# Is conjugation by gauge transformation of $G$-bundle contained in $\mathfrak{g}$?

Before painstakingly defining all these terms, let me ask my question in plain english: given a $$G$$-bundle, is every conjugate of a vector field by a gauge transformation an element of the Lie algebra?

For a little more context, the question arises from a calculation in Exercise 86 of Part II (Gauge Fields) of Gauge fields, knots, and gravity by Baez and Muniain. Spurred on by the comments, I have decided to spell out the terminology in full detail as follows.

Let $$M$$ be a smooth manifold, let $$G$$ be a Lie group, and let $$E\to M$$ be a $$G$$-bundle, viz. a vector bundle whose model fiber $$V$$ is a (finite dimensional) vector space equipped with a representation $$\rho\colon G\to \mathrm{GL}(V)$$ such that all bundle transition maps belong to $$\rho(G)$$. Denote the endormorphism bundle of $$E$$ by $$\mathrm{End}(E)$$ and let $$G(E)\subset \Gamma\bigl(\mathrm{End}(E)\bigr)$$ denote the gauge group of the bundle. Elements $$g\in G(E)$$ are sections of $$\mathrm{End}(E)$$ such that for every local trivialization $$\phi\colon \mathrm{End}(E)\restriction U\to U\times \mathrm{GL}(V)$$ and for every $$p\in U$$, it holds that $$\phi(g_p)\in \{p\}\times \rho(G)$$. The derivative $$d\rho\colon\mathfrak{g}\to \mathrm{End}(V)$$ furnishes an embedding of the Lie algebra of $$G$$ into the set of linear transformations of $$V$$. Paralleling the definition of $$G(E)$$, we define the infinitesimal gauge group $$\mathfrak{g}(E)$$ to be the set of sections $$\mathcal{G}\in\mathrm{End}(E)$$ such that for $$\phi(\mathcal{G}_p)\in \{p\}\times d\rho(\mathfrak{g})$$ for every $$p$$ and for every local trivialization $$\phi$$ around $$p$$. Finally, on any trivialization $$\phi$$ of $$\mathrm{End}(E)\restriction U$$ let $$D^{\phi}$$ denote the standard flat connection on the trivialization.

Question. For all vector fields $$X$$ and $$\rho,\phi, U$$ as above is $$\rho(g)^{-1}D^\phi_X\rho(g)\in \mathfrak{g}(E\restriction U)$$?

• Can you please change "$g$ be a smooth section of $\text{End}(E)$" to "$s$ be a smooth section of $\text{End}(E)$".. What does it mean to say $g(p)\in G$ for all $p\in M$? Jan 13, 2019 at 7:50
• A $G$-bundle (as defined in II.2 of the aforementioned book) consists of a bundle together with a representation of $G$ on its model fiber such that the bundle transition maps belong to (the image of) $G$. Thus it is meaningful to stipulate that a section of $\textrm{End}(E)$ take values in (the image of) $G$ (see page 214 in the aforementioned book for a discussion of this point). Jan 13, 2019 at 8:08
• What do you mean by $g^{-1} X g$? As you said, $g(p)$ is a linear (?) map between the fibers $E_p \to E_p$. How does such a map act on a tangent vector to $M$ at $p$? Do you assume that $E$ is a natural bundle so that you have a lift of vector fields on $M$ to vector fields on $E$? Jan 13, 2019 at 11:16
• In local coordinates around $p$ write $g=g^i_je_i\otimes e^j$ and let $Xg:= X(g^i_j)e_i\otimes e^j$. An implicit part of the question is to understand whether the condition $(g^{-1}Xg)_p\in\mathfrak{g}$ using the above definition is independent of the choice of chart. Jan 13, 2019 at 11:48
• I think if you add all you said in comments to question it will become easier for other to read.. you might get more attention.. Jan 16, 2019 at 9:47

I suggest an alternative definition of ''$$G$$-bundle'', not using trivializations, that may help you sort out the confusion. It appears in many places, e.g. Kobayashi Nomizu, Fundation of Dfferential Goemetry, vol 1, chap.2. Here is a quick summary.
Let $$E\to M$$ be a rank $$n$$ vector bundle and $$F\to M$$ the associated frame bundle. The fiber $$F_x$$ over a point $$x\in M$$ consists of all linear isomorphisms $$\varphi:\mathbb{R}^n\to E_x$$. The group $$\mathrm{GL}_n(\mathbb{R})$$ acts freely on $$F$$ on the right by $$\varphi\mapsto \varphi\circ g$$. Let $$G\subset \mathrm{GL}_n(\mathbb{R})$$ be a subgroup (instead of $$\rho(G)$$ of your question). A reduction of the structure group of $$F$$ to $$G$$ consists of a $$G$$-subbundle of $$F$$. That is, a submanifold $$F^G\subset F$$ such that each $$F^G\cap F_x$$ consists of a single $$G$$-orbit. The vector bundle $$E$$ is said to be a $$G$$-bundle, or equipped with a $$G$$-structure, if a reduction of the structure group of its frame bundle to $$G$$ has been chosen. Now you can show that $$E$$ admits a $$G$$-structure if and only if there exists a cover of $$M$$ by open subsets, over each of which $$E$$ can be trivialized, i.e. is isomorphic to the trivial vector bundle with fiber $$\mathbb{R}^n$$, and such that the transition functions between the trivializations take value in $$G$$.
You can make a rigorous definition of a $$G$$-structure on a vector bundle using trivializations alone, avoiding the frame bundle, but it's a bit tedious. The definition you gave in your question is not of a $$G$$-bundle but the condition of existence of a $$G$$-structure on a vector bundle. It is not the same. $$G(E)$$ does not make sense unless a $$G$$-structure has been chosen, it is not enough to know that there is one.
Anyway, using the above definition of a $$G$$-bundle, it is rather easy to sort out the rest. $$\mathrm{Aut}(E)$$ is the bundle whose fiber at a point $$x\in M$$ consists of all linear automorphisms of $$E_x$$. $$G(E)\subset\mathrm{ Aut}(E)$$, for a vector bundle $$E$$ with a given $$G$$-structure $$F^G\subset F$$, is defined as the bundle whose fiber at a point $$x\in M$$ consists of all linear automorphisms $$f:E_x\to E_x$$ such that $$\varphi\circ f\circ \varphi^{-1}\in G$$ for all $$\varphi\in F^G_x.$$ Similarly, $$\mathrm{End}(E)$$ is the vector bundle whose fiber at a point $$x\in M$$ consists of the linear transformations $$T:E_x\to E_x$$. Finally, $$\mathfrak{g}(E)\subset\mathrm{End}(E)$$ is the vector bundle whose fiber at a point $$x\in M$$ are the linear transformation $$T:E_x\to E_x$$ such that $$\varphi\circ T\circ \varphi^{-1}\in\mathfrak{g}$$ for all $$\varphi\in F^G_x,$$ where $$\mathfrak{g}\subset \mathrm{End}(\mathbb{R}^n)$$ is the Lie algebra of $$G$$.