Gauge covariant derivative and Leibniz rule

Question

Let's say I've got 2 different fields $a, b$ and I want to compute its covariant derivative $D_\mu = \partial_\mu + iA_\mu^a T^a$ where $\{A_\mu^a\}$ is the set of gauge fields and $\{T^a\}$ the algebra of the corresponding group under which $a$ and $b$ transform. Then,

$$ D_\mu(ab) = \partial_\mu(a)·b + a\partial_\mu b + iA_\mu a b = (D_\mu a)b + a\partial_\mu b = a(D_\mu b) + (\partial_\mu a)b, \quad A_\mu \equiv A_\mu^aT^a \tag1$$

So first of all we see an ambiguity because I can place $D_\mu$ either on $a$ or $b$. Nonetheless, we see something worse: the term in right hand side that goes with the partial derivative does not transform as the left hand side under the group because $(\partial_\mu a)b$ is not transformed into $U(\partial_\mu a)b$ while $D_\mu(ab)$ does.

An alternative expression for Eq. (1) would be to consider that Leibniz rule holds. In that case, left and right hand sides would transform in the same way, but how is that compatible with the definition $D_\mu = \partial_\mu + iA_\mu$ that leads to Eq. (1)?

Answer 1

If you have two tensor field on which the covariant derivative acts, the covariant derivative of their tensor product is

\begin{equation} \nabla (a \otimes b) = a \otimes \nabla b + b \otimes \nabla a \end{equation}

So that in coordinates, this is

\begin{eqnarray} D_\mu (a^\alpha b^\beta) &=& a^\alpha (\partial_\mu b^\beta + A_\mu b^\beta) + b^\beta (\partial_\mu a^\alpha + A_\mu a^\alpha)\\ &=& a^\alpha \partial_\mu b^\beta + b^\beta \partial_\mu a^\alpha + 2 A_\mu b^\beta a^\alpha \end{eqnarray}

The form is indeed slightly different from the original definition, but this is the general rule for covariant derivatives, similarly to how tensors of order 2 have two Christoffel symbols.

Edit : Sketch of a proof as of why :

Consider that our two fields transform under some gauge group :

\begin{eqnarray} a &\to& e^{\alpha(x)} a\\ b &\to& e^{\alpha(x)} b \end{eqnarray}

If you apply a derivative on their tensor product, you obtain

\begin{eqnarray} \partial_\mu (e^{\alpha(x)} a e^{\alpha(x)} b) &=& \partial_\mu (e^{2\alpha(x)} a b)\\ &=& 2 e^{2\alpha(x)}(\partial_\mu \alpha(x)) + e^{2\alpha(x)} \partial_\mu ( a b) \end{eqnarray}

Much like the usual derivatives for a gauge transformation, there is an extra term that prevents it from being invariant, but here that term has an extra factor of two. This is why you need a factor of two on your gauge field, to absorb the extra factor.