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I have a question about the meaning of Yang-Mills Functional.

It is stated everywhere that the Yang-Mills Functional is a measure of energy. But the formal definition of the Yang-Mills Functional is:

  • To have a manifold $M$ together with a smooth vector bundle $E\longrightarrow M$

  • To a given connection $A$ over $E\longrightarrow M$, the Yang-Mills Functional assigns the integral of the norm of the curvature of $A$.

I understand the mathematical background, but why does it represent energy? Is there any intuitive way to explain this link?

Any idea or suggestion is welcome.

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    $\begingroup$ The archetypal example of an energy functional is $E(u) = \int |\nabla u|^2$. If $u=u(x,y)$ measures the deviation of a taut membrane from perfect flatness (say by height), then $E(u)$ really does represent the potential energy stored in that particular configuration of the membrane. Many analogies flow naturally from this observation, including to this interpretation of the (euclidean signature!) Yang-Mills Functional. $\endgroup$
    – Igor Khavkine
    Mar 5, 2015 at 16:05
  • $\begingroup$ @IgorKhavkine In this case, how are a function and a connection related? $\endgroup$
    – Jjm
    Mar 6, 2015 at 8:53
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    $\begingroup$ A connection is a function, so the analogy is rather direct. More precisely, connections are in 1-1 correspondence with sections of a certain affine bundle. This is spelled out explicitly for the case of the Levi-Civita connection of (pseudo-)Riemannian geometry in the answers to this question. $\endgroup$
    – Igor Khavkine
    Mar 7, 2015 at 22:19

2 Answers 2

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The easiest way to see that the norm of the curvature corresponds to the energy is to consider the special case of an abelian U(1)-Yang-Mills theory (i.e. electrodynamics). If you write out the norm squared of the curvature in terms of the $E$ and $B$ fields you get the expression $E^2 + B^2$. This is exactly the familiar energy density of an electrodynamic field, see for example wikipedia.

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    $\begingroup$ You are mixing signatures, though. The norm of the curvature in lorentzian signature is proportional to the difference not the the sum: $B^2-E^2$. The energy density, obtained by a Legendre transform, is proportional to the sum $B^2 + E^2$, which happens to be the norm of the curvature in riemannian signature. $\endgroup$
    – José Figueroa-O'Farrill
    Mar 8, 2015 at 14:18
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    $\begingroup$ Yes of course you are right, my answer applies only in the Riemannian case. But as you noted, for Minkowski spacetime the norm of the curvature is not the Hamiltonian but the Lagrangian. Thus in this case the question does not make much sense. $\endgroup$
    – Tobias Diez
    Mar 8, 2015 at 16:02
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    $\begingroup$ I suspect that the answer to the OP might just be by analogy: the norm of the curvature in riemannian signature agrees with the energy of the lorentzian theory. $\endgroup$
    – José Figueroa-O'Farrill
    Mar 8, 2015 at 18:31
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As comment:

We can interpret Yang Mills functional via Hermitian Yang-Mills functional(Hermitian-Einstein functional) as follows when $M$ is a compact Kahler manifold

If we take YM functional and HYM functional as follows respectively

$YM(\nabla_A)=\int_X |F_A|^2\frac{\omega^n}{n!}$ and $HYM(\nabla_A)=\int_X|\Lambda_\omega F_A|^2\frac{\omega^n}{n!}$

Then

$$YM(\nabla_A)=HYM(\nabla_A)+\frac{4\pi^2}{(n-2)!}\int_X(2c_2(E)-c_1(E)^2)\wedge\omega^{n-2}$$

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