# Could extra dimensions be or become clustered?

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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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.