Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we place a monopole at the origin $\{{\bf 0}\}$, and the gauge field is well-definded in region $\mathbb R^3-\{0\}$ which is homomorphic to a sphere $S^2$.

Then the total manifold is $U(1)$ fibers attached to base $S^2$. We may ask how many types of "phase texture" are there on a sphere?

Then I use $\pi_2 (U(1)) = \pi_2(S^1)$ since we established the map $S^2\mapsto U(1) \sim S^1$.

But $\pi_2 (S^1) = 0$ not $\mathbb Z$! Where I got wrong? According to the famous book Topology and Geometry for Physicists, the correct formula should be $$ \pi_1(S^1) = \mathbb Z.$$ Why $\pi_1$ instead of $\pi_2$?

share|cite|improve this question
Did you mean $\mathbb R^3-\{0\}$ is homeomorphic to $S^2$? And if so, notice that the first space is non-compact (in particular it is not a bounded subset of $\mathbb R^3$) while $S^2$ is compact, so this statement alone is not true. – joshphysics Feb 1 '13 at 8:35
Maybe "homotopically equivalent to" would be more appropriate. – twistor59 Feb 1 '13 at 8:44
up vote 1 down vote accepted

What OP is classifying with $\pi_2 (G)$ [where the gauge group here is $G=U(1)$], is the globally$^1$ defined gauge transformations $g: M\to G$. This is typically not what we want to calculate.

When discussing the Dirac monopole physicists are instead interested in classifying inequivalent configurations of the dynamical variable of the theory, i.e. the gauge potential $A$. More precisely, in the Wu-Yang/bundle picture (which avoids the use of a Dirac string), we consider the associated vector bundle $$T^{*}M\otimes \mathfrak{g}~\to~ M,$$ where $\mathfrak{g}=\mathfrak{u}(1)$ is the corresponding Lie algebra. It is implicitly understood that the gauge potentials $A_{\alpha}: U_{\alpha}\to \mathfrak{g}$ are defined on local charts $U_{\alpha}\subset M$ with $\cup_{\alpha}U_{\alpha}=M$. Moreover it is implicitly understood that two local sections $A_{\alpha}: U_{\alpha}\to \mathfrak{g}$ and $A_{\beta}: U_{\beta}\to \mathfrak{g}$ are connected via local gauge transformations $g_{\alpha\beta}: U_{\alpha}\cap U_{\beta} \to G$ in the overlaps $U_{\alpha}\cap U_{\beta}$.

It turns out that we are therefore interested in counting maps from the equatorial overlap region $S^1$ to $G$, i.e. $\pi_1 (G)$, as twistor59 explains in his answer.


$^1$ The underlying space manifold for the Dirac monopole is $M=\mathbb{R}^3\backslash\{\bf 0\}$, which is homotopy equivalent (but not homeomorhic) to $S^2$. We have removed the origin, since the Dirac monopole is singular there. [The 't Hooft-Polyakov monopoles are regular in the full space $\mathbb{R}^3$, but we will not discuss these monopoles in this answer.]

share|cite|improve this answer

You want to specify how much "twist" there is in a $U(1)$ (i.e. effectively $S^1$) bundle over $S^2$. If you cover $S^2$ with a North and South patch, then the transition region is topologically $S^1$ so the bundle is classified by maps $S^1$ to $U(1)$, i.e. you want $\pi_1(U(1))=\pi_1(S^1) = \mathbb{Z}$

share|cite|improve this answer
I see. Does it apply to all closesd 2D surfaces with non-zero curvature? e.g. a $T^2$? In this case the TKNN invariant also reads $\pi_1 (U(1)) = \mathbb Z$ – ChenChao Feb 1 '13 at 9:08
$T^2$ means $S^1XS^1$? If so, then yes, I think in this case the circle bundles are classified by $H^2(M;\mathbb{Z})$ which is again the integers - the Euler class. – twistor59 Feb 1 '13 at 11:32
Yet, TKNN invariant are known to be first "Chern" numbers. – ChenChao Feb 1 '13 at 15:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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