Disclaimer: Nothing in this note is original.

The Maurer-Cartan form

Let $E\to M$ be a fiber bundle and let $G$ be the transition group. As usual, use the gothic letters ${\frak g}$ to denote the Lie algebra of $G$. On the bundle we can define connection and curvature. It is usually quite helpful to regard connection and curvature as Lie-algebra valued, or ${\frak g}$-valued object, to be specific ${\frak g}$-valued differential forms. So, what is ${\frak g}$-valued form?

Let $M$ be a $n$-dimensional manifold, and $\phi$ a locally defined exterior form on $M$, taking values in the fixed vector space ${\frak g}$. Note now we have switched to the geometrical viewpoint of a Lie group, where ${\frak g}$ is a vector space, and $G$ itself a manifold.

First we define a ${\frak g}$-valued 1-form on $G$ itself. By definitions, given a tangent vector on $G$, namely a vector in $TG$, a ${\frak g}$-valued 1-form should take this vector to an element of ${\frak g}$, which is a vector at the identity $e$. Now let’s see how this idea is realized.

Let $E_ {R}$ be the basis of ${\frak g}$ and $X_ {R}$ be the left invariant vector fields of $E_ {R}$, where $R=1,\dots,\text{rank}(G)$. In our convention we use the capital Latin letter to label the Lie algebra indices. Let $\sigma^{R}$ be left invariant 1-forms on $G$ forming a basis dual to $X_ {R}$. Then

\[\Omega := E_ {R}\otimes \sigma^{R}\]

defined by

\[\Omega(Y_ {g}) := E_ {R} \otimes \sigma^{R}(Y_ {g}) = E_ {R}Y^{R}\]

is a ${\frak g}$-valued 1-form. What $\Omega$ does is that it takes the vector $Y$ at $g\in G$ and left translate it back to the identity! This is the Maurer-Cartan 1-form on $G$.

Classically this would be written as

\[g^{-1}dg\]

which might look rather weird at the first look. Here $dg$ is better thought of as a vector at $T_ {g}M$ and $g^{-1}$ the induced map of $g^{-1}$ that sends a vector at $T_ {g}M$ to $T_ {e}M$, where $e$ is the identity of the group. We should write

\[\Omega = L(g^{-1})_ {\ast }\,\circ\,dg.\]

I admit all this is rather confusing, because by the look of it $dg$ should be a form not a tangent vector, but here it is best to think of $dg$ as a vector. I can’t think of a better way to interpret it.

For a matrix group, each $E_ {R}$ is just a generator matrix, we can apply the usual matrix algebra, including the matrix multiplication and inverse, to construct $\Omega$, interested readers can work it out for $SO(2)$ group for example.

Theorem. In any matrix group, $g^{-1}dg$ is a matrix with left invariant $1$-form entries.

The left invariantness is a little tricky, try to convince yourself that it is the case for $SO(2)$ group.

from 1-form to $p$-form

The most general ${\frak g}$-valued $p$-form on a manifold $M$ is

\[\phi := E_ {R}\otimes \phi^{R},\quad \phi^{R}\in \Omega^{p}(M).\]

Since the $E_ {R}$’s do not vary, it is natural to define the

\[d\phi := E_ {R}\otimes d\phi^{R}.\]

The multiplication of two such ${\frak g}$-valued p-forms are not so clear. We want the multiplication still to be a ${\frak g}$-valued p-form. Take ${\frak o}(n)$ for example, the Lie-algebra vectors are anti-symmetric matrices, but the product of two anti-symmetric matrices may not be anti-symmetric anymore, so we can’t adopt the naïve matrix multiplication. The solution is to take the Lie bracket of the Lie algebras, namely the commutator and define

\[[\phi,\psi] = [E_ {R}\otimes \phi^{R},E_ {S}\otimes \psi^{S}] := [E_ {R},E_ {S}]\otimes \phi^{R}\wedge \psi^{S}.\]

As an example, consider the Maurer-Cartan 1-form on $M=G$, we have

\[[\Omega,\Omega]=[E_ {R}\otimes \sigma^{R},E_ {S}\otimes \Omega^{S}] = [E_ {R},E_ {S}]\otimes \sigma^{R}\wedge \sigma^{S}=C_ {RS}^{T}E_ {T}\otimes \sigma^{R}\wedge \sigma^{S}.\]

From the Maurer-Cartan equation

\[d \sigma^{A} = -\sum_ {R<S} C^{A}_ {RS}\sigma^{R}\wedge \sigma^{S}\]

we have

\[d\Omega + \frac{1}{2} [\Omega,\Omega]=0\]

which is again called the Maurer-Cartan equation.

We have some immediate consequences from the definition, for $\phi \in\Omega^{p}$ and $\psi \in\Omega^{q}$

\[[\phi,\psi] = (-1)^{pq+1}[\psi,\phi]\]

and

\[d[\phi,\psi] = [d\phi,\psi]+(-1)^{p}[\phi,d\psi].\]

If we regard the Lie algebra vectors as matrices, then ${\frak g}$-valued forms are matrices with forms as entries, the product

\[\phi \wedge \psi = E_ {R}\phi^{R} \wedge E_ {S}\psi^{S}\]

is defined as matrix multiplication but using the exterior product of the entries. Then it can be shown that we have a relation between the Lie algebra product

\[[-,-]: {\frak g} \times {\frak g} \to {\frak g}\]

and the exterior product,

\[[\phi,\psi] = \phi \wedge \psi - (-1)^{pq}\psi \wedge \phi.\]

For example, for $p,q$ odd, we have

\[[\phi,\phi]=2\phi \wedge \phi.\]

Connection in a principal bundle

Let $E\to M$ be a real or complex vector bundle of rank $K$. Let $G$ be the structure group. By this we mean that, in each trivializing patch $U$ of $M$, there exists a collection of distinguished frames (e.g., orthonormal), which we may call $G$ frames, and such that any two such frames $e_ {U}$ and $e_ {V}$ in an overlap $U \cap V$ are related by $e_ {V} = e_ {U} c_ {UV}$ where $c_ {UV}\in G$.

A $G$-connections essentially means that given a distinguished frame $\mathbf{f}$, we can use the $G$ connection to parallel transport it along any curve, the transported frame at any point of any curve is still a distinguished frame. This requires the connection to be ${\frak g}$-valued (1-form). Then of course the curvature is also ${\frak g}$-valued,

\[\theta = d\omega+\omega \wedge \omega.\]

Under a change of frame, namely in the overlap of two trivializing patches $U$ and $V$, we require that (the compatibility condition) the following two operations to be equivalent (or in the terminology of category theory, to be commute)

  • parallel transport in $U$ first, then convert to $V$ patch;
  • convert to $V$ first, then parallel transport in $V$. This gives us the transition property of the connection and curvature under the change of patches,
\[\omega_ {V} = c_ {VU}(\omega _ {U} + d) c_ {UV}\]

and

\[\theta_ {V} = c_ {VU}\omega_ {U}c_ {UV}.\]

This should be familiar to readers.

Given a vector bundle $E$, consider the principal bundle $P$ of frames of sections of the vector bundle. The bundle $P\to M$ has for fiber $F$ the structure group $G$ and the transition function $c_ {UV}$ are the same as for $E$.

We can now define the connection $\omega^{\ast}$ on $P$. Note that $\omega^{\ast}$ is ${\frak g}$-valued connection on $P$ rather than $M$. However I must warn you that this is not the usual way to introduce the connection on the bundle.

  1. Introduce the local frame of $P$, denoted $e_ {U}$. It can be thought of as a section of $P$ over $U$.
  2. For a point $\mathbf{f}$ over $x\in U$ we can write it in the basis $e_ {U}$,
\[\mathbf{f} = e_ {U}g_ {U},\quad g_ {U}\in G.\]

But now the covariant derivative has two parts, one coming from $\nabla e_ {U}$, one from $\nabla g_ {U}$. We have

\[\begin{align} \nabla \mathbf{f} &= e_ {U}\omega g_ {U} + e_ {U} dg_ {U} = e_ {U} g_ {U} g_ {U}^{-1} \omega g_ {U} + e_ {U}g_ {U} g_ {U}^{-1} dg_ {U} \\ &= \mathbf{f}\otimes (g_ {U}^{-1}\omega g_ {U}+g_ {U}^{-1} dg_ {U}) \\ &=:\mathbf{f}\otimes \omega ^\ast . \end{align}\]

But we want to define a connection on $P$, in the meanwhile $\omega$ is a connection on the base manifold $M$, so we actually need $\pi ^\ast\omega$ instead of $\omega$ between $g_ {U}^{-1}$ and $g_ {U}$. Anyway we usually neglect $\pi ^\ast$ in $\pi ^\ast\omega$ since it is clear from the context.

The local frame $\mathbf{e}_ {U}$ gives us a trivialization of $\pi ^{-1}U$, namely a local product structure of $\pi ^{-1}U$. We can regard $g_ {U}$ and $x\in U$ as the “coordinates” of $\mathbf{f}$.

Given some curve on $P$, the velocity vector, or tangent vector is $(dx / dt, d g_ {U} / dt)$. When acted by $\omega ^\ast$, the $g_ {U}^{-1}\omega g_ {U}$ part acts on $dx / dt$ and yields

\[(g_ {U}^{-1} \omega g_ {U})\left( \frac{dx}{dt} \right) = g_ {U}^{-1} \omega\left( \frac{dx}{dt} \right)g_ {U}\in {\frak g}.\]

The $g_ {U}^{-1}dg_ {U}$ part acts on $d g_ {U} / dt$ yields

\[g_ {U}^{-1} dg_ {U}\left( \frac{dg_{U}}{dt} \right) = g_ {U}^{-1} \frac{dg_{U}}{dt} \in {\frak g}.\]

It can be shown the given two trivializing patches $U$ and $V$, although $\omega_ {U}\neq \omega_ {V}$ we still have

\[\omega ^\ast _ {U} = \omega ^\ast _ {V},\]

thus $\omega ^\ast$ defines not a local, but a global connection on P!

In $p$ we may then consider the distribution of $n$-planes transversal to the fibers, defined by

\[\omega ^\ast =0.\]

This distribution is called the horizontal distribution. Many books take this distribution as the starting point for their discussion of connections.

We then define the global ${\frak g}$-valued $2$-form connection

\[\theta ^\ast := d\omega ^\ast +\omega ^\ast \wedge \omega ^\ast = d\omega ^\ast + \frac{1}{2} [\omega ^\ast ,\omega ^\ast ] .\]

The last equality is true due to the fact that $\omega ^\ast$ is a one-form.

Note that unlike in the case of a tangent bundle of a surface, where the structure group $G$ is abelian, we cannot expect $\theta ^\ast$ to be globally exact!

Of course we can also push-forward $\theta ^\ast$ from $P$ to the base space $M$, we also have (the proof is omitted)

\[\theta ^\ast (x,g_ {U}) = g_ {U}^{-1} \pi ^\ast \theta_ {U}(x)g_ {U}.\]