On the definition of a connection 1-form

Question

Let $P\to M$ be a principal $G$-bundle. I'm reading Mathematical Gauge Theory by Mark Hamilton and he defines a connection $1$-form as $A\in \Omega^1(P,\mathfrak{g})$ satisfying

(i) $({\rm R}_g)^*A = {\rm Ad}_{g^{-1}}\circ A$ for all $g\in G$ and

(ii) $A(X^\#)=X$ for all $X\in \mathfrak{g}$.

Here ${\rm R}_g(p)=pg$ is the action map of $g$ and $X^\#\in \mathfrak{X}(P)$ is the action field of $X$. While condition (ii) seems very natural to me and I can sort of convince myself of needing $g^{-1}$ instead of $g$ on the right side of (i) (maybe because $G$ acts on $P$ by the right?), I'm not sure how to understand condition (i).

What does it mean?

Answer 1

One way you can understand condition (i) is by noting that condition (ii) already implies that it must be true on vertical vectors. This follows because $(R_g)_*X^\sharp = (\operatorname{Ad}_{g^{-1}}X)^\sharp$, and so $$ ((R_g)^*A) (X^\sharp) = A((R_g)_*X^\sharp) = A((\operatorname{Ad}_{g^{-1}}X)^\sharp) = \operatorname{Ad}_{g^{-1}}X = \operatorname{Ad}_{g^{-1}}[A(X^\sharp)], $$ i.e., $(R_g)^*A = \operatorname{Ad}_{g^{-1}}\circ A$ on vertical vectors $X^\sharp$. So (i) is a natural extension of this condition to all vectors in $P$.

A consequence of (ii) is that the horizontal distribution $H\subset TP$ corresponding to $A$, defined by $H := \ker A$, is invariant under the $G$-action. $H$ is complementary to the vertical distribution $V$, in the sense that $TP = H\oplus V$, and a connection can equivalently be defined as such a $G$-invariant complement.

If you were to start with $H$ as the fundamental definition of a connection, then defining the connection 1-form $A\in\Omega^1(P,\mathfrak{g})$ by $A(V+X^\sharp) := X$ for $V\in H$ and $X\in\mathfrak{g}$, condition (ii) would hold as a consequence of the $G$-invariance of $H$.

Answer 2

After studying a bit more, I'll just leave one more very clear interpretation of condition (ii): $G$ acts on itself by the left via conjugation, by $(a,g) \mapsto C_{g^{-1}}(a) = gag^{-1}$, meaning that it acts on $\mathfrak{g}$ via the derivative of the original action, i.e., $(X,g) \mapsto {\rm Ad}(g^{-1})X$ is a right action of $G$ on $\mathfrak{g}$. So saying that $A \in \Omega^1(P, \mathfrak{g})$ satisfies condition (ii) just means that $A$, seen as a map $A\colon TP \to \mathfrak{g}$, is $G$-equivariant.