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Let $\pi\colon P\to X$ be a locally trivial principal $G$-bundle over a Hausdorff paracompact space $X$, where $G$ is a topological group (we work in the category of topological spaces, as I do not think smooth structures are relevant to the question I am asking).

The group $\mathrm{Gau}(P)$ of gauge transformations is the group of automorphisms of $P$ covering the identity of $X$ (the group structure is given by composition, of course). A gauge transformation with compact support is an $f\in\mathrm{Gau}(P)$ which is the identity on $P\setminus\pi^{-1}(K)$, where $K$ is some compact subset of $X$. Clearly the set of gauge transformations with compact support forms a subgroup $\mathrm{Gau}_c(P)\subset \mathrm{Gau}(P)$.

I've been reading [1] and stumbled across a sentence (second page, first column) which in the above notation reads as:

If $G$ is not compact, then $\mathrm{Gau}_c(P)=\{\mathrm{id}_P\}$.

Alas, I could not come up with a proof. Any suggestion?

[1] Smoothness of the action of the gauge transformation group on connections, M. C. Abbati, R. Cirelli, A. Manià, and P. Michor, J. Math. Phys. 27, 2469 (1986), http://dx.doi.org/10.1063/1.527404

EDIT: I misinterpreted the paper. They actually mean that any gauge transformation which is a compactly supported homeomorphism $P\to P$ must be the identity. The statement as I put it is not true, as proved by Ben McKay.

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First you just look at what happens on one fiber: left (or right) translation of the structure group by a group element. No point of the fiber is fixed (if you act by a nonidentity element). This fiber is already not a compact set, and you are sliding its points around. If the set of nonfixed points in $P$ sits in a compact set $K$, then any fiber (being closed) intersects the nonfixed points in a relatively compact set inside $K$. But then the set of fixed points on the fiber must be nonempty, not being contained in $K$. But one fixed point ensures that all points of that fiber are fixed. So $f$ is the identity on each fiber, so the identity.

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Sorry for being dense, but is your $K$ a compact subset of $P$ or rather of $X$? I can't follow your argument... –  johndoe Jul 4 '13 at 12:40
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Oops, I thought $K$ was a compact subset of $P$. If $K$ is a compact subset of $X$, then of course there are lots of compactly supported gauge transformations. Take $P \to X$ trivial, and just make lots of compactly supported maps to $G$. So there are compactly supported gauge transformations, not the identity, precisely when the identity component of $G$ is not compact. –  Ben McKay Jul 4 '13 at 12:43
    
Oh I see, I misinterpreted the paper. They indeed mean $K$ in $P$ as you did. –  johndoe Jul 4 '13 at 12:49
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Now I'm embarrassed, what should I do with my question and your answer? –  johndoe Jul 4 '13 at 12:53

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