- #1
jostpuur
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Homework Statement
I'm supposed to prove, that when G is a Lie group, [itex]i:G\to G[/itex] is the inverse mapping [itex]i(g)=g^{-1}[/itex], then
[tex]
i_{*e} v = -v\quad\quad\forall \; v\in T_e G
[/tex]
where [itex]i_{*e}:T_e G \to T_e G[/itex] 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 [itex]\phi:M\to N[/itex] is a smooth function between two manifolds, then for each [itex]p\in M[/itex]
[tex]
\phi_{*p}:T_p M\to T_{\phi(p)} N,\quad\quad (\phi_{*p} v) f = v(f\circ \phi)\quad
\forall \; v\in T_p M \quad f:N\to\mathbb{R}
[/tex]
Quick googling revealed that this is often given other notations similar to derivative notations.
The Attempt at a Solution
We can choose some chart [itex]x:U\to\mathbb{R}^n[/itex], where [itex]U\subset G[/itex] is some environment of the [itex]e\in G[/itex], so that x(e)=0. After simply writing down some definitions, I was able to show that the result follows if we know
[tex]
\partial_j (x_k\circ i\circ x^{-1})(0) = -\delta_{jk}
[/tex]
Here [itex]x^{-1}:(x(U)\subset \mathbb{R}^n)\to U[/itex] is the inverse of the x, [itex]x_k:U\to\mathbb{R}[/itex] is a component mapping of the x, and [itex]\partial_j[/itex] 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 [itex]g\mapsto g^{-1}[/itex] 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)
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