In Gravity, Gauge Theories and Geometric Algebra by A. Lasenby, C. Doran and S. Gull develop a theory for gravity based on the use of a position-gauge field and a rotation-gauge field. The former is introduced in order to make gauge-invariant terms of the form $\nabla_x \phi (x)$, which under diffeomorphisms $x'=f(x)$ follow the transformation law:
$$\nabla_x \phi (x) \rightarrow\nabla_x \phi' (x) =\bar{\rm f}\left(\nabla_{x'} \phi (x')\right)$$
Later on in the document, the electromagnetic vector potential is introduced. From the classical gauge invariance property $A \rightarrow A+\nabla \phi$ (although it is presented a bit differently in the text, the concept is the same), concentrating on the term $\nabla \phi$ the authors affirm the vector potential should transform in the same way as a gradient does, with the conclusion:
$$A(x) \rightarrow A'(x) =\bar{\rm f}\left(A(x')\right)$$
Although it obviously makes sense that something which can be freely varied by adding a whatever gradient should indeed transform as a gradient, the mathematical procedure needed to prove this is not entirely clear to me. I present the few passages (which I'm not sure are completely correct) I tried to develop:
- the gauge invariance of vector potential may be written as:
$$A_1(x) - A_2(x) =\nabla_x \phi(x)$$
- applying the diffeomorphism yields:
$$A'_1(x) - A'_2(x) = \bar{\rm f}\left( \nabla _{x'} \phi (x') \right)$$
- even if I change the position dependance, the relation $A_1(x') - A_2(x') =\nabla_{x'} \phi(x')$ should still hold (as from the point of view of the math, it seems equivalent to simply putting an apostrophe near the $x$), so:
$$A'_1(x) - A'_2(x) = \bar{\rm f}\left( A_1(x') - A_2(x') \right)$$
- now, for every field $A$, the following relation should hold:
$$A'(x) - \bar{\rm f}\left(A(x') \right) = K$$
where K is a constant or field (whose exact dependance I was unable to derive). Assuming the previous passages are correct, the answer to my question should be then showing that $K$ is identically equal to $0$, which is exactly what I am missing.
EDIT: I forgot to specify the notation: $\underline{\rm f}$ is the differential of the diffeomorphism $f$, while $\bar{\rm f}$ is the adjoint of the differential
$ \newcommand\DD[2]{\frac{\mathrm d#1}{\mathrm d #2}} \newcommand\diff\underline \newcommand\R{\mathbb R} \newcommand\adj\overline \newcommand\tDD[2]{\mathrm d#1/\mathrm d#2} $
Let $M$ be a differentiable manifold, and let $f : M \to M$ be diffeomorphism. $f$ induces a transformation on the tangent bundle $TM$ by noting that a path $\gamma : [0,1] \to M$ transforms to $f\circ\gamma$. It follows by the chain rule that $$ \DD{}tf(\gamma(t)) = \diff f(\gamma(t);\,\tDD\gamma t). $$ Thus at a tangent vector $v = \tDD\gamma t$ at $x = \gamma(t)$ is "pushed" to $T_{f(x)}M$ via $$ v \mapsto \diff f(x; v). $$
Now consider a one-form $\alpha_x : T_xM \to \R$. Pushing this via $f$ gives $$ \alpha_x(v) \mapsto \alpha_{f(x)}(\diff f(x; v)) \implies \alpha_x \mapsto \alpha'_x = \alpha_{x'}\circ\diff f(x; {-}), $$ where $x' = f(x)$. If the vector potential $A_x$ is to be interpreted as "transforming like a one-form", then $A_x$ is vector such that $A_x = \alpha_x^\sharp$ where $\alpha_x$ is a one-form and $\sharp : T^*_xM \to T_xM$ is the isomorphism induced by a given metric. If $\{e_i\}_{i=1}^n$ is any basis at $x$ with reciprocal $\{e^i\}_{i=1}^n$ then $$ v\cdot A_x = \alpha_x(v),\quad \alpha_x^\sharp = \sum_{i=1}^ne^i\alpha_x(e_i). $$ If $\alpha_x$ transforms as above to $\alpha'_x$ then $$\begin{aligned} A_x \mapsto A'_x &= (\alpha'_x)^\sharp = \sum_{i=1}^ne^i\alpha'_x(e_i) = \sum_{i=1}^ne^i\alpha_{x'}(\diff f(x; e_i)) \\ &= \sum_{i=1}^ne^i(\diff f(x; e_i)\cdot A_{x'}) = \sum_{i=1}^ne^i(e_i\cdot\adj f(x; A_{x'})) \\ &= \adj f(x; A_{x'}). \end{aligned}$$
We can also do this without a basis using the following two facts. Let $\flat = \sharp^{-1}$ and let $L^* : W^* \to V^*$ denote the dual of a linear map $L : V \to W$.
First, let's note that the proper way of writing the transformation of $A$ should allow for gauge transformations; we should really write $$ A'_x + \nabla\phi(x) = \adj f(x; A_{x'}). $$ But then $$\begin{aligned} A'_x &= \adj f(x; A_{x'}) - \nabla\phi(x) \\ &= \adj f(x; A_{x'}) - \adj f\Bigl(x;\, \nabla'(\phi\circ f^{-1})(x')\Bigr) \\ &= \adj f\Bigl(x;\, A_{x'} - \nabla'(\phi\circ f^{-1})(x')\Bigr). \end{aligned}$$ Here we've used the shorthand $\nabla' = \nabla_{x'}$. This shows that such an $A'$ is in fact the transformation of $A$ in some gauge. Similarly, we can show that $A$ transforms properly to some gauge of $A'$: $$\begin{aligned} A'_x &= \adj f(x; A_{x'} + \nabla'\phi(x')) \\ &= \adj f(x; A_{x'}) + \adj f(x; \nabla'\phi(x')) \\ &= \adj f(x; A_{x'}) + \nabla(\phi\circ f)(x), \end{aligned}$$$$ \implies A'_x - \nabla(\phi\circ f)(x) = \adj f(x; A_{x'}) $$
So a good question to ask is: is this still possible if $A$ transforms like a vector? In that case, if $$ A'_x + \nabla\phi(x) = \diff f(x; A_{x'}) $$ then for all $f$ and $\phi$ there must be some $\psi$ such that $$ \diff f(x; \nabla'\psi(x')) = \nabla\phi(x). $$ I've not been able think of a counterexample yet.