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?


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)) $.

  • $\begingroup$ I don't see why $\frac{d}{dt}g \exp(tv_1)$is the definition of $(L_g)_* v_1$. Could you explain this? $\endgroup$ – Snow Dec 17 '13 at 4:56

First, $\mathfrak g$ is the Lie algebra. The mistake is in your calculation of the derivative. Although you can make sense of $gv_1$ when working with matrix groups, this doesn't make sense in general.

As @user116293 told you, in the setting of general manifolds, if $f\colon M\to N$ is a smooth map and $v\in T_pM$, one way to understand/define $f_*(v) = df_p(v)$ is to define a curve $\gamma(t)$ in $M$ with $\gamma(0)=p$ and $\gamma'(0) = v$ and then compute $(f\circ\gamma)'(0)\in T_{f(p)}N$. In your situation, $v=v_1$, $\gamma(t) = \exp(tv_1)$, and $f\colon G\to G$ is given by $f(x)=L_gx = gx$.

  • $\begingroup$ How do I calculate the derivative correctly? Also I didn't write $vg_1$. Do you mean $gv_1$? $\endgroup$ – Snow Dec 17 '13 at 5:00
  • $\begingroup$ Yes, sorry, edited. It was very late :) $\endgroup$ – Ted Shifrin Dec 17 '13 at 12:51

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