Worldvolume Flux, BIons, and F/D Bound States

D-branes are not just rigid hypersurfaces on which open strings end. They are dynamical solitons with gauge fields living on their worldvolume, and those gauge fields carry spacetime charges. This is one of the most useful lessons of D-brane physics: a lower-dimensional brane can be hidden inside a higher-dimensional brane as worldvolume flux.

The previous pages treated D$p$-branes as sources for Ramond—Ramond potentials $C_{p+1}$ and as objects whose open strings give a $U(N)$ gauge theory. The bridge between these statements is the Wess—Zumino coupling

$$S_{\rm WZ}=\mu_p\int_{\mathcal W_{p+1}} \sum_q P[C_q]\wedge e^{\mathcal F}, \qquad \mathcal F=P[B]+2\pi\alpha' F.$$

Here $\mathcal W_{p+1}$ is the D$p$-brane worldvolume, $P[\cdots]$ denotes pullback to it, $F=dA-iA\wedge A$ is the Chan—Paton field strength, and

$$\mu_p={1\over (2\pi)^p(\alpha')^{(p+1)/2}}$$

is the unit Ramond—Ramond charge. In a constant dilaton background the physical tension is

$$T_p={\mu_p\over g_s}.$$

The exponential in $S_{\rm WZ}$ is not a decorative shorthand. It is the statement that a D$p$-brane can carry D$(p-2)$, D$(p-4)$, and still lower charges when its worldvolume gauge bundle is topologically nontrivial.

Magnetic worldvolume flux induces lower-dimensional Ramond—Ramond charge. The first Chern class carries D$(p-2)$ charge, while instanton number on a D4-brane carries D0 charge.

The Wess—Zumino coupling and dissolved branes

Expanding the exponential gives

$$S_{\rm WZ}=\mu_p\int_{\mathcal W_{p+1}}\left( C_{p+1}+C_{p-1}\wedge \mathcal F+{1\over 2}C_{p-3}\wedge \mathcal F\wedge\mathcal F+\cdots \right).$$

The leading term tells us that a D$p$-brane is electrically charged under $C_{p+1}$. The next term tells us something subtler: if the D$p$-brane has magnetic flux through a two-cycle $\Sigma_2\subset \mathcal W_{p+1}$, then it carries D$(p-2)$ charge.

Set $B=0$ for the moment. Flux quantization says

$${1\over 2\pi}\int_{\Sigma_2} F=m\in \mathbb Z.$$

Then

$$\mu_p\int_{\mathcal W_{p+1}} C_{p-1}\wedge 2\pi\alpha' F = m\,\mu_{p-2}\int_{\mathcal W_{p-1}} C_{p-1},$$

because

$${\mu_{p-2}\over \mu_p}=(2\pi)^2\alpha'.$$

Thus one unit of magnetic flux carries one unit of D$(p-2)$ charge. The lower-dimensional brane is dissolved: it is not localized at a point in the two-cycle unless the flux is localized, but the conserved Ramond—Ramond charge is exactly the same.

This idea is the D-brane version of a very old field-theory fact: topological flux can behave like particle number. The novelty is that in string theory the topological number is literally a spacetime brane charge.

D2-branes with magnetic flux: dissolved D0-branes

The cleanest example is a D2-brane extended along $x^1,x^2$ with constant magnetic field $F_{12}$. Let

$$f=2\pi\alpha' F_{12}, \qquad V_2=\int dx^1dx^2.$$

The DBI energy is

$$E=T_2 V_2\sqrt{1+f^2}.$$

Flux quantization gives

$$F_{12}V_2=2\pi m, \qquad f={ (2\pi)^2\alpha' m\over V_2}.$$

Using $T_0/T_2=(2\pi)^2\alpha'$, the energy becomes

$$E=\sqrt{(T_2V_2)^2+(mT_0)^2}.$$

This is exactly the BPS mass formula for a bound state carrying D2 charge and $m$ units of D0 charge. Notice the important word bound. The mass is not $T_2V_2+mT_0$; instead it is the length of a charge vector in the central-charge plane. The D0 charge is not a collection of independent D0-branes sitting on the D2. It has been absorbed into the D2 as magnetic flux.

The same configuration has a simple T-dual interpretation. T-dualize along $x^2$. The magnetic field becomes the slope of a D1-brane in the $(x^1,\widetilde x^2)$ plane:

$$\tan\theta=2\pi\alpha' F_{12}.$$

So a fluxed D2 is dual to a tilted D1. The square-root energy above is then just the geometric length of the tilted brane times its tension. This is often the fastest way to remember why the DBI square root knows about BPS charge addition.

Higher Chern classes: D0-branes inside D4-branes

The next term in the Wess—Zumino expansion is

$${\mu_p\over 2}\int C_{p-3}\wedge (2\pi\alpha'F)\wedge(2\pi\alpha'F).$$

For a D4-brane this contains

$${\mu_4\over 2}\int_{\mathcal W_5} C_1\wedge (2\pi\alpha'F)\wedge(2\pi\alpha'F).$$

Therefore a gauge instanton on the four spatial directions of the D4-brane carries D0 charge. With standard trace conventions,

$$k={1\over 8\pi^2}\int_{\mathbb R^4}{\rm Tr}\,F\wedge F\in\mathbb Z$$

corresponds to $k$ dissolved D0-branes.

This fact is a cornerstone of the D-brane/gauge-theory dictionary. The moduli space of $k$ D0-branes bound to $N$ D4-branes is the moduli space of $k$ instantons in $U(N)$ gauge theory. In the open-string description, the D0—D0 and D0—D4 strings reproduce the ADHM variables; in the D4 worldvolume theory, the same physics is a smooth instanton gauge field.

For a stack of branes the WZ coupling is more properly written schematically as

$$S_{\rm WZ}=\mu_p\int {\rm Tr}\left(P[C]\wedge e^{P[B]+2\pi\alpha'F}\right),$$

with refinements involving curvature couplings and the non-Abelian pullback. For the present purpose, the essential point is already visible in the Abelian formula: Chern characters of the worldvolume gauge bundle are D-brane charges.

Electric flux and fundamental-string charge

Magnetic flux dissolves lower-dimensional D-branes. Electric flux dissolves fundamental strings.

Consider a D$p$-brane with an electric field $F_{0i}$. The DBI Lagrangian depends on $F_{0i}$, so the canonical electric displacement

$$\Pi^i={\partial \mathcal L_{\rm DBI}\over \partial F_{0i}}$$

is the conserved flux conjugate to the gauge potential $A_i$. By Gauss’s law, electric flux lines cannot simply end in the middle of the brane. If they do end, the endpoint must be charged under the worldvolume $U(1)$ gauge field. But the endpoint of an open fundamental string is precisely such a charge.

Thus a fundamental string ending on a D-brane appears in the D-brane gauge theory as an electric charge, and a bundle of dissolved F-strings appears as quantized electric flux.

For a D1-brane, this statement becomes especially sharp. Let

$$E=2\pi\alpha' F_{01}.$$

For one D-string in flat space with $B=C_0=0$, the DBI Lagrangian density is

$$\mathcal L=-T_{D1}\sqrt{1-E^2}, \qquad T_{D1}={1\over 2\pi\alpha' g_s}.$$

The electric displacement is

$$D={\partial \mathcal L\over \partial E} ={T_{D1}E\over \sqrt{1-E^2}}.$$

The Hamiltonian density is

$$\mathcal H=DE-\mathcal L =\sqrt{T_{D1}^2+D^2}.$$

Quantization of electric flux sets

$$D=pT_F, \qquad T_F={1\over 2\pi\alpha'}, \qquad p\in\mathbb Z.$$

Therefore

$$\mathcal H=T_F\sqrt{p^2+{1\over g_s^2}}.$$

This is the tension of a bound state of $p$ fundamental strings and one D-string. For $q$ coincident D-strings carrying $p$ units of electric flux, the BPS tension is

$$T_{(p,q)}={1\over 2\pi\alpha'}\sqrt{p^2+{q^2\over g_s^2}} \qquad (C_0=0).$$

The notation $(p,q)$ means $p$ units of NS—NS string charge and $q$ units of Ramond—Ramond string charge. A fundamental string is $(1,0)$, a D-string is $(0,1)$, and the simplest genuinely mixed object is a $(1,1)$ string.

BIon spikes: a string ending on a brane

The electric-flux picture has a beautiful spacetime avatar: the BIon spike. Consider a D3-brane with one transverse scalar $X(\boldsymbol x)$ and an electric field $E_i$ on the brane. A fundamental string ending on the brane pulls out a spike in the transverse direction. The spike is not an extra object added by hand; it is a classical solution of the DBI equations.

A fundamental string ending on a D3-brane is described on the brane by electric flux and a transverse scalar profile. The BPS condition ties the electric field to the slope of the spike.

The BPS equation takes the schematic form

$$2\pi\alpha' E_i=\pm \partial_i X.$$

Away from the endpoint, Gauss’s law implies

$$\nabla\cdot \boldsymbol\Pi=0,$$

so the scalar is harmonic. For a spherically symmetric spike on a D3-brane,

$$X(r)=X_\infty+{c\over r},$$

with $c$ proportional to the number $N_F$ of attached fundamental strings. The endpoint at $r=0$ is a source for electric flux; geometrically it is the place where the spike becomes a semi-infinite F-string.

The energy of the BPS configuration splits as

$$E=T_{D3}\,({\rm brane\ volume})+N_FT_F\,({\rm string\ length}),$$

up to the usual regularization of the infinite brane volume and infinite string length. This is the physical reason for the BPS equation: it rewrites the DBI energy as a sum of brane tension plus string tension, with no additional binding energy.

The magnetic analogue is also important. A D1-brane ending on a D3-brane is described by a magnetic monopole on the D3 worldvolume and a scalar spike obeying a Bogomolny equation

$$\boldsymbol B=\pm \nabla X.$$

This is the D-brane origin of the relation between monopoles in four-dimensional gauge theory and D-strings suspended between D3-branes.

The charge lattice of F/D bound states

For strings in type IIB theory, F-string and D-string charges form an integral lattice. At $C_0=0$ the tension of a $(p,q)$ string is

$$T_{(p,q)}=T_F\sqrt{p^2+{q^2\over g_s^2}}.$$

F1 and D1 charges form an integral lattice. A primitive vector $(p,q)$ labels a single half-BPS bound string; nonprimitive vectors describe multiple coincident copies of a primitive string.

A few comments prevent common misunderstandings.

First, the formula is a BPS formula. The charges enter through a central charge, so the tension is the magnitude of a charge vector, not the sum of constituent tensions.

Second, a primitive charge vector, $\gcd(p,q)=1$, represents a single stable bound string. If $(p,q)=n(p_0,q_0)$ with $n>1$, the configuration is at threshold for splitting into $n$ identical $(p_0,q_0)$ strings.

Third, the D1 electric-flux derivation is only one corner of a larger structure. Type IIB theory has an $SL(2,\mathbb Z)$ duality that rotates F1 and D1 charges into each other and combines the axion and dilaton into

$$\tau=C_0+ie^{-\Phi}.$$

We will treat that duality systematically next. For now, the essential result is already visible from the D-brane worldvolume: turning on electric flux on a D-string literally builds a string with both NS—NS and R—R charge.

A useful organizing principle

The examples above are not isolated tricks. They are manifestations of one organizing principle:

$$\boxed{\text{worldvolume topology and flux encode spacetime brane charges}.}$$

More concretely:

Worldvolume data	Spacetime interpretation
$\displaystyle {1\over 2\pi}\int F=m$ on a D$p$	$m$ units of D$(p-2)$ charge
$\displaystyle {1\over 8\pi^2}\int {\rm Tr}\,F\wedge F=k$ on a D4	$k$ units of D0 charge
electric flux on a D$p$	dissolved F-string charge
electric point source on a D$p$	endpoint of an F-string
magnetic monopole on a D3	endpoint of a D-string
non-Abelian instanton moduli	D0-branes bound to D4-branes

This principle is one of the reasons D-branes are so powerful. They convert questions about extended objects in spacetime into questions about gauge fields, topology, and solitons on a lower-dimensional worldvolume.

Exercises

Exercise 1: one unit of flux gives one D$(p-2)$-brane

Let a D$p$-brane wrap a two-cycle $\Sigma_2$ with $B=0$ and

$${1\over 2\pi}\int_{\Sigma_2}F=m.$$

Use the Wess—Zumino coupling to show that the configuration carries $m$ units of D$(p-2)$ charge.

Solution

The relevant WZ term is

$$\mu_p\int C_{p-1}\wedge 2\pi\alpha' F.$$

Integrating over $\Sigma_2$ gives

$$\mu_p(2\pi\alpha')\int_{\Sigma_2}F =\mu_p(2\pi\alpha')(2\pi m) =m\mu_p(2\pi)^2\alpha'.$$

Since

$$\mu_{p-2}=\mu_p(2\pi)^2\alpha',$$

the coupling becomes

$$m\mu_{p-2}\int C_{p-1},$$

which is precisely the coupling of $m$ D$(p-2)$-branes.

Exercise 2: the D2—D0 square-root mass formula

A D2-brane of area $V_2$ has constant magnetic field $F_{12}$ with

$$F_{12}V_2=2\pi m.$$

Show that the DBI energy is

$$E=\sqrt{(T_2V_2)^2+(mT_0)^2}.$$

Solution

The DBI energy is

$$E=T_2V_2\sqrt{1+(2\pi\alpha'F_{12})^2}.$$

Flux quantization gives

$$2\pi\alpha'F_{12}={(2\pi)^2\alpha'm\over V_2}.$$

Therefore

$$E^2=(T_2V_2)^2+\bigl[T_2V_2(2\pi\alpha'F_{12})\bigr]^2 =(T_2V_2)^2+\bigl[T_2(2\pi)^2\alpha'm\bigr]^2.$$

Using $T_0/T_2=(2\pi)^2\alpha'$, we get

$$E^2=(T_2V_2)^2+(mT_0)^2.$$

Exercise 3: flux as brane angle under T-duality

For an open string ending on a D2-brane with constant $F_{12}$, the boundary condition can be written schematically as

$$\partial_\sigma X^1+2\pi\alpha'F_{12}\partial_\tau X^2=0, \qquad \partial_\sigma X^2-2\pi\alpha'F_{12}\partial_\tau X^1=0.$$

T-dualize along $X^2$. Show that the dual object is a D1-brane tilted by an angle satisfying

$$\tan\theta=2\pi\alpha'F_{12}.$$

Solution

Under T-duality along $X^2$,

$$\partial_\tau X^2\longleftrightarrow \partial_\sigma \widetilde X^2, \qquad \partial_\sigma X^2\longleftrightarrow \partial_\tau \widetilde X^2.$$

The first boundary condition becomes

$$\partial_\sigma X^1+2\pi\alpha'F_{12}\partial_\sigma \widetilde X^2=0,$$

so the endpoint is constrained to move along a line in the $(X^1,\widetilde X^2)$ plane. Equivalently, the transverse combination is fixed. Up to orientation convention, the brane equation is

$$\widetilde X^2=(2\pi\alpha'F_{12})X^1+{\rm constant}.$$

Hence its slope is

$$\tan\theta=2\pi\alpha'F_{12}.$$

Exercise 4: D0 charge from a D4 instanton

A D4-brane has Wess—Zumino coupling

$${\mu_4\over 2}\int C_1\wedge (2\pi\alpha'F)\wedge(2\pi\alpha'F).$$

Explain why an instanton number

$$k={1\over 8\pi^2}\int {\rm Tr}\,F\wedge F$$

corresponds to $k$ units of D0-brane charge, up to the standard trace normalization.

Solution

The $C_1$ coupling measures D0-brane charge. The coefficient of $\int C_1$ is proportional to

$${\mu_4\over 2}(2\pi\alpha')^2\int {\rm Tr}\,F\wedge F.$$

Using

$$\int {\rm Tr}\,F\wedge F=8\pi^2 k,$$

this becomes

$$\mu_4(2\pi\alpha')^2(4\pi^2)k =\mu_4(2\pi)^4(\alpha')^2 k.$$

Since

$${\mu_0\over \mu_4}=(2\pi)^4(\alpha')^2,$$

the coupling is $k\mu_0\int C_1$. This is exactly the coupling of $k$ D0-branes.

Exercise 5: electric flux on a D-string

Start from

$$\mathcal L=-T_{D1}\sqrt{1-E^2},$$

where $E=2\pi\alpha'F_{01}$. Compute the Hamiltonian density in terms of the electric displacement $D=\partial\mathcal L/\partial E$.

Solution

The displacement is

$$D={T_{D1}E\over \sqrt{1-E^2}}.$$

Solving for $E$ gives

$$E^2={D^2\over T_{D1}^2+D^2}.$$

The Hamiltonian density is

$$\mathcal H=DE-\mathcal L ={T_{D1}\over \sqrt{1-E^2}} =\sqrt{T_{D1}^2+D^2}.$$

If $D=pT_F$, then

$$\mathcal H=T_F\sqrt{p^2+{1\over g_s^2}}.$$

This is the tension of a $(p,1)$ string at $C_0=0$.

Exercise 6: primitive and nonprimitive charge vectors

Use the tension formula

$$T_{(p,q)}=T_F\sqrt{p^2+{q^2\over g_s^2}}$$

to show that $(np_0,nq_0)$ has exactly $n$ times the tension of $(p_0,q_0)$.

Solution

Substitute $(p,q)=(np_0,nq_0)$:

$$T_{(np_0,nq_0)}=T_F\sqrt{n^2p_0^2+{n^2q_0^2\over g_s^2}} =nT_F\sqrt{p_0^2+{q_0^2\over g_s^2}} =nT_{(p_0,q_0)}.$$

So a nonprimitive vector is at threshold for splitting into $n$ copies of the primitive string. This is why primitive vectors label single stable bound strings.

Exercise 7: the BIon BPS equation

For a D3-brane with one scalar $X$ and electric field $E_i$, explain why the BPS equation

$$2\pi\alpha' E_i=\partial_i X$$

implies that the scalar profile is harmonic away from the string endpoint.

Solution

Away from sources, Gauss’s law is

$$\partial_i\Pi^i=0.$$

For a BPS configuration the electric displacement is parallel to the electric field, and the electric field is proportional to $\nabla X$. Thus Gauss’s law reduces schematically to

$$\nabla\cdot \nabla X=0,$$

or

$$\nabla^2X=0$$

away from the endpoint. On a D3-brane with spherical symmetry in the three spatial worldvolume directions, the harmonic solution is

$$X(r)=X_\infty+{c\over r}.$$

The singularity at $r=0$ is the endpoint of the fundamental string.