1
$\begingroup$

The Calabi-Yau manifold is a stable complex 3D (or real 6D) manifold on the geometry of which information about strings can be stored as fibre bundles of tensors, more or less like the electromagnetic four-vector can be seen as a vector on the small circle in Kaluza-Klein theory (corresponding to $\operatorname{U}(1)$ symmetry), represented in the ecxtended 5d metric tensor as eight extra symmetric off-diagonal terms.

The Calabi-Yau manifold incorporates the all three gauge symmetries in particle physics, i.e. $\operatorname{SU}(3)\operatorname{SU}(2)_{l}\operatorname{U}(1)_{Y}$. These gauge symmetries correspond to Lie groups and gauge fields. The gauge transformations leave spacetime untouched, unlike the the gauge transformations used to derive general relativity (corresponding to the Poincaré Poincaré group) which only touch upon spacetime itself (the symmetry demand being obviously that spacetime transformations leave the GR defined Lagrangian unchanged), so not an a particle field in spacetime.

How is this symmetry incorporated in the Calabi-Yau manifold, and how does that introduce graviton modes?

$\endgroup$
19
  • 1
    $\begingroup$ Why would it need to be "incorporated into the Calabi-Yau manifold"? The 10d theory you're compactifying is already a theory of (super)gravity. $\endgroup$
    – ACuriousMind
    Jul 14, 2022 at 17:14
  • $\begingroup$ @ACuriousMind The extra compactified dimensions can be associated with the three basic forces. Say a certain string mode corresponds to electric charge only, so we have an electron. The string vibrates in this way because of the shape of the manifold. How is the charge "mass" induced by the shape of the CY manifold? $\endgroup$
    – Gerald
    Jul 14, 2022 at 17:34
  • 2
    $\begingroup$ It isn't, and very little of what you are saying is compatible with my own technical understanding of string theory: For instance, non-Abelian Yang-Mills type gauge theories come from (coincident) D-branes, not from a string "vibrating because of the shape of the manifold", and for non-Abelian gauge enhancements we typically want singularities (so no smooth CY). $\endgroup$
    – ACuriousMind
    Jul 14, 2022 at 17:43
  • $\begingroup$ @ACuriousMind You mean intersecting branes between which strings can be attached? I'm a bit confused how a vibration causes mass. $\endgroup$
    – Gerald
    Jul 14, 2022 at 17:48
  • $\begingroup$ In general relativity on a spacetime manifold, the gauge group for gravity comes from the structure constants $f_{ab}^c$ the commutators $£_{\xi_a}\xi_b=f_{ab}^c\xi_c$ for a basis $\{\xi_d|d\in I\}$ ($I$ an index set) of the manifold's Killing vector fields. I'd be interested if e.g. @ACuriousMind shows how this relates to the situation in string theory. $\endgroup$
    – J.G.
    Jul 14, 2022 at 17:49

1 Answer 1

3
$\begingroup$
  1. The Yang-Mills type gauge groups of an effective theory in string theory do not arise solely from the compactification, but from "wrapping" D-branes (to which open strings can be attached) around non-trivial cycles in homology - that's the connection to the geometry of the compactification manifold. Non-Abelian gauge groups arise from coincident D-branes (for a discussion of why, see this question and its answers, and more generally this phenomenon is known as gauge enhancement), which in the geometry are typically modelled by taking some limit in which the compactification manifold becomes singular so that some of the distinct cycles "merge" at the singular point and the branes become coincident. Blowing up the singularity (i.e. returning to the non-singular version) corresponds to symmetry breaking (this is mostly an M-theory viewpoint).

  2. In contrast, gravity - which as you say is not gauge theory like the others, for an exact discussion of what is "gauge" about gravity see this answer of mine - is not generated by D-branes or singularities or anything like that: The 10-dimensional uncompactified low-energy effective description of string theory/M-theory, i.e. the theory that is being compactified when we talk about Calabi-Yau manifolds and whatnot, is already a theory of supergravity, because the spectrum of states of the superstring already contains a massless spin-2 particle, and by arguments very similar to how massless spin-1 bosons are always associated with Yang-Mills type gauge theories (see this answer of mine and Weinberg's book for the full argument for spin-1 and this answer and its links for the spin-2 argument), such a massless spin-2 boson is always the graviton of a theory that looks like gravity.

So there is no need to get gravitons from the compactification - they are always there from the outset (and this is arguably the reason people started thinking about string theory as a potential theory of quantum gravity in the first place after it was initially designed to explain QCD).

$\endgroup$
4
  • 1
    $\begingroup$ But how can there already be a spectrum if states in a non-compactified 10d space, in which there are no string states corresponding to other particles yet? Because it's a closed string state, which the graviton is? Or because of the branes in it? $\endgroup$
    – Gerald
    Jul 14, 2022 at 18:43
  • 1
    $\begingroup$ Is the connection with QCD the elastic strings that were envisioned between quarks? Surely these are no gravitons yet. $\endgroup$
    – Gerald
    Jul 14, 2022 at 19:00
  • 1
    $\begingroup$ @Gerald If you think that there is no spectrum in non-compactified 10d space, you're simply wrong. Any introductory text on string theory should derive both the non-compactified open and closed string spectrum. Can you give any references for the claims you are making? $\endgroup$
    – ACuriousMind
    Jul 14, 2022 at 20:15
  • $\begingroup$ "Any introductory text on string theory should derive both the non-compactified open and closed string spectrum" So in 10d non-compactified space there is a spectrum? $\endgroup$
    – Gerald
    Jul 15, 2022 at 8:10

Your Answer

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

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