_p$ on the
ground states in (\ref{groundstates}). Show that your choice satisfies $\Omega_p^2=1$ acting
on the ground states.
\newline\newline
The two strings and their orientifolding partners are depicted as follows:
\begin{center}
\includegraphics{14_4_a}
\end{center}
We expect the following action of $\Omega_p$ on the ground states in (\ref{groundstates}):
\begin{equation}
\Omega_p\ket{[ij]} = \ket{[\bar{j}\bar{i}]},\quad
\Omega_p\ket{[i\bar{j}]} = \ket{[\bar{j}i]},\quad
\Omega_p\ket{[\bar{i}j]} = \ket{[j\bar{i}]},\quad
\Omega_p\ket{[\bar{i}\bar{j}]} = \ket{[ji]}.
\end{equation}
It is obvious that $\Omega_p^2=1$ when acting on any of the ground states.
\newline\newline
{\bf{b.}} What are the possible interactions between strings in the four types of sectors built on
the states (\ref{groundstates})?
\newline\newline
The following interactions between strings can take place:
\begin{equation}
\begin{array}{cccc}
[ij]*[jk]=[ik] & [ij]*[j\bar{k}]=[i\bar{k}] & [i\bar{j}]*[\bar{j}k]=[ik] & [i\bar{j}]*[\bar{j}\bar{k}]=[i\bar{k}] \\
{} & {} & {} & {} \\
{[{\bar{i}}{j}]}*[jk]=[\bar{i}k] & [\bar{i}j]*[j\bar{k}]=[\bar{i}\bar{k}] & [\bar{i}\bar{j}]*[\bar{j}k]=[\bar{i}k] & [\bar{i}\bar{j}]*[\bar{j}\bar{k}]=[\bar{i}\bar{k}].
\end{array}
\end{equation}
However, the effective number of interactions is only half those listed, as the action of $\Omega_p$ induces an equivalence
class among them. Denoting the set of interactions by $\mathcal{S}$, we can form the quotient set $\mathcal{S}/\Omega_p$
that can be represented by
\begin{equation}
[ij]*[jk]=[ik] \qquad [i\bar{j}]*[\bar{j}k]=[ik] \qquad [\bar{i}j]*[jk]=[\bar{i}k] \qquad [\bar{i}\bar{j}]*[\bar{j}k]=[\bar{i}k].
\end{equation}
\newline\newline
{\bf{c.}} String products have a component along the ground state, as in
\begin{equation}
\ket{[i\bar{j}]} * \ket{[\bar{j}k]} = \ket{[ik]} + \cdots.
\end{equation}
Write the other possible ground state products. Test the consistency of your definition of $\Omega_p$ by acting with $\Omega_p$
on both sides of the equations giving ground state products. $\Omega_p$ acts on a product via
\begin{equation}\label{omegaact}
\Omega_p(\ket{A}*\ket{B})=(\Omega_P\ket{B})*(\Omega_p\ket{A}).
\end{equation}
\newline\newline
The possible ground state products are (denoting the equivalence relation described earlier by \approx)
\begin{eqnarray}
\ket{[ij]}*\ket{[jk]} \approx \ket{[\bar{k}\bar{j}]}*\ket{[\bar{j}\bar{i}]} &=& \ket{[ik]}+\ket{\bar{k}\bar{i}}\\
\ket{[i\bar{j}]}*\ket{[\bar{j}k]} \approx \ket{[\bar{k}j]}*\ket{[j\bar{i}]} &=& \ket{[ik]}+\ket{\bar{k}\bar
\end{eqnarray}
We act on the above equations using (\ref{omegaact}) to get
\begin{eqnarray}
\ket{[\bar{k}\bar{j}]}*\ket{[\bar{j}\bar{i}]} &=& \ket{[\bar{k}\bar{i}]} + \cdots \nonumber\\
\ket{[k\bar{j}]}*\ket{[\bar{j}\bar{i}]} &=& \ket{[k\bar{i}]} + \cdots \nonumber\\
\ket{[kj]}*\ket{[j\bar{i}]} &=& \ket{[k\bar{i}]} + \cdots \nonumber\\
\ket{[\bar{k}\bar{j}]}*\ket{[\bar{j}i]} &=& \ket{[\bar{k}i]} + \cdots \nonumber\\
\ket{[k\bar{j}]}*\ket{[\bar{j}i]} &=& \ket{[ki]} + \cdots \nonumber\\
\ket{[\bar{k}j]}*\ket{[ji]} &=& \ket{[\bar{k}i]} + \cdots \nonumber\\
\ket{[kj]}*\ket{[ji]} &=& \ket{[ki]} + \cdots,
\end{eqnarray}
which are consistent.
\newline\newline
{\bf{d.}} Find $\Omega_p$ action on $\alpha_n^i$ by using $\Omega_p\,X^i(\tau,\sigma)\,\Omega_p^{-1}
= X^i(\tau,\pi-\sigma)$. Find $\Omega_p$ action on $\alpha_n^a$ using 
\begin{equation}\label{omegaxdot}
\Omega_p\,\dot{X}^a(\tau,\sigma)\,\Omega_p^{-1} = -\dot{X}^a(\tau,\pi-\sigma).
\end{equation}
Verify that for any arbitrary product $R$ of oscillators of both types,
$\Omega_p\,R\,\Omega_p^{-1} = (-1)^{N_\perp}R$, where $N_\perp$ is the total number of $R$.
\newline\newline
We use the method of the previous homework problem on orientifolds and D-branes to get that
\begin{equation}
\Omega_p\,\alpha_n^i\,\Omega_p^{-1} = (-1)^n\alpha_n^i,
\end{equation}
where the $(-1)^n$ comes from a factor of $\cos(n\pi)$. To obtain the action of $\Omega_p$ on the transverse
oscillators, we use the expansion for $\dot{X}^a$
\begin{equation}
\dot{X}^a = -i\sqrt{2\alpha^\prime}\sum_{n\in\mathbb{Z}}\alpha_n^a e^{-in\tau}\sin n\sigma,
\end{equation}
which turns out to be the same for a string stretched between two parallel D-branes as for one D-brane.
We now apply relation (\ref{omegaxdot}) to get
\begin{equation}
\Omega_p\,\alpha_n^a\,\Omega_p^{-1} = (-1)^n\alpha_n^a.
\end{equation}
Thus each $\alpha_n^{\left(i,a\right)}$ contributes a $(-1)^n$, so we see that for a product of
oscillators of both kinds $R$ with number $N^\perp$, the action of $\Omega_p$ is given by
$(-1)^{N_\perp}$.
\newline\newline
{\bf{e.}} Describe the orientifold spectrum in terms of the spectrum of a single D$p$-brane.
\newline\newline
We can build a general state from any product $R$ of oscillators in the following manner:
\begin{equation}
\ket{\Psi_R} = \sum_{ij}\left(r_{ij}R\ket{[ij]}+r_{i\bar{j}}R\ket{[i\bar{j}]}+r_{\bar{i}j}R\ket{[\bar{i}j]}
+r_{\bar{i}\bar{j}}R\ket{[\bar{i}\bar{j}]}\right).
\end{equation}
Acting on both sides with $\Omega_p$, we obtain
\begin{equation}
\Omega_p\ket{\Psi_R} = \sum_{ij}(-1)^{N_\perp}\left(r_{ij}R\ket{[\bar{j}\bar{i}]}+r_{i\bar{j}}R\ket{[j\bar{i}]}
+r_{\bar{i}j}R\ket{[\bar{j}i]}+r_{\bar{i}\bar{j}}R\ket{[ji]}\right).
\end{equation}
If we require that our theory be $\Omega_p$ invariant, then we obtain the conditions
\begin{eqnarray}
r_{ij} &=& (-1)^{N_\perp}r_{\bar{j}\bar{i}} \nonumber\\
r_{i\bar{j}} &=& (-1)^{N_\perp}r_{j\bar{i}} \nonumber\\
r_{\bar{i}j} &=& (-1)^{N_\perp}r_{\bar{j}i}.
\end{eqnarray}
The first of these relations states that the matrix $r_{\bar{i}\bar{j}}$ is completely determined once
we fix the $N^2$ parameters in $r_{ij}$ (one is the $\pm$ transpose of the other). The other two conditions
state that the $r_{i\bar{j}}$ and $r_{\bar{i}j}$ matrices are either both symmetric or both antisymmetric.
For the case of $N_\perp$ odd, we see that the antisymmetric condition holds. This tells us that there
are in total $N^2 + 2\times(N/2)(N-1) = 2N(N-1)$ independent parameters that can be used to build states.
When $N_\perp$ is even, the symmetric condition holds and we then have $N^2 + 2\times(N/2)(N+1)=2N(N+1)$
independent parameters. This was what we wanted to show.
\clearpage
\begin{hwproblem}{15.3}
Properties of the string charge $\vec{Q}$
\end{hwproblem}
\newline
{\bf{a.}} Consider a string at some fixed time $t_0$ and a region $\mathcal{R}$ of space that contains
a portion of this string: the string enters the region $\mathcal{R}$ at a point $\vec{x}_i$ and leaves
$\mathcal{R}$ at a point $\vec{x}_f$ (assume there is no compactification of space). Calculate the
string charge $\vec{Q}=\int_{\mathcal{R}}\dd^d x \vec{j}^0$ contained in $\mathcal{R}$ at time $t_0$.
Use your result to show that the total string charge associated with a closed string is zero.
\newline\newline
We know that $\vec{j}^0$ is given by
\begin{equation}
\vec{j}^0 = \frac{1}{2}\int\dd\sigma\;\delta\left(\vec{x}-\vec{X}(t,\sigma)\right)\frac{\del\vec{X}(t,\sigma)}{\del\sigma}.
\end{equation}
Thus, $\vec{Q}$ is given by
\begin{eqnarray}
\vec{Q} &=& \int_{\mathcal{R}}\dd^d x\; \frac{1}{2}\int\dd\sigma\; \delta\left(\vec{x}-\vec{X}(t_0,\sigma)\right)
\frac{\del\vec{X}(t_0,\sigma)}{\del\sigma} \nonumber\\
{}&=& \frac{1}{2}\int\dd\sigma\int_{\mathcal{R}}\dd^d x\;\delta\left(\vec{x}-\vec{X}(t_0,\sigma)\right)
\frac{\del\vec{X}(t_0,\sigma)}{\del\sigma} \nonumber\\
{}&=& \frac{1}{2}\int\dd\sigma\;\frac{\del\vec{X}(t_0,\sigma)}{\del\sigma}\left\{
\begin{array}{rr}
1, & \vec{X}(t_0,\sigma)\in\mathcal{R} \\
0, & \vec{X}(t_0,\sigma)\notin\mathcal{R}
\end{array}\right\} \nonumber\\
{}&=& \frac{1}{2}\left(\vec{X}(t_0,\vec{x}_f)-\vec{X}(t_0,\vec{x}_i)\right).
\end{eqnarray}
Thus we see that the charge associated with a closed string is $\vec{0}$ because $\vec{x}_i$ and $\vec{x}_f$ are
identical when the region encloses the entire string.
\newline\newline
{\bf{b.}} A more abstract proof that $\vec{Q}$ is zero for any {\em{localized}} configuration of closed strings
requires showing that $\nabla\cdot\vec{j}^0=0$ implies $\int\dd^d x\,\vec{j}^0=0$.
\newline\newline
First we note that the following identity holds for any two functions $\phi$ and $\psi$:
\begin{equation}
\nabla\cdot(\phi\psi\vec{j}^0) = \phi\nabla\psi\cdot\vec{j}^0 + \psi\nabla\phi\cdot\vec{j}^0+\phi\psi\nabla\cdot\vec{j}^0.
\end{equation}
Integrating $\nabla\cdot(\phi\psi\vec{j}^0)$ over all space and using the divergence theorem, we have
\begin{equation}
\int_{\mathrm{space}}\dd^d x\,\nabla\cdot(\phi\psi\vec{j}^0) = \oint_{\del(\mathrm{space})}\phi\psi\vec{j}^0\cdot\vec{\dd\sigma}
= 0,
\end{equation}
where we used the fact that $\vec{j}^0$ is localized in space. We can then use our previously-derived identity, dropping
off the term containing $\nabla\cdot\vec{j}^0$, which vanishes, and using $\phi=1$ and $\psi=x^i$:
\begin{equation}
\int\dd^d x\,\frac{\del x^i}{\del x^j}\,\hat{e}^j\cdot\vec{j}^0 = \int\dd^d x\,\delta^i_j\,j^{0,j} = \int\dd^d x\,j^{0,i}=0,
\end{equation}
which holds for all $i$, so we've proven what we wanted to prove.
\newline\newline
{\bf{c.}} Assume now that one space coordinate $x$ is curled up into a circle of radius $R$, and consider
a closed string wrapped around this circle. Calculate the string charge $\vec{Q}$. Explain why the answer is
not zero, and compare with the result obtained in (15.72).
\newline\newline
The string charge is given by
\begin{equation}
\frac{1}{2}\left(\vec{X}(t_0,\vec{x}_f)-\vec{X}(t_0,\vec{x}_i)\right).
\end{equation}
Now this evaluates to $\frac{1}{2}2\pi R m=m\pi R$, where $m$ is the number of times the string is wrapped around the
compactified dimension. This is nonzero because there can be two different string coordinates associated with identified
points in space if some of the dimensions are compactified.
\clearpage
\begin{hwproblem}{17.3}
Charge carried by winding strings
\end{hwproblem}
\newline
The zero mode expansion of a string with winding number $\ell$ is given by
\begin{equation}
X^m(\tau,\sigma) = x^m(\tau),\qquad X(\tau,\sigma)=\ell R\sigma,
\end{equation}
where the index $m$ is used to denote all string coordinates except for $X^{25}\equiv X$. Consider the coupling term
of the Kalb-Ramond field to the string:
\begin{equation}
S=-\int\dd\tau\dd\sigma\,\frac{\del X^\mu}{\del\tau}\frac{\del X^\nu}{\del\sigma}B_{\mu\nu}(X).
\end{equation}
Calculate the terms in $S$ that couple $B_{m,25}=-B_{25,m}$ to the string trajectory $x^a(\tau)$.
\newline\newline
The term which couples $B_{m,25}$ to the string trajectory is given by
\begin{equation}
\left(\del_\tau X^m\del_\sigma X-\del_\tau X \del_\sigma X^m\right)B_{m,25},
\end{equation}
which simplifies to
\begin{equation}
\ell R\frac{\dd x^m}{\dd\tau}B_{m,25}.
\end{equation}
Note that this is strikingly similar to the term which couples a Maxwell field to the trajectory of a particle,
\begin{equation}
S_{\mathrm{EM}} = \frac{e}{c}\int_{\mathcal{P}}\dd\tau\,A_\mu\left(x(\tau)\right)\frac{\dd x^\mu}{\dd \tau}(\tau).
\end{equation}
Making the identification $(e/c)A_\mu\leftrightarrow (\ell R) B_{m,25}$ (up to a sign), we see that the electric
charge carried by the string is proportional to its winding number $\ell$.
\end{document}
