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".

  • $\begingroup$ For the exponential, you need to start thinking about bundles and simultaneous local descriptions, as in the answer below. For the "continuously connected to identity" part of the question, you can think in terms of physical states: "small" gauge transformations leave you in the vacuum, "large" ones take you over sphaleron barriers to different vacua. $\endgroup$
    – Demosthene
    Commented Feb 24, 2017 at 17:25
  • $\begingroup$ Minor comment to the post (v7): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Jan 14, 2023 at 16:33

2 Answers 2


Bundles and compactified spacetime

A gauge theory cannot be looked at purely locally, it has inherently global features one cannot see locally. The proper mathematical formalization of a Yang-Mills gauge theory is that the gauge field $A$ is a connection on a principal bundle $P\to M$ over spacetime $M$. However, in practice, it turns out that physicists don't actually want $M$ to be spacetime itself, but spacetime compactified.

We can see this most clearly in the construction of the BPST instanton on Euclidean $\mathbb{R}^4$: The gauge invariant trace of the field strength itself goes as $\mathrm{tr}(F)\propto \frac{\rho^2}{(x^2+\rho^2)^2}$ and is well-defined everywhere, falling off toward infinity. But if we consider the associated gauge potential $A$, one finds it is not well-defined everywhere, it goes as $A\propto \frac{x^3}{x^2(x^2+\rho^2)}$, which is singular for $x\to 0$, yet well-defined for $\lvert x\rvert\to\infty$ as $A(x)\to U(x)^{-1}\mathrm{d}U(x)$, where $U(x)$ is essentially the gauge transformation you wrote down in your question.

So we want $F$ as a physically allowed field strength, yet its corresponding $A$ is not well-defined on $\mathbb{R}^4$. The bundle viewpoint cannot help us because all bundles over Euclidean space are trivial, meaning $A$ must always be defined globally. However, if we pass to $S^4$ as the conformal compactification of $\mathbb{R}^4$ and identify one of the poles with "infinity" and the other with zero, then non-trivial bundles become possible, and we get two local descriptions on the northern and southern "hemispheres" which we can usually extend over the entire sphere except a single point. If the local description of $A$ extends over the entire sphere, then the principal bundle of the gauge theory is trivial.

But we already saw that the specific $A$ we chose does not extend to $x=0$, and in fact the topological invariant $\int\mathrm{tr}(F\wedge F)$ is non-zero, meaning the bundle is non-trivial, meaning $A$ cannot extend over the entire sphere. In particular, it is inherently impossible to find a $A$ that is well-defined at every $x\in\mathbb{R}^4$ and has a well-defined limit towards infinity that gives us the BPST instanton solution $F$.

So you have exactly two choices: Either we must consider gauge theory on $S^4$ instead on $\mathbb{R}^4$, or the BPST instantons - all instantons, in fact - are not actually allowed solutions of the gauge theory. Standard physics opts for the former, in light of instanton contributions to detectable things like the axial anomaly.

Large gauge transformations

Now that we know that we are looking at a principal bundle $P\to S^4$, a gauge transformation is a fiber-preserving automorphism $P\to P$, and it can happen that these are not homotopic to the identity map $P\to P$. As a toy example, consider the $\mathrm{U}(1)$-bundle $\mathrm{U}(1)\times S^1\to S^1$, which is the torus, and the gauge transformation $\mathrm{U}(1)\times S^1\to\mathrm{U}(1)\times S^1, (g,s)\mapsto (gs,s)$, which takes the canonical embedding $S^1\to \mathrm{U}(1)\times S^1$ and winds it once around the circle $\mathrm{U}(1)$. Since the winding number is a homotopy invariant, the image of the $S^1$, as a path, is not homotopic to the source and therefore this transformation is not homotopic to the identity. This is a large gauge transformation in the proper, mathematical sense, as defined in the Wikipedia article and discussed, for instance, in this answer by David Bar Moshe. I am actually not certain whether there are "true" large gauge transformations on $S^4$ in this sense, but I believe there aren't.

"False gauge transformations", or: Transition functions

By lacking the formal machinery of principal bundles, the physicist often confuses two different objects - the gauge transformations $P\to P$, which descend to functions $g_i : U_i\to G$ in the local description, and the transition function, which are gauge transformation-like functions $t_{ij} : U_i\cap U_j\to G$ that define the bundle in the local description and do not exist globally. Both the $g_i$ and the $t_{ij}$ fulfill certain compatibility conditions to be globally well-defined.

Now, if the physicist makes a gauge transformation, they usually only consider $\mathbb{R}^4$, meaning they implicitly set the gauge transformation on the other local description - the open set around $\infty$ - to be trivial. The compatibility condition then says that $g_i = t_{ij}$ on $U_i\cap U_j$. In the physicist's local description, this overlap is the sphere at infinity, i.e the behaviour of the gauge transformation as $\lvert x\rvert \to \infty$. So the condition that $U(x)\to I$ that confuses you, Itzykon, Zuber and probably countless others is nothing but the condition that the $U(x)$, given in this local description, actually lifts to a proper gauge transformation on the bundle $P\to S^4$.

A $U(x)$ that does not do this either needs to be complimented by its corresponding transformation in the other local description, or it changes the bundle, that is, the physicist has declared that it changed the transition function, and thereby (probably) the bundle. The $\mathrm{SU}(2)\cong S^3$ bundles over $S^4$ are classified by maps $S^3\to S^3$ "on the equator", in perfect analogy to $\mathrm{U}(1)$-bundles on $S^1$ as described in this answer of mine. And, as $x\gg c$, your $U^{(1)}(x)$ becomes a function $$ \frac{x}{\lvert x\rvert}\mapsto \exp\left(\frac{\mathrm{i}\tau_\mu x^\mu}{\lvert x\rvert}\right),$$ where $x/\lvert x\rvert$ is just a point on the unit sphere $S^3\subset\mathbb{R}^4$ and the r.h.s. lies in $\mathrm{SU}(2)\cong S^3$ naturally. So this is a map $S^3\to S^3$ whose homotopy class classifies the bundle, and it is not too hard to see that it winds the $S^3$ once around itself, in contrast to the constant map, so the implicit change it carries out on the transition function actually changes the bundle. It is not a gauge transformation for the theory on $S^4$ since it does so, not even a "large" one, but these transformations are also often called large gauge transformations. Note finally that it is also not an allowed gauge transformation on $\mathbb{R}^4$ since it is not smooth at $0$.

  • 3
    $\begingroup$ C.f. ncatlab.org/nlab/show/… $\endgroup$
    – jak
    Commented Mar 16, 2017 at 11:17
  • $\begingroup$ Do you mean that $\tau_\mu x^\mu = x^0 \mathbb{I} + \sum_{i=1}^3 x^i \tau_i$? Because in that case your expression for the limit of $U^{(1)}$ is not an element of $SU(2)$ because of its determinant. I think the expression $x^a \tau^a$ in the question is supposed to mean just $\sum_{i=1}^3 x^i \tau_i$, which is an element of $SU(2)$. But then the limit of $U^{(1)}$ is a map from $S^2$. $\endgroup$
    – Friedrich
    Commented Dec 21, 2017 at 18:15
  • $\begingroup$ @Friedrich I think the expression should just be $x^a \tau^a$ as in the question, but, no, it is not a map from $S ^2$ - note that three of the four components of a vector in $S^3$ already almost uniquely identify the point - the constraint of being a unit vector means that $x^0$ is determined up to a sign, so this is actually a map from a hemisphere of $S ^3$ to $S ^3$ (And extendible to a map $S ^3 \to S ^3$). However it's not as easy as I claim to see its winding number now, I think. $\endgroup$
    – ACuriousMind
    Commented Dec 22, 2017 at 12:41
  • $\begingroup$ Sorry for reviving this, but I don't understand why, instead of considering we "change the bundle" on the $S^4$, we can't just assume that we have a non-trivial $g^j$ on the open set around $\infty$. In other words, I don't understand why do we "implicitly" set the gauge transformation to the identity around $|x|\rightarrow \infty$. It seems we just shifted the goalpost of Zuber's question to this choice. $\endgroup$
    – Frotaur
    Commented Jul 19, 2022 at 10:45
  • $\begingroup$ @Frotaur: (let's use $i=1$ for the set around $0$ and $i=2$ for the one around $\infty$), The point is that setting $g_2 = 1$ means the bundle is completely specified by giving $g_1$ (the $U$ physicists talk about) and in particular that we can compute the topological charge $\lambda$ from looking at $t_{12} = g_1$ without having to actually talk about bundles. If $g_2 \neq 1$ then you can't compute the isomorphism class of the bundle just from $g_1$, and so we would have to always give both $g_1$ and $g_2$ instead of just $U$, but that would require actually talking about bundles. $\endgroup$
    – ACuriousMind
    Commented Jul 20, 2022 at 7:56

If you want a rather physical argument as to why you should have the extra condition $$ U(x) \rightarrow I \; \text{for} \; x \rightarrow \infty, $$ we refer to such a gauge transformation which asymptotically goes to identity as a trivial gauge transformation, because it acts trivially on the boundary of the spacetime considered (since it is the identity).

Suppose that this is not the case, that is that $U(\infty) \neq I$. In that case, in the path-integral formulation of QFT, such a pure gauge gives rise to non-zero boundary terms which affect directly the value of the action considered (just consider any integration by parts in the Lagrangian, in that case you cannot simply discard the extra boundary terms because the pure gauge is non-vanishing at the boundary). The boundary terms then have concrete physical implications. For instance in AdS/CFT, such non-vanishing pure gauges in the bulk correspond to global symmetries of the boundary CFT.

So it is not sufficient to connect the transformation to the identity: if it does not die off at infinity, then it will have physical consequences (on the boundary of the spacetime) and cannot be considered a trivial gauge transformation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.