Consider the multiplication map $\mu : G \times G \to G$ of a Lie group. So on the tangent level we have a map $T(G \times G) \to TG$. Making the proper identification $T(G\times G) \simeq TG \times TG$ and also identifying $TG$ with $G \times \mathfrak g$ through $\mathrm{triv}_L v_x \mapsto (x, \omega_G(v_x)) = (x , TL_{x^{-1}}\cdot v_x)$. Show that the map "$T\mu$" defined so that the following diagram commute is given by $$((x,v),(y,w))\mapsto (xy, TR_y \cdot v + TL_x \cdot w )$$ where the diagram is $$\require{AMScd}\begin{CD}T(G \times G) \simeq TG \times TG @>{T\mu}>> TG\\@VVV@VVV\\(G \times \mathfrak g)\times(G \times \mathfrak g)@>{"T\mu"}>> G \times \mathfrak g\end{CD}$$

Attempt: The idea was to simply find $T\mu = \mathrm{triv}_L \circ T\mu \circ (\mathrm{triv}_L \times \mathrm{triv}_L)^{-1}$ and this is what I have so far

$$(\mathrm{triv}_L \times \mathrm{triv}_L)^{-1} ((x,v),(y,w)) = (L_x ^ v, L_y^w) $$

then $$T\mu (L_x ^ v, L_y^w) = L_x ^ v + L_y^w \implies \mathrm{triv}_L (L_x ^ v + L_y^w) = (xy, TL_{(xy)^{-1}} (L_x^v + L_y^w))$$

Now working on the second coordinate of the last equation we obtain

$$\begin{align}TL_{(xy)^{-1}} (L_x^v + L_y^w) &= TL_{(xy)^{-1}} TL_{x}\cdot v_x + TL_{(xy)^{-1}}TL_{y}\cdot w_y\\&= TL_{y^{-1}x^{-1}}TL_{x}\cdot v_x + TL_{y^{-1}x^{-1}}TL_{y}\cdot w_y\\&=TL_{y^{-1}}\cdot v_x +TL_{y^{-1}}TL_{x^{-1}}TL_{y}\cdot w_y \end{align}$$

and I couldn't get past this.

Any ideas on how I should go on?

Note1: This is the Exercise 5.94 of Jeffrey Lee's Manifolds and Differential Geometry.

Note2: $L_x^v = T_e L_x \cdot v_x$ is the left-invariant vector field corresponding to $v_x \in TG$.


The formula for "$T\mu$" that you propose does not really make sense. If you take $(x,v)$ and $(y,v)$ in $G\times\mathfrak g$, then $TR_y\cdot v$ and $TL_x\cdot w$ are not elements in $\mathfrak g$, but in $T_yG$ and $T_xG$, respectively, so you cannot add them. Indeed, the formula you propose is the one for $T\mu:TG\times TG\to TG$. So you take $x,y\in G$, $\xi\in T_xG$ and $\eta\in T_yG$. Then $T_xR_y\cdot\xi$ and $T_yL_x\cdot \eta$ both lie in $T_{xy}G$, and $T_{(x,y)}\mu\cdot(\xi,\eta)= T_xR_y\cdot\xi+T_yL_x\cdot \eta$.

To get the formula in the left trivialization, you have to take $\xi=L^v_x$ and $\eta=L^w_y$. Then of course $T_yL_x\cdot L^w_y=L^w_{xy}$. For the other summand, you have to move $T_xR_y\cdot L^v_x$ back to the origin using a left translation. So this gives $T_{xy}L_{y^{-1}x^{-1}}\cdot T_xR_y\cdot L^v_x$ using that left and right translations commute and that $L_{y^{-1}x^{-1}}=L_{y^{-1}}L_{x^{-1}}$ you see that the result is $T_{y^{-1}}R_y\cdot T_eL_{y^{-1}}\cdot v=Ad(y^{-1})(v)$. Hence the correct formula in the left trivialization is $((x,v),(y,w))\mapsto (xy,Ad(y^{-1})(v)+w)$. This also agrees with the usual product formula for the left logarithmic derivative.

| cite | improve this answer | |
  • $\begingroup$ I had the impression that Aaron had problems seeing the result for $T\mu$. So consider $f: I\times I \rightarrow \mathbb{R}$, then $\frac{d}{dt}\big|_{t=0}f(t,t)=Df|_{t=0}\cdot(1,1)=\partial_1 f(0,0) + \partial_2 f(0,0)$. Applying the definition of $T\mu$, one exactly get $T_{(x,y)}\mu(\xi,\eta)=T_xR_y \xi+ T_yL_x\eta$ since $L_x=\mu(x,\cdot)$ and $R_y=\mu(\cdot,y)$. $\endgroup$ – Markus Heinrich May 25 '16 at 7:28
  • $\begingroup$ I must say that this Exercise was very misleading. But thanks to your observations I managed to prove everything accordingly. $\endgroup$ – Aaron Maroja May 25 '16 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.