# M-theory and Calabi-Yau compactifications

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The bundles like U(1) and SU(3) and so on, how are they related to Calabi-Yau compactifications? For example, how does one relate Calabi-Yau compactifications to SU(2)? does the Calabi-Yau compactification give rise to SU(2)?

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mitchell porter
First a detail in the title. d=6 Calabi-Yau spaces are used in d=10 string theory. In d=11 M-theory, to get analogous results, you use d=7 manifolds called G2 manifolds. (If you continue to d=12 F-theory, you use Calabi-Yau again, but now they are d=8.)

There are two classic ways to get gauge groups in string theory. One is to use a stack of branes, and then consider open strings that end on the branes. If there are N branes, the open strings will produce a gauge group of order N, e.g. U(N). See Clifford Johnson's "D-brane Primer" for more about this.

This way of generating a gauge group has no relationship to the Calabi-Yau at all, except in the sense that the brane stack inhabits a submanifold of the Calabi-Yau.

The other way is to start with a big gauge group intrinsic to the strings, and then break it in some way. For example, there are d=10 string theories with gauge groups E8xE8 or SO(32). 1980s string phenomenology revolved around identifying an SU(3) part of one of the E8 fields, with an SU(3)-valued connection on the Calabi-Yau, leaving an E6xE8 grand unified symmetry. The standard model gauge groups were then supposed to arise as subgroups of the E6.

There are still other ways to get gauge groups in string theory, e.g. the d=10 gauge group can be broken by "fluxes"; and @arivero proposes to use the classic Kaluza-Klein approach to obtain the non-chiral U(3) part of the standard model (obtaining chiral SU(2) from Kaluza-Klein is a problem, and is why it was abandoned).

But as I said, the main ways are (1) brane stacks and (2) embedding a spin connection in a stringy gauge group. Only the latter uses a bundle from the Calabi-Yau, and the bundle is identified with a part of the stringy gauge group that does not show up in phenomenology... Sorry if this isn't what you were expecting!

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Yes, over M we have a principal bundle X: The gauge group for example is SU(2). In string theory, we can have a spacetime of the form M × K where K is the compact manifold.

witten says that for example the index of the dirac operator on K has something to do with particle physics?

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Or any of the other gauge groups other than SU(2).

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we have with K the various groups like SU(2)? what is the connection between these?

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mitchell porter
In pre-string Kaluza-Klein theory - where one is dealing with quantum fields rather than strings - the gauge group arises from a symmetry of the compact manifold K. But as I mentioned, the gauge groups in string theory are obtained differently, since most of the K's considered there, don't have that kind of symmetry (continuous isometry group).

As for the Dirac operator, it's relevant for fermions. In the standard model, the observable fermions come from massless chiral fermions in different representations (righthanded are SU(2) singlets, lefthanded are SU(2) doublets) which are coupled via the Higgs field. Chiral fermions in conjugate representations can have a Dirac mass term and are expected to be heavy (admittedly this is a physical, not mathematical deduction, from the assumption that there is a "natural" interaction scale that is heavy).

For the Calabi-Yau, the index of the Dirac operator counts the difference between the number of lefthanded and righthanded spinorial harmonics that the manifold allows. In the 1980s heterotic models, that's the number of fermion families that will remain light because they don't pair up with a conjugate counterpart.

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Yes, I was discussing with my friend about how there is a principal bundle X over M.
what are these? is it like su(2)? and is m-theory in some sense the 'queen'? or is it the second, version of supergravity?

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To add to my question above, as I said we have a principal bundle I have read somewhere that, if the gauge group is G, and we have a representation of G on F, we have something called the associated bundle They also write, that the higgs field has something to do with this associated bundle. Does anyone know the details or references related to this? I also want to know why maps π-1(Ui) to Ui × F is called a local trivialisation where Ui is an open cover of M?

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bayakiv
And you do not want to know which geometric object has the symmetry (continuous group of isometries) of the group U(2)? If you want to know, then this is a torus laid (stretched) on a sphere so that its outer circle coincides with the northern polar circle, and inner circle with the southern one. In this case, the group U(2) is generated by the proper isometries of the torus and its rotations along the surface of the sphere.

bayakiv
In this case, the group U(2) is generated by the proper isometries of the torus and its rotations along the surface of the sphere.
By the way, the group SU(2) is generated by the shifts of the torus along its own circles of Villarso and its rotations along the sphere. Accordingly, to generate a group SU(3), it will be necessary to move the Willarso circles of the three-dimensional torus and rotate this torus along the surface of the three-dimensional sphere.

mitchell porter
There is a difference between the bundles of gauge theory, and the compact spaces of Kaluza-Klein and string/M theory. Those compact spaces are part of space-time, they are extra dimensions of space-time, and in particular the space-time metric is extended to describe them too. On the other hand, the various bundles of gauge theory are not extra space-time dimensions, they are geometric objects attached to the space-time points, and they are not described by the space-time metric.

There is also a difference between how Kaluza-Klein theory and how string/M theory obtain gauge groups. In Kaluza-Klein theory, the gauge groups come from the isometries of the compact space. In string theory, they mostly arise in other ways, as I suggested.

U(2) equals SU(2) x U(1), so it can arise as the isometry group of the manifold S^2 x S^1. But this is not the only way to obtain it.

@love_42, I find myself wondering if you are actually asking about "Higgs bundles". They are different from Higgs fields, though inspired by them.

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bayakiv
U(2) equals SU(2) x U(1), so it can arise as the isometry group of the manifold S^2 x S^1. But this is not the only way to obtain it.
So it is so, but in my example, the dimension of the manifold is smaller. In addition, in my example, the mass of an electron can probably be interpreted as the latitude of the polar circle.

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mitchell porter
I got confused about the isometry groups for a moment... @bayakiv is embedding one object in another and looking at the isometries of the combined object, I will discuss that in a moment. Normally (in Kaluza-Klein), one looks for a single compact manifold whose complete group of isometries is the desired gauge group.

I don't actually know a manifold whose isometry group is just SU(2). I mentioned S^2 but its isometry group is SO(3). CP1 comes close but its isometry group is actually SU(2)/Z2. But the way you might do it, is to find a manifold whose isometry group contains SU(2) as a subgroup, and then take a quotient of that manifold which breaks the larger symmetry, in a way such that only SU(2) remains.

As for product groups like U(2) or the standard model gauge group, you could get them by taking a product of manifolds whose symmetry groups are the factors in the group product. But one should also think holistically, and in the way previously mentioned: one could instead find a manifold with a larger group as isometry group, e.g. a manifold with a GUT group as isometry group, and then quotient the manifold just to get the SM group.

@bayakiv, I don't understand your scenarios. Are the original inner and outer circles of the torus pinned to those polar latitudes, or do you allow the torus to rotate around the torus's poloidal direction? (so that different toroidal circles of the torus, take turns being mapped to the polar latitudes). Also, is the torus-laid-on-the-sphere only allowed to rotate along the latitudes, or can it also tilt? And finally, no matter how I interpret the scenario, I don't see how you get U(2).

bayakiv
Maybe the following explanation will help you. The toroidal shell of a sphere is placed on a sphere between its polar circles, so if you draw on this shell a circle having a sphere radius that intersects two opposite points lying in two different polar circles, then one half of the circle is drawn on the outer side of the shell, and the other on internal. If we fix with the help of a needle an arbitrary point of the toroidal shell on the sphere, then it will still be able to rotate along the surface of the sphere, and if we also add the shell's own motion along Villarso circles (i.e., along the circles connecting opposite points of two polar circles ), then the complete group generated by these two groups is exactly the group SU(2 ). In turn, the extension of a group SU(2 ) to a group U(2 ) is due to the assumption of the complete group of proper motions of the torus and the complete group of rotation of the torus on the sphere.

bayakiv
In addition, in my example, the mass of an electron can probably be interpreted as the latitude of the polar circle.
It seems that I said this without thinking properly. Most likely, the latitude of the polar circle is responsible for the ratio of the coupling constants of the electromagnetic and weak interactions, and for the electron mass, another property of the toroidal shell should be chosen, for example, the length of the closed winding of the torus.

bayakiv
For @mitchell porter and those interested in the formation of lepton masses, I have one remarkable observation. If we assume that the lepton mass is proportional to the exponential of the pseudo-Euclidean length of the torus knot, then we get a fairly good correspondence of the electron to the torus knot (3,2), the muon to the torus knot (3,5), and the tau to the torus knot (3,7). Note here that the square of the pseudo-Euclidean torus knot length (p,q) is pq. However, it should be noted that the torus knot, which is responsible for the formation of mass, is formed not on the torus of internal symmetries, but on the torus of time.

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bayakiv
However, it should be noted that the torus knot, which is responsible for the formation of mass, is formed not on the torus of internal symmetries, but on the torus of time.
Here I again did not think, but after thinking I realized that torus knots should be knitted simultaneously on the torus of time and on the torus of internal symmetries.

bayakiv
It turns out that the ratio of the masses of leptons is described by the Koide formula and is equal to 2/3, but its meaning is hidden from me and therefore I will still return to my analogies, that is, to the representation of leptons by torus knots. So, let's try to fit the result of calculating the masses of leptons to the experimental data. Let the "mass" of the torus knot be equal to the exponent of the square of its pseudo-Euclidean length. Then if we take the Villarso circle node (1,1), the trefoil node (2,3) and the Cinquefoil node (2,5) as the electron, muon, and tau-lepton, then we will get "masses", the ratio of which will now be better match experimental data than what I presented earlier.

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bayakiv
It turns out that the ratio of the masses of leptons is described by the Koide formula and is equal to 2/3, but its meaning is hidden from me...
It seems I guessed where the secret meaning of Koide's formula is buried. If we take the sectoral (90 degrees) area of the golden spiral as the lepton mass, then in the limit we get that three successive sectors of the golden spiral satisfy a condition very similar to the Koide formula. Indeed, it is easy to prove that the limit of the ratio of the square of the sum of three consecutive Fibonacci numbers to the sum of their squares is equal to the square of the golden ratio. And each Fibonacci number is equal to the side of the square in which the sector of the golden spiral is inscribed. It remains only to correct the formula taking into account the dependence of the area of the sector of the golden spiral on the area of the corresponding square.

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bayakiv
it is easy to prove that the limit of the ratio of the square of the sum of three consecutive Fibonacci numbers to the sum of their squares is equal to the square of the golden ratio

Cyroos
Maybe the following explanation will help you. The toroidal shell of a sphere is placed on a sphere between its polar circles, so if you draw on this shell a circle having a sphere radius that intersects two opposite points lying in two different polar circles, then one half of the circle is drawn on the outer side of the shell, and the other on internal. If we fix with the help of a needle an arbitrary point of the toroidal shell on the sphere, then it will still be able to rotate along the surface of the sphere, and if we also add the shell's own motion along Villarso circles (i.e., along the circles connecting opposite points of two polar circles ), then the complete group generated by these two groups is exactly the group SU(2 ). In turn, the extension of a group SU(2 ) to a group U(2 ) is due to the assumption of the complete group of proper motions of the torus and the complete group of rotation of the torus on the sphere.
Villarso circles have the intrinsic property to make the toroidal shell on the sphere be placed between two arbitrary adjacent ribs scaled-down/up to symmetrical planes. If the rotational angle of the motion meridian circle does lie between two arbitrary adjacent ribs , then (quazi-)symmetry IS lost to a polygonal shape with three vertices attached by ridges to each other as well as to the indentation point. This spells that the vertical supports are located at the outer and inner equatorial circles of latitudinal bundle to transfer the binding fibers of the toroidal , in such a manner as to make the support reactive motions tangential to the shell middle surface.

bayakiv
@Cyroos Without drawing, we will not understand each other.

Cyroos
@Cyroos Without drawing, we will not understand each other.
can I adopt image(s) from internet ??

bayakiv
can I adopt image(s) from internet ??
I do not know. Probably modern visualization software are easy to cope with the problem. You just need to draw lines in different colors on the inner and outer side of the toroidal shell of a sphere without polar circles.

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Cyroos
I have made a word file as attached . ... .

sincerely
Cyroos

#### Attachments

• loaorbit.docx
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bayakiv
It is not surprising to depict Villarso's circles. And will you try to draw a toric 2,3-knot on the shell?

• Cyroos
Cyroos
It is not surprising to depict Villarso's circles. And will you try to draw a toric 2,3-knot on the shell?
You are very intelligent smart , bayakiv . .... this is comonly where the scenario leasds to . .. .However , in this case , we are dealing with topologization imaginiing plane topology rather than similar TO POINT-SET TOPOLOGY . . WE WANT NOT TO GO THAT WAY .. . We simply wanna bear it on our minds that toric knots rather than Villarso circles are the departure line of bifurcation.Thence :
At the focal point (F), the inherent property of the Vilarso circles is that the first rib practically draws the half-angle, while the second rib thus created transforms the first virtuality into an actual fork. Drawing the half-tilt symmetry is the most significant thing that we two have to do together. Scaling the latitude, then , would [ with very little doubt ] make it feasible for us both to soon get engaged with some of the thorniest issues in hydrostatics, Quantum Mechanics, and , later, the development of unimaginably tiny microfluidic chips in medicine( size less than 1 mm ). Quazi-symmetry would , of course , get us both somewhere. Nonetheless, Now that the Hand of Fate has brought our ideas and exchanges so close to one another , We two had MUCH better go right for nothing less than a full symmetry.

best Regards
prof Cyroos Sanaye

bayakiv
@Cyroos, If I understood you correctly, then in the geometric construction of the toroidal shell of the sphere you see not only the symmetry that generates the group SU(2), but also the quasi-symmetry associated with the latitude of the polar circles. I would like to add to this that we can still talk about the symmetries of the vector field of the toroidal shell, which has streamlines in the form of various torus knots (including Villarso's circles).

Best regards,
Igor V. Bayak

Cyroos
I would like to add to this that we can still talk about the symmetries of the vector field of the toroidal shell, which has streamlines in the form of various torus knots (including Villarso's circles).

Agreed. We shall continue discussing things in even more detail. I have to be off to the airport right now in order not to lose my flight. Make it a promise that we two shall continue together on this thread ( and arrive at its commercial benefits, too ) . . . .. . . .

Sincerely Yours
Reza Sanaye

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bayakiv
@Cyroos, I will be happy to continue working with you. But I have never been able to profit from my scientific interest.

Cyroos
I'll show U how to profit . . . ..

• weirdoguy and berkeman
Mentor
I'll show U how to profit . . . ..
Caution, @bayakiv -- This looks like a scam setup...

bayakiv
@berkeman You needlessly worry about me. I work for an enterprise subject to US sanctions, so is it worth fearing fraud when faced with the banditry of one state in relation to another? By the way, commerce is quite appropriate here, you can, for example, sell a vortex plasma thruster to an interested party.

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Cyroos
Dear Bayakiv !
i am no plotter / no scammer / no intriguer . ..
I just know how to put our in-common brand/branch of topology to best commercial use in Japan . . .
You decide for yourself to work with me in the long run OR not to . . .

Respect
Regards
Cyroos

• weirdoguy