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Let $G$ be a matrix Lie group and $\frak g$ be the corresponding Lie algebra. Furthermore, $X:I\rightarrow G$ is a smooth path in $G$.

I'd like to understand this statement better:

$$X^{-1} \cdot \frac{d X}{dt} \in \frak g.$$

Is there a proof for this from first principles? (e.g relying on matrix algebra, matrix exponential etc.)

Are there any illustrative, insightful examples?

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The basic point here is that left multiplication by $X^{-1}$ is a linear map, and thus its derivative also is left multiplication by $X^{-1}$. Now take a curve $c:I\to G$ with $c(0)=X$, so $c'(0)$ is tangent to $G$ at $X$. Then $X^{-1}\cdot c(t)$ is a curve in $G$ mapping $0$ to the unit matrix. Hence $\frac{d}{dt}|_{t=0}(X^{-1}\cdot c(t))$ is an element of $\mathfrak g$. But by linearity, this derivative is just $X^{-1}\cdot c'(0)$. Now you can do this in any point of a curve, which proves the claim.

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  • $\begingroup$ Ah, I see. The statement I read was a shorthand for $X^{-1} \frac{d}{dt} c(t)|_{t=0} \in {\frak g}$ with $c(0) = X$. Now, that makes all sense! $\endgroup$ – B0rk4 Mar 27 '17 at 4:54
  • $\begingroup$ I guess it is a bit more than a shorthand for that, since you can then do things for all $t$ and thus get $(X(t))^{-1}X'(t)$ as a curve in $\mathfrak g$. But the main argument can be done for fixed $t$. $\endgroup$ – Andreas Cap Mar 27 '17 at 6:38
  • $\begingroup$ True, I can pick my $c(a)$ freely (also for $a \neq 0$), since $X(a)$ is in $G$ for all $a \in I$. Thanks. $\endgroup$ – B0rk4 Mar 27 '17 at 7:00

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