pullback of Maurer-Cartan form for matrix lie groups

Question

I want to show that for a manifold $M$ and smooth map $\varphi:M\rightarrow GL(n,\mathbb{R})$ we have $\varphi^*\omega=\varphi^{-1} d\varphi$.

The Maurer-Cartan form is defined as for $X\in T_gG$ we have $\omega(X)=d_g(L_{g^{-1}})(X)$. My idea to approach this probblem was to first show that for a matrix Lie group the Maurer-Cartan form is given by $g^{-1}dg$ and then statment $\varphi^*\omega=\varphi^{-1} d\varphi$ should somehow follow.

First question: To show that $\omega(X)=d_g(L_{g^{-1}})(X)$ agrees with $g^{-1}dg$. For a matrix group $L_h$ is a linear operaion so $d_g(L_h)=L_h$ and then $\omega(X)=L_{g^{-1}}(X)=g^{-1}(X)$. But how do we get $dg$ into this?

Second question: Does $\varphi^*\omega=\varphi^{-1} d\varphi$ follow from $g^{-1}dg$ and if so how?

Answer 1

This can be done by some simple manipulation with your definitions:

\begin{eqnarray*} (\varphi^*\omega)_p(X_p) & = & \omega_{\varphi(p)}(\varphi_*X_p) \\ & = & d_{\varphi(p)}\left (L_{\varphi(p)^{-1}} \right ) d\varphi_p(X_p) \\ & = & \left . \frac{d}{dt} \right |_{t=0} L_{\varphi(p)^{-1}} \cdot \varphi(\gamma(t)) \\ & = & \varphi(p)^{-1}\left [\left . \frac{d}{dt} \right |_{t=0} \varphi(\gamma(t)) \right ] \\ & = & \varphi(p)^{-1}d\varphi_p(\gamma'(t)) \end{eqnarray*}

where we take $\gamma(t)$ to be the integral curve of $X$ with initial point $p$.

Answer 2

You can do this by writing everything in terms of the elements of a matrix $g \in GL(n,\mathbb{R})$.

First, recall that if you have a map $\phi: M \rightarrow N$ and a $1$-form $\omega = a_i(y)\,dy^i,$ then $$ \phi^*\omega = a_i(\phi(x))\,d(\phi(x))$$.

Let $g^i_j$ denote the element in the $i$-th row and $j$-th column of an element $g \in GL(n,\mathbb{R})$. Since $GL(n,\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$, the $g^i_j$, $1 \le i,j \le n$, are coordimates on $GL(n,\mathbb{R})$.

The Maurer-Cartan form is a matrix-valued $1$-form where the element in the $i$-th row and $j$-th column is $$ \omega^i_j = (g^{-1})^i_k\,dg^k_j. $$ Therefore, $$ \phi^*\omega^i_j = \phi^*((g^{-1})^i_k\,dg^k_j) = (\phi^{-1})^i_kd\phi^k_j = (\phi^{-1}\,d\phi)^i_j. $$