Is Maurer-Cartan form bi-invariant for $SO(n)$?

Question

If we parametrize $SO(2)=S^1$ by the angle $\theta$, then the left-invariant Maurer-Cartan form $\Theta=g^{-1}\, \mathrm{d}g$ is $$\begin{pmatrix} 0 & \mathrm{d}\theta \\ -\mathrm{d}\theta & 0 \end{pmatrix},$$ and is actually bi-invariant. Is the same true for $SO(n)$ in general?

I think it is not true because after some calculation I got this is equivalent to $\Theta(I_n)$ commuting with all $g\in SO(n)$, which forces $\Theta(0)=0$ if $n>2$.

Is my argument correct? Can anyone give a proof or disproof of the statement?

If the statement is not correct, how is Maurer-Cartan form related to the bi-invariant metric on compact Lie groups, say $SO(n)$?

Answer 1

You are right, a Maurer-Cartan form for $SO(n)$ is not bi-invariant in general.

For any $g\in SO(n)$, $$ g^{\top}g=I_n\Longrightarrow{\rm d}g^{\top}g+g^{\top}{\rm d}g=0\iff\left(g^{\top}{\rm d}g\right)^{\top}+g^{\top}{\rm d}g=0\iff g^{-1}{\rm d}g=-\left(g^{-1}{\rm d}g\right)^{\top}, $$ where the last step is due to $g^{-1}=g^{\top}$.

A Maurer-Cartan form is of course left-invariant, which, in this case, can be seen through $$ \left(L_a\right)^*g^{-1}{\rm d}g=\left(ag\right)^{-1}{\rm d}\left(ag\right)=g^{-1}a^{-1}a{\rm d}g=g^{-1}{\rm d}g. $$ By contrast, for right-translation, we have $$ \left(R_a\right)^*g^{-1}{\rm d}g=\left(ga\right)^{-1}{\rm d}\left(ga\right)=a^{-1}g^{-1}{\rm d}g\,a=a^{-1}\left(g^{-1}{\rm d}g\right)a. $$ The last term from above is not necessarily the same as $g^{-1}{\rm d}g$ even for $SO(3)$. For example, let $$ g=\left( \begin{array}{ccc} \cos\varphi&-\sin\varphi&0\\ \sin\varphi&\cos\varphi&0\\ 0&0&1 \end{array} \right)\in SO(3)\Longrightarrow g^{-1}{\rm d}g=\left( \begin{array}{ccc} 0&-1&0\\ 1&0&0\\ 0&0&0 \end{array} \right){\rm d}\varphi\in so(3). $$ Now consider $$ a=\left( \begin{array}{ccc} 1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&\cos\theta \end{array} \right)\in SO(3). $$ We have $$ a^{-1}\left(g^{-1}{\rm d}g\right)a=\left( \begin{array}{ccc} 0&-\cos\theta&\sin\theta\\ \cos\theta&0&0\\ -\sin\theta&0&0 \end{array} \right){\rm d}\varphi\not\equiv g^{-1}{\rm d}g. $$

For a compact Lie group $G$, it possesses a bi-invariant metric for its Lie algebra $\mathcal{G}$, i.e., $$ \left<\cdot,\cdot\right>:\mathcal{G}\times\mathcal{G}\to\mathbb{R}. $$ Yet note that $\mathcal{G}$ is isomorphic to $T_eG$, the tangent space of $G$ at its identity $e$. Thus the above metric is also a metric for $T_eG$, i.e., $$ \left<\cdot,\cdot\right>:T_eG\times T_eG\to\mathbb{R}. $$ Now thanks to the Maurer-Cartan form, this metric can be extended to a metric for $TG$, the whole tangent space of $G$, as follows. Let $$ \omega:TG\to\mathcal{G}\cong T_eG $$ be a Maurer-Cartan form on $G$. Intuitively, $$ \omega_g=\left(L_{g^{-1}}\right)_*,\quad\forall\,g\in G, $$ meaning that a Maurer-Cartan form maps a tangent vector in $T_gG$ to a tangent vector in $T_eG$ by left-translation. As such, $$ \left<\cdot,\cdot\right>\circ\omega_g:T_gG\times T_gG\to\mathbb{R},\quad\left(X_g,Y_g\right)\mapsto\left<\omega_g(X_g),\omega_g(Y_g)\right> $$ is a metric for $T_gG$, and in general, $$ \left<\cdot,\cdot\right>\circ\omega:TG\times TG\to\mathbb{R},\quad\left(X,Y\right)\mapsto\left<\omega(X),\omega(Y)\right> $$ which is a metric for $TG$. Since $\left<\cdot,\cdot\right>$ is bi-invariant (for $\mathcal{G}$), $\left<\cdot,\cdot\right>\circ\omega$ is bi-invariant (for $TG$).

Answer 2

A Maurer-Cartan left invariant form $\omega$ comes from a linear functional $l$ on $\mathfrak{g}$ that is then extended on all tangent spaces to make it left invariant, that is $\omega(X)=l(X_e)$. Now, $\omega$ is also right invariant if and only if $l$ is invariant under all the maps $Ad(g)$, $g\in G$. If $l\ne 0$, this cannot happen for any semisimple group $G$, since it has a trivial center, and the adjoint representation is self dual ( so, since there are no $X\in \mathfrak{g}$ invariant under all $Ad(g)$ - they have to lie in the center of $\mathfrak{g}$). However, for say the group $U(n)$ we have a bi-invariant Maurer-Cartan, obtained by extending the linear map $l(A)= Trace(A)$. $\omega$ gives a distribution and the kernels at each point are the tangent spaces to the cosets of the subgroup $SU(n)$. Similarly, we can get a bi-invariant Maurer-Cartan on $GL(n,\mathbb{R})$.