Trivial principal bundles and curvature.

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.

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.

• Thanks for your answer. Just one question: When you write $F_{\mathcal{M}}^{A}=dA+A\wedge A$, how do you view $A$? The connection form $A$ is a form in $\Omega^{1}(\mathcal{P},\mathfrak{g})$ and not on $\mathcal{M}$. What is true is that $F^{A}=\mathrm{d}A+A\wedge A$, but that equation is for $F^{A}\in\Omega^{2}(P,\mathfrak{g})$ and not $F_{\mathcal{M}}^{A}\in\Omega^{2}(\mathcal{M},\mathrm{Ad}(\mathcal{P}))$. Jul 12 at 13:57