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I've never even considered the possibility that a constant transformation would not qualify as a gauge transformation. But I'm reading a paper that seems to make exactly this distinction. In particular, the title of the paper itself begins with "Gauge Invariant". But their results clearly change under any Poincaré transformation (or more generally any BMS transformation). They even acknowledge this at one point deep in the paper. The context is gravitational radiation on $\mathscr{I}^+$.

Now, they have eliminated a more dynamical form of gauge invariance, which can vary from point to point. But I would say that at least somewhere early on in the paper, they should clarify that the "Gauge Invariant" they use in the title and throughout the paper only refers to those parts of the gauge freedom. In my opinion, an unqualified "gauge invariant" necessarily refers to all possible gauge transformations.

Am I being over-precise? Would most people normally understand that "gauge invariant" excludes constant gauge transformations?

Mike
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    $\uparrow$ which paper? – Qmechanic Feb 19 '15 at 22:23
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    Related: http://physics.stackexchange.com/q/13870/2451 and links therein. – Qmechanic Feb 19 '15 at 22:23
  • It's not yet posted; they sent it around to a list I'm on, so I can't share it. Sorry. – Mike Feb 19 '15 at 22:30
  • An $x$-independent/constant transformation is generically a non-geometric, coordinate-dependent notion. However, if it make sense in a restricted framework, say, within special relativity, it is often called a global transformation in physics speak. A global gauge transformation is a special case of a gauge transformation. – Qmechanic Jul 26 '17 at 08:25
  • FWIW, the paper was this one. After my objections, they at least softened their language in a later paper. In the latter, they included a footnote at one point claiming that since the quantity that is manifestly gauge dependent could be multiplied by vectors (they don't have access to) and result in a tensor, that quantity should be called "invariant". Suffice it to say: I disagree. – Mike Jul 28 '17 at 03:07

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There are such things are large gauge transformations, which I think are related to your question. For example, consider general relativity where the gauge invariance is diffeomorphism invariance. Typically gauge transformations are considered that leave the boundary invariant, but there are also large gauge transformations that for example rescale the boundary metric. As an example, $t \rightarrow 2t$ is technically a gauge transformation but it changes the asymptotic value of the metric. Similar statements hold in gauge theories as well.

Large gauge transformations can affect conserved charges, so although these are technically gauge transformations, they are often distinguished from "small" or what we might call ordinary gauge transformations.

Surgical Commander
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  • Interesting. I'd never heard about large vs. small. I'll have to do some more reading and thinking about this. – Mike Feb 19 '15 at 22:16
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    FWIW, "large" and "small" often mean "not homotopic" resp. "homotopic" to the identity transformation (where the homotopy is taken in the topology of the gauge group, naturally). – ACuriousMind Feb 19 '15 at 22:31
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In YM theory, you generally don't take the constant transformations to be gauge transformations, since the constant transformations are generated by the charge operator. If the charge operator generated gauge transformations, it would act trivially on all physical states, which would mean that you couldn't have charges.

user1504
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  • Wouldn't that just mean you can't have entire systems with a net charge - which is indeed mathematically the case? It doesn't mean you can't have any charge anywhere. – tparker Jul 26 '17 at 08:03
  • It would mean that any state in the physical Hilbert space -- 1 particle, 2 particle, pick your complicated configuration -- has charge 0. – user1504 Jul 28 '17 at 01:40
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    That statement is true! Any state in the physical Hilbert space does have charge $0$. If we make the reasonable assumption that the vacuum has charge 0 and is a physical state, then the current operator $J$ can only create particle-antiparticle pairs with no net charge, so any physical state has total charge $0$. – tparker Jul 28 '17 at 01:51
  • No one's ever figured out a way to mathematically define a gauge theory with net charge, because you need to be able to integrate by parts and discard boundary terms, so $\oint dn^\mu F_{\mu \nu}$ must go to zero at spatial infinity, which by Gauss's law can only happen if there is no net charge. – tparker Jul 28 '17 at 01:54
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I can't think of any sane definition of the phrase "gauge transformation" which would exclude a function that is identically equal to some constant $C$, but include a function that is equal to $C$ everywhere except for at a tiny region over in the Andromeda galaxy, where it takes on the value $C + \epsilon$ for a tiny shift $\epsilon$. So yes, a constant transformation still counts as a gauge transformation.

tparker
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