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The royal status of 11-dimensional supergravity

A few days ago, I discussed the maximally supersymmetric gauge theory, the Darth Vader theory, which is exceptional.

I mentioned that the \(\NNN=4\) gauge theory in \(d=4\) dimensions is a beautiful, fully consistent (at the quantum level) dimensional reduction of the 10-dimensional supersymmetric gauge theory which has many hidden virtues and symmetries. All these theories have 16 supercharges.

A portrait of eleven-dimensional supergravity. As Dilaton knows, "elf" means "eleven". ;-)

There's one more royal family that is, in some sense, comparably fundamental as the family descended from the ten-dimensional supersymmetric gauge theory. This additional family of theories has 32 real supercharges and not just 16 of them.

The master theory's spacetime has \(d=11\) dimensions rather than \(d=10\). It is not renormalizable and none of its compactifications is fully consistent at the quantum level, either. These theories are only meaningful at a classical level (or in some low-energy expansion). All of these theories contain gravity and they are limits of string/M-theory which had to be the case because string/M-theory is the only consistent quantum theory of gravity.

Even though the eleven-dimensional gravity is just a low-energy limit of the full string/M-theory, it (and its compactifications) already knows quite something about the wisdom of string/M-theory in its most symmetric incarnation.

First of all, the previous text looked at gauge theories. We saw that supersymmetry implied some pairing between fermionic and bosonic degrees of freedom. Because the fermions have to be organized as \(j=1/2\) spinors whose number of components grows exponentially, it's only possible to match them for \(d=3,4,6,10\). In particular, \(d=10\) was the maximum dimension in which a supersymmetric and pure gauge theory may exist.

Allowing gravity

Now we will allow massless fields with spin up to \(j\leq 2\). We know that Einstein's gravity predicts gravitational waves. Their energy has to be quantized i.e. these waves have to be composed of gravitons of energy \(E=\hbar\omega\). The spin of the gravitons is \(j=2\) essentially because the metric field \(g_{\mu\nu}\) has two indices.

If we linearize the metric tensor around a background, e.g. the flat Minkowski background, we obtain a dynamical spin-2 field. Much like spin-1 fields, it contains timelike components (in particular, components with an odd number of indices being timelike) which create negative-norm excitations. Much like in the Yang-Mills case, there has to be a gauge symmetry that decouples these dangerous excitations that could make the probabilities of many processes negative if they weren't killed.

For spin-1 fields, the possible gauge symmetries that do this job are Yang-Mills symmetries; we may choose the gauge group which also influences the number of components of the gauge fields (via the dimension of the gauge group). However, the gauge field of gravity – the metric tensor – has a spin that is greater by one. So the conserved charges – the generators of the symmetries – have to carry an additional Lorentz index. The generator of this symmetry can't be an "electric charge" or its non-Abelian generalization; it has to be a spacetime vector.

The only conserved vector that still allows realistic theories is the energy-momentum vector, i.e. the integral of the stress-energy tensor. Its conservation is linked to the translational symmetries of the spacetime via Emmy Noether's theorem; however, we are making this symmetry local or gauge so the translational symmetry is going to be promoted to all diffeomorphisms. If we required the conservation of yet another vector or even higher-spin tensors, the theory would be so constrained by the conservation laws (so many components) that it would essentially have to be non-interacting.

(Winding numbers of macroscopic strings and membranes may be a loophole and may be additional conserved quantities aside from the energy-momentum vector; this loophole is possible because pointlike excitations of the theory – and those are studied by the Coleman-Mandula theorem – carry vanishing values of these charges.)

If you think about these words, you will see that \(j=2\) is the maximum spin of a field for which you will be able to propose a reasonable gauge symmetry that kills the negative-norm excitations while allowing at least some interactions in the theory. Moreover, the \(j=2\) massless field has to be unique. There's only one translational symmetry of the spacetime so there's only one stress-energy tensor and it can only couple to one "gauge field" for this symmetry, the metric tensor.

The spin \(j=5/2\) would already give too many components.

Maximizing the number of supercharges

Fine. So we want to construct a supersymmetric theory of massless particles that has the maximum spacetime dimension and the maximum number of supercharges. Let me say in advance that we will be led to \(d=11\) dimensions and \(N=32\) supercharges by this maximization procedure. Why?

Imagine a supermultiplet of massless particles in such a theory. One-half of the supercharges i.e. \(N/2\) of them will annihilate the whole supermultiplet; all of the components will be invariant under one-half of the supercharges. The rest of the supercharges, \(N/2\) real supercharges, may be recombined into \(N/4\) raising operators and \(N/4\) lowering operators; the latter group may be composed of the Hermitian conjugates of the former group. Each of these complexified supersymmetry generators changes a chosen projection of the spin by \(\Delta j=\pm 1/2\).

We want to maximize the number of supercharges so we must allow the set of \(N/4\) raising operators to be able to climb from the minimum allowed component of the spin, e.g. \(j_{12}\), i.e. from \(j_{12}=-2\) (graviton), to the maximum allowed one, \(j_{12}=+2\), by those \(\Delta j=1/2\) steps. We therefore see that \[

\frac{N}{4} = \frac{(+2)-(-2)}{\frac{1}{2}}=8,\qquad N=32.

\] If the supersymmetry algebra is capable of climbing from one extreme component of the graviton to the opposite one, we must have \(N=32\) real supercharges. Now, recall that the Dirac spinor in \(d=10\) has \(2^{10/2}=32\) components so in \(d=10\), we actually find a 32-component spinor. However, the Dirac spinor is reducible, to the left-handed and right-handed chiral components, if you wish. That's not really a problem but it's a sign that we may go higher.

And indeed, the spinor in \(d=11\) has 32 real components. There's no chirality in \(d=11\) because eleven is an odd number. The \(d=11\) spinor simply becomes the Dirac spinor if you dimensionally reduce the theory by one dimension. So \(d=11\) is the right number of dimensions in which we should expect a nice maximally supersymmetric theory of gravity or supergravity.

Representation of the massless multiplet

So far I haven't said anything about interactions and even in this section, we will assume that we're considering the free limit of the theory only. The question we want to answer is how do the excitations of the massless gravitational supermultiplet transform under spacetime rotations.

In \(d=11\), the massless particles move in a direction. Their nonzero energy prevents us from mixing time with other coordinates while preserving the energy-momentum vector of the particle; the direction of the particle's motion removes one spatial dimension as well. So there are only \(d-2=9\) transverse dimensions that may be rotated into each other so that the energy-momentum vector is unchanged. (There are actually some light-like boosts as well but I only want to discuss compact groups here.)

So the little group is \(SO(9)\). As we have already said, the multiplet is annihilated by 16 supercharges while the remaining 16 "active" supercharges play the role of 8 raising and 8 lowering operators. The algebra satisfied by these 16 "active" Hermitian supercharges is \[

\{Q_a,Q_b\} = \delta_{ab} E, \qquad a,b\in\{1,2,\dots , 16\}

\] if we normalize the supercharges so that there's no extra coefficient. The Kronecker delta arises from a Dirac gamma matrix reduced to the simpler 16-dimensional space relevant for the little group. My point is that these 16 "active" supercharges transform as a spinor of the little group, \(SO(9)\), because they're a remnant of a spinor of \(SO(10,1)\), the Lorentz group of the 11-dimensional theory.

Do you know another way how to look at the anticommutators above? Well, you should. They're the same anticommutators as the algebra of Dirac gamma matrices:\[

\{\Gamma_a,\Gamma_b\} = \delta_{ab},\qquad a,b\in\{1,2,\dots , 16\}

\] up to the obvious change of the normalization. But we know what we get from the algebra above if we "quantize it", right? We get a 256-dimensional spinor of \(SO(16)\) or \(Spin(16)\), to be more accurate. It's 256-dimensional because each of the 8 raising operators (that exists besides the 8 lowering operators) may be either applied to the "ground state" or not. The mathematical problem we're solving here is completely \(Spin(16)\)-symmetric so we know that the states we get by "quantizing" the raising and lowering operators must produce a nice representation of \(Spin(16)\). Clearly, it has to be \(2^8\)-dimensional and it's nothing else than the spinor of \(SO(16)\).

Moreover, one may define the operator of chirality \(\Gamma_{17}\), to use a notation analogous to \(\Gamma_5\) in \(d=4\), that anticommutes with the sixteen \(\Gamma_a\) matrices. This operator makes it clear that the 256-dimensional representation is reducible; it decomposes at least to 128 components that have \(\Gamma_{17}=+1\) and 128 components with \(\Gamma_{17}=-1\).

If we return from the \(\Gamma_a\) notation to \(Q_a\), it's clear that \(\Gamma_{17}\) is nothing else than the operator remembering whether a state of the representation is bosonic or fermionic. That's the only conclusion from the fact that \(\Gamma_{17}\) anticommutes with all \(Q_a\) operators and those are fermionic ones.

We have just determined that the gravitational supermultiplet contains 128 bosonic and 128 fermionic states. We know that they transform as the full Dirac spinor (or the sum of the two chiral spinors) under \(SO(16)\). However, the little group \(SO(9)\) is "bizarrely" embedded into this \(SO(16)\) in such a way that the "vector" \({\bf 16}\) of \(SO(16)\) is the "spinor" \({\bf 16}\) of \(SO(9)\) whose vector is \({\bf 9}\), of course. So how do the 128-dimensional chiral spinors of \(SO(16)\) decompose under \(SO(9)\) which is a subgroup of \(SO(16)\)?

Getting the three fields of \(d=11\) SUGRA

Let's start with the fermionic ones. We should get a representation of \(Spin(9)\) which is 128-dimensional. This representation inevitably contains some states with \(j_{12}=3/2\), as expected for gravitinos. What are the representations of \(Spin(9)\) with this property?

Well, the answer is that there is a unique 128-dimensional representation of this kind and it is irreducible. Why? Take the tensor product of the vector and the spinor of \(Spin(9)\), \[

{\bf 9} \otimes {\bf 16}.

\] It's 144-dimensional which is too many for us. Is it irreducible? Well, it is not. We may require\[


\] That's the only linear, rotationally invariant condition we may demand from the spin-3/2 object \(\chi_{ia}\) which doesn't make it identically vanish. The condition above has a free \(b\) index so it eliminates sixteen components of the tensor with one vector index and one spinor index. So the number of independent surviving components is \(8\times 16=128\) rather than \(9\times 16=144\) and that's exactly what we need. The gravitino in \(d=11\) forms an irreducible 128-dimensional representation of the little group

What about the bosonic states?

They must contain the graviton which comes from transverse excitations of the metric tensor, \(h_{ij}\), where \(i,j=1,2,\dots,9\). This would have \(9\times 10/2\times 1=45\) components; count the number of squares in a symmetric matrix or a triangle that includes the diagonal. However, this 45-dimensional representation isn't irreducible. Again, we may demand an extra condition, namely tracelessness\[

\sum_{i=1}^9 h_{ii} = 0

\] and general relativity actually does imply that the physical states are traceless which means that the \(d=11\) graviton only has \[\frac{9\times 10}{2\times 1} - 1 = 44\] components. Where are the remaining 84 components?

I have already mentioned that the spin-2 states have to be unique. But the gravitational multiplet may still contain \(j=1\) and \(j=0\) states. Well, there have to be some additional \(j=1\) states as well. I originally wrote an explanation in terms of weights but it would be too annoyingly technical so I erased it and you will only be told the sketched result. The representations with \(j=1\) don't have to be just vectors; \(p\)-forms are fine, too. If you analyze the dimensions of the antisymmetric tensor representations, you will find out that \(C_{ijk}\) with three \(SO(9)\) indices has\[

\frac{9\times 8 \times 7}{3\times 2 \times 1} = 84

\] components which is exactly what you need. So the 128 bosonic states decompose to 44 states of the graviton and 84 states of the 3-form. Everything seems to work at the level of the little group and its representations.

Writing the Lorentz, diff-invariant Lagrangian

The \(d=11\) theory of supergravity should therefore contain the metric tensor \(g_{\mu\nu}\), a gravitino Rarita-Schwinger field \(\chi_{\mu\alpha}\), and a three-form \(C_{\lambda\mu\nu}\). Let's use the convention in which the metric and the three-form potential are dimensionless and the fermionic field has the dimension \({\rm mass}^{1/2}\). What is the Lagrangian?

You combine the usual Einstein-Hilbert action \[

\LL_{EH} = \frac{1}{16\pi G} R

\] i.e. the Ricci scalar with other terms. Note that its dimension is \({\rm mass}^2\) and when divided by Newton's constant which is \({\rm length}^9\), e.g. because \(A/4G\) should be the dimensionless entropy, we get \({\rm mass}^{11}\), as expected from an 11-dimensional Lagrangian density. Aside from the Einstein-Hilbert action, there will also be some nice Maxwell-like kinetic term for the three-form potential,\[

\LL_{EM} = \frac{C}{G} F_{\lambda\mu\nu\pi} F^{\lambda\mu\nu\pi}

\] where the 4-form \(F\) is gotten as the antisymmetrized derivative, \(F=*dC\), much like in the electromagnetic case. Finally, there have to be kinetic terms for the gravitino,\[

\LL_{RS} \sim \frac{C}{G} \chi^{\lambda \beta} (\Gamma\delta)_{\lambda}^{\mu\nu}{}^\alpha_\beta \partial_\nu \chi_{\mu\alpha}

\] where I don't want to write all the required combinations of contractions of the indices and their right relative normalizations so I have unified the gamma matrices and a Lorentz-vector Kronecker delta into a "hybrid" object. Just like the Dirac Lagrangian, this fermionic kinetic term is linear in derivatives. Note that all the terms in the Lagrangian have the right units; and all of them are proportional to \(1/G\).

You will find out that the theory above isn't supersymmetric. However, it's possible to systematically deduce all the additional interaction terms and the supersymmetry suddenly starts to hold. It's a kind of miracle. You will have to add gauge-fermion-like interactions \(F\psi\psi\), the Chern-Simons term \(C\wedge F\wedge F\) with the right coefficient (an 11-form), and a \(\chi^4\) term as well. But when you add all of them, the total action may be verified to be locally supersymmetric! It's also diffeomorphism symmetric and symmetric under \(\delta C = d\lambda\) where \(\lambda\) is a 2-form parameter of a gauge transformation for the 3-form that generalizes the electromagnetic gauge symmetry (but is still Abelian).

Why does the supersymmetry work?

It's a kind of a miracle that the supersymmetry holds when you adjust a few coefficients. I guess that only dozens of people in the world have verified the supersymmetry of the 11-dimensional supergravity's action explicitly, without any help of a computer, and roughly 100 people have done so with the help of a computer or some other improvements of their brains.

Is there an intuitive explanation why it works? I admit that, paradoxically enough, the most straightforward and heavy-algebra-free approach to prove that this limit exists could actually start from (type IIA) string theory. If any reader has an explanation why such an interacting theory with such a huge symmetry principle exists at all, an explanation will be highly welcome.

M2-branes, M5-branes

The field content of the theory includes a 3-form bosonic field. One may naturally include another term in the action\[

S = \int \dd \Sigma_3 C_{(3)}

\] that simply integrates the 3-form potential over some 3-dimensional manifold in the spacetime. Such privileged manifolds that do contribute to the action in this way may exist; they're of course the world volumes of the two-dimensional branes or membranes. The membranes in 11-dimensional supergravity or M-theory are known as M2-branes. They carry some specific "charge density" in the right direction. However, in electromagnetism, you know that you may also consider \(*F\) instead of \(F\) and study magnetic charges, too.

The same thing may be done here. The field strength \(F_{(4)}\) which is a 4-form may be Hodge dualized in 11 dimensions to a 7-form and this 7-form may be, at least in regions without sources, written as \(F_{(7)}=d\tilde C_{(6)}\). And this dual 6-potential may be integrated over 6-dimensional manifolds in spacetime: they're nothing else than the world volumes of M5-branes.

So the 11-dimensional theory predicts gravitational waves, generalizations of electromagnetic waves, black holes, M2-branes, and M5-branes, among other objects.

We're going to look at some dimensional reductions, perhaps oxidations, and dimensional chemistry. Song "Chemistry" is sung by Mr Xindl X.

Surprising richness of the compactifications

If you wanted the theory to be well-defined at the quantum level including the quantum corrections – which is the same as corrections that become strong at high energies – you would have to replace the 11-dimensional supergravity by its unique ultraviolet completion, M-theory. I don't want to do that in this article.

However, it's interesting to look at compactifications. The simplest compactifications are the toroidal ones, i.e. ones in which a couple of dimensions are periodically identified, i.e. points are identified with other points that are shifted by an element of a lattice.

\(\RR^n / \Gamma^n=T^n\) is a torus. If its size is comparable to the Planck length, the length scale at which the quantum M-theory corrections become really important, it's obvious that the compactification will be a compactification with extremely short, microscopic periodicities that look like zero from the viewpoint of long-distance probes. So if you only consider SUGRA and not the full M-theory, these compactifications will be just dimensional reductions and the size and angles in the torus will be physically inconsequential (for all effects at low energies).

And indeed, this fact will manifest itself as a noncompact symmetry of the dimensionally reduced supergravity theory: changing the shape and size of the torus is actually changing nothing about the low-energy physics. The dimensional reductions still have 32 real supercharges but they're organized as extended supersymmetry in the lower-dimensional spacetimes. You will find out that there is a noncompact version of the \(E_k\) symmetry if you dimensionally reduce the theory to \(11-k\) large dimensions of spacetime. That's true for \(k=6,7\) and, when you properly interpret some fields in 3 dimensions, even for \(k=8\). Well, one could argue that some of those statements may even be done for the Kač-Moody case \(k=9\) in \(d=2\) if not the hyperbolic case \(E_{10}\) in a 1-dimensional spacetime.

The groups \(E_5, E_4, E_3, E_2, E_1\) should be understood as \(SO(5,5)\), \(SL(5,\RR)\), \(SL(2,\RR)\times SL(3,\RR)\), \(SL(2,\RR)\), \(\RR\). If you study the full M-theory, using probes that can feel the Planckian physics, the shape of the torus matters and these noncompact symmetries are reduced to their discrete versions such as \(E_{7(7)}(\ZZ)\) in the case of the \(\NNN=8\) \(d=4\) supergravity, i.e. to the so-called U-duality groups. The quotient of the continuous group and its discrete subgroup spans the moduli space. Various charges and scalar fields etc. know about the noncompact group or its maximal compact subgroup.

It's a shocking set of mathematical facts. There exist mathematical explanations why the moduli spaces have to be quotients, why the exceptional groups are the only solutions, and so on. But I think that no one knows of any "truly conceptual" explanation why the exceptional groups appear in the discussion of a maximally supersymmetric theory of gravity which didn't start with any components that would resemble exceptional groups. The exceptional groups were spitted out as one of the amazingly surprising outputs.

All these mathematical facts become even more stunning if you study those theories beyond the low-energy approximation i.e. if you investigate the whole structure of string/M-theory. The 11-dimensional spacetime of M-theory, the completion of the 11-dimensional supergravity, is the maximum-dimensional spacetime in string/M-theory that may exist which is why people often say that it's a "more fundamental" limit than others. Of course, a more politically correct assertion is that it's just another limit, on par with many others such as the five 10-dimensional string "theories" (vacua). Still, by its having a higher number of dimensions, the description in terms of 11-dimensional theory is "more geometric" than others.

(F-theory has 12 dimensions in some counting but 2 of them have to be infinitesimal and they're a bit different than the remaining 10 dimensions. From some perspectives, F-theory is more geometrical and higher-dimensional than M-theory; from others, it's the other way around. This discussion is similar to the question whether mothers or fathers are more fundamental. There's no unique answer. After all, it's not just an analogy because M-theory stands for Mother while F-theory stands for Father. M-theory compactified on a circle easily produces type IIA string theory; F-theory is a toolkit to construct sophisticated nonperturbative type IIB string vacua.)

Technical: Speed up your MathJax \(\rm\LaTeX\) by a second per page, at least in Windows. Download these zipped OTF fonts, unzip them, and for each of the 22 OTF files, right-click Properties/General and "unblock". Then choose all the 22 OTF files and install them. It will display the same maths you're used from the web, but using your local copy of the fonts. (These are not STIX fonts which I consider uglier but the very same fonts you're used to.) Warning: the local \(\rm \TeX\) fonts may produce imperfect output when applied by Internet Explorer to pages with Greek letters. You should learn how to uninstall the MathJax fonts, too.
This text has been proofread once and quickly.

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