I am a little confused with making computations with Lie groups and Lie algebras, and would appreciate very much any help with the following questions.
In page 44 of my Brazilian edition of do Carmo's Riemannian Geometry, he states that if a Lie group $G$ has a bi-invariant metric, the inner product that the metric induces on the Lie algebra $\mathcal G$ satisfies
$$
\langle [U, X], V \rangle = – \langle U, [V, X] \rangle.
$$
He argues in the following way:
For every $a \in G$, the automorphism $R_{a^{-1}}L_a: G \longrightarrow G$ is a diffeomorphism that leaves $e$ fixed. Therefore, the differential $d(R_{a^{-1}}L_a) = Ad(a): \mathcal G \longrightarrow \mathcal G$ is a linear map.
First question: the differential above is at $e$, right?
He proceeds:
Explicitly,
$$
Ad(a)Y = dR_{a^{-1}} dL_a Y = dR_{a^{-1}}Y \quad \forall Y \in \mathcal G
$$
Second question: Here, the differential $dR_{a^{-1}}$ is computed at $a$, right? So it should be
$$
d(R_{a^{-1}})_a (dL_a)_e Y(e) = d(R_{a^{-1}})_aY(a).
$$
He now argues that if $x_t$ is the flow of $X \in \mathcal G$ then
$$
[Y, X] = \lim_{t \to 0} \frac1t (dx_t(Y) – Y).
$$
Third question: Shouldn't it be $$
[Y, X] = \lim_{t \to 0} \frac1t (dx_t(Y) – Y)x_t ?
$$
Following, he claims that since $X$ is left-invariant, then $L_y \circ x_t = x_t \circ L_y$
Why does it hold?
He then concludes the proof, but the remaining of it I think I can understand
Thanks in advance and kind regards.
Best Answer
First question: Yes, when dealing with Lie groups, if the base point of a differential is omitted you can assume that it is the identity.
Second question: This is a corollary of the first question: $${\rm d}(R_{a^{-1}}\circ L_a)_e(Y(e)) = {\rm d}(R_{a^{-1}})_a{\rm d}(L_a)_e(Y(e)) = {\rm d}(R_{a^{-1}})_a(Y(a)) \in \mathfrak{g}.$$
Third question: given two left-invariant fields $X$ and $Y$, $[Y,X] = -[X,Y]$ will be a left invariant field. If you regard $X$ and $Y$ as elements of $\mathfrak{g}$, then $[X,Y]$ will also be an element of $\mathfrak{g}$. A usual abuse of notation is, for $g \in G$ and $v \in T_aG$, to write $gv$ for ${\rm d}(L_g)_a(v)$ and $va$ for ${\rm d}(R_g)_a(v)$. The bracket $[X,Y]$ is supposed to be the Lie derivative $\mathcal{L}_XY$ of $Y$ in the direction of $X$, so you want to mimic the definition $$\lim_{t \to 0}\frac{Y(a+tX(a)) - Y(a)}{t},$$but you cannot subtract $Y(a+tX(a))$ and $Y(a)$ as they lie in different tangent spaces. The workaround is to drag $Y$ using the flow of $X$, so one gets $$[X,Y] = \lim_{t\to 0} \frac{ {\rm d}(x_{-t})_{x_t(a)}Y(x_t(a)) - Y(a)}{t}.$$Carry all the base points carefully and compare what you get with Wikipedia, for instance.
Last question: the flows of left invariant fields consist of right translations, and vice-versa. And left translations commute with right translations in general. For instance, if $X$ is left-invariant, you may check that $x_t(a) = a x_t(e)$ (and $x_t(e)$ earns the better name and notation $\exp(tX)$ or ${\rm e}^{tX}$ -- note that I use different fonts, $e$ for the unit element and ${\rm e}$ for the exponential).