3
$\begingroup$

In many textbooks the typical story is told that there are exactly 2 possible heterotic theories in 10 dimensions, the $E8\times E8$ as well as the $SO(32)$ theories. This is of course only true for supersymmetric theories. If one further demands stability, i.e. no tachyons the $O(16)\times O(16)$ theory is the only stable non-supersymmetric theory. But if one allows for tachyons much more is possible. In the recent arXiv paper https://arxiv.org/abs/2010.10521 by Kaidi some twisted theories are discussed using a gauging of the $\mathbb{Z}_2^5$, resulting in theories with $2^{5-n}$ tachyons where $n=1,\ldots,5$. These theories have gauge groups like $E8$, $(E7\times SU(2))^2$ or $O(8)\times O(24)$. In a footnote the author mentions that these are not all possible theories and one could further allow for torsion in the gauged $\mathbb{Z_2}$ groups. Also, this paper focuses purely on theories obtained from twisting the $SO(32)$ string theory, I would assume that similar twists exist for the $E8\times E8$ theory.

So my question is: Does there exist a classification of possible heterotic theories in 10 dimensions? Or if this is to broad, are there further known examples besides the ones mentioned above? Of course the theories should still be consistent in the sense that they are modular invariant and anomaly free. Especially interesting would be a theory with exactly 6 tachyons, which is not obtainable via gauging the $\mathbb{Z_2}$ symmetries.

$\endgroup$
2
  • $\begingroup$ Is the following helpful? arxiv.org/abs/hep-th/9112006 $\endgroup$ – Kevin Feb 4 at 16:14
  • $\begingroup$ It is for sure a interesting source, especially since it gives more examples. But I am not sure if it is actually a complete classification as it claims. It is claimed that there is only a single non-lattice theory. For example I do not see how the tachyonic $E_8$ theory appears in this classification. But I am also not really used to the notation they use so it could be included and I just fail to see how. $\endgroup$ – Rohbar Feb 4 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.