## March 26, 2008

### Bagger-Lambert

The low-energy theory on a (stack of) D2-brane(s) is a maximally-supersymmetric gauge theory in 2+1 dimensions. The Yang-Mill multiplet has a gauge field and 7 real scalars in the adjoint representation. At least if you are out on the Coulomb branch, where the gauge symmetry is Higgsed down to the Cartan, you can dualize the gauge fields to another scalar, which is circle-valued.

The M2-brane is obtained as the strong-coupling limit of this theory. The radius of the circles (one for each element of the Cartan subalgebra) go to infinity, and the $\mathrm{SO}\left(7\right)$ R-symmetry is promoted to $\mathrm{SO}\left(8\right)$. This strong-coupling limit is superconformal, but the above description is effective only for the free theory, where the M2-branes are separated (away from the origin, the moduli space looks like ${ℝ}^{8n}/{S}_{n}$). The theory of coincident M2-branes is an interacting SCFT which, so far, does not have a Lagrangian description. But there’s no theorem that rules out a Lagrangian description, so there may just be one.

Bagger and Lambert recently proposed a very interesting maximally supersymmetric interacting 2+1D Lagrangian field theory which, at least classically, seems to be superconformal. It does not arise as the dimensional reduction of some higher dimensional theory, and so it was missed in previous attempts at tackling this problem.

I never got around to blogging about Bagger and Lambert’s paper, but Bandres, Lipstein and Schwarz wrote a nice followup, which gives me an excuse to return to the subject.

#### Update:

Whoops! Even as was typing this, Mark van Raamsdonk came out with a paper making some of the points below. I’d better hurry up and post this, before there are yet-more followup papers to discuss.

Let me start with some notational preliminaries. The connected part of the Lorentz group in 2+1 dimensions is $\mathrm{SL}\left(2,ℝ\right)$, so we’ll use a Wess and Bagger-like notation for spinors1. Spinor indices are raised and lowered using the 2-component $ϵ$-symbol, ${ϵ}_{\alpha \beta }=-{ϵ}_{\beta \alpha }$. The $\gamma$-matrices are ${\sigma }_{\alpha \beta }^{\mu }={\sigma }_{\beta \alpha }^{\mu }$, and are real. After raising an index using ${ϵ}^{\alpha \beta }$, we can write $\left({\gamma }^{\mu }\right)^{\alpha }{}_{\beta }={ϵ}^{\alpha \delta }{\sigma }_{\delta \beta }^{\mu }$ in terms of Pauli matrices, with ${\gamma }^{0}=i{\sigma }_{2}$.

Also real are the $\mathrm{SO}\left(8\right)$ $\gamma$-matrices, which we can regard as a trilinear form $\Gamma :V\otimes S\otimes C\to ℝ$ where $V$, $S$, and $C$ are the three 8-dimensional real representations of $\mathrm{Spin}\left(8\right)$. Alternatively, letting “$I$” be an $\mathrm{SO}\left(8\right)$ vector index, we think of ${\Gamma }^{I}:C\to S$ We’ll also need $\begin{array}{rl}{\Gamma }^{IJ}& ={\stackrel{˜}{\Gamma }}^{\left[I}{\Gamma }^{J\right]}:C\to C\\ {\Gamma }^{IJK}& ={\Gamma }^{\left[I}{\stackrel{˜}{\Gamma }}^{J}{\Gamma }^{K\right]}:C\to S\end{array}$ where ${\stackrel{˜}{\Gamma }}^{I}={\left({\Gamma }^{I}\right)}^{t}:S\to C$.

What Bagger and Lambert do is introduce an auxiliary vector space, $W$ (really, a vector bundle, which we will take to be trivial). $W$ is endowed with a positive definite inner product, $\left(\cdot ,\cdot \right):{Sym}^{2}\left(W\right)\to ℝ$ and a skew-symmetric quadrilinear form, $f:{\wedge }^{4}W\to ℝ$ Alternatively, using the metric, we can regard $f$ as a trilinear product

(1)$\left[\cdot ,\cdot ,\cdot \right]:{\wedge }^{3}W\to W$

The fields of the model are $\begin{array}{rl}\varphi & \text{a scalar field taking values in}\phantom{\rule{thinmathspace}{0ex}}W\otimes V\\ {\psi }_{\alpha }& \text{a spinor field taking values in}\phantom{\rule{thinmathspace}{0ex}}W\otimes C\\ {A}_{\mu }& \text{a}\phantom{\rule{thinmathspace}{0ex}}G\subset \mathrm{SO}\left(W\right)\phantom{\rule{thinmathspace}{0ex}}\text{gauge connection}\\ \end{array}$

Now, $\mathrm{so}\left(W\right)\simeq {\wedge }^{2}W$ has the usual action on $W$,

(2)$\alpha \wedge \beta :X↦X+\alpha \left(\beta ,X\right)-\beta \left(\alpha ,X\right)$

But, because we have (1), we can contemplate an “exotic” action,

(3)$\alpha \wedge \beta :X↦X+\left[\alpha ,\beta ,X\right]$

We demand that $f$ be invariant under this action, which amounts to requiring

(4)$\left[\alpha ,\beta ,\left[X,Y,Z\right]\right]=\left[\left[\alpha ,\beta ,X\right],Y,Z\right]\right]+\left[X,\left[\alpha ,\beta ,Y\right],Z\right]+\left[X,Y,\left[\alpha ,\beta ,Z\right]\right]$

Then the antisymmetry of $f$ ensures that the inner product, $\left(\cdot ,\cdot \right)$ is also invariant.

Unfortunately, (3) is not really satisfactory. The action of $\mathrm{so}\left(W\right)$ must satisfy the the Lie-algebra relations,

(5)$\left[{\delta }_{\alpha \wedge \beta },{\delta }_{\gamma \wedge \delta }\right]=\left(\beta ,\gamma \right){\delta }_{\alpha \wedge \delta }+\left(\alpha ,\delta \right){\delta }_{\beta \wedge \gamma }-\left(\beta ,\delta \right){\delta }_{\alpha \wedge \gamma }-\left(\alpha ,\gamma \right){\delta }_{\beta \wedge \delta }$

so it would make sense, for instance, to gauge that symmetry. If we try to impose that (3) satisfy (5), this would require

(6)$\begin{array}{rl}\left[\left[\alpha ,\beta ,\gamma \right],\delta ,X\right]-\left[\left[\alpha ,\beta ,\delta \right],\gamma ,X\right]=& \left(\beta ,\gamma \right)\left[\alpha ,\delta ,X\right]+\left(\alpha ,\delta \right)\left[\beta ,\gamma ,X\right]\\ & -\left(\beta ,\delta \right)\left[\alpha ,\gamma ,X\right]-\left(\alpha ,\gamma \right)\left[\beta ,\delta ,X\right]\end{array}$

which does not hold in any of the known solutions.

The closest one comes is when $W$ is 4-dimensional. The quadrilinear form, $f$, is just the 4-index $ϵ$-symbol, and ${\wedge }^{2}W$ is decomposable into self-dual and anti-self-dual forms ${\wedge }^{2}W={\wedge }_{+}^{2}W\oplus {\wedge }_{-}^{2}W$ with respect to this form. The action of ${\wedge }_{-}^{2}W$ on $W$ is the one induced from (3), while the action of ${\wedge }_{+}^{2}W$ on $W$ is the one induced from $\alpha \wedge \beta :X↦X-\left[\alpha ,\beta ,X\right]$ These signs are made slightly more transparent by mapping ${ℝ}^{4}$ to the quaternions $X=\frac{1}{2}\left(\begin{array}{cc}{X}_{4}+i{X}_{3}& i{X}_{1}-{X}_{2}\\ i{X}_{1}+{X}_{2}& {X}_{4}-i{X}_{3}\end{array}\right)$ where we’ve represented the unit quaternions by $i$ times the Pauli matrices. In this notation, the inner product is $\frac{1}{2}\left(X,Y\right)=\mathrm{Tr}\left({X}^{†}Y\right)$ The matrix representing $\left[X,Y,Z\right]$ is $\begin{array}{rl}\left[X,Y,Z\right]\simeq & \frac{4}{3}\left(X{Y}^{†}Z+Y{Z}^{†}X+Z{X}^{†}Y\\ & -\left(Z{Y}^{†}X+Y{X}^{†}Z+X{Z}^{†}Y\right)\right)\end{array}$ In this notation, $\mathrm{SO}\left(4\right)\sim \mathrm{SU}\left(2{\right)}_{L}×\mathrm{SU}\left(2{\right)}_{R}$ acts by conjugation $X↦{g}_{L}X{g}_{R}^{-1}$ whereas (3) would have corresponded to $X↦{g}_{L}X{g}_{R}$, which fails to satisfy the group law for $\mathrm{SU}\left(2{\right)}_{R}$.

Anyway, introducing the covariant derivative2, ${D}_{\mu }X={\partial }_{\mu }X+{A}_{\mu }^{L}X-X{A}_{\mu }^{R}$ we can write the action as

(7)$S=k\left({S}_{m}+{S}_{\mathrm{CS}}\right)$

where $k\in ℤ$,

(8)$\begin{array}{rl}{S}_{m}=& \int {d}^{3}x\mathrm{Tr}\left(-\left({D}_{\mu }\Phi {\right)}^{†}\left({D}^{\mu }\Phi \right)+i{\Psi }^{†}{\sigma }^{\mu }{D}_{\mu }\Psi \\ & -\frac{2\pi i}{3}{\Psi }^{†}{\Gamma }^{\mathrm{IJ}}\left({\Phi }^{I}{\Phi }^{J†}\Psi +{\Phi }^{J}{\Psi }^{†}{\Phi }^{I}+\Psi {\Phi }^{I†}{\Phi }^{J}\right)\\ & -\frac{8{\pi }^{2}}{3}\left({\Phi }^{\left[I}{\Phi }^{J†}{\Phi }^{K\right]}{\Phi }^{\left[K†}{\Phi }^{J}{\Phi }^{I†\right]}\right)\right)\end{array}$

and

(9)${S}_{\mathrm{CS}}=\frac{1}{4\pi }\int \mathrm{Tr}\left({A}^{L}d{A}^{L}-\frac{2}{3}{A}^{L}\wedge {A}^{L}\wedge {A}^{L}-{A}^{R}d{A}^{R}+\frac{2}{3}{A}^{R}\wedge {A}^{R}\wedge {A}^{R}\right)$

is the Chern-Simons action at level-1 for $\mathrm{SU}\left(2{\right)}_{L}$ and level-($-1$) for $\mathrm{SU}\left(2{\right)}_{R}$.

The supercharges are spacetime spinors in the $S$ of $\mathrm{Spin}\left(8\right)$, and the supersymmetry variations

(10)$\begin{array}{rl}\delta {\Phi }^{I}& =i\eta {\Gamma }^{I}\Psi \\ \delta \Psi & =\eta {\Gamma }^{I}{\sigma }^{\mu }{D}_{\mu }{\varphi }^{I}+\frac{2\pi }{3}{\Phi }^{I}{\Phi }^{J†}{\varphi }^{K}\eta {\Gamma }^{IJK}\\ \delta {A}_{\mu }^{L}& =-i\pi \eta {\Gamma }^{I}{\sigma }_{\mu }\left({\Phi }^{I}{\Psi }^{†}-\Psi {\Phi }^{I†}\right)\\ \delta {A}_{\mu }^{R}& =-i\pi \eta {\Gamma }^{I}{\sigma }_{\mu }\left({\Psi }^{†}{\Phi }^{I}-{\Phi }^{I†}\Psi \right)\\ \end{array}$

While it’s probably unsurprising, Bandres et al verify that the rest of the $\mathrm{OSp}\left(8\mid 4\right)$ superconformal algebra holds at the classical level, as well. Perhaps a good challenge for our ERGE friends would be to check that (7) is superconformal at the quantum level.

Parity is implemented, in this theory, in a slightly nonstandard way: accompanying a reflection in one of the spatial coordinates, is an exchange of the two $\mathrm{SU}\left(2\right)$s and Hermitian conjugation on the matrices $\Phi$ and $\Psi$, $\begin{array}{c}{A}^{L}↔{A}^{R}\\ {\Phi }^{I}\to {\Phi }^{I†}\\ \Psi \to {\gamma }^{1}{\Psi }^{†}\end{array}$ A Majorana mass term, $\psi \psi$ is a pseudo-scalar in 2+1 dimensions which, as noted by Bandres et al accounts the requisite sign in the transformation under parity of the second line of (8).

Are there other realizations, where we demand that (3), (4), (5) hold for just some subalgebra, $𝔤\subset \mathrm{so}\left(W\right)$? An obvious guess would be to replace $W=ℍ$ by $W=ℍ\left(n\right)$, the space of $n×n$ matrices of quaternions. Bandres et al looked for other, more nontrivial, examples, but didn’t find any.

#### Update: Moduli Space

Van Raamsdonk points out that the moduli space — the space of zeroes of the scalar potential in (8), modulo $\mathrm{SO}\left(4\right)$ gauge transformations — of the Bagger-Lambert model is $\left({ℝ}^{8}×{ℝ}^{8}\right)/\mathrm{SO}\left(2\right)$. This is because a zero of the potential requires that all 8 of the ${\varphi }^{I}$ lie in a common 2-plane in $W$. Configurations which differ by a rotation within that 2-plane are gauge-equivalent, so the moduli space seems to be $V\otimes {ℝ}^{2}/\mathrm{SO}\left(2\right)$. Rotations in the orthogonal 2-plane comprise a residual unbroken $\mathrm{SO}\left(2\right)$ gauge symmetry, which does does not act on the moduli space.

This looks like a puzzle, because it’s hard to see how the bosonic spectrum (15 massless scalars and a nondynamical $U\left(1\right)$ gauge field) could be $N=8$ supersymmetric. Fortunately, Mukhi and Papageorgakis ride to the rescue. They show that, when one integrates out the massive modes, the $U\left(1\right)$ gauge field actually becomes dynamical3. A dynamical $U\left(1\right)$ gauge field can be dualized to a circle-valued scalar, so the full moduli space is $\frac{{ℝ}^{8}×{ℝ}^{8}}{\mathrm{SO}\left(2\right)}\phantom{\rule{thinmathspace}{0ex}}×\phantom{\rule{thinmathspace}{0ex}}{S}^{1}$ which is the desired answer for the moduli space of a pair of M2-branes. (If I were a little more energetic, I would attempt to show that this actually gets the discrete identifications right, and that the moduli space is $\left({ℝ}^{8}×{ℝ}^{8}\right)/{ℤ}_{2}$.)

1 In 3+1 dimensions, the Lorentz group is $\mathrm{SL}\left(2,ℂ\right)$, and we need to distinguish between the two distinct two-dimensional representations. Spinors in the $2$, ${\psi }_{\alpha }$, carry undotted indices. Their Hermitian conjugates, ${\overline{\psi }}_{\stackrel{˙}{\alpha }}$, carry dotted indices and transform in the $\overline{2}$. In 2+1 dimensions, there’s only one type of spinor, and we can impose a Majorana condition.

2 If you want to insist on using the previous notation, we can write this as ${D}_{\mu }X={\partial }_{\mu }X+\left[{A}_{\mu }^{L}-{A}_{\mu }^{R};X\right]$ where $\left[\omega ;X\right]:{\wedge }^{2}W\otimes W\to W$ is the action of 2-forms on $W$, induced from (3), and we explicitly separate out the self-dual and anti-self-dual pieces.

3 Note that, from the underlying parity-invariance of the theory, this $U\left(1\right)$ must have vanishing Chern-Simons coefficient, and so is a massless dynamical gauge field.

Posted by distler at March 26, 2008 11:37 AM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1642

### Re: Bagger-Lambert

Thanks for the very useful review.

The day before yesterday I had spent an hour trying to see whether (3), (4), (5) secretly encodes an ${L}_{\infty }$-algebra structure. So far I haven’t found a satisfactory solution.

The discussion in the appendix of J. Bagger, Comments on multiple M2-branes shows that (3)-(5) is equivalent to a vector space $A\oplus B$ equipped with a skew bracket which satisfies the Joacobi identity everywhere except on ${\wedge }^{3}A$. There the Jacobi identity plain fails (not “up to something”).

That looks odd. If that is all there is, I’d say there might be little point in talking about “3-algebras” in this context. Instead, it would seem that we are just talking about nothing but the volume form on ${ℝ}^{4}$.

But then of course, the way this arises from gauge transformations with not one but two gauge parameters suggests that there is more going on after all. I’ll need to think about this more.

Posted by: Urs Schreiber on March 27, 2008 4:44 AM | Permalink | Reply to this

### Lie algebra

Well, as I said, you can realize some $𝔤\subset {\wedge }^{2}W$, as a Lie algebra, acting on $W$ via (3). For instance, $\mathrm{su}\left(2{\right)}_{L}$ acts that way.

But an ${L}_{\infty }$ structure would have involved endowing $W$ itself with both a bilinear and a trilinear product. That’s not what seems to be going on here.

Posted by: Jacques Distler on March 27, 2008 8:57 AM | Permalink | PGP Sig | Reply to this

### Re: Bagger-Lambert

It’s been a long time that we have had the above exchange. But with Jim Stasheff I recently had another conversation about this which makes me come back here:

As Jim emphasized, the “3-algebra” appearing in Bagger-Lambert nothing but a Nambu bracket (for instance (18), (19) here).

That the Nambu bracket has something to do with membrane mechanics is an old idea, I suppose, for instance appearing here.

I haven’t tried to follow the development of the Bagger-Lambert idea since I posted the above comment. So I am wondering if anybody meanwhile has been making the connection between this new use of the Nambu bracket and that older one in membrane mechanics?

Posted by: Urs Schreiber on May 18, 2008 4:39 AM | Permalink | Reply to this

### Re: Bagger-Lambert

There’s alot of mathemaical work on Nambu algebras I was unaware of, so I’m starting a collection, e.g papers by Takhtajan et al and by Gautheron

jim

Posted by: jim stasheff on May 18, 2008 7:59 AM | Permalink | Reply to this

### Re: Bagger-Lambert

Hi Jacques,

Nice post. I think that the moduli space, including the modding out by ${S}_{2}$ works out. This is because there is a large gauge transformation that exchanges the real components in ${X}^{I}={x}_{4}^{I}+i{x}_{3}^{I}{\sigma }_{3}$. Namely choosing ${g}_{L}=i{\sigma }_{1}$ and ${g}_{R}=i{\sigma }_{2}$ does the trick. (Sorry for not using MathML)

Posted by: Joe Minahan on March 28, 2008 1:05 PM | Permalink | Reply to this

### Identification

(Sorry for not using MathML)

Itex, actually. I took the liberty of adding dollar signs around your equations, and changing the Text-filter to one of the itex-using filters.

Yeah, the ${ℤ}_{2}$ identification comes from an $\mathrm{SO}\left(4\right)$ transformation which is a reflection, simultaneously in the two 2-planes.

Equally obvious (I was just tweaking the ERGE guys) is that, if $N=8$ supersymmetry survives quantization, then this theory is superconformal at the quantum level.

The point is that there is only one independent coupling constant, $k$, sitting out in front of the whole Lagrangian. And $k$ is quantized, because it is the coefficient of a nonabelian Chern-Simons term. Hence it cannot be renormalized.

Posted by: Jacques Distler on March 28, 2008 2:19 PM | Permalink | PGP Sig | Reply to this

### Re: Bagger-Lambert

Hi,
Just thought I’d add. We (Tadrowski and Thompson and I) put a paper out last week which I think explains why SO(4) is sensible as the only possible option. We looked at the Bagger Lambert theory of open membranes and found that the target space of the boundary string is 6 dimensional only for SO(4). Since the boundary of a membrane has to be a fivebrane its natural then to limit to SO(4). These are exciting times for the membrane.

regards

David Berman

Posted by: David Berman on April 1, 2008 3:56 PM | Permalink | Reply to this