What is the intuition (geometrical interpretation) of $[\omega\wedge\omega]$ in Cartan's structure equation?

Question

The Cartan's equation in principal bundle:

$$ {\textrm{hor d}}\omega = \textrm{d}\omega + \frac{1}{2}[\omega\wedge\omega] $$

Is there an intuition or geometric interpretation of the second term, $[\omega\wedge\omega]$?

Answer 1

I don't know of a direct geometrically intuitive way to see this. However, one way to get to this is formula is to look for a way to make $d\omega$ a horizontal $2$-form that is tensorial with respect to the action of $G$.

Let $P$ be a principal $G$-bundle over $M$. For convenience, I will assume that the Lie group $G$ is a matrix group. Recall that $\omega$ is a $\mathfrak{g}$-valued $1$-form on the principal bundle $P$ such that given any section $s$ of the principal bundle $P$, $$ \nabla s = s^*\omega. $$ Given any section $\bar{s}$, there is a map $g: M \rightarrow G$ such that $\bar{s} = sg$. A straightforward calculation shows that $$ \bar{s}^*\omega = g^{-1}s^*\omega g + g^{-1}\,dg. $$ Clearly, $d\bar\omega$ depends on both $d\omega$ and $dg$. To get a geometric invariant, we want to find something that depends on $g$ but not on $dg$. A hint is given by the fact that the second term on the right satisfies the Maurer-Cartan equations $$ d\Omega + \Omega\wedge \Omega = 0. $$ On the other hand, the first term is defined using the adjoint action of $G$ on $\omega$. The vertical derivative will be the adjoint action of $\mathfrak{g}$. An explicit calculation shows that $$ d\bar\omega = g^{-1}d\omega\,g -\Omega\wedge \omega - \omega\wedge\Omega - \Omega\wedge \Omega. $$ At this point, it is not hard to see that "completing the square" eliminates the last 3 terms on the right: $$ d\bar\omega + \bar\omega\wedge\bar\omega = g^{-1}(d\omega + \omega\wedge\omega) $$ This means that the action of $G$ on the $2$-form $$ d\omega + \omega\wedge\omega $$ is zero-th order, i.e., it does not depend on the derivatives of $g \in G$. This implies the $2$-form is a well-defined tensor on $P$ and therefore a geometric invariant.

Answer 2

As the notation you use suggests, $d\omega+\tfrac12[\omega\wedge\omega]$ is the horizontal projection of $d\omega$, so $\tau:=-\tfrac12[\omega\wedge\omega]$ is the vertical projection of $d\omega$. This means that it vanishes upon insertion of one vector field that is horizontal and coincides with $d\omega$ on vertical vector fields. By definition, for vector fields $\xi$ and $\eta$, one has $\tau(\xi,\eta)=[\omega(\xi),\omega(\eta)]$, which readily shows that it vanishes if either $\xi$ or $\eta$ is horizontal. To check that it coincides with $d\omega$ on vertical vector fields, it suffices to insert a fundamental vector field. So for $A\in\mathfrak g$ consider the fundamental vector field $\zeta_A$, which by definition has the property that $\omega(\zeta_A)=A$. But the infinitesimal version of equivariancy of $\omega$ under the principal right action reads as $\mathcal L_{\zeta_A}\omega=-\text{ad}_A\circ\omega$ where $\mathcal L$ is the Lie derivative. Since $d(\omega(\zeta_A))=0$, $\mathcal L_{\zeta_A}\omega=i_{\zeta_A}d\omega$. But this exactly says that $d\omega(\zeta_A,\eta)=-(\text{ad}_A\circ \omega)(\eta)=-[A,\omega(\eta)]=-[\omega(\zeta_A),\omega(\eta)]$. Hence indeed $\tau$ coincides with $d\omega$ on vertical vector fields.