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Invariant connections on principal bundles and space-time symmetries of the Yang-Mills fields

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I am trying to understand the Wang's theorem (http://projecteuclid.org/download/pdf_1/euclid.nmj/1118800027) about invariant connections from the physical point of view. According with Kobayashi-Nomizu, the Wang's theorem reads:

THEOREM 11.5. If a connected Lie group $K$ is a fibre-transitive automorphism
group of a principal bundle $P( M, G)$  and if $J$ is the isotropy subgroup of
$K$ at $x_0 =\pi(u_0) $, then there is a  $1 : 1$ correspondence between the set of $K$-invariant
connections in the principal bundle $P$ and the set of linear mappings  $\Lambda: \kappa \rightarrow g$
which satisfies

$\Lambda(X) = \lambda(X)$  for $X\in  \chi$


$\Lambda(ad (j)(X)) = ad ( \lambda(j)) ( \Lambda(X))$ for  $j \in  J$ and $X \in \kappa$

$\Lambda(X) = \omega_{u_0}(\hat{X})$ for $X \in \kappa$

where $\hat{X}$ is the vector field on $P$ induced by $X$, $\kappa$ is the Lie algebra of the Lie group $K$, $\chi$ is the Lie algebra of Lie group $J$ ,  $ \lambda : J \rightarrow G$ is a homomorphism between Lie groups and   $ \lambda : \chi \rightarrow g$ is the induced homomorphism between the corresponding Lie algebras.

My questions are:

1.  How to prove the Wang's theorem for physicists .

2.  How to apply the Wang's theorem to the study of the space-time symmetries of the Yang-Mills fields.

3.  Let me know papers with applications in physics of the  Wang's theorem.

asked Aug 7, 2015 in Theoretical Physics by juancho (860 points) [ revision history ]
edited Aug 7, 2015 by juancho

2 Answers

+ 2 like - 0 dislike

1. Proving the Wang's theorem: (adapted from Kobayashi-Nomizu with step by step deductions)

First we prove that $ \lambda : J \rightarrow G$ is a homomorphism.

Given that $j_1 \in J$, $j_2 \in J$ and $ju_0 = u_0 \lambda(j)$ we have

$$u_0 \lambda(j_1j_2) = (j_1j_2)u_0 = j_1(j_2u_0)=j_1(u_0 \lambda(j_2))=(j_1u_0)\lambda(j_2)= (u_0\lambda(j_1))\lambda(j_2)$$

$$u_0 \lambda(j_1j_2) = u_0(\lambda(j_1)\lambda(j_2))$$

$$ \lambda(j_1j_2) = \lambda(j_1)\lambda(j_2)$$.

From $ju_0 = u_0 \lambda(j)$ and $eu_0=u_0e$ we deduce that $\lambda(e)=e$.

With $j_1 = j$ and $j_2 = j^{-1}$ we have that

$$\lambda(jj^{-1}) = \lambda(j)\lambda(j^{-1})$$

$$\lambda(e) = \lambda(j)\lambda(j^{-1})$$

$$e= \lambda(j)\lambda(j^{-1})$$

$$\lambda(j)^{-1}= \lambda(j^{-1})$$

We next prove that $\omega(X^*)$ is independent of the choice of $k$ and $a$. Let

$$u_0 = k u a = k_1 u a_1$$

where $k \in K$, $k_1 \in K$, $a \in G$ and $a_1 \in G$ . Then we have that $k^{-1}u_0a^{-1} = u$ and $k_1^{-1}u_0a_1^{-1} = u$; it is to say

$$k^{-1}u_0a^{-1} = k_1^{-1}u_0a_1^{-1} $$

which is equivalent to

$$k_1k^{-1}u_0 = u_0a_1^{-1}a$$.

Comparing the last equation with $ju_0 = u_0 \lambda (j)$ where $j \in J$ we deduce that

$$j = k_1k^{-1}  \in J$$

and

$$\lambda (j) = a_1^{-1}a$$.

From these last equations we obtain that

$$k_1 = jk$$

and

$$a_1=a\lambda (j)^{-1}=a \lambda (j^{-1}) $$.

Using all these we have

$$k_1 \circ R_{a_1}X^* = (jk)  \circ R_{a \lambda (j^{-1}) }X^*  $$

$$k_1 \circ R_{a_1}X^* = (jk)  \circ R_{ \lambda (j^{-1}) }( R_{a}X^* ) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( k \circ R_{a}X^* ) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( \hat{X}_{u_0} +A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j \circ R_{ \lambda (j^{-1}) }( \hat{X}_{u_0} )+j \circ R_{ \lambda (j^{-1}) }(A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0  \lambda (j^{-1})} )+j \circ R_{ \lambda (j^{-1}) }(A_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0  \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(jA_{u_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0  \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(A_{ju_0}^*) $$

$$k_1 \circ R_{a_1}X^* = j ( \hat{X}_{u_0  \lambda (j^{-1})} )+ R_{ \lambda (j^{-1}) }(A_{u_0\lambda(j)}^*) $$

$$k_1 \circ R_{a_1}X^* = \hat{Z}_{u_0}+ C^*_{u_0}$$

where

$$Z = ad(j)(X)$$

and

$$C = ad(\lambda (j) )(A)$$

Then we have that

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[\Lambda (ad(j)(X) )+ad(\lambda (j) )(A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(\lambda (j))(\Lambda(X))+ad(\lambda (j) )(A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(\lambda (j))(\Lambda(X)+A)] $$

$$ad(a_1)(\Lambda (Z) +C)= ad(a_1)[ad(a_1^{-1}a)(\Lambda(X)+A)] $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[ad(a_1^{-1}a)(\Lambda(X)+A)]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[(a_1^{-1}a)(\Lambda(X)+A)(a_1^{-1}a)^{-1}]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= a_1[(a_1^{-1}a)(\Lambda(X)+A)(a^{-1}a_1)]a_1^{-1} $$

$$ad(a_1)(\Lambda (Z) +C)= (a_1a_1^{-1})[a(\Lambda(X)+A)a^{-1}](a_1a_1^{-1} )$$

$$ad(a_1)(\Lambda (Z) +C)= e[a(\Lambda(X)+A)a^{-1}]e$$

$$ad(a_1)(\Lambda (Z) +C)= a(\Lambda(X)+A)a^{-1}$$

$$ad(a_1)(\Lambda (Z) +C)= ad(a)(\Lambda(X)+A)$$.

This proves  that $\omega(X^*)$ depends only on $X^*$.

Now, we prove that $\omega$ is a connection one-form. Let  $X^* \in T_u(P)$ and
$u_0 = kua$ . Let $b$ be an arbitrary element of the structure group $G$.  We write

$$Y^* = R_b X^* \in  T_v(P)$$

where $v = ub$ and for hence $u_0 =kua=ku(bb^{-1})a=k(ub)(b^{-1}a)= kv(b^{-1}a)$.  Then we have that

$$k \circ R_{b^{-1}a}Y^* = k \circ R_{b^{-1}a}R_b X^* = k \circ  R_{bb^{-1}a}X^*  = k \circ R_aX^*  =(\hat{X}_{u_0}+A^*_{u_0})$$.

From $k \circ R_aX^*  =\hat{X}_{u_0}+A^*_{u_0}$ we write the definition of $\omega$ as

$$\omega(X^*)=ad(a)(\Lambda (X) +A)$$

then from $k \circ R_{b^{-1}a}Y^* =\hat{X}_{u_0}+A^*_{u_0}$ we write that

$$\omega(Y^*)=ad(b^{-1}a)(\Lambda (X) +A).$$

Then we have

$$\omega(R_b X^*)=\omega(Y^*)=ad(b^{-1}a)(\Lambda (X) +A)$$

$$\omega(R_b X^*)= (b^{-1}a)(\Lambda (X) +A)(b^{-1}a)^{-1}$$

$$\omega(R_b X^*)= (b^{-1}a)(\Lambda (X) +A)(a^{-1}b)$$

$$\omega(R_b X^*)= b^{-1}[a(\Lambda (X) +A)a^{-1}]b$$

$$\omega(R_b X^*)= ad(b^{-1})[a(\Lambda (X) +A)a^{-1}]$$

$$\omega(R_b X^*)= ad(b^{-1})[ad(a)(\Lambda (X) +A)]$$

$$\omega(R_b X^*)= ad(b^{-1})\omega(X^*).$$

Finally, we prove that $\omega$  is invariant by $K$. Let  $X^* \in T_u(P)$ and $k_1$  be an arbitrary element of $K$. Then $k_1X^* \in T_{k_1u},(P)$ and we have that


$$\omega(k_1 X^*)=k_1(ad(a)(\Lambda (X) +A))$$

$$\omega(k_1 X^*)=ad(ak_1)(k_1(\Lambda (X) )+k_1(A))$$

$$\omega(k_1 X^*)=ad(ak_1)[ad(k_1^{-1})(\Lambda (X) )+ad(k_1^{-1})(A)]$$

$$\omega(k_1 X^*)=ad(ak_1)[ad(k_1^{-1})(\Lambda (X) )]+ad(ak_1)[ad(k_1^{-1})(A)]$$

$$\omega(k_1 X^*)=(ak_1)[ad(k_1^{-1})(\Lambda (X) )](ak_1)^{-1}+(ak_1)[ad(k_1^{-1})(A)](ak_1)^{-1}$$

$$\omega(k_1 X^*)=(ak_1)[k_1^{-1}\Lambda (X) k_1](k_1^{-1}a^{-1})+(ak_1)[k_1^{-1}Ak_1](k_1^{-1}a^{-1})$$

$$\omega(k_1 X^*)=a(k_1k_1^{-1})\Lambda (X)(k_1k_1^{-1})a^{-1}+a(k_1k_1^{-1})A(k_1k_1^{-1})a^{-1}$$

$$\omega(k_1 X^*)=ae\Lambda (X)ea^{-1}+aeAea^{-1}$$

$$\omega(k_1 X^*)=a\Lambda (X)a^{-1}+aAa^{-1}$$

$$\omega(k_1 X^*)=a[\Lambda (X)+A]a^{-1}$$

$$\omega(k_1 X^*)=ad(a)(\Lambda (X)+A)$$

$$\omega(k_1 X^*)=\omega( X^*)$$

answered Aug 7, 2015 by juancho (860 points) [ revision history ]
edited Aug 14, 2015 by juancho
+ 2 like - 0 dislike

2.  Example of invariant connection.

Wang`s theorem can be applied to construct spherically symmetric  $SU(2)$ connections over $S^2$.  In this case the isotropy subgroup is $U(1)$.  We start searching for a flat $SU(2)$ connection of the form

$$\hat{A} = \hat{\tau}_1d \theta+(f(\theta) \hat{\tau}_2+g(\theta) \hat{\tau}_3)d\phi$$

where $\hat{\tau}_k = \frac{\hat{\sigma_k}}{2i}$ being $\hat{\sigma_k}$ the Pauli matrices.  The matrices $\hat{\tau}_k$ are the generators of the Lie algebra $su(2)$ with the commutation relations

$$[\hat{\tau}_1,\hat{\tau}_2] = \hat{\tau}_3$$

$$[\hat{\tau}_1,\hat{\tau}_3] = -\hat{\tau}_2$$

and being $\hat{\tau}_3$ is the generator of the Lie algebra $u(1)$.

The corresponding Yang-Mills field is given by

$$\hat{F} = d\hat{A} + \frac{1}{2}[\hat{A},\hat{A}] =d\hat{A} + \hat{A}  \wedge \hat{A} $$

Then we have

$$\hat{F} = d[ \hat{\tau}_1d \theta+(f(\theta) \hat{\tau}_2+g(\theta) \hat{\tau}_3)d\phi] +$$

$$[ \hat{\tau}_1d \theta+(f(\theta) \hat{\tau}_2+g(\theta) \hat{\tau}_3)d\phi] \wedge [ \hat{\tau}_1d \theta+(f(\theta) \hat{\tau}_2+g(\theta) \hat{\tau}_3)d\phi] $$

which is expanded as

$$\hat{F} = \frac{df}{d\theta }\hat{\tau}_2 d\theta \wedge  d\phi +\frac{dg}{d\theta }\hat{\tau}_3 d\theta \wedge  d\phi +f\hat{\tau}_1 \hat{\tau}_2 d\theta \wedge  d\phi+g\hat{\tau}_1 \hat{\tau}_3 d\theta \wedge  d\phi+ $$

$$f\hat{\tau}_2 \hat{\tau}_1 d\phi \wedge  d\theta +g\hat{\tau}_3 \hat{\tau}_1 d\phi \wedge  d\theta $$

Then we have

$$\hat{F} = \frac{df}{d\theta }\hat{\tau}_2 d\theta \wedge  d\phi +\frac{dg}{d\theta }\hat{\tau}_3 d\theta \wedge  d\phi +f\hat{\tau}_1 \hat{\tau}_2 d\theta \wedge  d\phi+g\hat{\tau}_1 \hat{\tau}_3 d\theta \wedge  d\phi-$$

$$f\hat{\tau}_2 \hat{\tau}_1 d\theta \wedge  d\phi -g\hat{\tau}_3 \hat{\tau}_1 d\theta \wedge  d\phi $$

which can be rewritten as

$$\hat{F}= [ \frac{df}{d\theta }\hat{\tau}_2  +\frac{dg}{d\theta }\hat{\tau}_3  +f(\hat{\tau}_1 \hat{\tau}_2-\hat{\tau}_1 \hat{\tau}_2) +g(\hat{\tau}_1 \hat{\tau}_3 -\hat{\tau}_3 \hat{\tau}_1)]d\theta \wedge  d\phi$$

it is to say

$$\hat{F}= ( \frac{df}{d\theta }\hat{\tau}_2  +\frac{dg}{d\theta }\hat{\tau}_3  +f[\hat{\tau}_1 ,\hat{\tau}_2] +g[\hat{\tau}_1, \hat{\tau}_3] )d\theta \wedge  d\phi$$

Using the commutation relations we obtain

$$\hat{F}= ( \frac{df}{d\theta }\hat{\tau}_2  +\frac{dg}{d\theta }\hat{\tau}_3  +f\hat{\tau}_3 -g\hat{\tau}_2 )d\theta \wedge  d\phi$$

which is rewritten as

$$\hat{F}= [(\frac{df}{d\theta }-g)\hat{\tau}_2  +(\frac{dg}{d\theta }+f)\hat{\tau}_3  ]d\theta \wedge  d\phi$$

The connection is flat when $\hat{F} = 0$ and then

$$0= [(\frac{df}{d\theta }-g)\hat{\tau}_2  +(\frac{dg}{d\theta }+f)\hat{\tau}_3  ]d\theta \wedge  d\phi$$

which is equivalent to

$$\frac{df}{d\theta }-g = 0$$

$$\frac{dg}{d\theta }+f =0$$

The solution of such system of coupled linear ordinary differential equations  is

$$f \left( \theta \right) =C_{{1}}\sin \left( \theta \right) +C_{{2}}
\cos \left( \theta \right)$$

$$g \left( \theta \right) =C_{{1}}\cos \left( \theta \right) -C_{{2}}
\sin \left( \theta \right)$$

Then the invariant flat connection takes the form

$$\hat{A} = \hat{\tau}_1d \theta+[(C_{{1}}\sin \left( \theta \right) +C_{{2}}
\cos \left( \theta \right)) \hat{\tau}_2+(C_{{1}}\cos \left( \theta \right) -C_{{2}}
\sin \left( \theta \right)) \hat{\tau}_3]d\phi$$

Now we construct a non-flat invariant connection of the form

$$\hat{A} = \hat{\Lambda}_1d \theta+(f(\theta) \hat{\Lambda}_2+g(\theta) \hat{\Lambda}_3)d\phi$$

where the Wang map $\Lambda  : su(2) \rightarrow su(2)$ is determined by $ \hat{\Lambda}_k = \Lambda(\hat{\tau}_k)$ with

$$[\hat{\Lambda}_1,\hat{\Lambda}_3] = -\hat{\Lambda}_2$$

Then the corresponding Yang-Mills field takes the form

$$\hat{F}= ( \frac{df}{d\theta }\hat{\Lambda}_2  +\frac{dg}{d\theta }\hat{\Lambda}_3  +f[\hat{\Lambda}_1 ,\hat{\Lambda}_2] +g[\hat{\Lambda}_1, \hat{\Lambda}_3] )d\theta \wedge  d\phi$$

which is reduced to

$$\hat{F}= ( \frac{df}{d\theta }\hat{\Lambda}_2  +\frac{dg}{d\theta }\hat{\Lambda}_3  +f[\hat{\Lambda}_1 ,\hat{\Lambda}_2] -g\hat{\Lambda}_2)d\theta \wedge  d\phi$$

Now using that $\frac{df}{d\theta }=g $ and $\frac{dg}{d\theta }=-f$, the Yang-Mills field acquires the form

$$\hat{F}= ( g\hat{\Lambda}_2  -f\hat{\Lambda}_3  +f[\hat{\Lambda}_1 ,\hat{\Lambda}_2] -g\hat{\Lambda}_2)d\theta \wedge  d\phi$$

which is reduced to

$$\hat{F}= (   [\hat{\Lambda}_1 ,\hat{\Lambda}_2] - \hat{\Lambda}_3)fd\theta \wedge  d\phi$$

Finally using that $f \left( \theta \right) =C_{{1}}\sin \left( \theta \right) +C_{{2}}
\cos \left( \theta \right)$ we obtain

$$\hat{F}= (   [\hat{\Lambda}_1 ,\hat{\Lambda}_2] - \hat{\Lambda}_3)(C_{{1}}\sin \left( \theta \right) +C_{{2}}
\cos \left( \theta \right))d\theta \wedge  d\phi$$

From this last equation we observe that when the Wang map $\Lambda  : su(2) \rightarrow su(2)$ is a Lie algebra homomorphism, it is to say, when $[\hat{\Lambda}_1 ,\hat{\Lambda}_2] =\hat{\Lambda}_3 $; then $\hat{F}=0$ and for hence the invariant connection is flat.

answered Aug 10, 2015 by juancho (860 points) [ revision history ]
edited Aug 11, 2015 by juancho

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