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In one of my books (the great Baez & Munian's "Gauge fields, knots and gravity"), the vector potential is defined as a $End(E)$ valued 1-form, with $End(E)$ endomorphisms of the fiber $E$. So, with $e_i$ as basis of sections in $E$, the basis of sections in $End(E)$ is $e_i \otimes e^j$, then $$A(x) = A_{\mu j}^{i} (e_i \otimes e^j)\wedge dx^{\mu}.$$ Fine. Covariant derivative of a section is then:

$$(D_{\mu} s)^i = \partial_{\mu}s^i + A_{\mu j}^{i}s^j$$

But in QFT books I find that vector potential is defined as $A(x) = A_{\mu}^{i} \tau_{i} dx^{\mu}$, with $\tau$ the generator of the group (that is, basis of sections), and covariant derivative as:

$$(D_{\mu} s)^i = \partial_{\mu}s^i + \tau^{ai}{}_{j}A^{a}_{\mu}s^j$$

(summation in repeated indexes). I don't know how to reconcile these two definitions. The vector potential should have 3 indexes: one for the spacetime (base manifold) and two for the $End(E)$, so why are QFT books showing a vector potential with one spacetime and only one index for the fiber?

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This is because the generators $\tau_a$ of the groups are themselves matrices, they carry two indices: $\tau_a = (\tau_a)_j^i (e_i \otimes e^j)$. In other words, what you call $$A_{\mu j}^i (e_i \otimes e^j)$$ in the first part of your question should be identified with $$A_\mu^a \tau_a$$ in the second part. The transition between the two expressions can be made as follows: $$A_\mu = A_{\mu j}^i (e_i \otimes e^j)= A_\mu^a (\tau_a)_j^i (e_i \otimes e^j) \, . $$

To summarize, you can describe your gauge field with two or three indices, the relation between the two notations being $$A_{\mu j}^i = A_\mu^a (\tau_a)_j^i \, . $$

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    $\begingroup$ Thanks. It seems you're right. It surprises me a lot that a very good book like the one by Baez & Muniain doesn't make explicit that $A$ involves a sum $A^a \tau_a$. I'll accept your answer in a few hours to see if someone wants to say something more. $\endgroup$ – David Aug 23 '17 at 22:39
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    $\begingroup$ Yes indeed, this book doesn't seem to say explicitly that the gauge field can be expressed in a basis of Lie algebra generators. But you can find this in standard "physicists'" textbook, for instance Peskin and Schroeder. $\endgroup$ – Antoine Aug 25 '17 at 9:57
  • $\begingroup$ Yes, but the "standard QFT textbooks" miss the diff.geom. approach altogether :( $\endgroup$ – DanielC Aug 29 '17 at 22:47
  • $\begingroup$ One book used by physicists which has a strong diff.geom. approach is Nakahara's Geometry, Topology and Physics $\endgroup$ – Antoine Aug 30 '17 at 16:26

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