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How can instanton solution to Yang-Mills theory with gauge group $SU(3)$ or $SU(N)$ be obtained? For $SU(2)$ it is explained in textbooks but what about more general color gauge groups?

EDIT: How does $A^\mu$ look like for SU(3) or SU(N). Wikipedia only gives for SU(2). It is given by for SU(2)

$$A^\mu_a = 2/g \frac{\eta^a_{\mu\nu}(x-z)}{(x-z)^2+\rho^2}$$

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    $\begingroup$ Wikipedia tells you how to do it. $\endgroup$
    – twistor59
    Commented May 8, 2013 at 14:08
  • $\begingroup$ @Qmechanic why did you put a book tag on this? I rather thought he is looking for a reference and that the answer will rather be found in a paper than in a textbook ...? Certainly the question will be killed now with this tag even though it has 5 upvotes and 2 stars which means that many people would be interested in seing an answer here :-/ $\endgroup$
    – Dilaton
    Commented May 8, 2013 at 18:50
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    $\begingroup$ I think the upvotes are because people would genuinely like to hear about instantons themselves (as opposed to just be handed a list of references and links). On that note, @Raj, it would be good if you (or somebody else?) could reformulate your question to ask an actual physics question (as opposed to just asking for references). $\endgroup$
    – Qmechanic ♦
    Commented May 8, 2013 at 20:28
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    $\begingroup$ Sorry if posting like answer. I forgot to register. My question is: how does the potential $A^\mu$ look like if SU(3) or SU(N) is considered? Wikipedia only shows for SU(2). $\endgroup$
    – Raj
    Commented May 8, 2013 at 21:04
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    $\begingroup$ Hi @Raj. If you would like to merge your two accounts, see physics.stackexchange.com/help/user-merge $\endgroup$
    – Qmechanic ♦
    Commented May 8, 2013 at 21:43

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TO have an instanton solution, you need to map the (euclideanized) "spacetime at infinity" to the group manifold. In the case of SU(2), both the spacetime at infinity and the group manifold are $S^3$ and instantons are characterized by the integers. I hope you understand that much, at least for SU(2).

If you're interested in 4d instantons, they are characterized by $H_3(M_G)$ where $M_G$ is the group manifold -- since the spacetime at infinity is $S^3$. So, for every (homologically distinct) non-contractible 3-cycle of the group manifold, you can find an instanton. As the Wikipedia link given by @twistor says, the gauge fields corresponding to directions on that 3-cycle will have the same profile as the SU(2) instanton and the other gauge fields will have a trivial configuration (of course, up to a gauge transformation). Essentially, you're seeking the possible embeddings of SU(2) inside your gauge group and then making instantons out of those SU(2) subgroups.

If you understand that, the generalization to arbitrary number of dimensions should be straightforward.

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  • $\begingroup$ Is SU(2) embeddable into SU(N), $N\geq 3$? $\endgroup$
    – firtree
    Commented May 9, 2013 at 5:08
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    $\begingroup$ 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. $\endgroup$
    – Siva
    Commented May 9, 2013 at 8:40
  • $\begingroup$ Should it not be characterized by the third homotopy group $\pi_3$ rather than the third homology group $H_3$? $\endgroup$
    – Ruben Verresen
    Commented Feb 12, 2016 at 2:38
  • $\begingroup$ @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? $\endgroup$
    – Siva
    Commented Feb 14, 2016 at 6:06
  • $\begingroup$ @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 $\pi_3(G)$. $\endgroup$
    – Ruben Verresen
    Commented Feb 14, 2016 at 16:11

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