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  What defines a large gauge transformation, really?

+ 3 like - 0 dislike

Usually, one defines [large gauge transformations](http://en.wikipedia.org/wiki/Large_gauge_transformation) as those elements of $SU(2)$ that can't be smoothly transformed to the identity transformation. The group $SU(2)$ is simply connected and thus I'm wondering why there are transformations that are not connected to the identity. (Another way to frame this, is to say that large gauge transformations can not be built from infinitesimal ones.)

An explicit example of a large gauge transformation is

$  U^{\left(  1\right)  }\left(  \vec{x}\right)  =\exp\left(  \frac{i\pi
x^{a}\tau^{a}}{\sqrt{x^{2}+c^{2}}}\right) $

How can I see explicitly that it is impossible to transform this transformation to the identity transformation?

I can define

$U^\lambda(\vec x) =  \exp\left( \lambda  \frac{i\pi
x^{a}\tau^{a}}{\sqrt{x^{2}+c^{2}}}\right) $

and certainly

$ U^{\lambda=0}(\vec x) = I $

$ U^{\lambda=1}(\vec x) = U^{\left(  1\right)  }\left(  \vec{x}\right) $

Thus I have found a smooth map $S^3 \to SU(2)$ that transforms $U^{\left(  1\right)  }\left(  \vec{x}\right)$ into the identity transformation. So, in what sense is it not connected to identity transformation?

Framed differently: in what sense is it true that $U^{\lambda=1}(\vec x)$ and $U^{\lambda=0}(\vec x)$ aren't homotopic, although the map $U^\lambda(\vec x)$ exists? My guess is that at as we vary $\lambda$ from $0$ to $1$, we somehow leave the target space $SU(2)$, but I'm not sure how I can see this.

In addition, if we can write the large gauge transformation as an exponential, doesn't this does mean explicitly that we get a finite large gauge transformation, from infinitesimal ones?

According to this paper, the defining feature of large gauge transformations is that the function in the exponent $\omega(x)$ is singular at some point. Is this singularity the reason that we can't transform large gauge transformations "everywhere" to the identity transformations? And if yes, how can we see this?

I got another idea from this paper. There, the authors state that its not enough that we find a map $U^\lambda(\vec x)$, with the properties mentioned above, but additionally this map must have the following limit
$  U^\lambda(\vec x) \to I \quad \text{ for } x\to \infty \quad \forall \lambda. $
Obviously, this is not correct for my map $U^\lambda(\vec x)$. However, I don't understand why we have here this extra condition.

Edit 2: As mentioned above, there only exists no smooth map between $U^{\lambda=1}(\vec x)$ and $U^{\lambda=0}(\vec x)$, if we restrict ourselves to those gauge transformations that satisfy

$ U(x) \to I  \quad \text{ for } x\to \infty. $

The mystery therefore is, why we do this. It seems, I'm not the only one puzzled by this, because Itzykson and Zuber write in their QFT book:

"there is actually no very convincing argument to justify this

asked Feb 28, 2017 in Theoretical Physics by JakobS (110 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

The requirement that at large $x$ the unitary transform tends to the identity is a precise specification of the meaning of ''small'' in ''small gauge transformation''. It means that only a local region is significantly affected by the transformations. On the other hand, "large gauge transformations'' have significant effects outside any bounded region - that's why they are called large. 

For a connected gauge group such as SU(2), all gauge transformations can be built up gradually from infinitesimal ones by solving a differential equation. But the group of gauge transformations is not the gauge group $SU(2)$ but the direct sum of one copy of SU(2) at each space-time point. And a small gauge transformation must be built up not just from arbitrary infinitesimal transforms but (because of the label ''small'') from local infinitesimal transforms that are the identity outside a bounded region. This only allows one to produce gauge transformation that behave in the large-$x$ limit as the identity transformation.

The reason for making the distinction is that gauge fixing usually fixes only the small gauge transformations and leads to so-called Gribov copies related to large gauge transformations.

answered Feb 28, 2017 by Arnold Neumaier (15,787 points) [ revision history ]

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