Show that $g^{-1}\frac{dg}{dt} \in \mathfrak{g}$

Question

Let's $G$ be a Lie Group with its respective Lie Algebra $\mathfrak{g}$. I would like to show the next statement:

If $g(t)$ is a smooth curve in $G$, then $g^{-1} \frac{dg(t)}{dt} \in \mathfrak{g}$

For simplicity, I am supposing matrix Lie groups, so exponential map is simply exponential matrix.

My work:

There exists some smooth curve $X(t) \in \mathfrak{g}$, such that $g(t) = \exp{(X(t))}$. Let's be $\{ G_1 \dots G_n\}$ a base of $\mathfrak{g}$ such that $X(t) = \sum_{i=1}^n x_i(t) G_i$.

\begin{equation} \begin{array}{rl} g^{-1} \frac{dg}{dt} = & g^{-1} \frac{d}{dt} \left( \exp (\sum_{i=1}^n x_i(t) G_i) \right) \\ = & g^{-1} \frac{d}{dt} \left( \sum_{k=0}^{\infty} \frac{(\sum_{i=1}^n x_i(t) G_i)^k}{k!} \right) \\ = & g^{-1} \sum_{j=1}^{n} \left\{ \frac{\partial }{\partial x_j(t)} \left( \sum_{k=0}^{\infty} \frac{(\sum_{i=1}^n x_i(t) G_i)^k}{k!} \right) \frac{d x_j(t)}{dt} \right\}\\ = & g^{-1} \sum_{j=1}^{n} \left\{ \sum_{k=0}^{\infty} \left( \frac{k X(t)^{k-1}}{k!}\right) G_j \frac{d x_j(t)}{dt} \right\}\\ = & g^{-1} \sum_{j=1}^{n} \left\{ \exp(X(t)) G_j \frac{d x_j(t)}{dt} \right\}\\ = & \sum_{j=1}^{n} \left\{ G_j \frac{d x_j(t)}{dt} \right\} \in \mathfrak{g} \end{array} \end{equation}

Questions: I am not sure at all that this development is right, so I have the next questions:

1. Is this development right? If not, how I should proceed?
2. Regarding existence of $X(t) \in \mathfrak{g}$ s.t. $g(t) = \exp(X(t))$, Is it because of the smoothness of the exponential map?
3. How much different would be the generalization to other Lie groups?

N.B. I found this statement in "Stochastic Models, Information Theory, and Lie Groups, Volume 2" (available online)

I really appreciate any help

Answer 1

The statement as written doesn't make sense for an arbitrary Lie group. On the left side, $g^{-1}$ (or, more accurately, $g(t)^{-1}$) is an element of the group, while $dg(t)/dt$ is an element of the tangent space $T_{g(t)}G$, so it makes no sense to multiply them together, let alone to ask whether their product is an element of the Lie algebra of $G$, which is the space of left-invariant vector fields on $G$.

However, if $G$ is a matrix group (i.e., a Lie subgroup of $GL(n,\mathbb R)$ for some $n$), then there is a way to interpret this statement so that it makes sense, and in fact is true.

First of all, our hypothesis implies that $G$ is a submanifold of the vector space $M(n,\mathbb R)$ of all real $n\times n$ matrices. It's a standard fact that because $M(n,\mathbb R)$ is a vector space, its tangent space at any point can be identified with $M(n,\mathbb R)$ itself. And because $G$ is a submanifold, its tangent space at any point can be identified with a certain linear subspace of $M(n,\mathbb R)$. Furthermore, it's a standard argument in Lie theory that the Lie algebra $\mathfrak g$ can be canonically identified with $T_IG$, the tangent space to $G$ at the identity matrix.

Now for any $g\in G$, consider the left multiplication map $L_g\colon G\to G$ given by $L_g(h) = gh$. It is a diffeomorphism with inverse $L_{g^{-1}}$. Therefore, its differential at the identity, $d(L_g)|_I\colon T_I G \to T_g G$ is a linear isomorphism, with inverse $d(L_{g^{-1}})|_g\colon T_g G \to T_I G$. So if we identify $T_I G$ with $\mathfrak g$ as described above, we can think of $d(L_{g^{-1}})|_g$ as a linear isomorphism from $T_g G$ to $\mathfrak g$.

We can compute this linear isomorphism explicitly. Let $\mathscr L_{g^{-1}}\colon M(n,\mathbb R)\to M(n,\mathbb R)$ denote the left multiplication map acting on all of $M(n,\mathbb R)$. This is a linear map, so in terms of the usual identification of tangent spaces of $M(n,\mathbb R)$ with $M(n,\mathbb R)$ itself, the differential $d(\mathscr L_{g^{-1}})|_g$ is the same linear map: that is, in terms of matrices, $d(\mathscr L_{g^{-1}})|_g(X) = g^{-1}X$ for any $X\in T_gG$. Moreover, since $L_{g^{-1}}\colon G\to G$ is the restriction of $\mathscr L_{g^{-1}}$, its differential is the restriction to $T_gG$ of $d(\mathscr L_{g^{-1}})|_g$, i.e., left multiplication by the matrix $g^{-1}$. So the upshot is that if we consider $T_gG$ as a linear subspace of $M(n,\mathbb R)$, then left multiplication by the matrix $g^{-1}$ is an isomorphism from $T_g G$ to $T_I G \cong \mathfrak g$.

Now suppose $J$ is an interval and $g\colon J\to G$ is a smooth curve. For any $t\in J$, the velocity vector $dg(t)/dt$ is an element of the tangent space $T_{g(t)}G$, and the discussion above shows that when we left multiply it by the matrix $g(t)^{-1}$, we get an element of $T_I G \cong \mathfrak g$.