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\hfill {\tt hep-th/0301220}
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{\Large\bf
Low energy dynamics of self-dual $A_1$ strings
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{\large\bf
Andreas Gustavsson
}\\
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{\it Institute of Theoretical  Physics,
Chalmers University of Technology, \\
S-412 96 G\"{o}teborg, Sweden}\\
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{\tt f93angu@fy.chalmers.se}
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\noindent{\bf Abstract:} 
We examine the interrelation between the (2,0) supersymmetric six dimensional effective action for the $A_1$ theory, and the corresponding low-energy theory for the collective coordinates associated to selfdual BPS strings. We argue that this low energy theory is a two-dimensional N = 4 supersymmetric sigma model.
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\section{Introduction}
We still do not have a microscopic understanding of the interacting $(2,0)$ supersymmetric quantum theories in six dimensions, although many years now have passed since they where first discovered by Witten in 1995 \cite{Witten}. But whatever the microscopic theory is, we may always consider the low-energy effective action obtained by integrating out all massive degrees of freedom, provided that we have given a non-zero vacuum expectation value to say the fifth scalar field in the $(2,0)$-supermultiplet, so that the selfdual strings and all other subtle degrees of freedom become massive. We will in this letter only consider the $A_1$ effective theory in which we have just one $(2,0)$-supermultiplet, which on-shell consists of a selfdual two-form gauge potential $B^+$ with selfdual field strength $H^+ = dB^+$, five scalars $\phi^a$, and a symplectic Majorana spinor $\psi$ (we do not know the off-shell multiplet if any such exists).

The low energy dynamics of supersymmetric magnetic monopoles in superYang-Mills theories has been examined in \cite{Gauntlett} for $N=2$ SYM and in \cite{Blum} for $N=4$ SYM. A natural next step would therefore be to examine the low energy dynamics of supersymmetric selfdual strings in the $(2,0)$ supersymmetric theories in six dimensions. We will however not give a complete derivation of their low energy dynamics. This is so because we have not determined the full structure of the six-dimensional $(2,0)$-supersymmetric effective action yet.

\section{Effective action and collective coordinates}
We will start out from the following terms in the effective action for our $A_1$ theory in a flat Minkowski spacetime,
\bea
&&\frac{1}{2\pi}\int \(-\frac{1}{2}H \wedge *H - d\phi^a \wedge *d\phi^a + H \wedge A(\phi) \right.\cr
&&\left. + i\psi^T c\Omega \gamma^{M}\partial_M \psi + \frac{2\pi}{6V_4}i\psi^T c\Omega \epsilon_{abcde}\frac{1}{(\phi^a \phi_a)^{\frac{5}{2}}}\phi^a\partial_M \phi^c\partial_N \phi^d\partial_P \phi^d \gamma^{MNP} \sigma^e \psi + ... \). \label{toy}
\eea
Here $H = dB + A(\phi)$ and $A(\phi)$ is the pullback of the M theory three form to the five brane. $V_4 = \frac{8\pi^2}{3}$. We will choose the same gauge for $A(\phi)$ as in \cite{Gustavsson}. In particular $A_{0MN} = 0$. \footnote{This action should be invariant under $A \rightarrow A + d\Lambda$, $B\rightarrow B - \Lambda$. In order to achieve this gauge invariance we must add the term $-\tilde{A}$ where $d\tilde{A} = *dA - A \wedge dA$ as explained in \cite{Aharony}. This term should thus be among the `$+...$'.} As Lorentz indices we use $M = 0,...,5$. By $+...$ we mean terms needed to make the action $N=(2,0)$ supersymmetric, in the sense that the equations of motion derived from it are supersymmetric. (Of course $H$ is not in the supermultiplet, only its selfdual part is.) This action is also invariant under the Lorentz group $SO(1,5)$ and under a global R-symmetry group $SO(5)_R$. \footnote{To make the $SO(5)_R$ symmetry manifest we must make an integration by parts to get $B\wedge F(\phi)$.} As $SO(5)$ vector indices we use $a,b,... = 1,...,5$ and as $SO(5)$ Dirac-spinor indices we use $i,j,...$. As $SO(1,5)$ Weyl spinor indices we use $\alpha, \beta,....$. The spinors transform in $(4,4)$ of $SO(1,5) \times SO(5)_R$ and are constrained by a symplectic Majorana condition
\be
\bar{\psi} = \psi^T c \Omega
\ee
where $\bar{\psi} := \psi^{\dag} \gamma^0$ and where $c$ and $\Omega$ are charge conjugation matrices of $SO(1,5)$ and $SO(5)$ respectively. The gamma matrices are denoted by $\gamma^M$ and $\sigma^a$ for these two groups respectively. We choose the conventions so that ${\gamma^{M}}^T = -c\gamma^M c^{-1}$ and ${\sigma_a}^T = \Omega \sigma_a \Omega^{-1}$. These objects we will be born with the index structures $\Omega_{ij}$, $c^{\alpha \beta}$, $\psi_{\alpha}^i$, $(\gamma^M)_{\alpha}{}^{\beta}$, $(\sigma_a)^i{}_j$.

The moduli spaces we will consider are those of $k=1,2,...$ parallel BPS saturated selfdual strings. These moduli spaces should be given by ${\cal{M}}_k \times {\bf{R}}^4$ where the center of mass moduli space is ${\bf{R}}^4$ and corresponds to the four transverse directions to the (center of mass of the) strings. It seems very plausible that the dimension of the moduli space is $4 k$ (corresponding, in the classical picture of widely separated strings, to the four transverse coordinates of each string). But to prove this one has to count e.g the number of fermionic zero modes in a configuration with total charge $2\pi k$.
Parallel BPS-saturated strings are necessarily straight. To describe their motion tangent the moduli space we thus need just one parameter, which we will take to be the time-coordinate. But the generic string configuration is not a BPS configuration, and to describe such a configuration we also need a parameter running along the strings, which we take to be $x^5$. The tangent space to a BPS string configuration should thus be a two dimensional plane in the configuration space of all physically permitted configurations. This should be the natural generalization of \cite{Manton}. If we let $X^i$ ($i = 1,...,\dim {\cal{M}}_k$) denote the bosonic moduli parameters on ${\cal{M}}_k$ and assume that the BPS strings are aligned in the $x^5$ direction, then we make the following ansatz,
\bea
\phi^a(x^0,...,x^5) & = & \phi^a(x^1,...,x^4,X^i(x^0,x^5))\cr
B^+_{MN}(x^0,...,x^5) & = & B^+_{MN}(x^1,...,x^4,X^i(x^0,x^5))
\eea
for the bosonic fields. When we let the moduli parameters depend on $x^0, x^5$, they will be called collective coordinates. We will use the indices $\mu, \nu, ... = 0,5$ for the string world-sheets, and $I = 1,2,3,4$ for the transverse coordinates.

For a single string, $k=1$, the fermionic zero modes corresponds to broken supersymmetries $u_-$ which are constant spinors such that $\gamma^{05}\sigma_5 u_{\pm} = \pm u_{\pm}$. The zero modes are given by
\bea
\psi_0 = \gamma^{I} \partial_I \phi^5 u_-.
\eea 
That these really are zero modes is justified in the appendix. We could promote the moduli $u_-$ to fermionic collective coordinates $u_-$ just by letting them depend on $x^0,x^5$ and expand the fermion field as 
\be
\psi = \gamma^I \partial_I \phi^5(x^I,X^I(x^0,x^5)) u_-(x^0,x^5).
\ee
However we want to label the fermionic and the bosonic collective coordinates with the same label. That is, for $k = 1$ we want to convert $u_-$ into $\lambda^I$. $u_-$ transforms in the representation $(-\frac{1}{2},2,2') \oplus (+\frac{1}{2},2',2)$ of $SO(1,1) \times SO(4) \times SO(4)_R$. We may find one such $u_-$ such that $\lambda^I : = \gamma^I u_-$ constitute four linearly independent spinors. These will be four two-component Majorana spinors under $SO(1,1)$. But we may choose other representations for the $SO(4)$ gamma matrices. If we transform $\gamma^I \rightarrow \gamma^J J_{J}{}^I$ then the Clifford algerbra is preserved if and only if $J_{I}{}^{I'} J_{J}{}^{J'} \delta_{I'J'} = \delta_{IJ}$. Demanding $J$ to have real entries (which is needed in order to preserve the hermiticity property of the gamma matrices) we find that $J$ is a unit quaternion. We thus have the freedom to make the redefinition $\lambda^I \rightarrow \lambda^J J_J{}^I$ of the collective coordinates. 

For a general $k$ we make the following ansatz for the collective coordinates,
\be
\psi(x^M) = \partial_i \phi^5(x^I,X^i(x^0,x^5)) \lambda^i(x^0,x^5)
\ee
or with $\lambda^i$ replaced by ${\cal{J}}_j{}^i \lambda^j$ where ${\cal{J}}_i{}^j$ is the quarternion induced by $J_I{}^J$ and is given by ${\cal{J}}_i{}^j = G^{jk}\int d^4 x \partial_i B_{I5}\partial_k B^{J5}J_J{}^I$ where $G_{ij}$ is the metric on the moduli space (see (\ref{metric})). This is showed by using completeness and orthonormality relations for the massless zero modes together with the infinitely many massive modes.

\section{Low energy dynamics}
We get the action for the collective coordinates by expanding the fields about the BPS configuration, inserting this expansion into the 6d Wilsonian low-energy effective action (\ref{toy}) and integrating over the transverse coordinates. As explained in \cite{Harvey} this is an expansion in inverse length or in $n = n_{\partial_0} + \frac{n_f}{2}$ where $n_{\partial_0}$ is the number of time derivatives and $n_f$ the number of fermions, if we let $X^i$ have mass dimension zero and $\lambda^i$ mass dimension $\frac{1}{2}$. To zeroth order in $n$ we have the BPS solution in \cite{Gustavsson} for widely separated strings. Actually we will not need the exact form of the Higgs profile for our considerations (which we actually do not know since we do not know the exact form of the effective action). We will just need the the results that $\phi^A = 0$ ($A=1,...4$)and that we for the four transverse components have $H = *_4 d\phi^5$. We also need that $\int_{S^3}H = \int_{S^3}A$ for any sphere in the transverse space (which follows from $dH =dA$ upon using the Stokes' theorem and the facts that both $H$ and $A$ (as we have chosen it) are globally defined all over the six-dimensional spacetime. 

When inserting the $n=0$ BPS solution, $H$ becomes self-dual and hence $H\wedge *H = 0$. In our gauge we get $\int H \wedge A = \int_{R^2 \times R_+} H \int_{S^3} A = \int_{R^2 \times R_+} *H \int_{S^3} H = \int d\phi^5 \wedge dx^0\wedge dx^5 \int_{S^3} H = \int dx^0 dx^5 \int d\phi^5 \wedge H = \int dx^0 dx^5 QD$ where $Q = 2\pi k$ and $D=\phi^5(\infty)$. Similarly $\int_{R^4} d\phi^5 \wedge *_4 d\phi^5 = \int d\phi^5 \wedge H$. The total tension of the strings is $2QD$ with our conventions. The fermionic interaction term involving two fermions becomes zero due to a factor $\gamma_I\int d^4 x x^I = 0$. Using these results when inserting the collective coordinate expansions into (\ref{toy}), we get the action
\be
\frac{1}{2\pi}\int dx^0 dx^5 \(G_{ij}\(\{X^k\}\) \(\eta^{\mu\nu}\partial_{\mu} X^i \partial_{\nu} X^j + i{\lambda^i}^T c\Omega \gamma^{\mu} D_{\mu} \lambda^j\) - 4\pi k D + ...\). \label{low}
\ee
Here
\be
G_{ij} = \int d^4 x \partial_i \phi^5 \partial_j \phi^5\label{metric}
\ee 
is the metric on the moduli space and
\be
D_{\mu} \lambda^i = \partial_{\mu} \lambda^i + \Gamma^i_{kl}\partial_{\mu} X^k \lambda^l
\ee
where
\be
\Gamma^i_{jk} = G^{il}\int d^4 x (\partial_j \partial_k \phi^5) \partial_l \phi^5.
\ee
It is easily seen that $\Gamma^i_{jk}$ is the Christoffel symbol associated with the metric $G_{ij}$, so $D_{\mu}$ is a covariant derivative. The terms in $+...$ in (\ref{low}) arise from those in $+...$ in (\ref{toy}).

The unbroken supersymmetry $\epsilon_+$ relates the bosonic and fermionic collective coordinates. The bosonic zero modes may be obtained by an unbroken supersymmetry variation as
\bea
\delta \phi^a & = & \epsilon^T_+ c\Omega \sigma^a \psi_0\cr
& = & \epsilon^T_+ c\Omega \sigma^a \lambda^i \partial_i \phi^5.
\eea
Expanding $\delta \phi^a = (\delta X^i) \partial_i \phi^a$, we see that in a BPS configuration where $\partial_i \phi^A = 0$ the right hand side must also vanish. Indeed it does identically (which can be seen by inserting $\gamma^{05}\sigma^5$ and first let it act on $\lambda^i_-$ and then on $\epsilon_+$). Taking $a=5$ we see that the bosonic collective coordiniates $X^i$ are related to the fermionic collective coordinates $\lambda^i$ as
\be
\delta_{\epsilon} X^i = \epsilon^T_+ c\Omega \lambda^i.
\ee
Similarly, by inserting the expansions
\bea
\phi & = & \phi(X=const) + \delta X^i \partial_i \phi(X=const) + Ordo((\delta X)^2)\cr
B & = & B(X=const) + \delta X^i \partial_i B(X=const) + Ordo((\delta X)^2)
\eea
into the (unbroken) supersymmetry variation
\be
\delta_{\epsilon} \psi  =  -i\(\frac{1}{24}\gamma^{MNP}H^+_{MNP} + \frac{1}{2}\gamma^{M}\sigma_a \partial_{M}\phi^a\)\epsilon_+
\ee
and using that in the BPS configuration (that is for the fields at $X=const$) this variation vanishes, we get only a contribution from term involving derivatives of $\delta X^i$. To first order this is
\be
\delta_{\epsilon} \psi = -i\gamma^{\mu}\sigma_a (\partial_{\mu} X^i) \partial_i \phi^a \epsilon_+.
\ee
Noting that $\partial_i \phi^A = 0$, we get
\be
\delta_{\epsilon}(\partial_i \phi^5 \lambda^i) = -i\sigma_5 \gamma^{\mu} \partial_{\mu} X^i \partial_i \phi^5 \epsilon_+.
\ee
Multiplying by $\partial_j \phi^5$ and integrating over the transverse coordinates $x^i$ we get
\be
\delta_{\epsilon} \lambda^i = -i\gamma^{\mu} \partial_{\mu}X^i \epsilon_+ - \Gamma^i_{jk} \delta X^k \lambda^j.
\ee
The 2d spinors are two-component Majorana spinors. These are the supersymmetries of a non-linear sigma model. Notice that $\epsilon$ is a two-component Majorana spinor under $SO(1,1)$. The $SO(4)\times SO(4)_R$ spinor indices on $\epsilon$ and $\lambda^i$ are always trivially contracted and we can forget about them from the two-dimensional point of view. 

We have four supersymmetries obtained by substituting ${\cal{J}}_j{}^i \lambda^j$ for $\lambda^i$. So the 2d action should be the $N=4$ supersymmetric sigma model. The action we obtained is only part of the $N=4$ sypersymmetric sigma model action. As shown in \cite{Alvarez-Gaume} the quaternions ${\cal{J}}_i{}^j$ comprise the three complex structures of a hyper-Kahler manifold. So we conclude that the moduli space ${\cal{M}}_k$ is a hyper-Kahler manifold. 

We should not be surprised that we did not get all the terms up to order $n = 2$ of the $N = 4$ sigma model because the 6d action we started with was missing terms too, and was therefore not completely (2,0) supersymmetric. We miss one term at order $n = 2$, namely the term proportional to \cite{Alvarez-Gaume}
\bea
R_{ijkl}\bar{\lambda}^i \lambda^j \bar{\lambda}^k \lambda^l & = & 2\(\frac{\partial^2 G_{ik}}{\partial X^j \partial X^l}
+ G_{mn}\Gamma^m_{ik}\Gamma^n_{jl}\) \bar{\lambda}^i \lambda^j \bar{\lambda}^k \lambda^l\cr
& = & 2\(\int \partial_i \partial_l \phi^5 \partial_j \partial_k \phi^5 
+ G^{pq} \int d^4 x \partial_p \phi \partial_i \partial_k \phi^5  \int d^4 y \partial_q \phi^5 \partial_j \partial_l \phi^5 \) \bar{\lambda}^i \lambda^j \bar{\lambda}^k \lambda^l\cr
& = & 2\(\int \partial_i \partial_l \phi^5 \partial_j \partial_k \phi^5 
 + \int \partial_i \partial_k \phi^5 \partial_j \partial_l \phi^5\)\bar{\lambda}^i \lambda^j \bar{\lambda}^k \lambda^l \label{terms}
\eea
where in the last step we have used the completeness and orthonormality conditions of the modes to rewrite the second term.

These terms obviously must come from terms in the 6d action which involve four fermions. The mass dimension the 6d Lagrangian is $6$, the dimension of $\psi$ is $\frac{5}{2}$ and the dimension of $\phi$ is $2$. The only Lorentz invariant combination which involves only scalars and fermions and with the correct dimension and which does not involve derivatives seems to be $\frac{1}{\phi_a \phi^a} \bar{\psi} \psi \bar{\psi} \psi$. It is easily seen that by an integration by parts this terms gives rise to the terms in (\ref{terms}), but we then have to cancel the other unwanted terms by something else. If we allow derivative terms we get infinitely many possibilities, and presumably one can find a combination of such terms that exactly gives the first term in (\ref{terms}). The lifting procedure to get a 6d action that reduces to a 2d action is not unique. For instance the quadratic fermion interaction term could not be seen in the 2d action as it vanished when we inserted the $n = 0$ solution into it. But maybe these additional terms become uniquely determined if we also require the 6d action to be (2,0) supersymmetric.




\newpage
\appendix

\section{Fermionic zero modes}
We notice that $\int_{S^3}H^+ = \int_{S^3}A$ (for our particular gauge, see section $3$ and \cite{Gustavsson}) for any three-sphere in the transverse space, implies that $H^+ = A$ when viewed as three forms in the transverse space. (We do not get that $H = A$ in the six dimensional spacetime.) We may take a supersymmetry variation of both sides, $\delta_{\epsilon} H^+ = \delta_{\epsilon} A$ while rememebering that this equation is only $SO(4)$ covariant, not $SO(1,5)$ covariant. But we can easily find an $SO(1,5)$ covariant equation that reduces to it. This is how we obtained the interaction term which involves two fermions in the action (\ref{toy}). The corresponding $SO(4)\times SO(5)_R$ covariant equation reads
\be
\gamma^{I}\partial_{I} \psi = -\frac{2\pi}{6V_4}\frac{1}{r^5}\epsilon_{abcde}\phi^a \partial_I \phi^b \partial_J \phi^c \partial_K \phi^d \sigma^e \gamma^{05} \psi
\ee
where will let $\rho = \sqrt{\phi^A \phi_A}$ ($A=1,...,4$) and $r=\sqrt{\rho^2 + (\phi^5)^2}$. 

For $k = 1$ the map $x^A \mapsto \phi^A$ has winding number one, and $\phi^A = \rho(R) \frac{x^A}{R}$. We can then write the equation of motion as
\be
\gamma^{I}\partial_{I} \psi = \frac{2\pi}{V_4}\(\frac{\rho^4}{r^5}\frac{x^I}{R^4}\gamma_I\gamma^{05} \sigma^5 + ...\)\psi
\ee
Here $+...$ are terms proportional to $\gamma_A\gamma^{05}\sigma^A \psi$ and to $x_A\gamma^{05}\sigma^A \psi$.

Inserting that $\psi_{\pm} = \gamma^{I}\partial_I \phi^5 u_{\pm}$ where the BPS solution (\cite{Gustavsson}) is given by $\phi^5 = (D - \frac{|Q|}{2V_3 R^2})\theta(R - R_0)$ with $Q = 2\pi$ and $R_0 = \sqrt{|Q|/(2V_3 D)}$, we get the l.h.s.
\be
\gamma^{J} \partial_{J}(\gamma^I \partial_I \phi^5 u_{\pm}) = (\partial^I \partial_I \phi^5) u_{\pm} = \frac{|Q|}{V_3 {R_0}^3}\delta(R - R_0)u_{\pm}
\ee
To compute the r.h.s. we need the following results,
\bea
\lim_{\rho \rightarrow 0}\frac{\rho^4}{r^5} & = & \frac{4}{3}\delta(\phi^5)\cr
\delta(\phi^5) & = & \frac{\delta(R - R_0)}{\frac{d\phi^5}{dR}(R_0)} = \frac{V_3{R_0}^3}{|Q|} \delta(R-R_0)
\eea
The terms $+...$ should combine into something that is proportional to $\delta'(\phi^5)\phi^5 + \delta(\phi^5) = \(\delta(\phi^5)\phi^5\)'$  (with $'$ denoting $d/d\phi^5$) and thus vanishes identically. We then get the r.h.s.
\bea
- \frac{2\pi}{V_4}\frac{\rho^4}{r^5}\frac{x^I}{R} \gamma_I \gamma^{05} \sigma^5 \psi_{\pm} & = & \mp \frac{2\pi}{V_4}\frac{4}{3}\delta(\phi^5)\frac{x^I}{R^4}\gamma_I \psi_{\pm}\cr
& = & \mp \frac{|Q|}{V_3 {R_0}^3}\delta(R - R_0) u_{\pm}
\eea
The r.h.s. is then equal to the l.h.s. only for the choice $u_-$. These are thus fermionic moduli for $k=1$.

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\newpage
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