7
$\begingroup$

Suppose the winding number $n$ of the Yang-Mills instanton configuration. It ia given by the expression $$ \tag 1 n = \frac{1}{16\pi^2}\int \limits_{S^4} d^{4}x\text{tr}\big[F_{ij}\tilde{F}^{ij}\big], $$ where the integration is over 4-sphere $S^4$. I've met two different approaches of calculation of the right hand-side of $(1)$.

Approach 1 - the transition function

The sphere $S^4$ is divided on two semispheres $H^+, H^-$, whose manifolds are $$ \partial H^+ = S^3, \ \partial H^- = -S^3 $$ Next, let's use the identity $F\tilde{F} = dK$ on each semisphere. We have definite local bundle patches $$ H^{+}\times SU(n), \ H^{-}\times SU(n) \ \ \text{with coordinates} \ \ \{x,f_{+}\},\{x,f_-\} $$ Along the equatorial intersection $H^{+}\cap H^{-}= S^3$ the coordinates $f_{\pm}$are related by the transition function $g_{-+}$ $$ f_+ = g_{-+} f_- $$ By using the formalism above, the integral $(1)$ can be rewritten in the form $$ \tag 2 n = \int \limits_{H^+}dK_{+} + \int \limits_{H^-}dK_{-} = \int \limits_{S^3}\text{tr}\big[(g_{-+}\partial g_{-+}^{-1})^3\big] $$

Approach 2 - the gauge $A_0 = 0$

One can fix the gauge $A_0 = 0$. Then at unfinities $t\to \pm \infty$ the 4-potential $A_i$ tends to pure gauge $$ A_{i}(\mathbf x , t \to \pm \infty) = g_{\pm}\partial_i {g}_{\pm}^{-1}, $$ and on the spatial boundaries $A_{i}(\mathbf x \to \infty, t) = 0$. Then (see the comment section) $$ \tag 3 n = \int d\sigma_{\mu}K^{\mu} = \int\limits_{S^3}\text{tr}\big[(g_{+}\partial g^{-1}_{+})^{3}\big] - \int\limits_{S^3}\text{tr}\big[(g_{-}\partial g^{-1}_{-})^{3}\big] $$

The question. How these approaches are related to each other? Precisely, it seems that $(2)$ is obtained in gauge fixing independent way, while $(3)$ is obtained in the fixed gauge. Therefore, is $(2)$ reduced to $(3)$ in the temporary gauge $A_{0} = 0$? Or these expressions aren't equivalent?

$\endgroup$
5
  • $\begingroup$ The second approach is formally very unclear. On $S^4$, what exactly do you mean by "at $t\to\pm\infty$" or "on $\mathbf{x}\to\infty$"? Also, did you forget the ³ exponents in eq. (3)? What is the integral of $K$ in the second term in eq. (3) over and how is the final expression obtained? Where did you find this approach? $\endgroup$
    – ACuriousMind
    Jan 19 '17 at 15:26
  • $\begingroup$ @ACuriousMind : It seems that I don't understand the derivation of $(3)$ in a clear manner. First, it seems that I am incorrect in the statement that it is obtained in the compactified $R^4 \to S^{4}$ space. Maybe the hypersurfaces $\sigma_{\mu}$ are 3-dimensional surfaces in the finite euclidean space (i.e., in the box). $\endgroup$
    – Name YYY
    Jan 19 '17 at 15:45
  • $\begingroup$ @ACuriousMind : Second, I forgot about the exponent in Eq. $(3)$. Third, sources of derivations heuristically represent the space-time by a cylinder, whose upper and lower caps being time-like hypersurfaces of constant $t$, and lateral caps being spatial hypersurfaces. The condition $A(\mathbf r \to \infty ) = 0$ means that the integral $(3)$ on spatial hypersurfaces $\sigma_{i}, i\neq 0$ is equal to zero, while the integrals on two time-like hypersurfaces with $t\to \pm \infty$ give the two summands in $(3)$. The one of sources where I seen this approach is Weinberg's QFT Vol. 2, sec. 23.5 $\endgroup$
    – Name YYY
    Jan 19 '17 at 15:47
  • $\begingroup$ @ACuriousMind : The time-like hypersurfaces are 3-spheres, since the gauge field tends to pure gauges on it, and the gauge element $g\to 1$ for $\mathbf x \to \infty$. $\endgroup$
    – Name YYY
    Jan 19 '17 at 15:51
  • 1
    $\begingroup$ Aha! Now you've said it yourself: eq. (3) is on a cylinder $[-T,T]\times S^3$, not on the 4-sphere $S^4$. That's the reason the two equations are different, and why they cannot be equivalent. $\endgroup$
    – ACuriousMind
    Jan 19 '17 at 15:54
1
$\begingroup$

The expressions eq. (2) and eq. (3) are both correct, but not equivalent, because they are given in different settings:

  1. Eq. (2) is the winding number on a sphere $S^4$, i.e. the one-point compactification of Minkowski space $\mathbb{R}^{1,3}$. The $g_{+-}$ is a transition function corresponding to the single choice necessary to glue the local gauge potentials on the hemispheres to a global gauge potential on $S^4$. One should note that, technically, the cocycle construction of the principal bundle would require the intersection of the hemispheres to be open in $S^4$, i.e. they would have to overlap on an $S^3\times(-\epsilon,\epsilon)$ for $\epsilon > 0$. Eq. (2) is what results at $\epsilon\to 0$, at which point we do not technically fulfill the conditions for the cocycle construction anymore, yet the value of $n$ remains unchanged because its dependence on $\epsilon$ is smooth and $n$ takes only discrete values, so it is constant in $\epsilon$.

  2. Eq, (3) is the winding number of an instantonic field configuration on a cylinder $S^3\times[-T,T]$ for $T > 0$. Here, $g_+$ and $g_-$ are gauge functions such that $A(\pm T) = g_\pm^{-1}\mathrm{d} g_\pm$ (again, the physicist likes to take the limit $T\to\infty$). The crucial difference is that the $g_\pm$ are now no transition functions, but related to the gauge potential value. Also, we do not need to take the temporal gauge here, we can obtain eq. (3) without any gauge fixing at all: $$ n \propto \int_{S^3\times [-T,T]} F\wedge F = \int_{S^3\times [-T,T]} \mathrm{d}K = \int_{S^3\times\{-T\}}K(g_-) - \int_{S^3\times\{T\}}K(g_+)$$ where the sign appears because the $S^n$ at the end of a cylinder are oppositely oriented and $K(g_\pm)$ denotes the Chern-Simons 3-form for the respective gauge field configuration $A_\pm$.

$\endgroup$
1
  • $\begingroup$ What would be the winding number of SO(4) instantons $\endgroup$
    – J. Hristov
    Aug 15 '21 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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