What exactly is an orbifold? I've come across orbifolds on several occasions and I know they are important to string theory, but what is an orbifold? I've seen some very technical mathematical definitions, but I was wondering if there was a more basic/intuitive definition. Also, what is the physical interpretation of an orbifold?

  • $\begingroup$ For comparison so that answerers know what you're looking for here, what would you consider a "basic" or "intuitive" definition of a manifold? $\endgroup$
    – ACuriousMind
    Commented Jun 12, 2020 at 15:35
  • 1
    $\begingroup$ I would say a "basic/intuitive" definition is that a (smooth) manifold is locally diffeomorphic to Euclidean space. Though, I am familiar with the definition of manifolds in terms of coordinate charts, transition functions, etc. $\endgroup$
    – user267286
    Commented Jun 12, 2020 at 15:39
  • 1
    $\begingroup$ Due diligence? Like this? $\endgroup$ Commented Jun 12, 2020 at 16:08
  • $\begingroup$ Just to clarify Ahaha's "basic/intuitive" definition of a smooth manifold: consider an n-dimensional space, $V_n\,,$ and a point p of it and a small enough region around p that can be covered with a finite number of "open sets" (which are kind of like open balls). If that region can be mapped to $\mathbb{R}^n$, i.e. n-dimensional Euclidean space, by a function which is differentiable to all orders (this results in smoothness) then one is on the way to calling $V_n$ a smooth manifold. There are some technical details being left out in this description. $\endgroup$ Commented Jun 15, 2020 at 20:07
  • $\begingroup$ I should add that the mapping of the region around $p$ from $V_n$ to $\mathbb{R}^n$ needs to be one-to-one, onto and invertible. There are still some technical details omitted. $\endgroup$ Commented Jun 15, 2020 at 21:27

1 Answer 1


From the mathematical perspective: Orbifolds are locally quotients of differentiable manifolds by finite groups. They are smooth except for very simple, finite and tractable class of singularities.

The amazement is that despite its simplicity, orbifolds share strong similarities with much more general spaces such as stacks. It is also relatively easy (and suprisingly beautiful) to give explicit constructions of very important algebraic data such as coherent sheaves and use them to verify and impressive class of very deep phenomena such as equivalences of derived categories of coherent sheaves or the McKay correspondence.

Now, why are orbifolds so important for string theory? The answer is that orbifolds provide a wide class of examples of singular spacetimes at which strings can propagate in a demonstrably consistent way (see the classic Strings On Orbifolds).

A basic expectation is that a truly quantum theory of gravity should be able to deal with situations were the curvature of spacetime is very high (even planckian) as was the curvature of the very early universe.

You can read about propagation of strings even in elementary string theory textbooks such as the one of Zwiebach. You can also learn how branes provide physical mechanisms for singularity resolution or how string theory successfully deal with topology change in spacetime and how all the latter can be used to exactly compute black hole degeneracies or to provide phenomenologicaly realistic scenarios were famous no-go theorems are circumvented.

The understanding of the physics of black hole and cosmological singularities is one of the most greatest goals in theoretical physics. Even is possible that full quantum gravity can be understood purely in terms of high-curvature spacetime fluctuations (Wheeler's spacetime foam) and strings propagating on orbifolds are a beautiful example of how string theory is guiding us towards the achieve of those dreams.

Edit: I forgot to tell that ADE like singularities can be explicity defined as "Branes". For example, an $A_{N}$ singularity in type IIB superstring theory can be seen to be equivalent to n sparated M5-branes after a lift to M-theory . The dictionary can be found in Branes and toric geometry.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.