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

$$i_{*e} v = -v\quad\quad\forall \; v\in T_e G$$

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 $p\in M$

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

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

$$\partial_j (x_k\circ i\circ x^{-1})(0) = -\delta_{jk}$$

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

$$i_{*e} v = -v$$

would be to show

$$vf = -v(f\circ i)$$

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

$$\dot{g}_0 f = D_t (f\circ g)(t)\Big|_{t=0}$$

and the right side is

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

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

$$0 = D_t f\big(g(t)\cdot (i\circ g)(t)\big)\Big|_{t=0} = \cdots$$

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

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

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

$$t\mapsto \gamma^1(t)\cdot \gamma^2(t)$$

by multiplying the images.

Is the equation

$$\dot{(\gamma^1\cdot\gamma^2)}_0 = \dot{\gamma}^1_0 + \dot{\gamma}^2_0$$

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.