# The tangent map of multiplication - Maurer-Cartan form

Question:

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.
• 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 $R_y=\mu(\cdot,y)$. – Markus Heinrich May 25 '16 at 7:28