Timeline for Yang-Mills instanton
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2016 at 16:40 | comment | added | Ruben Verresen | A more 'maths-y' way: the set of topologically distinct $G$-bundles on a manifold X is equivalent to the set of homotopically distinct maps $X \to BG$, denoted as $[X,BG]$ (this is the classification theorem for vector bundles, where $BG$ is the classifying space of $G$). So we see that the instanton numbers on a spacetime $S^4$ are classified by $[S^4,BG] = \pi_4(BG)$. It is then a general fact from mathematics that $\pi_4(BG) = \pi_3(G)$. `Physically' this last equation says that the second Chern number on spacetime (left) coincides with the flat Chern-Simons action at infinity (right). | |
| Feb 14, 2016 at 16:29 | comment | added | Ruben Verresen | The above is however not really rigorous: it shows that (1) the integral of the Chern-Simons form at infinity is an integer (since it has to be equal to the second Chern number on (compactified) spacetime), and (2) that the connection at infinity is determined by a map $S^3 \to G$. Hence it is very suggestive that the integral of the Chern-Simons form is measuring the index of the map $S^3 \to G$, but the above doesn't prove it. | |
| Feb 14, 2016 at 16:11 | comment | added | Ruben Verresen | @Siva: The usual `physics-y' way of looking at: suppose you have your instanton such that the action coincides with the second Chern number $\int F \wedge F$, then at the boundary on infinity this equals $\int_{S^3} \omega$ where by Stokes $d\omega = F \wedge F$. This means $\omega$ is the Chern-Simons form $\omega = A\wedge d A + \frac{2}{3} A^3$. But since we demand our curvature to be zero at infinity, our connection must be purely gauge $A = g^{-1}dg$. So the connection is determined by this $g$ as a function of $S^3$, i.e. a map $S^3 \to G$. But that is by definition $\pi_3(G)$. | |
| Feb 14, 2016 at 6:06 | comment | added | Siva | @RubenVerresen: I think you might be right. Do you have a physical explanation for why the relevant objects are homotopies instead of homologies, when talking about instantons on various manifolds? | |
| Feb 12, 2016 at 2:38 | comment | added | Ruben Verresen | Should it not be characterized by the third homotopy group $\pi_3$ rather than the third homology group $H_3$? | |
| May 9, 2013 at 8:40 | comment | added | Siva | Yes. If you know some group theory, the Dynkin diagram of SU(2) is a single "dot" and for any SU(N) is a bunch of "dots" with some lines between them. The way I see it, roughly, you could place the SU(2) dot at each of the sites of the Dynkin diagrams of any larger group. In fact, there might be multiple embeddings. Maybe someone could give a link dealing with this. | |
| May 9, 2013 at 5:08 | comment | added | firtree | Is SU(2) embeddable into SU(N), $N\geq 3$? | |
| May 9, 2013 at 0:41 | history | answered | Siva | CC BY-SA 3.0 |