In superstring theory, it says that they wrap 16 dimensions on a torus given by $\mathbb{R}^{16}$ divided by a SO(32) or $E_8 \times E_8$ lattice and this gives a gauge group of the same name.

But in Kaluza-klein theory it is the isometry group of the compact dimensions that gives the gauge group.

Doesn't a 16 dimensional torus have isometry group of $U(1)^n$ since it is the product of $S_1$?

Or is the reduction of the 16 dimensions different to usual Kaluza-Klein compactification?

I'm struggling to see how compactification on an $E_8$ torus would give an $E_8$ gauge group especially as $E_8$ has no 8 dimensional representation? If anything it could only be the isometry of a 248 dimensional surface...


2 Answers 2


You are talking about the two ways to construct a consistent 10-dimensional heterotic string theory. The original paper "Heterotic string theory: I. The free heterotic string" by Gross, Harvey, Martinec and Rohm is rather accessible and describes the detailed construction. I'll address your specific question of how the compactification on a torus manages to generate such "large" groups as $\mathrm{SO}(32)$ and $E_8\times E_8$ without giving the details.

You are correct that we usually would expect only a Kaluza-Klein style $\mathrm{U}(1)^{16}$ from compactifying on a 16-dimensional torus. However, this gauge group as a gauge group for a 10d SUGRA theory is forbidden since the gravitational and gauge anomalies don't cancel, so not only does this construction not yield the known heterotic string, it doesn't yield a consistent effective theory at all!

The crucial point is "choosing the right torus", that is, the ratios of the radii of the 16 circles must be specifically chosen to yield a consistent theory. The usual way to encode this choice is to think of the $T^{16}$ as $\mathbb{R}^{16}/\Gamma$, where $\Gamma$ is a discrete sixteen-dimensional lattice. Now one examines the bosonic part of the spectrum of a closed heterotic string on this background. It turns out that the excitations that correspond to the string being "wound" around the compactified dimensions become massless at special choices of $\Gamma$. Those excitations, together with the usual massless Kaluza-Klein modes, now together transform as the adjoint of a larger gauge group $G$ closely related to the lattice $\Gamma$, which turns out to be the root lattice of $G$. Now the dimensions of the geometry and the groups match in the way that the dimension of the lattice/torus corresponds to the rank (not the dimension) of the group, so the 16-dimensional torus has no issue generating the 496 dimensional $E_8\times E_8$.

Further considerations concerning the consistency of the interacting string theory heavily constrain the lattice $\Gamma$ to be integral, self-dual, and even. In 16 dimensions, the only two such lattices that exist are those associated with $E_8\times E_8$ and $\mathrm{SO}(32)$. As an interesting side note, it has recently (2010) been shown by Adams, deWolfe and Taylor in "String universality in ten dimensions" that the other two choices of gauge groups, in particular $\mathrm{U}(1)^{496}$, do not possess a consistent Green-Schwarz mechanism and are anomalous, so these two groups are really the only allowed gauge groups for a 10D SUGRA $\mathcal{N}=1$ gauge theory.

  • $\begingroup$ Ahhh... its becoming clearer.... So I'm guessing the Kaluza Klein modes (which would normally give $U(1)^{16}$ would correspond to the 16 neutral bosons. And then the winding modes around the torus give the 240+240 charges. And then further guessing if the strings were wound round N times it would give particles with N times the charge. As soon as you said 'winding modes' it all made sense! $\endgroup$
    – user84158
    Sep 14, 2016 at 21:17
  • $\begingroup$ And I guess the integral part is to do with the strings having a preferred length at which they vibrate in the same way electron orbits are at a specific distance from the nucleus? $\endgroup$
    – user84158
    Sep 14, 2016 at 21:30

I think you are thinking of toroidal compactification. For the bosonic string we consider $O(26-d,10-d, \mathbb R)$ for the set of transformations that compactify to the $10$ dimensional supergravity. This is for $26-d$ vertex operators $\partial X^\mu\psi^i$ and $10-d$ vertex operators $\partial X^i\psi^\mu$. The lattice of roots is assumed to be given by some discrete group $\Gamma$ and for $g~\in~O(26-d,10-d)$ we can build a heterotic string theory. For the moduli space $$ {\cal M}~=~\frac{O(26-d,10-d,\mathbb r)}{O(26-d,\mathbb R)\times O(10-d,\mathbb R)\times O(26-d,10-d,\mathbb Z)} $$ $$ =~\frac{g}{g_1\times g_2\times O(26-d,10-d,\mathbb Z)}, $$ where $O(26-d,10-d,\mathbb Z)$ is the T-duality group of the modular or Mobius group of linear fractional transformations. This solution satisfies $g_1gg_2\Gamma~\simeq~g\Gamma$. We then have $10-d$ Kaluza-Klein bosons and $26-d$

For $d~=~5$ this is the compactification of the $26$ dimensional bosonic string and for $d-2$ this is over the $SO(32)$ string when constrained to the special orthogonal group. This then gives $$ SO(32)\times U(1)^{36-2d}~\simeq~E_8\times E_8\times U(1)^{36-2d}. $$ The torus is $36-2d$ dimensional, which for $d~=~10$ is the $\mathbb T^{16}$

  • $\begingroup$ I cannot tell what this answer is trying to say - what does "set of transformations that compactify to the 10-dimensional supergravity" mean? The moduli space of what are you writing down here, and what as this to do with the question? Why does a "lattice of roots" appear? What does the equation $g_1g g_2\Gamma = g\Gamma$ mean? There's a lot of technical terms here but I cannot tell what you're trying to say or how this answers the question. $\endgroup$
    – ACuriousMind
    Sep 14, 2016 at 17:34

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