Lie algebra question

**Fredrik** · Aug19-07, 02:19 AM

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 $LaTeX Code: \\phi:M\\rightarrow N$ is a diffeomorphism. Then we can define a function

$LaTeX Code: \\phi(p)_*:T_pM\\rightarrow T_{\\phi(p)}N$

for each $LaTeX Code: p\\in M$ .

A Lie group is both a group and a manifold. We can use any member g of a group G to construct two diffeomorphisms $LaTeX Code: \\rho_g$ and $LaTeX Code: \\lambda_g$ that map G onto itself:

$LaTeX Code: \\rho_g(h)=hg$
$LaTeX Code: \\lambda_g(h)=gh$

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 $LaTeX Code: \\mathfrak{g}=T_eG$

We can use either right or left multiplication to map the Lie algebra onto the tangent space at any other point g:

$LaTeX Code: \\rho_g(e)_*:\\mathfrak{g}\\rightarrow T_gG$
$LaTeX Code: \\lambda_g(e)_*:\\mathfrak{g}\\rightarrow T_gG$

Let's simplify the notation a bit:

$LaTeX Code: \\rho_g(e)_*(L)=Lg$
$LaTeX Code: \\lambda_g(e)_*(L)=gL$

We can use these maps to construct two vector fields $LaTeX Code: X_L^\\rho$ and $LaTeX Code: X_L^\\lambda$ for each vector L in the Lie algebra:

$LaTeX Code: X_L^\\rho|_g=Lg$
$LaTeX Code: X_L^\\lambda|_g=gL$

Either of these two vector fields can be used to define a Lie bracket on the Lie Algebra:

$LaTeX Code: [K,L]=[X_K^\\rho,X_L^\\rho]_e$
$LaTeX Code: [K,L]=[X_K^\\lambda,X_L^\\lambda]_e$

(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:

$LaTeX Code: [K,L](f)=K(X_L^\\rho f )-L(X_K^\\rho f)$
$LaTeX Code: [K,L](f)=K(X_L^\\lambda f)-L(X_K^\\lambda f)$

$LaTeX Code: (X_L^\\rho f)(h)=L(f\\circ\\rho_h)$
$LaTeX Code: (X_K^\\lambda f)(h)=L(f\\circ\\rho_h)$

$LaTeX Code: (f\\circ\\rho_h)(k)=f(kh)$
$LaTeX Code: (f\\circ\\lambda_h)(k)=f(hk)$

What am I missing? I have a feeling it's something simple.

**quidamschwarz** · Aug30-07, 07:37 AM

Hi Fredrik,
I also tried to prove the equivalence of the left and right Lie bracket. But unfortunately it’s just not true – the right Lie bracket is the reverse of the left one. You can find a note on

http://planetmath.org/encyclopedia/LieGroup.html

The main idea is to consider the inversion map

$LaTeX Code: \\phi:G\\to G,\\,g\\mapsto g^{-1}\\;.$

Using your notation and

for a tangentvector at $LaTeX Code: g\\in G$ we have

$LaTeX Code: \\phi(p)_*(X_g) = -g^{-1}\\,X_g\\,g^{-1}\\;,$

and for your right-invariant vectorfield $LaTeX Code: X_L^\\rho|_g=Lg$ we get

$LaTeX Code: (\\phi_*X_L^\\rho)|_g := \\phi(g^{-1})_*(X_L^\\rho|_{g{-1}}) = -g\\,Lg^{-1}g = -gL = -X_L^\\lambda|_g = X_{-L}^\\lambda|_g$

(Here we used the $LaTeX Code: \\phi$ -transformation $LaTeX Code: \\phi_*X$ of a vectorfield

. In the next step we also need, that the $LaTeX Code: \\phi$ -transformation of vectorfields is a Liealgebra-homomorphism of the set of vectorfields, i.e. $LaTeX Code: [\\phi_*X,\\,\\phi_*Y] = \\phi_*([X,Y])$ )

Therefore:

$LaTeX Code: [K,L] _\\lambda :=[X_K^\\lambda,X_L^\\lambda]_e = [-\\phi_*X_K^\\rho,-\\phi_*X_L\\rho]_e = (\\phi_*([X_K^\\rho,X_L^\\rho]))_e = -[X_K^\\rho,X_L^\\rho]_e=:-[K,L]_\\rho = [L,K]_\\rho \\;,$

i.e. the left Lie bracket is the negative of the right Lie bracket, or you can say the left Lie bracket is the reverse of the right Lie bracket.
The result is somewhat surprising (at least to me), because this means, that if you have a Lie group, which is a matrix Lie group, the Lie algebra bracket

is related to the left Lie bracket, not to the right, even though it first looked somewhat symmetric…

A question:
I don’t understand your ‘observations’ except the last two. There you have an unusual mixture of Lie algebra elemente like

and

with vectorfields and functions. What does $LaTeX Code: L(f\\circ\\rho_h)$ or $LaTeX Code: K(X_L^\\rho f )$ mean?

**Fredrik** · Aug30-07, 04:33 PM

Thank you for the reply. I haven't been here for a week because it seemed that no one was going to answer.

Originally Posted by quidamschwarz

What does $LaTeX Code: L(f\\circ\\rho_h)$ or $LaTeX Code: K(X_L^\\rho f )$ mean?

$LaTeX Code: L(f\\circ\\rho_h)$ is just the real number you get when $LaTeX Code: L\\in\\mathfrak{g}=T_eG$ acts on the function $LaTeX Code: f\\circ\\rho_h:G\\rightarrow\\mathbb{R}$ .

$LaTeX Code: X_L^\\rho f$ is the map from G into $LaTeX Code: \\mathbb{R}$ defined by $LaTeX Code: g\\mapsto X_L^\\rho|_g f$ .

This is the notation I would use to define the commutator [X,Y] of two vector fields X and Y:

This holds for all g, so the same fact can be expressed this way instead:

This equation says exactly the same thing, except that here I'm expressing the identity using vector fields (such as

) instead of tangent vectors (such as

).

I hope that helps. If there's anything else you want me to explain, just let me know. You probably don't need to pay any attention to the results I called "observations" though. Those are just results I obtained while trying to show that the two definitions of the Lie Bracket are equivalent, and they may not be useful at all.

I have changed my mind about where it's appropriate to put the asterisk in an expression involving the "push-forward" function. E.g. I think it makes more sense to write $LaTeX Code: \\phi_*(g)X_g$ than $LaTeX Code: \\phi(g)_*X_g$ , but from now on I'll just write $LaTeX Code: \\phi_*X_g$ . I believe that's what most people do, and it isn't very helpful to keep the "g" anyway.

Originally Posted by quidamschwarz

$LaTeX Code: \\phi(p)_*(X_g) = -g^{-1}\\,X_g\\,g^{-1}$

I have tried to show this, but it seems to me that it can't be true. Maybe I misunderstood something. The left-hand side acting on a function f is

$LaTeX Code: \\phi_*X_g(f)=X_g(f\\circ\\phi)$

But the right-hand side of your identity acting on the same function f is

$LaTeX Code: (-g^{-1}\\,X_g\\,g^{-1})(f)=-X_g(f\\circ\\lambda_{g^{-1}}\\circ\\rho_{g^{-1}})$

The same vector, except for the sign, acting on a different function...I don't see how the results can be the same.

Originally Posted by quidamschwarz

$LaTeX Code: (\\phi_*X_L^\\rho)|_g = \\cdots = X_{-L}^\\lambda|_g$

I tried to show this without using the other identity, and this is the closest I've been able to get:

$LaTeX Code: \\phi_*X_L^\\rho|_g(f) = X_L^\\rho|_g(f\\circ\\phi) = \\rho_{g*}L(f\\circ\\phi) = L(f\\circ\\phi\\circ\\rho_g)$

$LaTeX Code: =L(f\\circ\\lambda_{g^{-1}}\\circ\\phi) = \\phi_*L(f\\circ\\lambda_{g^{-1}}) = (\\lambda_{g^{-1}*}(\\phi_*L))(f) = X_{\\phi_*L}^\\lambda|_{g^{-1}}(f)$

**quidamschwarz** · Aug31-07, 04:11 AM

Thanks for reminding me to think of

and

as operators instead just as tangentvectors. For a moment I thought only vectorfields can act on functions, so I got confused.

The left-hand side acting on a function f is $LaTeX Code: \\phi_*X_g(f)=X_g(f\\circ\\phi)$ .

Here you have to keep track of the position you’re taking the derivatives:

$LaTeX Code: (\\phi_*X)_g(f) := X_{\\phi^{-1}(g)}(f\\circ\\phi)$

On the LHS we take the derivative of

in

. Since on the RHS $LaTeX Code: \\phi$ maps

first to $LaTeX Code: \\phi(g)$ we have to use $LaTeX Code: X_{\\phi^{-1}(g)}$ instead of

.
You might better see this by considering $LaTeX Code: \\phi_*X$ as a vectorfield not as an operator: $LaTeX Code: \\phi$ is a diffeomorphism of

and the derivative $LaTeX Code: d_g\\phi = \\phi_*|_g$ maps

into $LaTeX Code: T_{\\phi(g)}G$ , so define

$LaTeX Code: (\\phi_*X)_g := d_{\\phi^{-1}(g)}\\phi(X_{\\phi^{-1}(g)})$

to receive a vector tangent at $LaTeX Code: g\\in G$ . The two definitions coincide as you can check e.g. in local coordinates.
(All of this clearly can be done for an arbitrary manifold

instead of the Liegroup

.)

Now to prove $LaTeX Code: \\phi(p)_*(X_g) = -g^{-1}\\,X_g\\,g^{-1}$ for the inversion map $LaTeX Code: \\phi(g) = g^{-1}$ we have to compute $LaTeX Code: d_{\\phi^{-1}(g)}\\phi$ . For this consider

$LaTeX Code: m\\circ(id,\\phi): G\\to G\\times G\\to G: g\\mapsto (g,g^{-1})\\mapsto g\\,g^{-1} \\,$

where $LaTeX Code: m(g,h) = g\\,h$ is just the product map of

. Since $LaTeX Code: g\\,g^{-1} = e$ we have $LaTeX Code: m\\circ(id,\\phi) = \\text{const}$ , so

$LaTeX Code: 0=d_g\\big(m\\circ(id,\\phi)\\big)(X_g) = \\big(d_{(g,g^{-1})}m\\big)\\circ\\big(d_g(id,\\phi)\\big)(X_g) = d_{(g,g^{-1})}m\\big(X_g,\\,d_g\\phi(X_g)\\big) = d_g\\rho_{g^{-1}}(X_g) + d_{g^{-1}}\\lambda_g\\big(d_g\\phi(X_g) \\big) = X_g\\,g^{-1} + g\\, d_g\\phi(X_g)\\;.$

Here I used the chainrule and that the differential of the product map can be written as

$LaTeX Code: d_{(g,h)}m: T_gG\\times T_hG\\to T_{gh}G: (X_g,\\,Y_h)\\mapsto d_g\\rho_h(X_g) + d_h\\lambda_g(Y_h) = X_g\\,g + h\\,Y_h\\;.$

With this we finally get

$LaTeX Code: d_g\\phi(X_g) = -g^{-1}\\,X_g\\,g^{-1} \\;.$

So far I haven't tried to prove this by using the 'operator-viewpoint'. It might be easier.

**Fredrik** · Sep3-07, 12:24 PM

I still don't get it, but it's not your fault. I just don't really understand product manifolds very well. I'll have to work through some of the basics thoroughly before I'll be able to understand this.

When I said that $LaTeX Code: \\phi_*X_g=-g^{-1}X_g g^{-1}$ can't be true, I didn't consider the possibility that the left-hand side was supposed to be $LaTeX Code: (\\phi_*X)_g$ . I didn't even think about how to define $LaTeX Code: \\phi_*$ acting on a vector field until after I had finished writing post #3.

It wasn't too hard to figure out how to define it and use the definition and to prove the identity $LaTeX Code: [\\phi_*X,\\,\\phi_*Y] = \\phi_*([X,Y])$ that you used earlier, so at least I understand something.

You have probably given me enough information to understand this. Now I just need to make the effort to think hard about some of the things I still don't understand. I appreciate all the help you've given me.

Edit: I have now understood the things I didn't understand before about product manifolds, so I hope I'm going to figure out the rest tomorrow. I also read #2 again, and I see that you made it clear that $LaTeX Code: \\phi_*$ was acting on a vector field. I don't know how I could have missed that before.