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  Differential Geometry of the Standard Model

+ 4 like - 0 dislike
1519 views

I am trying to understand the mathematical structure of the standard model according to the spirit of the Nakahara`s book  "Geometry, Topology and Physics" .  In few words the Nakahara`s spirit consists in to convert in physics the theorems in "Foundations of Differential Geometry Vol 1" by Kobayashi and Nomizu.   In this sense I think that the proposition 6.3 of Kobayashi-Nomizu gives a very important foundation of the mathematical structure of the standard model from the point of view of the modern differential geometry.   

Proposition 6.3  Let ΓP a connection in the principal bundle P(M,G)  and let  ΓQ a connection in the principal bundle Q(M,H).  Let P×Q the principal bundle over M×M with structure group G×H.  Let P+Q the restriction of P×Q to the diagonal ΔM of M×M . Since ΔM and M are diffeomorphic with each other in a natural way, we consider P+Q as a principal  bundle over  M with group G×H.. The restriction of the projection  P×QP to the projection P+QP, denoted by  hP, is a homomorphism with the corresponding natural homomorphism hG:G×HG. Similarly, for  hQ:P+QQ and hH:G×HH.  Then

(a) There is a unique connection  Γ  in the principal bundle  P+Q  such that  the  homomorphisms hP:P+QP and hQ:P+QQ maps Γ  into  ΓP  and ΓQ respectively.

(b)  If   ω, ωP and ωQ  are the connection one- forms and Ω, ΩP and ΩQ, are the curvature  two-forms of Γ, ΓP and ΓQ respectively, then

ω=hPωP+hQωQ

Ω=hPΩP+hQΩQ

My questions are: 

1. How to prove the proposition 6.3?

2. How to apply the proposition 6.3 to the Standard Model?

asked Jul 26, 2015 in Theoretical Physics by juancho (1,130 points) [ revision history ]
edited Jul 26, 2015 by juancho

The proposition says in words simply that given gauge fields with structure group G and H, then there is the joint gauge field with structure group G x H. So if you have an SU(3)-connection modelling the strong force (say in some instanton sector) and an SU(2)xU(1) field for the electroweak force, then you may combine them to a single SU(3)xSU(2)xU(1) gauge field.

@UrsSchreiber, many thanks for you comment, it is very illustrative.  According with your comment in the standard model with gauge group SU(3)xSU(2)xU(1)  the proposition 6.3 implies that

ωSU(3)×SU(2)×U(1)=hSU(3)ωSU(3)+hSU(2)ωSU(2)+hU(1)ωU(1)

.

Please let me know which must  be the general expressions for: hSU(3)ωSU(3), hSU(2)ωSU(2) and hU(1)ωU(1).

The precise form of ωSU(3) etc. is not fixed (beyond the Cartan conditions it needs to satisfy) but is what encodes the gauge field itself. That's the conent of what it means to say that gauge fields in physics are mathematically modeled by principal bundles with connection.

Famous choices for the connection are instanton configurations. Look for that term, and you'll find plenty of explicit and relevant examples-

@UrsSchreiber, my apologies I want mean the general expressions.  Many thanks.

2 Answers

+ 3 like - 0 dislike

2. Application of the proposition 6.3 to the Standard Model(Thanks to Urs Schreiber).

Let M a four-dimensional space-time manifold, let ΓC   a connection in the principal bundle for the color denoted C(M,SU(3)C),  let ΓL   a connection in the principal bundle for the left weak isospin denoted L(M,SU(2)L) and let ΓY   a connection in the principal bundle for the hypercharge denoted Y(M,U(1)Y)). Let C×L×Y  the principal bundle over M×M×M with structure group SU(3)C×SU(2)L×U(1)Y. Let C+L+Y the restriction of  C×L×Y to the diagonal ΔM of M×M×M .  Since ΔM and M  are diffeomorphic with each other in a natural way, we consider C+L+Y  as a principal  bundle over  M  with  structure group SU(3)C×SU(2)L×U(1)Y. The restriction of the projection C×Y×LC to the projection C+L+YC, denoted  by hC, is a homomorphism with the corresponding natural homomorphism hSU(3)C:SU(3)C×SU(2)L×U(1)YSU(3)C

.

Similarly  for hL:C+L+YL with

hSU(2)L:SU(3)C×SU(2)L×U(1)YSU(2)L

;

and  hY:C+L+YY with

hU(1)Y:SU(3)C×SU(2)L×U(1)YU(1)Y

,

Then,

(a) There is a unique connection  Γ  in the principal bundle for the standard model chromo-electroweak  C+L+Y  such that  the  homomorphisms

hC:C+L+YC

,

hL:C+L+YL

and

hY:C+L+YY

maps Γ  into ΓC , ΓL and ΓY respectively.

(b)  If   ωωC , ωL and ωY  are the connection one- forms and Ω, ΩC, ΩL and ΩY, are the curvature  two-forms of Γ, ΓC, ΓL and ΓY respectively, then

ω=hCωC+hLωL+hYωY

Ω=hCΩC+hLΩL+hYΩY


More explicitly we have :

ωμ=(hCωC)μ+(hLωL)μ+(hYωY)μ

ωμ=hC(ωC,μ)+hL(ωL,μ)+hY(ωY,μ)

ωμ=hC(gCˆλα2Aαμ)+hL(gLˆσβ2Bβμ)+hY(gLY2Cμ)

where gC, gL and gY are the coupling constants; ˆλα are the eight Gell-Mann matrices which are the generators of the Lie group SU(3), ˆσβ are the three Pauli matrices which are the generators of the Lie group SU(2) and Y is the generator of the Lie group U(1). Aαμ, Bβμ and Cμ are the gauge potentials for the vector bosons.

Then we have:

ωμ=gChC(ˆλα)2Aαμ+gLhL(ˆσβ)2Bβμ+gYhY(Y)2Cμ

ωμ=gChSU(3)C(ˆλα)2Aαμ+gLhSU(2)L(ˆσβ)2Bβμ+gYhU(1)Y(Y)2Cμ

ωμ=gCˆλαI22Aαμ+gLI3ˆσβ2Bβμ+gYI3I2Y2Cμ

ωμ=gCˆλαI22Aαμ+gLI3ˆσβ2Bβμ+gYI6Y2Cμ

where I2 is the two-dimensional identity matrix, I3 is the three-dimensional identity matrix and I6 is the six-dimensional identity matrix.

The covariant derivative for a field which is a 0-form in the fundamental representation of SU(3)C×SU(2)L×U(1)Y has the form

Dμ=I6μ+gCˆλαI22Aαμ+gLI3ˆσβ2Bβμ+gYI6Y2Cμ

The covariant derivative for a field which is 0-form in the adjoint representation of  SU(3)C×SU(2)L×U(1)Y has the form

DμΦ=I6μΦ+[gCˆλαI22Aαμ+gLI3ˆσβ2Bβμ+gYI6Y2Cμ,Φ]

where

Φ=Φα1ˆλαI22+Φβ2I3ˆσβ2+Φ3I6Y2

answered Jul 28, 2015 by juancho (1,130 points) [ revision history ]
edited Jul 29, 2015 by juancho
+ 2 like - 0 dislike

1.  Proof of the proposition 6.3:

(b) First we prove that  ω((Ra)X)=ad(a1)ω(X) where aG×H and X(P+Q)u with uP+Q.  We assume that a=aG×aH where aGG , aHH and then hG(a)=aG and hH(a)=aH .  Also we assume that hP(X)=XP and hQ(X)=XQ. and then ωP((RaG)XP)=ad(a1G)ωP(XP) and ωQ((RaH)XQ)=ad(a1H)ωQ(XQ) . Then we have that

ω((Ra)X)=(hPωP+hQωQ)((Ra)X)

ω((Ra)X)=(hPωP)((Ra)X)+(hQωQ)((Ra)X)

ω((Ra)X)=ωP(hP((Ra)X))+ωQ(hQ((Ra)X))

ω((Ra)X)=ωP((RhG(a))hP(X))+ωQ((RhH(a))hQ(X))

ω((Ra)X)=ωP((RhG(aG×aH))hP(X))+ωQ((RhH(aG×aH))hQ(X))

ω((Ra)X)=ωP((RaG)hP(X))+ωQ((RaH)hQ(X))

ω((Ra)X)=ωP((RaG)XP)+ωQ((RaH)XQ)

ω((Ra)X)=ad(a1G)ωP(XP)+ad(a1H)ωQ(XQ)

From other side we have that

ad(a1)ω(X)=aω(X)a1

ad(a1)ω(X)=(aG×aH)ω(X)(aG×aH)1

ad(a1)ω(X)=(aG×aH)ω(X)(a1G×a1H)

ad(a1)ω(X)=(aG×aH)[(hPωP+hQωQ)(X)](a1G×a1H)

ad(a1)ω(X)=(aG×aH)[(hPωP)(X)+(hQωQ)(X)](a1G×a1H)

ad(a1)ω(X)=(aG×aH)[ωP(hP(X))+ωQ(hQ(X))](a1G×a1H)

ad(a1)ω(X)=(aG×aH)[ωP(XP)+ωQ(XQ)](a1G×a1H)

ad(a1)ω(X)=(aG×aH)ωP(XP)(a1G×a1H)+(aG×aH)ωQ(XQ)(a1G×a1H)

ad(a1)ω(X)=(aGωP(XP)a1G)(aHa1H)+(aGa1G)(aHωQ(XQ)a1H )

ad(a1)ω(X)=(aGωP(XP)a1G)(eH)+(eG)(aHωQ(XQ)a1H )

ad(a1)ω(X)=aGωP(XP)a1G+aHωQ(XQ)a1H

ad(a1)ω(X)=ad(a1G)ωP(XP)+ad(a1H)ωQ(XQ)

.

Second, we prove that ω(A)=A as follows.

ω(A)=(hPωP+hQωQ)(A)

ω(A)=(hPωP)(A)+(hQωQ)(A)

ω(A)=ωP(hP(A))+ωQ(hQ(A))

ω(A)=ωP(AP)+ωQ(AQ)

ω(A)=AP+AQ

ω(A)=A

Finally, we prove that  Ω=hPΩP+hQΩQ.

Ω=dω+12[ω,ω]

Ω=d(hPωP+hQωQ)+12[(hPωP+hQωQ),(hPωP+hQωQ)]

Ω=d(hPωP)+d(hQωQ)+12[hPωP,hPωP]+12[hPωP,hQωQ]+

12[hQωQ,hPωP]+12[hQωQ,hQωQ]

Ω=d(hPωP)+d(hQωQ)+12[hPωP,hPωP]+0+0+12[hQωQ,hQωQ]

Ω=d(hPωP)+d(hQωQ)+12[hPωP,hPωP]+12[hQωQ,hQωQ]

Ω=d(hPωP)+d(hQωQ)+12hP([ωP,ωP])+12hQ([ωQ,ωQ])

Ω=hP(dωP)+hQ(dωQ)+12hP([ωP,ωP])+12hQ([ωQ,ωQ])

Ω=hP(dωP+12[ωP,ωP])+hQ(dωQ++12[ωQ,ωQ])

Ω=hPΩP+hQΩQ

.

answered Jul 27, 2015 by juancho (1,130 points) [ revision history ]
edited Jul 27, 2015 by juancho

I think it's better to put it into one single answer?

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