Gauge-invariance of Lagrangians

Question

I am rereading David Bleecker's Gauge Theory and Variational Principles, and I have realized I don't understand something.

The offending part is in 3.3 (page 50-52), however I am reproducing the formalism here in my own notation.

Preliminaries: Here $M$ is a smooth manifold ("spacetime") and $P\rightarrow M$ is a principal fibre bundle with structure group $G$. Let $V$ be a finite dimensional real or complex vector space, and let $\rho:G\rightarrow\text{GL}(V)$ be a representation.

The notation $C(P,V,\rho)$ refers to smooth functions $\psi:P\rightarrow V$ that satisfies the equivariance property $\psi(ug)=\rho(g^{-1})\psi(u)$ for $u\in P$ and $g\in G$. The notation $\Omega^k(P,V,\rho)$ refers to differential $k$-forms on $P$ with values in $V$ that are 1) horizontal 2) equivariant in the sense of $(r_g)^\ast\xi=\rho(g^{-1})\circ\xi$ (here $r_g$ is the right action of $G$ on $P$).

The notation $C(P,G)$ refers to smooth maps $\tau:P\rightarrow G$ such that $\tau(ug)=\mathbf{Ad}_{g^{-1}}\tau(u)=g^{-1}\tau(u)g$.

A gauge transformation is a vertical automorphism of $P$, eg. it is a diffeomorphism $f:P\rightarrow P$ with $f(ug)=f(u)g$ and it covers the identitiy (eg. $\pi(f(u))=\pi(u)$). The group of gauge transformations of $P$ is denoted as $\text{GA}(P)$. There is a one-to-one correspondance between elements of $\text{GA}(P)$ and $C(P,G)$. For any $\tau\in C(P,G)$ we have an $f\in\text{GA}(P)$ defined as $f(u)=u\tau(u)$, and for any $f\in\text{GA}(P)$, we have a $\tau\in C(P,G)$ defined as $f(u)=u\tau(u)$.

Lagrangians: The space of 1-jets of maps from $P$ to $V$ are defined as $$ J(P,V)=\left\{(u,v,\theta):\ u\in P,\ v\in V,\ \theta:T_uP\rightarrow V\ \text{is linear} \right\}. $$ This is a smooth manifold in a natural way.

A Lagrangian is a map $L:J(P,V)\rightarrow \mathbb R$ such that $$ L(u,v,\theta)=L(ug,\rho(g^{-1})v,\rho(g^{-1})\circ\theta\circ (r_{g^{-1}})_\ast). $$

It is shown in the text that a Lagrangian determines a function $\mathcal L:C(P,V,\rho)\rightarrow C(M)$ by $$ \mathcal L(\psi)(x)=L(u,\psi(u),d\psi|_u) $$ with $x=\pi(u)$.

A Lagrangian is $G$-invariant if $L(u,v,\theta)=L(u,\rho(g)v,\rho(g)\circ\theta)$.

A Lagrangian is gauge-invariant if for any $f\in\text{GA}(P)$ and $\psi\in C(P,V,\rho)$ we have $ \mathcal L(\psi)=\mathcal L(f^{-1\ast}\psi)$.

It is then shown in the text that $G$-invariance does not imply gauge invariance.

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Question 1: If a Lagrangian $L:J(P,V)\rightarrow\mathbb R$ is defined by the definition I have given above, then, as I have stated, $L(u,\psi(u),d\psi|_u)$ depends only on the base point $x\in M$, but not on the fiber point $u\in P_x$.

Considering that a "choice of gauge" at $x\in M$ is a fiber point $u\in P_x$, it seems to me that the bare definition of the Lagrangian already implies a sort of gauge-invariance (at least a "global" gauge invariance), since the value of the Lagrangian is insensitive to displacements along the fibers.

So what is the actual difference between a "Lagrangian" and a "$G$-invariant Lagrangian"? What would be an example of a Lagrangian that isn't $G$-invariant?

Question 2: Assuming Question 1 gets sorted out, I assume the difference between a $G$-invariant Lagrangian and a gauge-invariant Lagrangian is basically the same thing as the difference between a "globally $G$-invariant Lagrangian" and a "locally $G$-invariant Lagrangian" in the "traditional" (local tensor calculus-based) approach, where in the traditional language, "globally invariant" means that the $G$-transformation is constant, while "locally invariant" means that the $G$-transformation depends on the base space points.

Is this correct?

Question 3: I find Bleecker calling a vertical automorphism of $P$ a "gauge transformation" fairly odd, because previous he has referred to a local trivialization (LT) of $P$ as a choice of gauge. And of course it is intuitively clear that a LT is precisely what corresponds to a choice of gauge in the traditional local formalism.

However this would imply to me that a gauge transformation is actually a transition function between two LTs. How are elements of $\text{GA}(P)$ related to transition functions? Especially that it seems to me that a transition function is more closely related to what in the usual traditional physics literature is called a gauge transformation.

Answer 1

1. Yes, $G$ invariance means what is commonly understood as 'global gauge invariance'. This can be seen at most examples found in physics Literature where local gauge invariance forces you to substitute $\partial_{\mu} \to D_{\mu}$ which, in Bleecker's language, reads as $d \to D_{\omega}$ (where $\omega$ is the connection form). The Klein Gordon Lagragian, for instance, is $g$ invariant under a suitable $U(1)$ representation, but without the aforementioned substitution, it's not gauge invariant. As an example, I think the following should work: Let $M=\{*\}, G=GL_n(\mathbb{R})$ ($n=1$ is not excluded) and $P=GL_n(\mathbb{R})$. Take $V=\mathbb{R}$ and the representation to be the determinant: $$\rho(A)x := \det(A)x, x \in \mathbb{R}.$$ Observe that we can identify $T^*_AGL_n(\mathbb{R}) \cong T_AGL_n(\mathbb{R}) \cong \mathbb{R}^{n \times n}$. More precisely, the first isomorphism is given by $\theta^* \mapsto \theta,$ where $\theta^*(\theta) = 1$, and the second isomorphism is canonical (not that it really matters, but it's nice). In particular, if $\theta^* \in T^*_AGL_n(\mathbb{R})$ then, under this isomorphism, we have $$\det(B)^{-1} \theta^* \circ B^{-1} \mapsto \det(B) B \theta$$ (as $\det(B)^{-1} \theta^* \circ B^{-1} (\det(B) B \theta) = \theta^* B^{-1}B \theta = 1$). Now, the differential of the right translation in $GL_n(\mathbb{R})$ is just the right translation in $\mathbb{R}^{n \times n}$ (as $GL_n(\mathbb{R}) \subset \mathbb{R}^{n \times n}$ is open) and thus, if we set $$L(A,\theta^*) = \det(A^{-1})^{n+1}\det(\theta)$$ we compute: $$L(AB, \det(B^{-1}) \theta^* \circ B^{-1}) = \det(A^{-1})^{n+1} \det(B^{-1})^{n+1} \det(B)^n\det(\theta B)$$ $$= \det(A^{-1})^{n+1} \det(\theta).$$ But the Lagragian is not $GL_n(\mathbb{R})-$invariant, as $\det(A^{-1})^{n+1} \det(B)^n \det(\theta) \neq \det(A^{-1})^{n+1} \det(\theta)$ if $\det(B) \neq \pm 1$. So, in general, the Lagragian may depend on the choice of gauge, only when applied to 'nice' functions this dependence vanishes.
2. Yes. (An example is actually given in $\S$4 of Bleecker's book$-$take $M$ to be flat and look at the properties of the Lagragian if instead of '$D_{A}$' only '$d$' is taken, and pull it back viá a local section.)
3. Given a gauge transformation $f:P \to P$, you can proof that there exists a smooth function $\mu:P \to G$ such that $f(p)=p\mu(p)$ (where $\mu(pg)= g^{-1} \mu(p) g$). If you take two sections $s_i:U \to P; i=1,2$ then we may as well assume that we can trivialise over $U$ and, without loss of generality, write $s_i(x)=(x, g_i(x))$. Moreover, in this trivialisation we have $p\mu(p)=(x,g \mu(p))$, and thus a smooth function which satisfies $g_1(x)\mu((x,g_1(x)))=g_2(x)$ would determine your sought gauge transformation $f$ (at least locally), which satisfies $f(s_1(x))=s_2(x)$.