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String theory - for example - requires extra spatial dimension. Say for example in 10 dimensional string theory, what theoretically prevents clustering of the extra 6 dimensions in 2 timeless 3 dimensional (infinite) spaces?

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up vote 6 down vote accepted

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.

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.

I will assume that the word clustering in the question (v2) essentially refers to if $(K^6,g^{(6)})$ could be a product $K^6=K^3\times L^3$ with metric $g^{(6)}=g^{(3)}\oplus h^{(3)}$ of two $3$-dimensional manifolds $(K^3,g^{(3)})$ and $(L^3,h^{(3)})$?

Again, to have avoided experimental detection, the two $3$-dimensional manifolds $K^3$ and $L^3$ must both be compact. Now, another bit of traditional string wisdom is, that to have unbroken $N=1$ supersymmetry i $4$ 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 question.

On the other hand, the biggest holonomy group that a $3$-dimensional Riemannian manifold can have, 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.

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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. –  Ron Maimon Sep 1 '11 at 4:10
    
Nice answer, +2 (to compensate for Ron s accidental downvote) :-) –  Dilaton Nov 30 '11 at 19:20
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