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  Differential Geometry of the Standard Model

+ 4 like - 0 dislike
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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 $\Gamma_{{P}}$ a connection in the principal bundle $P \left( M,G \right)$  and let  $\Gamma_{{Q}}$ a connection in the principal bundle $Q \left( M,H \right)$.  Let $P \times Q$ the principal bundle over $M \times M$ with structure group $G \times H$.  Let $P + Q$ the restriction of $P \times Q$ to the diagonal $ \Delta M$ of $M \times M$ . Since $ \Delta M$ and $M$ are diffeomorphic with each other in a natural way, we consider $ P + Q$ as a principal  bundle over  $M$ with group $G \times   H$.. The restriction of the projection  $P \times Q  \rightarrow P$ to the projection $ P + Q \rightarrow P $, denoted by  $h_P$, is a homomorphism with the corresponding natural homomorphism $h_G : G \times H \rightarrow G$. Similarly, for  $h_Q : P + Q \rightarrow Q$ and $ h_H : G \times H  \rightarrow H$.  Then

(a) There is a unique connection  $\Gamma$  in the principal bundle  $P + Q$  such that  the  homomorphisms $h_P  : P + Q \rightarrow P$ and $h_Q : P + Q \rightarrow Q$ maps $\Gamma$  into  $\Gamma_{{P}}$  and $\Gamma_{{Q}}$ respectively.

(b)  If   $\omega$, $\omega_P$ and $\omega_Q$  are the connection one- forms and $\Omega$, $\Omega_P$ and $\Omega_Q$, are the curvature  two-forms of $\Gamma$, $\Gamma_P$ and $\Gamma_Q$ respectively, then

$$\omega = h^*_P \omega_P + h^*_Q \omega_Q $$

$$\Omega = h^*_P \Omega_P + h^*_Q \Omega_Q $$

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 (1,030 points) [ revision history ]
edited Jul 27, 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

$$\omega_{SU(3)\times SU(2) \times U(1)} = h^*_{SU(3)} \omega_{SU(3)} + h^*_{SU(2)} \omega_{SU(2)}+h^*_{U(1)} \omega_{U(1)}$$.

Please let me know which must  be the general expressions for: $h^*_{SU(3)} \omega_{SU(3)}$, $h^*_{SU(2)} \omega_{SU(2)}$ and $h^*_{U(1)} \omega_{U(1)}$.

The precise form of \(\omega_{SU(3)}\) 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

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

Let $M$ a four-dimensional space-time manifold, let $\Gamma_{{C}}$   a connection in the principal bundle for the color denoted $C \left( M,SU(3)_C \right)$,  let $\Gamma_{{L}}$   a connection in the principal bundle for the left weak isospin denoted $L \left( M,SU(2)_L \right)$ and let $\Gamma_{{Y}}$   a connection in the principal bundle for the hypercharge denoted $Y \left( M,U(1)_Y) \right)$. Let $C \times L \times Y$  the principal bundle over $M \times M \times M$ with structure group $SU(3)_C \times SU(2)_L \times U(1)_Y$. Let $C + L+ Y$ the restriction of  $C \times L \times Y$ to the diagonal $\Delta M$ of $M \times M \times M$ .  Since $\Delta M$ 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 $SU(3)_C \times SU(2)_L \times U(1)_Y$. The restriction of the projection $C  \times Y \times L  \rightarrow C$ to the projection $C +L + Y \rightarrow C$, denoted  by $h_C$, is a homomorphism with the corresponding natural homomorphism $$h_{SU(3)_C} :  SU(3)_C \times SU(2)_L \times U(1)_Y  \rightarrow SU(3)_C$$.

Similarly  for $h_L : C + L + Y \rightarrow L$ with

$$h_{SU(2)_L} :  SU(3)_C \times SU(2)_L \times U(1)_Y  \rightarrow SU(2)_L$$;

and  $h_Y : C + L + Y \rightarrow Y$ with

$$h_{U(1)_Y} :  SU(3)_C \times SU(2)_L \times U(1)_Y  \rightarrow U(1)_Y$$,

Then,

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

$$h_C  : C + L + Y \rightarrow C$$,

$$h_L  : C + L+ Y \rightarrow L$$

and

$$h_Y :C+  L + Y \rightarrow Y$$

maps $\Gamma$  into $\Gamma_{{C}}$ , $\Gamma_{{L}}$ and $\Gamma_{{Y}}$ respectively.

(b)  If   $\omega$, $\omega_C$ , $\omega_L$ and $\omega_Y$  are the connection one- forms and $\Omega$, $\Omega_C$, $\Omega_L$ and $\Omega_Y$, are the curvature  two-forms of $\Gamma$, $\Gamma_C$, $\Gamma_L$ and $\Gamma_Y$ respectively, then

$$\omega =  h^*_C \omega_C+h^*_L \omega_L + h^*_Y \omega_Y $$

$$\Omega = h^*_C\Omega_C +  h^*_L \Omega_L + h^*_Y \Omega_Y $$


More explicitly we have :

$$\omega_{\mu} =  (h^*_C \omega_C)_{\mu} +(h^*_L \omega_L )_{\mu} + (h^*_Y \omega_Y )_{\mu} $$

$$\omega_{\mu} =  h^*_C( \omega_{C, \mu})+h^*_L( \omega_{L, \mu}) +h^*_Y( \omega_{Y, \mu}) $$

$$\omega_{\mu} =  h^*_C( g_C\frac{\hat{\lambda}_{\alpha}}{2}A_{\mu}^{\alpha})+h^*_L(g_L\frac{\hat{\sigma}_{\beta}}{2}B_{\mu}^{\beta}) +h^*_Y(g_L\frac{Y}{2}C_{\mu}) $$

where $g_C$, $g_L$ and $g_Y$ are the coupling constants; $\hat{\lambda}_{\alpha}$ are the eight Gell-Mann matrices which are the generators of the Lie group $SU(3)$, $\hat{\sigma}_{\beta}$ are the three Pauli matrices which are the generators of the Lie group $SU(2)$ and $Y$ is the generator of the Lie group $U(1)$. $A_{\mu}^{\alpha}$, $B_{\mu}^{\beta}$ and $C_{\mu}$ are the gauge potentials for the vector bosons.

Then we have:

$$\omega_{\mu} = g_C\frac{ h^*_C(\hat{\lambda}_{\alpha})}{2}A_{\mu}^{\alpha}+g_L\frac{h^*_L(\hat{\sigma}_{\beta})}{2}B_{\mu}^{\beta} +g_Y\frac{h^*_Y(Y)}{2}C_{\mu} $$

$$\omega_{\mu} = g_C\frac{ h^*_{SU(3)_C}(\hat{\lambda}_{\alpha})}{2}A_{\mu}^{\alpha}+g_L\frac{h^*_{SU(2)_L}(\hat{\sigma}_{\beta})}{2}B_{\mu}^{\beta} +g_Y\frac{h^*_{U(1)_Y}(Y)}{2}C_{\mu} $$

$$\omega_{\mu} = g_C\frac{ \hat{\lambda}_{\alpha} \otimes I_2} {2}A_{\mu}^{\alpha}+g_L\frac{ I_3  \otimes\hat{\sigma}_{\beta}}{2}B_{\mu}^{\beta} +g_Y\frac{ I_3 \otimes I_2  Y}{2}C_{\mu} $$

$$\omega_{\mu} = g_C\frac{ \hat{\lambda}_{\alpha} \otimes I_2} {2}A_{\mu}^{\alpha}+g_L\frac{ I_3  \otimes\hat{\sigma}_{\beta}}{2}B_{\mu}^{\beta} +g_Y\frac{ I_6 Y}{2}C_{\mu} $$

where $I_2$ is the two-dimensional identity matrix, $I_3$ is the three-dimensional identity matrix and $I_6$ is the six-dimensional identity matrix.

The covariant derivative for a field which is a 0-form in the fundamental representation of $SU(3)_C \times SU(2)_L \times U(1)_Y$ has the form

$$D_{\mu} = I_6 \partial_{\mu} + g_C\frac{ \hat{\lambda}_{\alpha} \otimes I_2} {2}A_{\mu}^{\alpha}+g_L\frac{ I_3  \otimes\hat{\sigma}_{\beta}}{2}B_{\mu}^{\beta} +g_Y\frac{ I_6 Y}{2}C_{\mu} $$

The covariant derivative for a field which is 0-form in the adjoint representation of  $SU(3)_C \times SU(2)_L \times U(1)_Y$ has the form

$$D_{\mu}\Phi = I_6 \partial_{\mu}\Phi + [g_C\frac{ \hat{\lambda}_{\alpha} \otimes I_2} {2}A_{\mu}^{\alpha}+g_L\frac{ I_3  \otimes\hat{\sigma}_{\beta}}{2}B_{\mu}^{\beta} +g_Y\frac{ I_6 Y}{2}C_{\mu}, \Phi ]$$

where

$$\Phi  = \Phi _1^{\alpha}\frac{ \hat{\lambda}_{\alpha} \otimes I_2}{2} + \Phi _2^{\beta}\frac{ I_3  \otimes\hat{\sigma}_{\beta}}{2}+ \Phi _3\frac{ I_6 Y}{2}$$

answered Jul 28, 2015 by juancho (1,030 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  $\omega((R_a)_* X) = ad(a^{-1})\omega(X)$ where $a \in G \times H$ and $X \in (P+Q)_u$ with $u \in P + Q$.  We assume that $a = a_G \times a_H$ where $a_G  \in  G$ , $a_H  \in  H$ and then $h_G (a) = a_G$ and $h_H (a) = a_H$ .  Also we assume that $h_P (X) = X_P$ and $h_Q(X) = X _Q$. and then $\omega_P((R_{a_G})_* X_P) = ad(a_{G}^{-1})\omega_P(X_P)$ and $\omega_Q((R_{a_H})_* X_Q) = ad(a_{H}^{-1})\omega_Q(X_Q)$ . Then we have that

$$\omega((R_a)_* X)  = (h^*_P \omega_P + h^*_Q \omega_Q)((R_a)_* X)$$

$$\omega((R_a)_* X)  = (h^*_P \omega_P)((R_a)_* X) + (h^*_Q \omega_Q)((R_a)_* X)$$

$$\omega((R_a)_* X)  =  \omega_P(h_P((R_a)_* X))+  \omega_Q(h_Q((R_a)_* X))$$

$$\omega((R_a)_* X)  =  \omega_P((R_{h_G(a)})_* h_P(X))+  \omega_Q((R_{h_H(a)})_* h_Q(X))$$

$$\omega((R_a)_* X)  =  \omega_P((R_{h_G(a_G \times a_H)})_* h_P(X))+  \omega_Q((R_{h_H(a_G \times a_H)})_* h_Q(X))$$

$$\omega((R_a)_* X)  =  \omega_P((R_{a_G })_* h_P(X))+  \omega_Q((R_{a_H})_* h_Q(X))$$

$$\omega((R_a)_* X)  =  \omega_P((R_{a_G })_* X_P)+  \omega_Q((R_{a_H})_* X_Q)$$

$$\omega((R_a)_* X)  =  ad(a_{G}^{-1})\omega_P(X_P) +ad(a_{H}^{-1})\omega_Q(X_Q) $$

From other side we have that

$$ad(a^{-1})\omega(X)= a\omega(X)a^{-1}$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)\omega(X)(a_G \times a_H)^{-1}$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)\omega(X)(a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[(h^*_P \omega_P + h^*_Q \omega_Q)(X)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[(h^*_P \omega_P )(X)+ (h^*_Q \omega_Q)(X)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[ \omega_P (h_P(X))+ \omega_Q(h_Q (X))](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H)[ \omega_P (X_P)+ \omega_Q( X_Q)](a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \times a_H) \omega_P (X_P) (a_G^{-1} \times a_H^{-1}) + (a_G \times a_H)\omega_Q( X_Q)(a_G^{-1} \times a_H^{-1})$$

$$ad(a^{-1})\omega(X)= (a_G \omega_P (X_P)a_G^{-1}) ( a_H  a_H^{-1} )  + (a_G a_G^{-1})(a_H\omega_Q( X_Q)a_H^{-1} \ )$$

$$ad(a^{-1})\omega(X)= (a_G \omega_P (X_P)a_G^{-1}) ( e_H )  + (e_G)(a_H\omega_Q( X_Q)a_H^{-1} \ )$$

$$ad(a^{-1})\omega(X)= a_G \omega_P (X_P)a_G^{-1}  + a_H\omega_Q( X_Q)a_H^{-1} $$

$$ad(a^{-1})\omega(X)= ad(a_{G}^{-1})\omega_P(X_P) +ad(a_{H}^{-1})\omega_Q(X_Q) $$.

Second, we prove that $\omega(A^*) = A$ as follows.

$$\omega(A^*)= (h^*_P \omega_P + h^*_Q \omega_Q)(A^*)$$

$$\omega(A^*)= (h^*_P \omega_P)(A^*) + (h^*_Q \omega_Q)(A^*)$$

$$\omega(A^*)=  \omega_P(h_P(A^*)) + \omega_Q( h_Q(A^*))$$

$$\omega(A^*)=  \omega_P(A^*_P) + \omega_Q( A^*_Q)$$

$$\omega(A^*)=  A_P+ A_Q$$

$$\omega(A^*)=  A$$

Finally, we prove that  $\Omega = h^*_P \Omega_P + h^*_Q \Omega_Q$.

$$\Omega = d\omega+\frac{1}{2} [\omega,\omega]$$

$$\Omega = d(h^*_P \omega_P + h^*_Q \omega_Q)+\frac{1}{2} [(h^*_P \omega_P + h^*_Q \omega_Q),(h^*_P \omega_P + h^*_Q \omega_Q)]$$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ] +\frac{1}{2} [h^*_P \omega_P , h^*_Q \omega_Q] +$$

$$\frac{1}{2} [h^*_Q \omega_Q , h^*_P \omega_P] +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ] +0 +0 +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2} [h^*_P \omega_P , h^*_P \omega_P ]  +\frac{1}{2} [h^*_Q \omega_Q , h^*_Q \omega_Q ] $$

$$\Omega = d(h^*_P \omega_P) + d(h^*_Q \omega_Q)+\frac{1}{2}h^*_P ( [ \omega_P , \omega_P ] ) +\frac{1}{2}h^*_Q( [ \omega_Q ,  \omega_Q ]) $$

$$\Omega =h^*_P (d \omega_P) + h^*_Q (d\omega_Q)+\frac{1}{2}h^*_P ( [ \omega_P , \omega_P ] ) +\frac{1}{2}h^*_Q( [ \omega_Q ,  \omega_Q ]) $$

$$\Omega =h^*_P (d \omega_P+\frac{1}{2} [ \omega_P , \omega_P ]) + h^*_Q (d\omega_Q+  +\frac{1}{2} [ \omega_Q ,  \omega_Q ]) $$

$$\Omega = h^*_P \Omega_P + h^*_Q \Omega_Q$$.

answered Jul 27, 2015 by juancho (1,030 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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