Taylor expansion of a gauge field

Question

Suppose I'm in flat spacetime, and I have some $C^\infty$ $SU(N)$ gauge field coefficient function $A_\mu^a(x)$, and I'd like to know what the value of this function is at a spacetime point $y$ not very far from $x$. Does $$A^a_\mu(y)=e^{(y-x)\cdot\partial}A^a_\mu(x)$$ give the correct result? I.e. is the connection for the gauge index trivial like it is for the spacetime index?

More out of general interest, if the above is correct, how does it generalize? I'd imagine for a curved spacetime one has $$A^a_\mu(y)=e^{(y-x)\cdot\nabla}A^a_\mu(x),$$ where $\nabla_\mu$ is the covariant derivative with metric compatible Christoffel symbol; is that correct?

Is there a physical setup in which the gauge fibre bundle(?) is curved and then one has a covariant derivative involving the color index?

EDIT: Some more context. Suppose I have a quark field $\psi(x)$ that transforms under a local gauge transformation as $\psi(x)\rightarrow V(x)\psi(x)$, where $V(x) = \exp(i\alpha^a(x)t^a)$, $\alpha^a(x)\in\mathbb R$ and $t^a\in SU(N)$. The QCD fermionic action generalizes from the free fermionic action in the way one would expect: the partial derivative becomes a covariant derivative, $\partial_\mu\rightarrow D_\mu$, where $D_\mu \equiv \partial_\mu -igA_\mu$.

Now if I want $\psi(x+\epsilon)$, I'd figure that the usual Taylor expansion in terms of partial derivatives would be generalized to covariant derivatives. To wit, I'd expect that $$\psi(x+\epsilon) = \psi(x)+\epsilon^\mu \partial_\mu \psi(x) + \mathcal O(\epsilon^2) \rightarrow \psi(x+\epsilon) = \psi(x)+\epsilon^\mu D_\mu \psi(x) + \mathcal O(\epsilon^2).$$ The latter equation is consistent with Eq. 7.25 in Greiner's QCD book.

However, let's consider the transformation properties of $\psi(x+\epsilon)$. Under a gauge transformation we have that $$\psi(x+\epsilon) \rightarrow V(x+\epsilon)\psi(x+\epsilon) = \bigl( V(x) + \epsilon^\mu\partial_\mu V(x)\bigr)\bigl(\psi(x) + \epsilon^\mu\partial_\mu \psi(x) \bigr) + \mathcal O (\epsilon^2) \\ = V(x)\psi(x) + \epsilon^\mu \bigl( \partial_\mu V(x)\bigr)\psi(x) + V(x)\epsilon^\mu\partial_\mu\psi(x) + \mathcal O (\epsilon^2).$$ Now consider the transformation of $$\psi(x+\epsilon) = \psi(x)+\epsilon^\mu\partial_\mu\psi(x) + \mathcal O(\epsilon^2) \\ \rightarrow V(x)\psi(x) + \epsilon^\mu\partial_\mu\bigl(V(x)\psi(x)\bigr) + \mathcal O(\epsilon^2) \\ = V(x)\psi(x) + \epsilon^\mu \bigl( \partial_\mu V(x)\bigr)\psi(x) + V(x)\epsilon^\mu\partial_\mu\psi(x) + \mathcal O (\epsilon^2).$$

On the other hand, if we take as Greiner says $\psi(x+\epsilon) = \psi(x)+\epsilon^\mu D_\mu\psi(x) + \mathcal O(\epsilon^2)$, then we'll pick up extra $A_\mu$ terms that don't cancel.

(Parenthetically, even if we decided to change all partial derivatives to covariant derivatives in the above, we still don't get the right transformation properties. To wit suppose that $$\psi(x+\epsilon)\rightarrow V(x+\epsilon)\psi(x+\epsilon) \\ = \bigl[ V(x)+\epsilon^\mu D_\mu V(x) \bigr] \bigl[ \psi(x) + \epsilon^\mu D_\mu\psi(x) \bigr] + \mathcal O(\epsilon^2) \\ = V(x)\psi(x) + \epsilon^\mu D_\mu\bigl( V(x)\psi(x) \bigr) + \mathcal O(\epsilon^2).$$ At the same time, $$\psi(x+\epsilon) = \psi(x) + \epsilon^\mu D_\mu\psi(x) + \mathcal O(\epsilon^2) \\ \rightarrow V(x)\psi(x) + \epsilon^\mu V(x)D_\mu\psi(x)+\mathcal O(\epsilon^2),$$ and clearly the above two expressions are different.)

Perhaps, then, a better way to phrase the above question is: why doesn't the Taylor expansion for a gauge theory field not generalize from partial derivatives to covariant derivatives as one would naively expect (and as written in at least one book)?

Answer 1

Fiber bundles can have nontrivial topology, but for this question, we might as well work in a patch where it's trivial. Also, this answer uses classical fields to avoid possible technical issues when dealing with field operators (or distributions) on a Hilbert space.

For any analytic function $f(x)$, we have the identity $$ f(x+c)=e^{c\cdot\partial}f(x) \tag{1} $$ where $\partial$ is the derivative with respect to $x$ and $c$ is independent of $x$. We might be tempted to write this as $$ f(y)=e^{(y-x)\cdot\partial}f(x), \tag{2} $$ but that's not correct unless we interpret it like this: $$ f(y)=\Big[e^{(y-x_0)\cdot\partial}f(x)\Big]_{x_0=x}. \tag{3} $$ Either way, it doesn't matter whether $f$ is a component of a gauge field (in flat or curved spacetime) or a component of a spinor field (in a gauge theory or not). In other words, it doesn't matter what the function $f$ represents. Equations (1) and (3) are identities that hold for any analytic function $f$. By the way, regarding the distinction between analytic and smooth, see https://en.wikipedia.org/wiki/Non-analytic_smooth_function.

In contrast, covariant derivatives are used to define new fields that preserve certain properties, which can facilitate the construction of gauge-invariant lagrangians. Consider a spinor field $\psi$ coupled to an abelian gauge field $A$. I'm using an abelian gauge field in this example so that we don't need to worry about path-ordering. Suppose we define a new field $\psi'$ by $$ \psi'(x+c)=\psi(x+c)-\exp\left(i\int_x^{x+c} dy\cdot A(y)\right)\psi(x). \tag{4} $$ Under a gauge transform \begin{gather} \psi(x)\to e^{i\theta(x)}\psi(x) \\ A_\mu(x)\to A_\mu(x)+\partial_\mu\theta(x), \tag{5} \end{gather} the field $\psi'$ transforms as $$ \psi'(x)\to e^{i\theta(x)}\psi'(x), \tag{6} $$ just like $\psi$ does. For infinitesimal $c$, equation (4) becomes \begin{align} \psi'(x+c) &=\psi(x)+c\cdot\partial\psi(x)-ic\cdot A\psi(x) + O(c^2) \\ &=\psi(x)+c\cdot D\psi(x) + O(c^2) \tag{7} \end{align} with $D_\mu=\partial_\mu-iA_\mu$. By the way, the finite version (4) is used in lattice gauge theory, where derivatives are replaced by finite differences.

I don't have a copy of Greiner's book on hand, but I'm guessing it's using the same symbol $\psi$ for both functions, writing $\psi$ instead of $\psi'$ on the left-hand side of (7). Yeah, that's careless, but it's also common: people often use the same symbol for different-but-related things. It's a compromise to avoid a proliferation of primes, tildes, hats, subscripts/superscripts, and other decorations when several different related quantities are involved. I don't know if that's what Greiner's book is doing here, but it's a plausible guess.