I'm using Instantons and four manifolds by Freed and Uhlenbeck to write a seminar on Yang-Mills instantons. I used Kobayashi and Nomizu's Foundations of differential geometry to fill the gap I had on principal bundles and connections in order to understand the first paragraphs of the second chapter of Instantons and four manifolds where Yang-Mills theory is described. On page 30 however it is stated that given a principal bundle $P(M,G)$ on a 4-manifold $M$ and an associated vector bundle $\eta$ the covariant derivative (or connection or Gauge potential) is expressed locally as $d+A_\alpha$ where $$A_\alpha: U_\alpha\rightarrow T^*(U_\alpha)\otimes\mathfrak{g}$$ where $\mathfrak{g}$ is the Lie algebra of $G$. I can't find references for this. I think it has to do with linear connections on vector bundle and so I tried looking in KN and the only thing I could find is proposition 2.9 where there is a decomposition of derivations on tensor fields in terms of covariant differentiation and linear endomorphism of $T_x(M)$. Can anybody give me a reference for why the covariant derivative has that local expression?
1 Answer
The associated vector bundle $P \times_G \mathfrak{g} \to M$ is a vector bundle. Now, a connection $1$-form $A \in \mathcal{A}(P)$ can also be specified equivalently as a collection of local $1$-forms $$ \{A_i \in \Omega^1(U_i,\mathfrak{g}) \}_{i \in I} $$ associated to the trivialization cover $\cup_{i \in I} U_i=M$. The expressions $A_i$ is just $s_i^*A$, i.e. $A$ in local gauge $s:U_i \to P$. We require that $A_i$ transforms like a connection under the $G$-valued transition functions of the principal $G$-bunde $P \to M$. This means that $$ A_j=Ad^{-1}_{g_{ij}} \circ A_j + g_{ij}^*\mu $$ for $g_{ij}:U_i \cap U_j \to G$ the transition functions of $P \to M$.
Now, any connection on a vector bundle can be specified by such a collection of Lie algebra (of the structure group of that vector bundle) valued $1$-forms $A_i$, that transform like a connection w.r.t to the transition functions taking values in the structure group of that vector bundle. But now the structure group and transition functions of the associated vector bundle are the same as for $P \to M$, hence the $A_i$s have the values in the "correct Lie algebra" and they also transform like a connection. Hence they define a connection on the associated vector bundle. Lets see why real quick: The corresponding local expressions $$ \nabla_A|_{U_i}=d+A_i $$ can be checked to be independent of the choice of frame of the associated vector bundle and hence defines a (global) connection $\nabla_A$ on that bundle. For that, we need to transform $A_i$ like a connection and note that the change of frame is precisely given by the transition functions $g_{ij}$.
TL;DR You start with the local expression and show that it is indepenent of your local trivialization and hence defines the global expression. This global expression is the one you wanted.
Edit: Two frames are related by $$ s_i=g_{ij}s_j $$ and you can verify, that on the overlap you have $$ \nabla_{A_i}s_i=\nabla{A_i}g_{ij}s_j=\nabla_{A_j}s_j. $$ You can also take a look at Hamiltons "Mathematical Gauge Theory", it contains a lengthy discussion of these topics from multiple angles.
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$\begingroup$ Renaming $\nabla_A$ as $\nabla$, $Ad(g_{ij}^{-1})\circ\nabla_{\vert U_i}+g_{ij}^*\mu=Ad(g_{ij}^{-1})\circ d+Ad(g_{ij}^{-1})\circ A_i+g_{ij}^*\mu=d+A_j$. This would give the independence of choice of frame(?), although I can't really justify why $Ad(g_{ij}^{-1})\circ d=d$ $\endgroup$ Commented Aug 24, 2023 at 13:15
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$\begingroup$ Inspired by mathoverflow.net/questions/229242/… could this also be the proposition at page 168 of bookstore.ams.org/gsm-25 (which is given without proof...)? $\endgroup$ Commented Aug 24, 2023 at 13:46
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1$\begingroup$ Edited the response a bit and added a literature recommendation. $\endgroup$ Commented Aug 24, 2023 at 22:07
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1$\begingroup$ It's theorem 5.9.2 in the book you recommended. Thank you $\endgroup$ Commented Aug 25, 2023 at 10:37
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$\begingroup$ You are welcome. Sorry if my answer confused you at first. $\endgroup$ Commented Aug 25, 2023 at 11:18