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A What defines a large gauge transformation, really?

  1. Feb 27, 2017 #1
    Usually, one defines large gauge transformations 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

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

    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?

    Edit:
    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 restriction".
     
  2. jcsd
  3. Feb 28, 2017 #2

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    From the definition of large gauge transformation in the Wikipedia article that you linked, any map from the base manifold into ##SU(2)## that is not null homotopic is a large gauge transformation.

    For instance, since the third homotopy group of ##SU(3)## is infinite cyclic, if the base manifold is ##S^3## then the group of large gauge transformations is infinite cyclic.
     
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