Inverse in lie group, tangent space

[jostpuur]

Homework Statement

I'm supposed to prove, that when G is a Lie group, i:G\to G is the inverse mapping i(g)=g^{-1}, then

<br /> i_{*e} v = -v\quad\quad\forall \; v\in T_e G<br />

where i_{*e}:T_e G \to T_e G is the tangent mapping.

Homework Equations

I'm not sure how standard the tangent mapping notation or terminology is, so here's how it is defined on our course. If \phi:M\to N is a smooth function between two manifolds, then for each p\in M

<br /> \phi_{*p}:T_p M\to T_{\phi(p)} N,\quad\quad (\phi_{*p} v) f = v(f\circ \phi)\quad<br /> \forall \; v\in T_p M \quad f:N\to\mathbb{R}<br />

Quick googling revealed that this is often given other notations similar to derivative notations.

The Attempt at a Solution

We can choose some chart x:U\to\mathbb{R}^n, where U\subset G is some environment of the e\in G, so that x(e)=0. After simply writing down some definitions, I was able to show that the result follows if we know

<br /> \partial_j (x_k\circ i\circ x^{-1})(0) = -\delta_{jk}<br />

Here x^{-1}:(x(U)\subset \mathbb{R}^n)\to U is the inverse of the x, x_k:U\to\mathbb{R} is a component mapping of the x, and \partial_j is a normal partial derivative in an Euclidean space.

This is where I ran out of ideas. I have no clue how the group structure gets related to what happens in the chart images. I know that g\mapsto g^{-1} is supposed to be smooth, but this merely justifies the existence of the partial derivative in the above equation.

(btw. I will not return any solution anywhere. I'm preparing to an exam by doing old exercises)

 

[Dick]

You can think of tangent vectors as represented by differentiable curves through e, the group identity. So pick g(s) to be a representative of v, i.e. g(0)=e, g'(0)=v. Now g(s)*i(g(s))=e. Differentiate both sides, put s=0 and get t+i_*(t)=0.

 

[jostpuur]

Your hint certainly pushed in the right direction, but I still have some problems with this. One way to prove

<br /> i_{*e} v = -v<br />

would be to show

<br /> vf = -v(f\circ i)<br />

for arbitrary smooth f:U\to\mathbb{R} on some environment U\subset G of e\in G.

If g:I\to G has been chosen so that 0\in I, g(0)=e, and \dot{g}_0 = v, then the left side is

<br /> \dot{g}_0 f = D_t (f\circ g)(t)\Big|_{t=0}<br />

and the right side is

<br /> -\dot{g}_0 (f\circ i) = -D_t(f\circ i\circ g)(t)\Big|_{t=0}.<br />

According to you hint, it looks like I should start computing something like this

<br /> 0 = D_t f\big(g(t)\cdot (i\circ g)(t)\big)\Big|_{t=0} = \cdots<br />

But I don't know how to do anything to that. How do you get the derivation inside the f?

I would know how to compute

<br /> D_t\Big((f\circ g)(t)\; (f\circ i\circ g)(t)\Big)\Big|_{t=0} =\cdots = f(e)\big( vf + (i_{*e} v) f\big)<br />

for example, but for this there does not seem to be any obvious reason why this would be zero.

 

[Dick]

If you think of the Lie group as a matrix group, then the argument I gave is not just a hint, it IS the proof. I'm having more than a little trouble wrapping my head around your more abstract formalism.

 

[jostpuur]

If you think of the Lie group as a matrix group, then the argument I gave is not just a hint, it IS the proof.

I can see this.

I'm having more than a little trouble wrapping my head around your more abstract formalism.

This is the rigor general formalism. In general Lie group is just a differentiable manifold with a group structure. Tangent vectors are linear mappings that map smooth functions into real numbers.

In fact the problem comes down to this:

If \gamma^1 and \gamma^2 are some smooth paths I\to G on some Lie group, so that \gamma^1(0)=\gamma^2(0)=e, then we get a new smooth path

<br /> t\mapsto \gamma^1(t)\cdot \gamma^2(t)<br />

by multiplying the images.

Is the equation

<br /> \dot{(\gamma^1\cdot\gamma^2)}_0 = \dot{\gamma}^1_0 + \dot{\gamma}^2_0<br />

true? (okey, that dot looks horrible, when its trying to be above both of the gammas...)

It seems to be, but I don't know the proof. Once this is clear, the previous problem becomes solved as you showed.