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In my last comment I forgot to write the (n-j) before |x_j| and |s(x_j)|. These two signs only differ by n(n-1)/2 on V^n, so they both work for the decalage isomorphism. But I was trying to say \sum (n-j)|x_j| is consistent with the convention (f,g)(v,w)=(-1)^{|g||v|}f(v)g(w), while \sum (n-j)|s(x_j)| is not.
I don’t think the sign should be \sum |sx_i|. It should be \sum |x_i|. Let’s write (s(x_1),...,s(x_n)) as (s_1(x_1),...,s_n(x_n)), where the subscript of s_j only means it is applied to x_j. The we move these s_j to obtain (s_1,s_2,...,s_n)(x_1,x_2,...,x_n), which really produces a sign \sum |x_j|. The sign \sum |s(x_j)| that you claimed should come with (s_n,...,s_2,s_1)(x_1,x_2,...,x_n), which I don't think is the case.
Thank you Bertram (sorry for my late response). The question I asked was a lemma which turns out to be incorrect, however the main theorem in that section is correct. The extension of gauge transformation that I wanted could be obtained from similar arguments as in Uhlenbeck’s paper. To justify that procedure, it is easy to see that there is no obstruction since the local gauges are constructed from exponential map from a bounded subset of the Lie algebra. Nevertheless the extension can be directly constructed by an inductive argument.
