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

+ 2 like - 0 dislike
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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 x0=π(u0), 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  Λ:κg
which satisfies

Λ(X)=λ(X)  for Xχ


Λ(ad(j)(X))=ad(λ(j))(Λ(X)) for  jJ and Xκ

Λ(X)=ωu0(ˆX) for Xκ

where ˆX is the vector field on P induced by X, κ is the Lie algebra of the Lie group K, χ is the Lie algebra of Lie group Jλ:JG is a homomorphism between Lie groups and   λ:χg 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 (1,130 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 λ:JG is a homomorphism.

Given that j1J, j2J and ju0=u0λ(j) we have

u0λ(j1j2)=(j1j2)u0=j1(j2u0)=j1(u0λ(j2))=(j1u0)λ(j2)=(u0λ(j1))λ(j2)

u0λ(j1j2)=u0(λ(j1)λ(j2))

λ(j1j2)=λ(j1)λ(j2)

.

From ju0=u0λ(j) and eu0=u0e we deduce that λ(e)=e.

With j1=j and j2=j1 we have that

λ(jj1)=λ(j)λ(j1)

λ(e)=λ(j)λ(j1)

e=λ(j)λ(j1)

λ(j)1=λ(j1)

We next prove that ω(X) is independent of the choice of k and a. Let

u0=kua=k1ua1

where kK, k1K, aG and a1G . Then we have that k1u0a1=u and k11u0a11=u; it is to say

k1u0a1=k11u0a11

which is equivalent to

k1k1u0=u0a11a

.

Comparing the last equation with ju0=u0λ(j) where jJ we deduce that

j=k1k1J

and

λ(j)=a11a

.

From these last equations we obtain that

k1=jk

and

a1=aλ(j)1=aλ(j1)

.

Using all these we have

k1Ra1X=(jk)Raλ(j1)X

k1Ra1X=(jk)Rλ(j1)(RaX)

k1Ra1X=jRλ(j1)(kRaX)

k1Ra1X=jRλ(j1)(ˆXu0+Au0)

k1Ra1X=jRλ(j1)(ˆXu0)+jRλ(j1)(Au0)

k1Ra1X=j(ˆXu0λ(j1))+jRλ(j1)(Au0)

k1Ra1X=j(ˆXu0λ(j1))+Rλ(j1)(jAu0)

k1Ra1X=j(ˆXu0λ(j1))+Rλ(j1)(Aju0)

k1Ra1X=j(ˆXu0λ(j1))+Rλ(j1)(Au0λ(j))

k1Ra1X=ˆZu0+Cu0

where

Z=ad(j)(X)

and

C=ad(λ(j))(A)

Then we have that

ad(a1)(Λ(Z)+C)=ad(a1)[Λ(ad(j)(X))+ad(λ(j))(A)]

ad(a1)(Λ(Z)+C)=ad(a1)[ad(λ(j))(Λ(X))+ad(λ(j))(A)]

ad(a1)(Λ(Z)+C)=ad(a1)[ad(λ(j))(Λ(X)+A)]

ad(a1)(Λ(Z)+C)=ad(a1)[ad(a11a)(Λ(X)+A)]

ad(a1)(Λ(Z)+C)=a1[ad(a11a)(Λ(X)+A)]a11

ad(a1)(Λ(Z)+C)=a1[(a11a)(Λ(X)+A)(a11a)1]a11

ad(a1)(Λ(Z)+C)=a1[(a11a)(Λ(X)+A)(a1a1)]a11

ad(a1)(Λ(Z)+C)=(a1a11)[a(Λ(X)+A)a1](a1a11)

ad(a1)(Λ(Z)+C)=e[a(Λ(X)+A)a1]e

ad(a1)(Λ(Z)+C)=a(Λ(X)+A)a1

ad(a1)(Λ(Z)+C)=ad(a)(Λ(X)+A)

.

This proves  that ω(X) depends only on X.

Now, we prove that ω is a connection one-form. Let  XTu(P) and
u0=kua . Let b be an arbitrary element of the structure group G.  We write

Y=RbXTv(P)

where v=ub and for hence u0=kua=ku(bb1)a=k(ub)(b1a)=kv(b1a).  Then we have that

kRb1aY=kRb1aRbX=kRbb1aX=kRaX=(ˆXu0+Au0)

.

From kRaX=ˆXu0+Au0 we write the definition of ω as

ω(X)=ad(a)(Λ(X)+A)

then from kRb1aY=ˆXu0+Au0 we write that

ω(Y)=ad(b1a)(Λ(X)+A).

Then we have

ω(RbX)=ω(Y)=ad(b1a)(Λ(X)+A)

ω(RbX)=(b1a)(Λ(X)+A)(b1a)1

ω(RbX)=(b1a)(Λ(X)+A)(a1b)

ω(RbX)=b1[a(Λ(X)+A)a1]b

ω(RbX)=ad(b1)[a(Λ(X)+A)a1]

ω(RbX)=ad(b1)[ad(a)(Λ(X)+A)]

ω(RbX)=ad(b1)ω(X).

Finally, we prove that ω  is invariant by K. Let  XTu(P) and k1  be an arbitrary element of K. Then k1XTk1u,(P) and we have that


ω(k1X)=k1(ad(a)(Λ(X)+A))

ω(k1X)=ad(ak1)(k1(Λ(X))+k1(A))

ω(k1X)=ad(ak1)[ad(k11)(Λ(X))+ad(k11)(A)]

ω(k1X)=ad(ak1)[ad(k11)(Λ(X))]+ad(ak1)[ad(k11)(A)]

ω(k1X)=(ak1)[ad(k11)(Λ(X))](ak1)1+(ak1)[ad(k11)(A)](ak1)1

ω(k1X)=(ak1)[k11Λ(X)k1](k11a1)+(ak1)[k11Ak1](k11a1)

ω(k1X)=a(k1k11)Λ(X)(k1k11)a1+a(k1k11)A(k1k11)a1

ω(k1X)=aeΛ(X)ea1+aeAea1

ω(k1X)=aΛ(X)a1+aAa1

ω(k1X)=a[Λ(X)+A]a1

ω(k1X)=ad(a)(Λ(X)+A)

ω(k1X)=ω(X)

answered Aug 7, 2015 by juancho (1,130 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 S2.  In this case the isotropy subgroup is U(1).  We start searching for a flat SU(2) connection of the form

ˆA=ˆτ1dθ+(f(θ)ˆτ2+g(θ)ˆτ3)dϕ

where ˆτk=^σk2i being ^σk the Pauli matrices.  The matrices ˆτk are the generators of the Lie algebra su(2) with the commutation relations

[ˆτ1,ˆτ2]=ˆτ3

[ˆτ1,ˆτ3]=ˆτ2

and being ˆτ3 is the generator of the Lie algebra u(1).

The corresponding Yang-Mills field is given by

ˆF=dˆA+12[ˆA,ˆA]=dˆA+ˆAˆA

Then we have

ˆF=d[ˆτ1dθ+(f(θ)ˆτ2+g(θ)ˆτ3)dϕ]+

[ˆτ1dθ+(f(θ)ˆτ2+g(θ)ˆτ3)dϕ][ˆτ1dθ+(f(θ)ˆτ2+g(θ)ˆτ3)dϕ]

which is expanded as

ˆF=dfdθˆτ2dθdϕ+dgdθˆτ3dθdϕ+fˆτ1ˆτ2dθdϕ+gˆτ1ˆτ3dθdϕ+

fˆτ2ˆτ1dϕdθ+gˆτ3ˆτ1dϕdθ

Then we have

ˆF=dfdθˆτ2dθdϕ+dgdθˆτ3dθdϕ+fˆτ1ˆτ2dθdϕ+gˆτ1ˆτ3dθdϕ

fˆτ2ˆτ1dθdϕgˆτ3ˆτ1dθdϕ

which can be rewritten as

ˆF=[dfdθˆτ2+dgdθˆτ3+f(ˆτ1ˆτ2ˆτ1ˆτ2)+g(ˆτ1ˆτ3ˆτ3ˆτ1)]dθdϕ

it is to say

ˆF=(dfdθˆτ2+dgdθˆτ3+f[ˆτ1,ˆτ2]+g[ˆτ1,ˆτ3])dθdϕ

Using the commutation relations we obtain

ˆF=(dfdθˆτ2+dgdθˆτ3+fˆτ3gˆτ2)dθdϕ

which is rewritten as

ˆF=[(dfdθg)ˆτ2+(dgdθ+f)ˆτ3]dθdϕ

The connection is flat when ˆF=0 and then

0=[(dfdθg)ˆτ2+(dgdθ+f)ˆτ3]dθdϕ

which is equivalent to

dfdθg=0

dgdθ+f=0

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

f(θ)=C1sin(θ)+C2cos(θ)

g(θ)=C1cos(θ)C2sin(θ)

Then the invariant flat connection takes the form

ˆA=ˆτ1dθ+[(C1sin(θ)+C2cos(θ))ˆτ2+(C1cos(θ)C2sin(θ))ˆτ3]dϕ

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

ˆA=ˆΛ1dθ+(f(θ)ˆΛ2+g(θ)ˆΛ3)dϕ

where the Wang map Λ:su(2)su(2) is determined by ˆΛk=Λ(ˆτk) with

[ˆΛ1,ˆΛ3]=ˆΛ2

Then the corresponding Yang-Mills field takes the form

ˆF=(dfdθˆΛ2+dgdθˆΛ3+f[ˆΛ1,ˆΛ2]+g[ˆΛ1,ˆΛ3])dθdϕ

which is reduced to

ˆF=(dfdθˆΛ2+dgdθˆΛ3+f[ˆΛ1,ˆΛ2]gˆΛ2)dθdϕ

Now using that dfdθ=g and dgdθ=f, the Yang-Mills field acquires the form

ˆF=(gˆΛ2fˆΛ3+f[ˆΛ1,ˆΛ2]gˆΛ2)dθdϕ

which is reduced to

ˆF=([ˆΛ1,ˆΛ2]ˆΛ3)fdθdϕ

Finally using that f(θ)=C1sin(θ)+C2cos(θ) we obtain

ˆF=([ˆΛ1,ˆΛ2]ˆΛ3)(C1sin(θ)+C2cos(θ))dθdϕ

From this last equation we observe that when the Wang map Λ:su(2)su(2) is a Lie algebra homomorphism, it is to say, when [ˆΛ1,ˆΛ2]=ˆΛ3; then ˆF=0 and for hence the invariant connection is flat.

answered Aug 10, 2015 by juancho (1,130 points) [ revision history ]
edited Aug 10, 2015 by juancho

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