Trivial principal bundles and curvature.

Question

Let $\mathcal{M}$ be a smooth manifold, $G$ a Lie group with Lie algebra $\mathfrak{g}$ and $\mathcal{P}\xrightarrow{\pi}\mathcal{M}$ a principal bundle. If $A\in\Omega^{1}(\mathcal{P},\mathfrak{g})$ is a connection form, we can define its curvature $F^{A}\in\Omega^{2}(\mathcal{P},\mathfrak{g})$. Now, it is a general fact that, since $F^{A}$ is "horizontal and of type Ad'', it can be identified with an element $F^{A}_{\mathcal{M}}\in\Omega^{2}(\mathcal{M},\mathrm{Ad}(\mathcal{P}))$, where $\mathrm{Ad}(\mathcal{P}):=\mathcal{P}\times_{\mathrm{Ad}}\mathfrak{g}$ denotes the adjoint bundle.

Now, let us assume that $\mathcal{P}$ is the trivial $G$-bundle, i.e. $\mathcal{P}\cong\mathcal{M}\times G$. As a consequence, also the adjoint bundle is the trivial vector bundle, i.e. $\mathrm{Ad}(\mathcal{P})\cong\mathcal{M}\times\mathfrak{g}$ (correct me if I am wrong). In particular, this implies that $F_{\mathcal{M}}^{A}\in\Omega^{2}(\mathcal{M},\mathfrak{g})$. Now, since $\mathcal{P}$ is trivial, there is a global section $s\in\Gamma^{\infty}(\mathcal{P})$ and we can define $F_{s}:=s^{\ast}F^{A}\in\Omega^{2}(\mathcal{M},\mathfrak{g})$.

Is there any relation between $F_{\mathcal{M}}^{A}$ and $F_{s}$, both of which are elements of $\Omega^{2}(\mathcal{M},\mathfrak{g})$?

If $G$ is abelian, then the answer is clearly yes, since in this case, one can easily show that $F_{s}$ is independent of the choice of $s$, by the transformation law of $F^{A}$ under gauge transformations. However, in the non-abelian case, it is not clear.

Answer 1

In the non-abelian case, the curvature forms $F_{\mathcal{M}}^{A}$ and $F_{s}$ are generally not equal. This is because the connection form $A$ and the section $s$ transform differently under gauge transformations, leading to distinct curvatures.

Let's analyze the situation more carefully. In the case of a trivial principal $G$-bundle $\mathcal{P} \cong \mathcal{M} \times G$, the connection form $A$ can be written as $A = g^{-1}dg$ for some gauge transformation $g: \mathcal{P} \rightarrow G$. Note that the choice of $g$ determines the connection form $A$ uniquely.

Under a gauge transformation $h: \mathcal{M} \rightarrow G$, the section $s$ and the connection form $A$ transform as follows:

\begin{align*} s &\mapsto s' = h \cdot s, \\ A &\mapsto A' = h^{-1}Ah + h^{-1}dh. \end{align*}

Now, let's consider the curvatures. The curvature $F_{\mathcal{M}}^{A}$ associated with $A$ is given by $F_{\mathcal{M}}^{A} = dA + A \wedge A$, where $d$ denotes the exterior derivative on $\mathcal{M}$.

On the other hand, we can define the curvature $F_s$ associated with the section $s$ by pulling back the connection form $A$ using $s$. In other words, $F_s = s^*F^{A}$, where $F^{A}$ is the curvature on $\mathcal{P}$. This can be expressed as $F_s = ds^{-1} \wedge ds + s^*A \wedge s^*A$.

Now, let's compare $F_{\mathcal{M}}^{A}$ and $F_s$:

\begin{align*} F_{\mathcal{M}}^{A} &= dA + A \wedge A, \\ F_s &= ds^{-1} \wedge ds + s^*A \wedge s^A. \end{align*}

The difference between the two curvatures is given by:

\begin{align*} F_s - F_{\mathcal{M}}^{A} &= ds^{-1} \wedge ds + s^*A \wedge s^*A - (dA + A \wedge A) \\ &= ds^{-1} \wedge ds + s^*A \wedge s^*A - (g^{-1}dg + g^{-1}dg \wedge g^{-1}dg) \\ &= ds^{-1} \wedge ds + s^*A \wedge s^A - g^{-1}dg - g^{-1}dg \wedge g^{-1}dg. \end{align*}

The expression $ds^{-1} \wedge ds$ arises due to the transformation of the inverse of the section $s$, while the term $g^{-1}dg \wedge g^{-1}dg$ arises due to the transformation of the connection form $A$.

Therefore, in general, $F_s$ and $F_{\mathcal{M}}^{A}$ are not equal, as there is an additional term $ds^{-1} \wedge ds - g^{-1}dg \wedge g^{-1}dg$ in their difference. This additional term accounts for the non-trivial transformation properties of the section $s$ and the connection form $A$ under gauge transformations.