M-Theory Origin of Type IIA Branes

Type IIA string theory has a remarkable strong-coupling limit. It does not merely become a more strongly interacting ten-dimensional string theory. Instead, an eleventh spacetime dimension opens up. The type IIA coupling $g_s$ is reinterpreted as the radius of a compact circle, and the nonperturbative branes of type IIA become ordinary geometric or membrane/fivebrane objects in eleven dimensions.

This is one of the cleanest places where D-branes stop looking mysterious. Their unusual $1/g_s$ tensions, Ramond—Ramond charges, and duality properties are not arbitrary features added to perturbative string theory. They are consequences of reducing eleven-dimensional supergravity and its two basic branes on a circle.

We use

$$\ell_s=\sqrt{\alpha'}, \qquad x^{11}\sim x^{11}+2\pi R_{11},$$

and denote the eleven-dimensional Planck length by $\ell_{11}$. The central relations are

$$\boxed{R_{11}=g_s\ell_s}, \qquad \boxed{\ell_{11}^3=g_s\ell_s^3}.$$

Thus weakly coupled type IIA, $g_s\ll1$, is M-theory on a very small circle. Strongly coupled type IIA, $g_s\gg1$, is eleven-dimensional M-theory on a large circle.

The elementary eleven-dimensional ingredients are the metric, the three-form $C_3^{(11)}$, the M2-brane, the M5-brane, gravitational waves, and KK monopoles. Compactifying on $S^1_{11}$ produces the perturbative and nonperturbative branes of type IIA string theory.

The D0 tower forces an eleventh dimension

The quickest route to the M-theory circle is the D0-brane. The type IIA D$p$-brane tension is

$$T_{Dp}={1\over (2\pi)^p g_s\ell_s^{p+1}}.$$

For $p=0$ this gives a particle of mass

$$M_{D0}=T_{D0}={1\over g_s\ell_s}.$$

A bound state of $n$ D0-branes has BPS mass

$$M_n={n\over g_s\ell_s}.$$

This looks exactly like a Kaluza—Klein momentum spectrum on a circle,

$$M_n={n\over R_{11}},$$

provided

$$R_{11}=g_s\ell_s.$$

This is a sharp argument, not just a mnemonic. The D0-branes are BPS, so their masses are protected. As $g_s$ increases, the whole tower of D0 bound states becomes lighter. A tower of particles with masses $n/R$ is the characteristic signature of a compact extra dimension of radius $R$. The strong-coupling limit of type IIA therefore reveals an eleventh dimension.

The D0-brane tower has the spectrum of KK momentum along the M-theory circle. As $g_s$ grows, $R_{11}=g_s\ell_s$ grows and the eleventh dimension becomes macroscopic.

The second relation, $\ell_{11}^3=g_s\ell_s^3$, follows from comparing Newton constants. In ten-dimensional string frame,

$$2\kappa_{10}^2=(2\pi)^7g_s^2\ell_s^8,$$

while in eleven dimensions

$$2\kappa_{11}^2=(2\pi)^8\ell_{11}^9.$$

Compactifying eleven-dimensional gravity on a circle gives

$$2\kappa_{11}^2=(2\pi R_{11})\,2\kappa_{10}^2.$$

Substituting $R_{11}=g_s\ell_s$ yields

$$\ell_{11}^9=g_s^3\ell_s^9, \qquad \ell_{11}=g_s^{1/3}\ell_s.$$

So the string length and the eleven-dimensional Planck length are not the same physical scale except at $g_s\sim1$.

Eleven-dimensional supergravity and its reduction

The low-energy limit of M-theory is eleven-dimensional supergravity. Its bosonic fields are just

$$G_{MN}, \qquad C_3^{(11)}, \qquad F_4=dC_3^{(11)}.$$

There is no dilaton in eleven dimensions. The bosonic action is

$$S_{11}={1\over 2\kappa_{11}^2}\int d^{11}x\sqrt{-G} \left(R-{1\over 2\cdot 4!}F_{MNPQ}F^{MNPQ}\right) -{1\over 12\kappa_{11}^2}\int C_3^{(11)}\wedge F_4\wedge F_4.$$

The absence of a dilaton is conceptually important. The type IIA coupling is not a local scalar that existed independently in eleven dimensions. It is the asymptotic size of the compact circle.

Let $x^{11}$ be the circle coordinate. A convenient string-frame reduction ansatz is

$$\boxed{ ds_{11}^2=e^{-2\varphi/3}ds_{10,\mathrm{str}}^2 +e^{4\varphi/3}\left(dx^{11}+C_1\right)^2 }$$

where

$$\varphi=\Phi-\Phi_\infty, \qquad \Phi_\infty=\log g_s.$$

This convention normalizes the circle to have physical radius $R_{11}$ at infinity. The ten-dimensional fields appearing here are the string-frame metric $g_{\mu\nu}$, the dilaton $\Phi$, and the R—R one-form $C_1$. Thus the R—R one-form of type IIA is geometrical: it is the Kaluza—Klein gauge field associated with translations around the M-theory circle.

The eleven-dimensional three-form decomposes as

$$\boxed{ C_3^{(11)}=C_3+B_2\wedge dx^{11} }$$

up to the harmless normalization of the circle coordinate. Therefore the type IIA R—R three-form $C_3$ and the NS—NS two-form $B_2$ have a common eleven-dimensional origin.

This already explains a lot:

$$\begin{array}{ccl} \text{KK momentum along }S^1_{11} &\longleftrightarrow& \text{D0-brane},\\ \text{KK magnetic monopole of }S^1_{11} &\longleftrightarrow& \text{D6-brane},\\ \text{M2 coupling to }C_3^{(11)} &\longleftrightarrow& \text{F1 or D2},\\ \text{M5 magnetic under }C_3^{(11)} &\longleftrightarrow& \text{NS5 or D4}. \end{array}$$

The full brane dictionary is obtained by asking whether the M-brane wraps the compact circle.

M2-branes: the common parent of F1 and D2

The M2-brane is electrically charged under $C_3^{(11)}$. Its tension is

$$\boxed{ T_{M2}={1\over (2\pi)^2\ell_{11}^3}. }$$

The corresponding BPS supergravity solution is

$$ds_{11}^2=H_2^{-2/3}dx_{0,1,2}^2+H_2^{1/3}dx_\perp^2,$$

with

$$F_4=d(H_2^{-1})\wedge dx^0\wedge dx^1\wedge dx^2, \qquad H_2(r)=1+{R_2^6\over r^6}.$$

The transverse space is eight-dimensional, so the harmonic function falls as $r^{-6}$. The near-horizon geometry of many coincident M2-branes is $AdS_4\times S^7$, but for the present purpose the important fact is simpler: the M2-brane has two possible reductions to ten dimensions.

M2 not wrapped on the circle gives a D2-brane

If the M2-brane lies entirely in the ten noncompact directions and the circle is transverse to it, it becomes a D2-brane. This is natural from the coupling. The unwrapped M2 couples to the component of $C_3^{(11)}$ with no leg along $x^{11}$, namely the type IIA R—R potential $C_3$. A D2-brane is precisely the object electrically charged under $C_3$.

The tension check is exact:

$$T_{M2}={1\over (2\pi)^2\ell_{11}^3} ={1\over (2\pi)^2 g_s\ell_s^3} =T_{D2}.$$

The $1/g_s$ in the D2-brane tension is therefore not mysterious. It is the statement that the eleven-dimensional Planck length is related to the string length by $\ell_{11}^3=g_s\ell_s^3$.

M2 wrapped on the circle gives the fundamental string

If one spatial direction of the M2-brane wraps $S^1_{11}$, the remaining one spatial direction is a string in ten dimensions. Its coupling comes from the component of $C_3^{(11)}$ with one leg along the circle,

$$C_3^{(11)}\supset B_2\wedge dx^{11}.$$

Thus the wrapped M2 couples electrically to $B_2$, so it is the fundamental type IIA string.

The tension is the wrapped membrane tension:

$$T_{F1}=(2\pi R_{11})T_{M2} =(2\pi g_s\ell_s){1\over (2\pi)^2g_s\ell_s^3} ={1\over 2\pi\ell_s^2}.$$

This is exactly the fundamental string tension. The perturbative string of type IIA is literally an M2-brane wrapped around the eleventh dimension.

M5-branes: the common parent of D4 and NS5

The M5-brane is magnetically charged under $C_3^{(11)}$. Equivalently, it couples electrically to the dual six-form potential $C_6^{(11)}$, with

$$dC_6^{(11)}=*_{11}F_4+\text{Chern--Simons terms}.$$

Its tension is

$$\boxed{ T_{M5}={1\over (2\pi)^5\ell_{11}^6}. }$$

The corresponding BPS solution is

$$ds_{11}^2=H_5^{-1/3}dx_{0,\ldots,5}^2+H_5^{2/3}dx_\perp^2,$$

where the transverse space is five-dimensional and

$$F_4=*_{\mathbb R^5}dH_5, \qquad H_5(r)=1+{R_5^3\over r^3}.$$

The M5-brane has a chiral two-form gauge field on its worldvolume with self-dual three-form field strength. This is the seed of several deep facts about D4-branes, five-dimensional super-Yang—Mills theory, and the six-dimensional $(2,0)$ theory.

M5 wrapped on the circle gives a D4-brane

If the M5-brane wraps $S^1_{11}$, the remaining object has four spatial dimensions in type IIA. It is a D4-brane. Its tension is

$$(2\pi R_{11})T_{M5} =(2\pi g_s\ell_s){1\over (2\pi)^5g_s^2\ell_s^6} ={1\over (2\pi)^4g_s\ell_s^5} =T_{D4}.$$

The D4-brane is electrically charged under the type IIA R—R five-form potential $C_5$, or magnetically charged under $C_3$. This is exactly what one obtains by reducing the M5-brane coupling.

This relation is especially important dynamically. The worldvolume theory on $N$ D4-branes is five-dimensional maximally supersymmetric Yang—Mills theory. It is nonrenormalizable by power counting, but its ultraviolet completion is the six-dimensional $(2,0)$ theory on $N$ M5-branes compactified on the M-theory circle.

M5 not wrapped on the circle gives an NS5-brane

If the M5-brane does not wrap the circle, it remains a fivebrane in ten dimensions. It is the type IIA NS5-brane. Its tension is

$$T_{M5}={1\over (2\pi)^5\ell_{11}^6} ={1\over (2\pi)^5g_s^2\ell_s^6} =T_{NS5}.$$

The $1/g_s^2$ scaling is the hallmark of a solitonic NS-sector object. The NS5-brane is magnetic under the NS—NS two-form $B_2$, just as the unwrapped M5-brane is magnetic under the component of $C_3^{(11)}$ with one leg on the M-theory circle.

This unifies two objects that look very different in perturbative type IIA: the D4-brane with tension $1/g_s$ and the NS5-brane with tension $1/g_s^2$ are simply the wrapped and unwrapped reductions of the same M5-brane.

D0 and D6 are geometry

The M2/M5 dictionary accounts for F1, D2, D4, and NS5. The D0 and D6 branes are different: they come from the eleven-dimensional metric itself.

D0-branes are KK momentum

A ten-dimensional D0-brane is a particle. In eleven dimensions it is a graviton carrying momentum $p_{11}$ around the circle. Momentum quantization gives

$$p_{11}={n\over R_{11}}.$$

Thus the $n$-D0 BPS bound state is the $n$th KK momentum mode. This explains why D0 charge is carried by the R—R one-form $C_1$: the field $C_1$ is the Kaluza—Klein gauge field from the metric component $G_{\mu,11}$.

The ten-dimensional coupling is

$$S_{D0}\supset \mu_0\int C_1.$$

In eleven dimensions this is simply the minimal coupling of a particle to the KK gauge field associated with circle momentum.

D6-branes are KK monopoles

A D6-brane is magnetically charged under the same one-form $C_1$. Since $C_1$ is a Kaluza—Klein gauge field, its magnetic source is a Kaluza—Klein monopole. The eleven-dimensional geometry is a Taub—NUT space fibered over the three transverse directions to the D6-brane:

$$\boxed{ ds_{11}^2=ds_{6,1}^2+H(r)d\vec r^{\,2} +H(r)^{-1}\left(dx^{11}+\vec A\cdot d\vec r\right)^2 }$$

with

$$\nabla\times\vec A=\nabla H, \qquad H(r)=1+{N R_{11}\over 2r}$$

up to conventional normalizations of the circle coordinate. The circle is nontrivially fibered over the linking two-sphere in the transverse $\mathbb R^3$. Its first Chern class is the D6 charge,

$${1\over 2\pi}\int_{S^2}F_2=N, \qquad F_2=dC_1.$$

For a single D6-brane, the eleven-dimensional Taub—NUT geometry is smooth at the core: the $S^1_{11}$ fiber shrinks smoothly, just as the angular circle in polar coordinates shrinks at the origin. The apparent D6-brane singularity in ten-dimensional supergravity is largely an artifact of reducing along a circle whose fiber degenerates.

The D6-brane is the magnetic dual of the D0-brane. In eleven dimensions it is a Taub—NUT geometry: the M-theory circle is fibered over the transverse $\mathbb R^3$ and shrinks at the monopole core.

This geometric interpretation also explains the electric-magnetic pairing

$$\text{D0} \quad\leftrightarrow\quad \text{D6}.$$

The D0 is electric charge for the KK gauge field $C_1$, while the D6 is magnetic charge for the same field. In ten dimensions a one-form gauge potential couples electrically to particles and magnetically to sixbranes, exactly as the D-brane charge rule requires.

The complete IIA/M-theory brane dictionary

The dictionary can be summarized as follows:

M-theory object	Relation to $S^1_{11}$	Type IIA object	Ten-dimensional charge
M2	wrapped	F1	electric under $B_2$
M2	unwrapped	D2	electric under $C_3$
M5	wrapped	D4	electric under $C_5$, magnetic under $C_3$
M5	unwrapped	NS5	magnetic under $B_2$
graviton	momentum along $S^1_{11}$	D0	electric under $C_1$
KK monopole	Taub—NUT circle	D6	magnetic under $C_1$

The pattern is beautifully economical. Eleven-dimensional M-theory has no fundamental string and no D-branes as elementary inputs. After compactification, the wrapped M2 becomes the fundamental string, the unwrapped M2 becomes the D2, the wrapped M5 becomes the D4, the unwrapped M5 becomes the NS5, and pure geometry gives D0/D6.

The type IIA D8-brane is not included in this ordinary circle compactification. D8-branes are domain walls in massive type IIA supergravity, where the Romans mass $F_0$ jumps across the brane. They do not arise from reducing conventional eleven-dimensional supergravity on a smooth circle in the same direct way as the branes above.

How the tension scalings become obvious

The different powers of $g_s$ in type IIA are often the first clue that a hidden geometric origin exists. Using

$$R_{11}=g_s\ell_s, \qquad \ell_{11}^3=g_s\ell_s^3,$$

we find

$$\begin{aligned} \text{wrapped M2:}\qquad (2\pi R_{11})T_{M2} &={1\over 2\pi\ell_s^2}=T_{F1},\\ \text{unwrapped M2:}\qquad T_{M2} &={1\over (2\pi)^2g_s\ell_s^3}=T_{D2},\\ \text{wrapped M5:}\qquad (2\pi R_{11})T_{M5} &={1\over (2\pi)^4g_s\ell_s^5}=T_{D4},\\ \text{unwrapped M5:}\qquad T_{M5} &={1\over (2\pi)^5g_s^2\ell_s^6}=T_{NS5},\\ \text{KK momentum:}\qquad {1\over R_{11}} &={1\over g_s\ell_s}=T_{D0}. \end{aligned}$$

The only entry in this list whose tension is not obtained by simply multiplying by a wrapped length is the D6-brane, because it is not a wrapped M-brane. It is a KK monopole. Its ten-dimensional tension still has the correct D-brane scaling,

$$T_{D6}={1\over (2\pi)^6g_s\ell_s^7},$$

but its eleven-dimensional interpretation is gravitational rather than membrane-like.

Local radius and when the M-theory lift is needed

In backgrounds with a varying dilaton, the M-theory circle has a varying local size. From the metric reduction ansatz,

$$R_{11}^{\mathrm{local}}(x)=R_{11}\,e^{2(\Phi(x)-\Phi_\infty)/3}.$$

This is a practical diagnostic. If a type IIA supergravity solution develops strong coupling in some region, $e^\Phi\gg1$, then the M-theory circle becomes large there and the ten-dimensional description is not the right one. The correct weakly curved description may be eleven-dimensional.

This explains several standard lifts:

$$\begin{array}{ccl} \text{D2-brane at strong coupling} &\longrightarrow& \text{M2-brane},\\ \text{D4-brane UV completion} &\longrightarrow& \text{M5-brane on }S^1,\\ \text{D0-brane quantum mechanics at strong coupling} &\longrightarrow& \text{eleven-dimensional momentum dynamics},\\ \text{D6-brane core} &\longrightarrow& \text{Taub--NUT geometry}. \end{array}$$

A good rule of thumb is this: type IIA is a ten-dimensional description only when the M-theory circle is small compared with the scales being probed. When the circle is large, the same physics is better described as eleven-dimensional.

Supersymmetry and chirality

Eleven-dimensional supergravity has one Majorana supercharge with $32$ real components. Compactification on a circle preserves all of them. In ten dimensions these become the two Majorana—Weyl supercharges of type IIA, which have opposite chirality.

This explains why the strong-coupling limit of type IIA is eleven-dimensional, while type IIB has a different nonperturbative structure. Type IIB has two supercharges of the same chirality and no simple decompactification to eleven-dimensional supergravity on a circle. Instead, its nonperturbative symmetry is the $SL(2,\mathbb Z)$ S-duality discussed earlier.

The M2- and M5-branes each preserve half of the eleven-dimensional supersymmetry. Their reductions therefore produce half-BPS type IIA branes. The D0, D2, D4, D6, F1, and NS5 all fit into this same BPS structure, even though perturbatively they look like very different objects.

Exercises

Exercise 1. D0-branes as KK momentum

Starting from the D0-brane tension $T_{D0}=1/(g_s\ell_s)$, show that a BPS bound state of $n$ D0-branes has the spectrum of KK momentum on a circle. What is the radius of that circle?

Solution

The BPS mass of an $n$-D0 bound state is additive in the central charge:

$$M_n=nT_{D0}={n\over g_s\ell_s}.$$

A particle carrying $n$ units of momentum on a circle of radius $R$ has

$$p_{11}={n\over R}.$$

Identifying $M_n=p_{11}$ gives

$$R=g_s\ell_s.$$

Thus the D0-brane tower is the Kaluza—Klein tower of an eleventh dimension with radius $R_{11}=g_s\ell_s$.

Exercise 2. Derive $\ell_{11}^3=g_s\ell_s^3$

Use

$$2\kappa_{10}^2=(2\pi)^7g_s^2\ell_s^8, \qquad 2\kappa_{11}^2=(2\pi)^8\ell_{11}^9,$$

and the circle reduction relation

$$2\kappa_{11}^2=(2\pi R_{11})2\kappa_{10}^2$$

to derive the relation between $\ell_{11}$ and $\ell_s$.

Solution

Substituting the two gravitational couplings gives

$$(2\pi)^8\ell_{11}^9 =(2\pi R_{11})(2\pi)^7g_s^2\ell_s^8.$$

Canceling $(2\pi)^8$ gives

$$\ell_{11}^9=R_{11}g_s^2\ell_s^8.$$

Using $R_{11}=g_s\ell_s$,

$$\ell_{11}^9=g_s^3\ell_s^9.$$

Taking the cube root of both sides yields

$$\ell_{11}^3=g_s\ell_s^3, \qquad \ell_{11}=g_s^{1/3}\ell_s.$$

Exercise 3. M2 reductions

Using

$$T_{M2}={1\over (2\pi)^2\ell_{11}^3}, \qquad R_{11}=g_s\ell_s, \qquad \ell_{11}^3=g_s\ell_s^3,$$

show that an unwrapped M2-brane is a D2-brane and a wrapped M2-brane is a fundamental string.

Solution

For an unwrapped M2-brane, the ten-dimensional tension is simply

$$T_{M2}={1\over (2\pi)^2g_s\ell_s^3}.$$

This equals the type IIA D2-brane tension

$$T_{D2}={1\over (2\pi)^2g_s\ell_s^3}.$$

For an M2-brane wrapped on the circle, the effective string tension is the membrane tension times the circumference:

$$T=(2\pi R_{11})T_{M2} =(2\pi g_s\ell_s){1\over (2\pi)^2g_s\ell_s^3} ={1\over 2\pi\ell_s^2}.$$

This is exactly the fundamental string tension $T_{F1}$.

Exercise 4. M5 reductions

Show that a wrapped M5-brane gives a D4-brane and an unwrapped M5-brane gives an NS5-brane by comparing tensions.

Solution

The M5-brane tension is

$$T_{M5}={1\over (2\pi)^5\ell_{11}^6}.$$

Since $\ell_{11}^3=g_s\ell_s^3$, we have

$$\ell_{11}^6=g_s^2\ell_s^6.$$

For an unwrapped M5-brane,

$$T_{M5}={1\over (2\pi)^5g_s^2\ell_s^6}=T_{NS5}.$$

For an M5-brane wrapped on the circle,

$$(2\pi R_{11})T_{M5} =(2\pi g_s\ell_s){1\over (2\pi)^5g_s^2\ell_s^6} ={1\over (2\pi)^4g_s\ell_s^5}=T_{D4}.$$

Thus the wrapped M5 is a D4-brane and the unwrapped M5 is an NS5-brane.

Exercise 5. Field decomposition and brane charges

Use

$$C_3^{(11)}=C_3+B_2\wedge dx^{11}$$

to explain why the unwrapped M2 is a D2-brane and the wrapped M2 is a fundamental string.

Solution

An M2-brane couples electrically to $C_3^{(11)}$:

$$S_{M2}\supset T_{M2}\int_{M2}C_3^{(11)}.$$

If the M2-brane is unwrapped, its worldvolume has no leg along $dx^{11}$. It therefore couples to the component $C_3$, which is the type IIA R—R three-form. The electric source for $C_3$ is a D2-brane.

If the M2-brane wraps the circle, one worldvolume direction is $x^{11}$. Pulling back $B_2\wedge dx^{11}$ and integrating over the circle leaves a coupling

$$\int_{\text{string}} B_2.$$

This is the defining electric coupling of the fundamental string. Hence the wrapped M2 is the type IIA F1 string.

Exercise 6. D0 and D6 as electric and magnetic KK charges

Explain why D0- and D6-branes are electric/magnetic duals from the eleven-dimensional viewpoint.

Solution

The type IIA R—R one-form $C_1$ is the Kaluza—Klein gauge field coming from the metric component $G_{\mu,11}$. Momentum along the circle is electric charge under this KK gauge field. Therefore a D0-brane, which couples to $C_1$, is an electric KK charge.

The magnetic source for a KK gauge field is a KK monopole. In eleven dimensions the D6-brane is a Taub—NUT geometry in which the M-theory circle is fibered over the transverse $\mathbb R^3$. The magnetic charge is the first Chern class of the circle bundle:

$${1\over 2\pi}\int_{S^2}F_2=N, \qquad F_2=dC_1.$$

Thus D0 and D6 are electric and magnetic objects for the same gauge field $C_1$.

Exercise 7. The local M-theory radius

Suppose a type IIA background has dilaton $\Phi(x)$ with asymptotic value $\Phi_\infty$. Use the metric ansatz to argue that the local M-theory circle radius scales as

$$R_{11}^{\mathrm{local}}(x)=R_{11}e^{2(\Phi(x)-\Phi_\infty)/3}.$$

What happens in a region where $e^{\Phi(x)}$ becomes large?

Solution

The reduction ansatz is

$$ds_{11}^2=e^{-2\varphi/3}ds_{10,\mathrm{str}}^2 +e^{4\varphi/3}(dx^{11}+C_1)^2, \qquad \varphi=\Phi-\Phi_\infty.$$

The proper length of the circle is controlled by the square root of the coefficient of $(dx^{11})^2$. Therefore the local radius is multiplied by

$$\sqrt{e^{4\varphi/3}}=e^{2\varphi/3}.$$

Thus

$$R_{11}^{\mathrm{local}}(x)=R_{11}e^{2(\Phi(x)-\Phi_\infty)/3}.$$

If $e^{\Phi(x)}$ becomes large, then the M-theory circle becomes locally large. A ten-dimensional type IIA description is then no longer the most natural description; the background should be lifted to eleven dimensions.