# Principal bundle automorphism generating global gauge transformations

Consider a principal $G$-bundle $P$ with connection form $\omega$. An automorphism $f$ of $P$ is by definition a (smooth) $G$-equivariant map: $f(p \cdot g) =f(p) \cdot g$ for all $p\in P$ and $g\in G$. If $s$ is a local cross section over $U$, then $s^\prime\equiv f ^{-1}\circ s$ is also a cross section over $U$ and there exists a unique map $g:U\rightarrow G$ such that $s^\prime (x)=s (x)\cdot g(x)$. If $A=s^* \omega$, $A^\prime= s^\prime \omega$ then $$A^\prime = g ^{-1} A g + g^{-1} \mathrm{d}g.$$

My question is: which bundle automorphisms generate the global gauge transformations $A^\prime=g^{-1}Ag$, with $g\in G$ constant?

Because of the property $$R_g ^*\omega= g ^{-1}\omega g,$$ I would have thought that they are generated by right multiplication by a group element but this transformation is not, somehow surprisingly for me, a bundle automorphism as, denoting by $f$ the right multiplication by $h\in G$, $$f(p)\cdot g= p\cdot hg \neq f(p\cdot g) =p\cdot gh.$$ Am I getting something wrong?

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I'm confused by several things. I don't see what the connection form has to do with anything. I don't know why you seem to assume that gauge transformations are somehow given by constants in $G$. Anyway, yes, right multiplication by a group element is not a bundle automorphism. – Seub Jan 10 '14 at 19:17
I am sorry if I have not been clear, that's probably because I myself am quite confused. I am not assuming that gauge transformations are given by constants in $G$, but rather that 1)Gauge transformations are generated by local or global automorphisms of the bundle. 2)There are bundle automorphisms which generate the particular gauge transformation $A\rightarrow g^{-1}Ag$ with $g$ constant. Given this I was trying to understand which are these automorphisms. – GFR Jan 10 '14 at 20:55

Let me rephrase your question first to see if I understand what you are asking:

Suppose we are given a principal $G$-bundle $P \longrightarrow M$, a connection one-form $\omega$ on $P$, and an open neighborhood $U \subset M$ over which we have locally trivialized $P$ via local cross-sections $s$ and $s'$. Write $A$ and $A'$ for the local gauge fields for $\omega$ in $U$ with respect to the trivializations given by $s$ and $s'$, respectively. If $$A' = g^{-1} A g$$ on all of $U$ for some $g \in G$, then what is the automorphism $f: P \longrightarrow P$ such that $s' = f^{-1} \circ s$?

Now that the question has been restated, let me tell you why (my interpretation of) your question is ill-posed.

The problem is that you only have given us local gauge fields $A$ and $A'$ over one open neighborhood $U \subset M$, while to define a global gauge transformation relating $A$ and $A'$, we need to know what the local gauge fields are on all open neighborhoods in some local trivialization of $P$, or in other words we need to know what $A$ and $A'$ are as elements of $\Omega^1_M(\mathrm{Ad}(P))$. Since we don't know what $A$ and $A'$ are globally, there is nothing we can do.

Now if $P$ is trivial, so that we can take $U = M$, we can indeed find the corresponding global gauge transformation $f$, since in this case we know $A$ and $A'$ on all of $M$.

Let me first describe a general way to relate global gauge transformations with local gauge transformations. Let $f: P \longrightarrow P$ be a given bundle automorphism, and suppose we have a local trivialization $\{(U_\alpha, \psi_\alpha)\}$ of $P$. We can think of these trivializations as given by local sections $s_\alpha: U_\alpha \longrightarrow \pi^{-1}(U_\alpha)$. For any $m \in U_\alpha$ and $p \in \pi^{-1}(m)$, there is a unique element $g_\alpha(p) \in G$ such that $p = s_\alpha(m).g_\alpha(p)$. Now define a map $$\bar{\phi}_\alpha: \pi^{-1}(U_\alpha) \longrightarrow G,$$ $$\bar{\phi}_\alpha(p) = g_\alpha(f(p))g_\alpha(p)^{-1}.$$ One can easily check that $\bar{\phi}_\alpha(p.g) = \bar{\phi}_\alpha(p)$, so that $\bar{\phi}_\alpha$ is constant on the fibers of $P$ and hence descends to a map $\phi_\alpha: U_\alpha \longrightarrow G$. It can be verified that the $\phi_\alpha$ patch together to form a section $\phi \in \Omega^0_M(\mathrm{Ad}(P))$, and furthermore the map $f \mapsto \phi$ gives a bijective correspondence between the group of gauge transformations $\mathscr{G}$ and $\Omega^0_M(\mathrm{Ad}(P))$. We also have that if $\omega$ has gauge field $A \in \Omega^0_M(\mathrm{Ad}(P))$ and $f^\ast \omega$ has gauge field $A' \in \Omega^0_M(\mathrm{Ad}(P))$, then $$A^\prime_\alpha = \phi_\alpha^{-1}A\phi_\alpha - \phi_\alpha^{-1}d\phi_\alpha.$$

Now let us return to the question at hand. In the case of the trivial bundle, there is only one $U_\alpha$ so we will drop the subscript. You are interested in the local gauge transformation $\phi(m) = g$ for some $g \in G$. You want to know what $f \in \mathscr{G}$ corresponds to this $\phi$. You can easily check that $f$ is given by left multiplication by $g$: $$f(m, h) = (m, gh).$$ In the above formula for $f$ we are imagining $P$ as $M \times G$, i.e. using the standard trivialization of the trivial principal $G$-bundle. The formula changes if we use a different choice of trivialization. But in any case, you can still show that the local gauge transformation associated to this $f$ is $\phi \equiv g$, so that \begin{align*} A' & = \phi^{-1} A \phi - \phi^{-1} d\phi \\ & = g^{-1} A g - g^{-1} dg. \end{align*}

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Dear Henry, thanks, let me see if I understand your answer. I have tried to study it carefully and I have one question. Basically you have considered two different trivialisations on the same open cover, related by $s_\alpha^\prime(x)=s_\alpha (x)\bar\phi_\alpha(p)$ and seen what are the expressions for the gauge potentials in the two trivializations. Now if $A_\alpha=s_\alpha^*\omega$ I think that $A^\prime_\alpha= (f^{-1}\circ s_\alpha)^* \omega= (s^\prime_\alpha)^*\omega=\phi_\alpha^{-1}A_\alpha +\phi_\alpha^{-1}d\phi$. – GFR Jan 15 '14 at 19:24
Note thatI get a plus in the second term for $A^\prime$ and that I defined $A^\prime =s^*\left( \left (f^{-1}\right )^*\omega \right )$, while you seem to have defined it as $A^\prime =s^*\left( f^*\omega \right )$. Is what I am saying correct/equivalent to what you wrote? – GFR Jan 15 '14 at 19:29
Very helpful answer, Henry! But could you elaborate on why $A_{\alpha}' = \phi_{\alpha}^{-1} A \phi_{\alpha} - \phi_{\alpha}^{-1} d\phi_{\alpha}$? GFR seems to have a second opinion about it. – PhysicsMath May 17 at 18:21

Once you've chosen a cross-section $s$, your bundle becomes trivialized. So perhaps it will be easier to consider the case of a trivial bundle $P = U \times G$.

Now the automorphisms of $G$ (thought of as a principle hom. space via right mult.) are exactly $G$ (exacting via left translations).

Thus the automorphisms of the bundle $U \times G$ are given by maps $U \to G$; we then left multiply by these. (Your guess of right mult. went wrong because $G$ is not commutative; but left mult. is okay!)

Note that since the cross-section $s$ corresponds to the identity section of $U \times G$ (since we used $s$ to obtain our trivialiation), left and right multiplication on $s$ (i.e. on the identity) coincide, so in your description of how to obtain $s'$ from $s$, you could have used left mult. instead.

If we take our map $U \to G$ to be constant, we get what I think you are looking for.

Note that the above discussion only makes sense in the context of a trivial bundle. Typically the automorphisms given by left. mult. won't extend to subsets of the base over which the bundle is non-trivial (unless it happens that $G$ is abelian, so that left and right mult. coincide), and (again, unless $G$ is abelian) they are dependent on the particular choice of trivialization.

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