To compactify 2 open dimensions to a torus, the method of identification written down for this example as

$$ (x,y) \sim (x+2\pi R,y) $$

$$ (x,y) \sim (x, y+2\pi R) $$

can be applied.

What are the methods to compactify 6 open dimensions to a Calaby-Yau manifold and how exactly do these methods work?


What you wrote down is the quotient of the 2-dimensional translation group by a discrete subgroup. But by far not every closed manifold arises as a quotient of groups this way.

One should be aware that the term "compactification" in physics is used not so much to refer to what in mathematics is called compactification of non-compact spaces. For instace one-point compactification in the mathematical sense turns the real line line into the circle. (However it also turns the plane into the 2-sphere, not into the torus.)

Instead, what is meant by "compactification" in physics is that you just choose a closed (and hence compact) manifold $Q$, then choose spacetime $X$ to be a $Q$-fiber bundle over space base space (often assumed to be just a product $X = Q \times Y$), and then describe the Kaluza-Klein mechanism for passing from physics on $X = Q \times Y$ to effective physics on just $Y$.

In particular for Calabi-Yau "compactifications" you just choose $Q$ to be a Calabi-Yau manifold, and then consider the Kaluza-Klein mechanism on spacetimes which are $Q$-fiber bundles. You don't actually obtain these spacetimes as compactifications of non-compact spacetimes in the sense of mathematics.

(Well one could consider that problem, but this is not what is generally meant by "compactification" in physics.)

  • 1
    $\begingroup$ Thanks for this nice clarification, going to follow the links now :-) $\endgroup$ – Dilaton Jul 14 '13 at 18:20
  • 2
    $\begingroup$ I thought that "$Q$" was the Calabi-Yau manifold, not $Y$. For instance $X_{10} = CY_6 * M_{3,1}$. So I would have thought of $CY$-fiber bundles with a Minkowski base space $M_{3,1}$ $\endgroup$ – Trimok Jul 15 '13 at 9:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.