Principal Bundles, Chern Classes, and Abelian Instantons

I am attempting to make sense of Chern classes of principal bundles and associated bundles. Let $\pi: P \to M$ be a principal $G$-bundle with $M$ a four-dimensional Riemannian manifold. One canonical vector bundle associated to $P$ and the adjoint representation is the adjoint bundle $\text{ad}P = P \times_{\text{ad}} \mathfrak{g}$ where the fibers are identified with $\mathfrak{g}$ and the transition functions are given by the adjoint action.

I understand characteristic classes in the context of vector bundles, so I'm satisfied to say that the adjoint bundle has topological invariants $\text{rank}(\mathfrak{g})$, $c_{1}(\text{ad}P)$, and $c_{2}(\text{ad}P)$. Roughly speaking, one can think of these as measuring "obstructions" to $\text{ad}P$ being trivial. More specifically, we have

$$c_{1}(\text{ad}P) = \frac{i}{2 \pi} \big[\text{Tr}F_{A}\big] \,\,\,\,\,\, \text{and} \,\,\,\,\,\, c_{2}(\text{ad}P) = -\frac{1}{4 \pi^{2}}\big[ \text{Tr}(F_{A} \wedge F_{A})\big].$$

My first question is: does it make sense to talk about Chern classes of principal bundles themselves, or when someone speaks of such a thing, are they implicitly referring to the classes of the adjoint bundles? For example, the above Chern classes measure the obstruction of $\text{ad}P$ from being trivial. Are there such classes which measure the failure of $P$ itself from being trivial?

My second question is somewhat tangential but still related. In algebraic geometry, on a complex surface $S$, one is interested in the instanton moduli space $\mathcal{M}_{N, c_{1}, c_{2}}(S)$ where $N, c_{1}, c_{2}$ are the rank, and first and second Chern classes of the bundles. These are the discrete moduli one fixes. In order to get a nice moduli space, one often allows certain sheaves with those fixed invariants.

For example, to formalize instantons with invariants $(N, c_{1}, c_{2})$ on $\mathbb{C}^{2} = \mathbb{R}^{4}$, one gets a compact moduli space by considering torsion-free sheaves of rank $N$ and those Chern classes on $\mathbb{P}^{2}$ with a framing at infinity. In the case $N=1$, $c_{1}=0$, and $c_{2}=n$ (the instanton number) the moduli space of $U(1)$ instantons on $\mathbb{C}^{2}$ is given by the Hilbert scheme:

$$\text{Hilb}^{n}(\mathbb{C}^{2}).$$

My second question is: how can $U(1)$ instantons have a non-zero second Chern class? I understand that rank one sheaves can have a second Chern class, of course, but if $P$ is a $U(1)$ principal bundle, then $\text{ad}P$ cannot have a second Chern class.

I would say that $U(1)$ instantons are trivial. So if this giant Hilbert scheme is the moduli space of instantons it seems like you're not just throwing in a few degenerate objects, you're throwing in everything! Any idea what's going on here?

For the first question I think you have some misunderstandings. Firstly Chern classes are defined for complex vector bundles - or equivalently principal $U(n)$-bundles. If $G$ is an arbitrary compact Lie group then the Chern classes of a principal $G$-bundle will be understood in terms of some representation $\rho:G\rightarrow U(n)$ (these always exist). For instance when $G=SU(n)$ the representation $\rho$ is the canonical inclusion.

Now if you are given a principal $G$-bundle $p:P\rightarrow X$ over a suitable space $X$ for some topological group $G$, and a representation $\rho:G\rightarrow U(n)$ then you can form the associated vector bundle $P\times_\rho\mathbb{C}^n\rightarrow X$ with total space

$P\times_\rho\mathbb{C}^n=P\times \mathbb{C}^n/[(e,v)\sim(eg,\rho(g)^{-1}v)]$

where $e\in P$, $v\in\mathbb{C}^n$ and $g\in G$. You can now ask for characteristic classes of this vector bundle, and in particular its Chern classes (and there are an infinte number of Chern classes, not just two).

Note that the Chern classes of this bundle are not the characteristic classes of the principal bundle $P$, but are the characteristic classes of some principal $U(n)$-bundle $Q\rightarrow X$ that admits a $(G,\rho)$-reduction of structure to $P$.

The principal bundle $P$ will have its own set of characteristic classes deriving from the cohomology of the classifying space $BG$ of $G$.

Note in particular that the vector bundle $P\times_\rho\mathbb{C}^n$ is not the adjoint bundle $ad(P)$, which is another vector bundle whose characteristic classes have even less to do with those of $P$. Consider for instance the canonical line bundle $\eta$ over $\mathbb{C}P^1=S^2$. Since $U(1)$ is abelian we have that $ad(\eta)\cong S^2\times \mathbb{R}$ is trivial (recall that the Lie algebra of $U(1)$ is $i\mathbb{R}$). However $\eta$ itself is not trivial and is in fact associated to the Hopf $S^1$-bundle $S^3\rightarrow S^2$ which has first Chern class $1\in H^2(S^2)$.

In summary it makes equal sense to ask for the characteristic classes of a principal bundle or the characteristic classes of a vector bundle with structure group $G$ if the representation $\rho$ is understood.

As for the second question, instantons cannot have a Chern class. Instantons are generally understood as certain field configurations. Given for instance as a suitable connection on a principal of vector bundle. I think what you are referring to is the instanton charge, which in the case above is defined to be the Chern class of the bundle to which the instanton is related. As I understand it a $U(1)$-instanton can not have non-zero second Chern class. It can have non-zero first Chern class. In fact for rank $N=1$ we have $c_1=n$ and $c_2=0$.

I wasn't quite clear above. In the following I'll limit attention of $U(n)$-bundles. There is a one-to-one correspondence between (complex) vector bundles over $X$ and $U(n)$-bundles over $X$. This happens in one direction through the construction I gave above, and in the other by taking the frame bundle of a given vector bundle.

Thus, given a complex vector bundle $E\rightarrow X$ of rank $n$ we have a principal $U(n)$-bundle $Fr(E)\rightarrow X$ and an isomorphism of vector bundles $Fr(E)\times_{U(n)}\mathbb{C}^n\cong E$ over $X$. We can then take a connection on either bundle and ask for the Chern classes it gives. To see that the classes defined by each are the same recall firstly that a connection on a principal bundle induces one on any associated bundle. So a connection $A$ on the principal bundle $Fr(E)$ induces one, $\nabla_A$, on the vector bundle $Fr(E)\times_{U(n)}\mathbb{C}^n\cong E$. Moreover it is seen that both connections $A$ and $\nabla_A$ yield identical curvature 2-forms. Thus

$c_i(E)^{\nabla_A}=c_i(Fr(E))^A\in H^*(X)$

where we add a superscript to denote which connection we use to define the classes.

Now recall that the Chern classes are independent of the choice of connection. It follows that any other connection $A'$ on $Fr(E)$ yields identical Chern classes

$c_i(Fr(E))^A=c_i(Fr(E))^{A'}\in H^*(X)$.

By the same reasoning any other connection $\nabla'$ on your original vector bundle $E$ yields identical Chern classes

$c_i(E)^{\nabla'}=c_i(E)^{\nabla_A}\in H^*(X)$.

The point is that it doesn't matter which bundle $E$ or $Fr(E)$ we use to define the Chern classes: they are the same. The one-to-one correspondence between vector bundles and principal bundles carries over to their characteristic classes and you can consider the Chern classes to 'belong' to either bundle.

As for your comments regarding the Yang-Mills functional. Firstly I believe what you say about it being bounded by the second Chern class is only relevant to Yang-Mills theory over a 4-dimensional base. It is certainly not true for Yang-Mills on a surface, or 8-manifold, say. As for what bundle, it is the one you are working with, which should be granted by deeper context. By the above, it does not matter it is a vector bundle or the corresponding principal bundle.

And if you are talking about some $G$-bundle then yes, there must be an implicit unitary representation. The Chern classes are the characteristic classes belonging to $U(n)$. They can be defined as certain cohomology classes in the cohomology of the classifying space $BU(n)$. You cannot ask for the Chern classes of a $G_2$-bundle, say, which will have its own set of characteristic classes. If you are working with a vector bundle which has a $G$-structure then this $G$-structure provides a unitary representation for $G$.

*Edit 2: Clarification of further comments.

The Chern classes may be defined as certain classes in the cohomology of the classifying space $BU(n)$ of $U(n)$. The integral cohomology of $BU(n)$ is the polynomial ring $H^*(BU(n))\cong \mathbb{Z}[c_1,\dots c_n]$ on classes $c_i\in H^{2i}(BU(n))$. The classes $c_i$ are the universal Chern classes: the Chern classes belonging to the universal $U(n)$-bundle $EU(n)\rightarrow BU(n)$ with contractible total space (you can think of $BU(n)$ as the Grassmanian of $n$-planes in $\mathbb{C}^\infty$).

Then if $f:X\rightarrow BU(n)$ is a suitable classifying map for the bundle $E\rightarrow X$ (principal $U(n)$ or complex vector bundle, we have already decided these are the same things in a certain sense), in that $f^*EU(n)\cong E$, you can define the Chern classes of $E$ to be

$c_i(E)=f^*(c_i)\in H^{2i}(X)$.

This is the point of view in algebraic topology, and its truly quite marvelous that this same information can be extracted by studying the geometric data contained in a connection on $E$.

It may have been misleading when I said that there were an infinte number of Chern classes. On the one hand it's true if you consider the infinite unitary group which has $H^*(BU(\infty))\cong\mathbb{Z}[c_1,\dots,c_n,\dots]$, and you can certainly embed a rank $n$ vector bundle $E$ into the rank $n+1$ vector bundle $E\oplus\epsilon^1$ and ask for its Chern classes. However whilst you can take $n$ all the way to infinity in this manner, there will only be at most $n$ non-trivial Chern classes for a rank $n$ bundle. There are an infinite number you can ask for, but only $n$ interesting ones.

Now more generally you may be interested not in the individual Chern classes but in certain polynomial combinations of them, and there are indeed an infinite number of these you could potentially ask for.

Now the final point is that over a 4-manifold $X$ there will only be two potentially non-trivial Chern classes. This is simply because $H^*(X)$ vanishes in dimensions greater than $4$. Thus we have $c_1(E)\in H^2(X)$ and $c_2(E)\in H^4(X)$ and anything else lives in higher degrees and is thus trivial.

An application of this? If $E\rightarrow X$ is a rank $n$ complex vector bundle over a $4$-manifold with $n>2$, then there is a rank $2$ vector bundle $E'\rightarrow X$ and a bundle morphism $E\cong E'\oplus\epsilon^{n-2}$.

• Thank you, indeed I seem to have some misunderstandings. So if you take $A$ a connection on a principal bundle $P$, you get the curvature two-form $F_{A}$ which you can then use to define for example, the first and second Chern classes, as I wrote in my question. But what bundle exactly are these the Chern classes of? It is well-known that the Yang-Mills functional is bounded from below by (up to numerical factors) the second Chern class. But again, the second Chern class of what bundle? Are we implicitly assuming we have a unitary representation of $G$, and taking classes of that bundle? – Benighted Aug 26 '17 at 20:06
• I've added the details for your questions to my original answer. It was far to long for a comment. I hope it is helpful. – Tyrone Aug 26 '17 at 21:12
• Thanks so much! Your edit was extremely enlightening. I think part of my confusion stemmed from a background in physics where they are mostly studying Yang-Mills theory with a $U(n)$ or $SU(n)$ structure group. So the representation is immediate, and they omit more general discussion. And yes, my comment about the YM functional was special to four-manifolds. Thanks again. – Benighted Aug 27 '17 at 18:58
• Quick question: in the context of your edit, if $X$ is a four-manifold, will you get more that two Chern classes? You mentioned initially that there will be infinitely many Chern classes of an associated complex vector bundle. – Benighted Aug 27 '17 at 19:11
• So I've written a little more. It's looking like quite an essay at this point! I'll try to keep anything further to the comments. I'm a algebraic topologist so I am much more used to the viewpoint that I wrote about this last time. You've hopefully had some expose to classifying spaces in the physics texts and you'll understand a little better why they are important. – Tyrone Aug 27 '17 at 21:20