17:51
I'm getting a bit ahead, but I've got a question on the tangent space.
I'll set up a few propositions first.
Proposition 1: For any point $p$ in a manifold $M$ and any tangent vector $X_p \in T_pM$, there are $\varepsilon > 0$ and a smooth curve $c : (-\varepsilon, \varepsilon) \to M$ such that $c(0) = p$ and $c'(0) = X_p$.
We define a smooth curve on a manifold $M$ to be a smooth map $c : (a, b) \to M$ and for $t_0 \in (a, b)$, we define the velocity vector to be $$c'(t_0) = c_* \bigg( \frac{d}{dt} \bigg \vert_{t_0} \bigg) \in T_{c(t_0)}M.$$
Proposition 2: Let $c : (a, b) \to M$ be a smooth curve, and let $(U, (x^1, \ldots, x^n))$ be a coordinate chart about $c(t)$. Write $c^i = x^i \circ c$ for the $i$th coordinate of $c$ in the chart. Then $c'(t)$ is given by $$c'(t) = \sum_{i = 1}^n \dot c^i(t) \frac{\partial}{\partial x^i} \bigg \vert_{c(t)}.$$
Proposition 3: Let $F : N \to M$ be a smooth map of manifolds, $p \in N$ and $X_p \in T_pN$. if $c$ is a smooth curve starting at $p$ in $N$ with velocity $X_p$ at $p$, then $$F_*(X_p) = \frac{d}{dt} \bigg \vert_0 F \circ c.$$
Now to the problem: I want to calculate the differential of left multiplication by a matrix on the general linear group over $\mathbb R$. Or more precisely, I want to understand a line in a proof.
The Statement: If $g \in \text{GL}_n(\mathbb R)$, let $l_g : \text{GL}_n(\mathbb R) \to \text{GL}_n(\mathbb R)$ be defined by $B \mapsto g \cdot B$.
Notice that $\text{GL}_n(\mathbb R) \subseteq \mathbb R^{n^2}$ and so is naturally a submanifold of $\mathbb R^{n^2}$.
This tells us that $T_g(\text{GL}_n(\mathbb R)) \cong T_g(\mathbb R^{n^2}) \cong \mathbb R^{n^2}$.
I want to show that $(l_g)_* : T_I(\text{GL}_n(\mathbb R)) \to T_g(\text{GL}_n(\mathbb R))$ is also multiplication by $g$.
Here's a proof I've come across and am trying to decipher:
The Proof: Let $X \in T_I(\text{GL}_n(\mathbb R))$. By Proposition 1, there exist $\varepsilon > 0$ and a smooth curve $c : (-\varepsilon, \varepsilon) \to \text{GL}_n(\mathbb R)$ such that $c(0) = I$ and $c'(0) = X$.
It follows that $l_g \circ c = g \cdot c$ (simply matrix multiplication).
By Proposition 3: $$(l_g)_*(X) = \frac{d}{dt} \bigg \vert_{t = 0} l_g \circ c = \frac{d}{dt} \bigg \vert_{t = 0} g \cdot c = g c'(0) = gX.$$
In this computation, $\displaystyle \frac{d}{dt} \bigg \vert_{t = 0} g \cdot c = g \cdot c'(0)$ by $\mathbb R$-linearity and Proposition 2.
It's this last line I can't understand. I don't see why it's true, why $\mathbb R$-linearity comes into play and how Proposition 2 is used.
