Could the 6 extra dimensions in superstring theory be a product of two manifolds?

  • $\begingroup$ QMechanic, please look to the question: physics.stackexchange.com/q/255944 $\endgroup$
    – Dmitry
    May 17 '16 at 8:43
  • 1
    $\begingroup$ @Dmitry: Tip: To ping a user in a comment write '@username'. It is not enough to write 'username'. $\endgroup$
    – Qmechanic
    Jan 11 '17 at 16:34

I) I will here only comment on the traditional superstring theory story, say, from the first superstring revolution in the 1980s, and leave it to others to include more recent developments.

II) Traditionally, the $10$-dimensional target space $(M^{10},g^{(10)})$ with a metric $g^{(10)}$ is viewed as a product $$M^{10}~=~M^4 \times K^6$$ with metric $g^{(10)}=g^{(4)}\oplus g^{(6)}$, where $(M^4,g^{(4)})$ is the $4$-dimensional spacetime with a $4$-metric $g^{(4)}$, which we see and observe; and $(K^6,g^{(6)})$ is a compact $6$-dimensional Riemannian manifold, whose characteristic length scales are so small that it has avoided experimental detection so far.

Let us mention for later that the biggest holonomy group a $6$-dimensional Riemannian manifold can have, is the $15$-dimensional Lie group $O(6)$, which is locally isomorphic to $SU(4)$.

III) Let us now reformulate OP's question as follows.

Could the compact manifold $(K^6,g^{(6)})$ be a product $$K^6~=~K^{6-n}\times L^n$$ with metric $g^{(6)}=g^{(6-n)}\oplus h^{(n)}$ of a $(6\!-\!n)$-dimensional manifold $(K^{6-n},g^{(6-n)})$ and a $n$-dimensional manifold $(L^n,h^{(n)})$, where $n=1,2,3$?

I will argue below that that is not possible.

IV) Again, to have avoided experimental detection, the two manifolds $K^{6-n}$ and $L^n$ must both be compact. Now, another bit of traditional string wisdom is, that to have unbroken ${\cal N}=1$ supersymmetry in $4$ spacetime dimensions, the holonomy group of $(K^6,g^{(6)})$ must be the $8$-dimensional Lie group $SU(3)$, see e.g., Green, Schwarz and Witten, Superstring theory, chap. 15. See also this Phys.SE question.

Case $n=3$: The maximal holonomy group of a $3$-dimensional Riemannian manifold is the $3$-dimensional Lie group $O(3)$, so $K^6=K^3\times L^3$ can at most have holonomy group $O(3)\times O(3)$, which is $6$-dimensional, and therefore too small to be $SU(3)$. Hence a product manifold $K^6=K^3\times L^3$ is ruled out.

Case $n=2$: A similar argument rules out a product manifold of the form $K^6=K^4\times L^2$, because the corresponding maximal holonomy group $O(4)\times O(2)$ is only $7$-dimensional.

Case $n=1$: Finally, a product manifold of the form $K^6=K^5\times L^1$ is ruled out because $SU(3)$ is not$^1$ a subgroup of $O(5)$.


$^1$$SU(3)$ is not a subgroup of $O(5)$. Maximal subgroups of $SO(5)$ are isomorphic to the $6$-dimensional Lie group $SO(4)$.

Alternatively: If the Lie algebra $su(3)$ is a subalgebra of $so(5)$, then the complexification $sl(3)=A_2$ must be a subalgebra of $so(5,\mathbb{C})=B_2$, but the root system of $A_2$ does not fit inside the root system of $B_2$. Contradiction.

  • $\begingroup$ I downvoted this for a stupid reason, and realized my mistake too late--- the downvote is already locked in. My apologies, the argument is completely fine. $\endgroup$
    – Ron Maimon
    Sep 1 '11 at 4:10
  • $\begingroup$ Nice answer, +2 (to compensate for Ron s accidental downvote) :-) $\endgroup$
    – Dilaton
    Nov 30 '11 at 19:20

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