Left and right fundamental differential form on Lie group

Suppose G is a Lie group. R_a and L_a are right and left translations. Assume e_i is a base of T_eG, the tangent space at the unit element e. X_i denote the right-invariant vector field obtained by the right translation and \tilde{X}_i by left translation. Let w^i and \tilde{w}^i denote the dual of X_i and \tilde{X}_i. Then we have the Maurer-Cartan equation

\displaystyle \begin{cases}dw^i=-\frac{1}{2}\sum\limits_{l,k=1}^n c^i_{lk}w^j\wedge w^k\\ c^i_{lk}+c^i_{kl}=0\end{cases}

Here c^i_{lk} are constants. Also we have the structure equation with respect to \tilde{w}^i. We see that there is a relation between \tilde{c}^i_{lk} and c^i_{lk}, that is

\tilde{c}^i_{lk}=-{c}^i_{lk}

\textbf{Proof: } Choose a local coordinate system (U,x^i) at e. Then

\displaystyle dy^i=\frac{\partial \phi^i(x,y)}{\partial x^j}\bigg|_{x=e}w^j

where \phi(x,y)=x\cdot y. Apply exterior differentiation on both sides

\displaystyle 0=\frac{\partial^2 \phi^i(x,y)}{\partial x^j\partial y^p}\bigg|_{x=e}dy^p\wedge w^j+\frac{\partial \phi^i(x,y)}{\partial x^j}\bigg|_{x=e}dw^j

Letting y=e, we get

\displaystyle dw^i=\frac{\partial^2 \phi^i(x,y)}{\partial x^k\partial y^l}\bigg|_{(e,e)}w^l\wedge w^k=-\frac{1}{2} \left(\frac{\partial^2 \phi^i(x,y)}{\partial x^l\partial y^k}\bigg|_{(e,e)}-\frac{\partial^2 \phi^i(x,y)}{\partial x^k\partial y^l}\bigg|_{(e,e)}\right)w^l\wedge w^k

Then it is easy to see \tilde{c}^i_{lk}=-{c}^i_{lk}.

\textbf{Remark: } see Chern’s lectures on differential geometry. p182

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