8
$\begingroup$

So I have a good appreciation of the non-abelian Yang-Mills story. We take a Riemannian manifold $M$ and a symmetry group $G$ giving rise to an infinite-dimensional gauge group $\mathcal{G}$. In addition, we take a principal $G$-bundle $P$ on $M$, with $\mathcal{A}$ the infinite-dimensional affine space of connections on $P$. Remarkably, $\mathcal{A}/\mathcal{G}$ is a finite dimensional space (not in general, but for example in Yang-Mills theory on Riemann surfaces, it's finite dimensional), and the Yang-Mills functional $\rm{YM}(\cdot)$ is a map from this space to $\mathbb{R}_{\geq 0}$, whose critical points satisfy the Yang-Mills equations. Then of course, we have the instantons which are self-dual and anti-self dual connections. This part of the story I understand well (I hope!) but I could use some assistance when it comes to the Hermitian Yang-Mills story.

Here, people start discussing a holomorphic vector bundle $E$ over $M$. Well, certainly this isn't the $G$-bundle $P$, and I can only guess that it's meant to be the associated vector bundle to $P$? But doesn't this require picking a particular representation of the group $G$? Moreover, the Hermitian Yang-Mills equation is a second order PDE for the curvature $F$, which seems to be the curvature of a connection on $E$. How does this relate to curvatures of connections in $\mathcal{A}$, if at all? I've heard Kobayashi-Hitchin relates instantons to stable holomorphic vector bundles, so I think they must be related, but I'm not sure how.

$\endgroup$
3
  • 2
    $\begingroup$ Yes, you need a representation. Usually, you pick the adjoint representation. The curvature is related in the obvious way since you first have a Lie algebra valued form which have a basis. $\endgroup$
    – user40276
    Commented Aug 4, 2016 at 10:37
  • $\begingroup$ Where do you find this result for $\mathcal{A}/\mathcal{G}$? $\endgroup$
    – 54321user
    Commented Aug 6, 2016 at 16:58
  • 1
    $\begingroup$ Sorry, I think maybe I was too general. If you look on page 3 here (cds.cern.ch/record/260640/files/9403042.pdf) apparently $\mathcal{A}/\mathcal{G}$ is finite dimensional in Yang-Mills theory on Riemann surfaces, and a few other settings. In general, it appears not to be finite dimensional. Sorry, I forgot I was mostly thinking about YM on curves! $\endgroup$
    – Benighted
    Commented Aug 6, 2016 at 17:02

1 Answer 1

7
$\begingroup$

As user40276 said, you need a representation to make this constructions. If you want to consider additional fields, you need several of those. I'm certainly no expert in this field, but I try to shed some light on representation-based constructions and hopefully this lifts some of your confusion.

Given a representation $\rho: G \rightarrow Aut(V)$ of $G$ on some vector space $V$, the associated vector bundle $E$ is constructed via $E=P\times_\rho V := \; ^{\textstyle{P \times V}}\big/_{\textstyle{\sim}}$, where $(p,v)\sim(p\cdot g,\rho^{-1}(g)\cdot v)$ for some $g\in G$.

A vector-valued form $\alpha\in\Omega^k(P,V)$ is called $G$-equivariant iff $(R_g)^*\alpha = \rho(g^{-1})\cdot\alpha$ holds $\forall g\in G$ (here $R_g$ is the right action of $G$ on $P$). Given a principle connection $\omega$ on $P$, we can define an exterior covariant derivative on equivariant forms as $$ D\alpha = d\alpha + \omega \wedge_\rho \alpha, $$ where the wedge product is defined as usual as anti-symmetrised product, where the "multiplication" is now taken to be the point-wise action of $\rho_*\circ\omega: TM\rightarrow End(V)$.

As an example, consider the adjoint representation $Ad: G\rightarrow Aut(\mathfrak g)$. Then $\omega$ is an equivariant 1-form with respect to the adjoint representation and the exterior covariant derivative is $$ D\alpha = d\alpha + [\omega \wedge \alpha], $$ which shows Cartan's structure equation for the curvature: $\Omega = D\omega = d\omega + [\omega \wedge \omega]$.

Note that an equivariant $k$-form $\alpha$ can identified with a vector bundle-valued form $\bar\alpha\in\Omega^k(M,E)$ via the fiber-wise identification $i_p: V\rightarrow E_{\pi(p)}, v\mapsto [p,v]$ and the horizontal lifting of vector fields on $M$. Concretely, we have $$ \bar\alpha_x(X_1,\dots,X_k)=i_p\circ\alpha_p(\tilde X_1,\dots\tilde X_k), $$ where the tilde denotes horizontal lifts of vector fields and $x=\pi(p)\in M$.

Then, a connection $\omega$ on $P$ induces an affine connection $\nabla^\omega$ on $E$ via the relation $$ \nabla^\omega \bar\alpha = \overline{D\alpha}. $$ Of course, $\nabla^\omega$ gives rise to a curvature tensor $R^\omega\in\Omega^2(M,End(E))$.

Let us again consider the adjoint representation. The associated bundle is called the adjoint bundle $Ad(P)$. In this case we find two curvature tensors: First, the bar map identifies equivariant forms of adjoint type, like $\omega$ and $\Omega=D\omega$ with $Ad(P)$-valued forms on $M$, so we get $F^\omega:=\bar{\Omega}\in\Omega^2(M,Ad(P))$. Second, we have the induced affine connection $\nabla^\omega$ on $Ad(P)$ and hence we have a curvature tensor $R^\omega\in\Omega^2(M,End(Ad(P)))$. These two are actually the same objects since $ad=Ad_*: \mathfrak g \rightarrow End(\mathfrak g)$ lifts to a bundle map $Ad(P)\rightarrow End(Ad(P))$.

For pure Yang-Mills theory, we need a $G$-equivariant inner product on $\mathfrak g$, i.e. $\langle Ad_gX, Ad_g Y\rangle = \langle X, Y\rangle$, or in other words, a bundle metric on $Ad(P)$. In this way we can define wedge products on $\Omega(M,Ad(P))$ and a Hodge dual and build up the Yang-Mills functional.

Scalar fields are modeled as sections of an associated vector bundle $E$ with respect to a representation of $G$. Covariant differentiation works as introduced above. Spinor fields are a bit more complicated, they are sections of $S\otimes E$, where $S$ is a spinor bundle over $M$, i.e. an associated vector bundle of the spin frame bundle over $M$. In both cases, we again need a bundle metric to build up bilinear forms for the Lagrangian.

$\endgroup$
2
  • $\begingroup$ Thanks for the fabulous answer! What exactly is $\omega$ in the equation, $D \alpha = d \alpha + \omega \wedge_{\rho} \alpha$? Is it only dependent on data on the base manifold, or is it bundle dependent too? $\endgroup$
    – Benighted
    Commented Aug 6, 2016 at 17:07
  • 1
    $\begingroup$ Sorry, I just noticed that I mentioned it nowhere. $\omega$ is a principal connection on the principle bundle $P$. I updated my post. $\endgroup$ Commented Aug 8, 2016 at 6:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .