Let $G$ be a matrix Lie group. Let $v$ be a left invariant vector field on $G$ and $v_1 \in \frak g$, where $\frak g$ is a Lie group of $G$. Let $v_1$ be its value at the identity. We define $\phi_t :G \to G$ by $\phi_t(g)=g \exp(t v_1)$.
I want to show that
$$ \frac{d}{dt} \phi_t(g)|_{t=0} = v_g $$ for all $g \in G$.
I calculated that
$$ \frac{d}{dt} \phi_t(g)|_{t=0} = g v_1 \exp(tv_1) |_{t=0}=gv_1$$
I want to show that $g v_1=v_g$.
Now $v_g=(L_g)_* v_1$, where $L_g$ is a left multiplication map by $g\in G$.
Evaluated at $f \in C^{\infty}(G)$, we have
$$v_g(f)=(L_g)_*v_1(f)=v_1(L_g^* f)$$.
On the other hand we have
$$ g v_1(f)=L_g v_1 (f)$$.
But I could not show that these are equal.
Did I make any mistake in the calculation? How can I prove they are equal?
Best Answer
Since for $ h \in G $, $ L_g(h) = g h $, and since $ exp(t v_1) $ is a curve with velocity $ v_1 $, the line $ \frac{d}{dt} \phi_t(g)|_{t=0} = \frac{d}{dt} g \exp(tv_1) $ is precisely the definition of the pushforward of $ v_1 $ by $ L_g $,
Note that you have assumed that $ G $ is embedded in a matrix group $ GL_n $ for some $ n $, so that the group operation is matrix multiplication and the exponential is literally matrix exponentiation. You are also so choosing embedding $ T_g G $ in $ T_g GL_n \approx End(\mathbb{R}^n) $. This is why you can further simplify $L_g v_1 $ to write $ g v_1 \in T_g G $.
Edit: Remember that you can think of the tangent space of a manifold $ M $ at $ p $ as equivalence classes of curves passing through $ p $. In this picture, if a vector $ v \in T_p M $ is represented by a curve $ \gamma(t) $ with $ \gamma(0) = p, \frac{d\gamma}{dt} (0) = v $, and if $ f: M \rightarrow M $, then the pushforward is given by composition: $ f_* v $ is represented by $ f(\gamma(t)) $.