# Gauge-invariance of Lagrangians

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.

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.

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 $$f(s_1(x))=s_2(x)$$.