Orientation of left and right translations on a Lie Group I read that if $G$ is a connected Lie Group, then for every $g \in G$, the left and right translations ($L_g$ and $R_g$, respectively), preserve orientation. The reason given is that $L_g$ and $R_g$ can be deformed through diffeomorphisms of $G$ to the identity map on $G$.
This last part (tell me if I'm wrong) means that it exists diffeomorphisms $f_i: G \to G (i=1,2)$ such that both
$$f_i \circ L_g: G \to G \text{ and } f_2 \circ R_g: G \to G$$
are the identity map on G. In fact, we can take $f_1 = L_{g^{-1}}$ and $f_2 = R_{g^{-1}}$. From here, it follows that (differentiating and applying the chain rule)
$$(D_{g^{-1}h}L_g) \circ (D_{h} L_{g^-1}) = I$$
where $I$ is the identity matrix of order $\dim G$ (and similarly for $R_g$). Taking the determinant,
$$\det(D_{g^{-1}h}L_g)  \det(D_{h} L_{g^-1}) = 1.$$
I know that, since $L_g$ is a diffeomorphism, it either preserves orientation everywhere or reverses orientation everywhere. From the equation above, it's obvious that $L_g$ and $L_{g^{-1}}$ cannot be such that one preserves and the other reserves the orientation. But how do we know that they both preserve orientation?
Note
My definition for left and right translations is the following: for $g \in G$
$$L_g: G \to G, h \mapsto gh \text{ and } R_g: G \to G, h \mapsto hg.$$
(It is not explicit here, but $gh = g \cdot h$ where $\cdot$ is the operation on $G$).
I read that on the 2nd page of
http://www.math.toronto.edu/karshon/grad/2009-10/2010-01-11.pdf
 A: This is to expand on Jack Lee's comment. It really more or less follows from the (canonical) definition of an orientation of a Lie group.
Given a Lie group and let $\mathcal O_e = [v_1, \cdots, v_n]$ be an orientation on $T_eG$, here $(v_1, \cdots, v_n)$ is an basis of $T_eG$. Then one can define, for each $g\in G$ an orientation $\mathcal O_g$ on $T_gG$ by
$$\mathcal O_g := [ d(L_g)_e v_1, \cdots, d(L_g)_e v_n],$$
where $L_g$ is the left multiplication. Since the mapping
$$ G\times T_eG \to TG, \ \ (g, v) \mapsto d(L_g)_e v$$
is smooth, $\mathcal O$ is an orientation on $G$.
It is easy to check that $L_g$ preserves the orientation $\mathcal O$: for each $h\in G$, using $L_g\circ L_h = L_{gh}$, the equivalence class
$$ \mathcal O_h = [d(L_h)_e v_1, \cdots, d(L_h)_e v_n]$$
is sent to
\begin{align}
[d(L_g)_hd(L_h)_e v_1, \cdots, d(L_g)_hd(L_h)_e v_n] &= [d(L_{gh})_e v_1, \cdots, d(L_{gh})_e v_n] = \mathcal O_{gh}.
\end{align}
Of course, we can replace $(v_1, \cdots, v_n)$ with $(-v_1, v_2, \cdots, v_n)$ and construct another orientation $\widetilde {\mathcal O}$ on $G$, and the same argument implies that $L_g$ is orientation preserving with respect to this new orientation.
Now we use connectedness: Since $G$ is connected, $G$ has exactly two orientations. Thus the orientation on $G$ is either $\mathcal O$ or $\widetilde{\mathcal O}$ and wehave shown that $L_g$ is orientation preserving.
The same argument for $R_g$. We can replace $L_g$ with $R_g$ in the above constructions to construct the two orientations on $G$ and show that $R_g$ preserves both orientations. But these two new orientations are exactly $\{ \mathcal O, \widetilde{\mathcal O}\}$ since $G$ has exactly two orientations. Thus $R_g$ is also orientation-preserving with respect to $\mathcal O$ or $\widetilde{\mathcal O}$.
