- #1
- 10,876
- 423
I'm reading about gauge theory and the text goes through some stuff about Lie groups and algebras rather quickly. I tried to prove one of the things they state without proof and got stuck.
Suppose that M and N are manifolds and \phi:M\rightarrow N is a diffeomorphism. Then we can define a function
\phi(p)_*:T_pM\rightarrow T_{\phi(p)}N
for each p\in M.
A Lie group is both a group and a manifold. We can use any member g of a group G to construct two diffeomorphisms \rho_g and \lambda_g that map G onto itself:
\rho_g(h)=hg
\lambda_g(h)=gh
The Lie algebra associated with the Lie Group is defined as the tangent space at the identity element, with a Lie bracket that will be defined below. Let's use the notation \mathfrak{g}=T_eG
We can use either right or left multiplication to map the Lie algebra onto the tangent space at any other point g:
\rho_g(e)_*:\mathfrak{g}\rightarrow T_gG
\lambda_g(e)_*:\mathfrak{g}\rightarrow T_gG
Let's simplify the notation a bit:
\rho_g(e)_*(L)=Lg
\lambda_g(e)_*(L)=gL
We can use these maps to construct two vector fields X_L^\rho and X_L^\lambda for each vector L in the Lie algebra:
X_L^\rho|_g=Lg
X_L^\lambda|_g=gL
Either of these two vector fields can be used to define a Lie bracket on the Lie Algebra:
[K,L]=[X_K^\rho,X_L^\rho]_e
[K,L]=[X_K^\lambda,X_L^\lambda]_e
(I assume that anyone who can help me with this already knows the definition of the commutator of two vector fields, which is used on the right).
The claim I haven't been able to prove is that these two definitions of the Lie bracket are equivalent, i.e. that it doesn't matter if we define it using right or left multiplication. So my question is, can someone help me prove that?
A few observations:
[K,L](f)=K(X_L^\rho f )-L(X_K^\rho f)
[K,L](f)=K(X_L^\lambda f)-L(X_K^\lambda f)
(X_L^\rho f)(h)=L(f\circ\rho_h)
(X_K^\lambda f)(h)=L(f\circ\rho_h)
(f\circ\rho_h)(k)=f(kh)
(f\circ\lambda_h)(k)=f(hk)
What am I missing? I have a feeling it's something simple.
Suppose that M and N are manifolds and \phi:M\rightarrow N is a diffeomorphism. Then we can define a function
\phi(p)_*:T_pM\rightarrow T_{\phi(p)}N
for each p\in M.
A Lie group is both a group and a manifold. We can use any member g of a group G to construct two diffeomorphisms \rho_g and \lambda_g that map G onto itself:
\rho_g(h)=hg
\lambda_g(h)=gh
The Lie algebra associated with the Lie Group is defined as the tangent space at the identity element, with a Lie bracket that will be defined below. Let's use the notation \mathfrak{g}=T_eG
We can use either right or left multiplication to map the Lie algebra onto the tangent space at any other point g:
\rho_g(e)_*:\mathfrak{g}\rightarrow T_gG
\lambda_g(e)_*:\mathfrak{g}\rightarrow T_gG
Let's simplify the notation a bit:
\rho_g(e)_*(L)=Lg
\lambda_g(e)_*(L)=gL
We can use these maps to construct two vector fields X_L^\rho and X_L^\lambda for each vector L in the Lie algebra:
X_L^\rho|_g=Lg
X_L^\lambda|_g=gL
Either of these two vector fields can be used to define a Lie bracket on the Lie Algebra:
[K,L]=[X_K^\rho,X_L^\rho]_e
[K,L]=[X_K^\lambda,X_L^\lambda]_e
(I assume that anyone who can help me with this already knows the definition of the commutator of two vector fields, which is used on the right).
The claim I haven't been able to prove is that these two definitions of the Lie bracket are equivalent, i.e. that it doesn't matter if we define it using right or left multiplication. So my question is, can someone help me prove that?
A few observations:
[K,L](f)=K(X_L^\rho f )-L(X_K^\rho f)
[K,L](f)=K(X_L^\lambda f)-L(X_K^\lambda f)
(X_L^\rho f)(h)=L(f\circ\rho_h)
(X_K^\lambda f)(h)=L(f\circ\rho_h)
(f\circ\rho_h)(k)=f(kh)
(f\circ\lambda_h)(k)=f(hk)
What am I missing? I have a feeling it's something simple.
Last edited:
