Holonomy group is closed or not?

Question

Let $P\to M$ be a $\mathrm{SU}(2)$-principal bundle over a closed connected manifold $M$. Let $A$ be a flat connection on $P$, fix $x\in M$ and let $H:=\mathrm{Hol}_{x,A}(\pi_1(M,x))<\mathrm{Aut}(P_x)\cong \mathrm{SU}(2)$ be $A$'s holonomy based at $x$ where $\mathrm{Hol}_{x,A}$ is the holonomy map from the based loop space to $\mathrm{Aut}(P_x)$. Since $\pi_1(M,x)$ is countable, $H$ is countable and hence a proper subgroup of $\mathrm{SU}(2)$.

Question : Is $H$ necessarily closed in $\mathrm{SU}(2)$ ?

• According to [DK] at top of p.132, $H$ is necessarily closed. Why ?

• According to [KN-I] at p.73, thm. 4.2, $H$ is a Lie subgroup of $\mathrm{SU}(2)$, and hence $H$ is necessarily closed.

• According to this link at middle of p.14, the restricted holonomy group (which here is trivial since $A$ is flat) is necessarily closed but the (non restricted) holonomy group $H$ is not necessarily closed. Why ? Is there a problem with thm. 4.2 in [KN-I] ?

Remark 1 : Since $H$ is proper in $\mathrm{SU}(2)$, we know that "if $H$ is dense in $\mathrm{SU}(2)$, then $H$ not a closed subgroup of $\mathrm{SU}(2)$". I already know that proper dense subgroups of $\mathrm{SU}(2)$ exist ; can $H$ be one of them ?

Remark 2 : The aim of this question is to clarify what is an irreducible flat connection. I found many different definitions :

1. "$A$ is reducible if $H$ lies in some proper subgroup of $\mathrm{SU}(2)$". See [DK] at bottom of p.131.
2. "$A$ is reducible if the set $Q\subset P$ corresponding to all points in $P$ horizontally joinable from a fixed $p_0\in P_x$ is a (strict) structural reduction $Q\to M$ with structural group $H<G$ of the $G$-bundle $P\to M$" (here $G=\mathrm{SU}(2)$). See [KN-I] at top of p.82.
3. "$A$ is reducible if the centralizer $C_G(H)$ is strictly bigger than the center $Z(G)$" (here again $G=\mathrm{SU}(2)$, so $Z(\mathrm{SU}(2)) = \{-1,1\}$). Since $C_G(H)$ is isomorphic to the stabiliser $\mathcal{G}_A$ of $A$ in the gauge group $\mathcal{G}$, this present definition of reducibility is equivalent to "A is reducible if $\mathcal{G}_A$ is strictly bigger than $Z(G)$". Then, since the only possible discrete centralizer $C_G(H)$ is minimal $\{-1,1\}$, for $G=\mathrm{SU}(2)$, this present definition of reducibility is also equivalent to "A is reducible if the kernel $\ker \mathrm{d}_A$ of the exterior covariant derivative $\mathrm{d}_A:\Omega^0(M;\mathrm{Ad}P)\to \Omega^1(M;\mathrm{Ad}P)$ on the space of $\mathrm{Ad}P$-valued differential 0-forms is not injective". Lastly, this present definition of irreducibility is equivalent to "A is reducible if $H$ acts reducibly on $\mathbb C^2$". These equivalent definitions of $A$ being reducible ($C_G(H)\ne Z(G)$, $\mathcal{G}_A \ne Z(G)$, $\ker \mathrm{d}_A|_{\Omega^0}\ne 0$ and $H$ acts reducibly on $\mathbb{C}^2$) seem to be the mainstream definitions of reducibility.

According to definition (1), since $H$ lies in itself which is proper in $\mathrm{SU}(2)$, every flat connection is reducible. This is problematic (because of the amount of papers about moduli spaces of irreducible $\mathrm{SU}(2)$ flat connections). So definition (1) is not a good one. Is [DK] wrong then ? Yes and no, because they start their "definition" with "In general..." which is certainly not "always".

Now, if $H$ can be, and is, a dense subgroup of $\mathrm{SU}(2)$ (i.e. $H$ not closed), then :

• $A$ would be irreducible according to definition (2) because a dense submanifold $Q\subset P$ is not a submanifold and hence not a structural reduction in the classical way. Or maybe $A$ could be reducible if we extend the notion of structural reduction to dense subgroups of the structural group. I don't know if this notion exists or not.
• $A$ would be irreducible according to definition (3) (because $C_G(H)$ is minimal for $H$ dense in $\mathrm{SU}(2)$).

If $H$ cannot be, and hence is not, a dense subgroup of $\mathrm{SU}(2)$, then :

• $A$ is reducible, according to definition (2), to a $H$-bundle $Q$ inside $P$.
• $A$ could be reducible or not according to definition (3). Indeed, if $H=\{1,-1\}$, then $C_G(H)=G=\mathrm{SU}(2)$ which is strictly bigger than the center $Z(G) = \{1,-1\}$, then $A$ would be reducible. But if $H$ is the binary icosahedral group inside $\mathrm{SU}(2)$, then $C_G(H)=\{1,-1\}$, and hence $A$ would be irreducible.

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p.s. this question is a sequel to this question.

p.p.s maybe the answer to my question lies in the details of the proof of thm 4.2 in [KN-I]. Still, there seems to be a bit of confusion around the notion of flat irreducible connections.

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[DK] : The Geometry of Four-Manifolds (Donaldson, Kronheimer)

[KN-I] : Foundations of Geometry, Vol. I (Kobayashi, Nomizu)

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Conclusion : According to the answers I got to my question, it seems that $H$ can be dense in $\mathrm{SU}(2)$. This is what I thought, but was unsure because it would contradict both Kobayashi-Nomizu and Donaldson-Kronheimer. Now, what happens to the definitions of reducibility of connections ? Definition (1) from [DK] is out of the game, may $H$ be dense or not in $\mathrm{SU}(2)$. Definitions (2) from [KN-I] and definition (3) tells us that $"H$ dense implies $A$ irreducible". I'm fine with that. But there is still an ambiguity between (2) and (3) where if $H$ is the binary icosahedral group, then $A$ is reducible according to (2) but irreducible according to (3). That's another story and leads to a new question (that you can find here).

Answer 1

The holonomy group need not be closed. The simplest example I know is when $M=S^1$. Take an element of infinite order $g\in SU(2)$ and define the representation $$ \rho: {\mathbb Z}=\pi_1(S^1)\to SU(2) $$ by sending the generator $1\in {\mathbb Z}$ to $g$. Let $P\to M$ be the associated (with this representation) principal flat fiber bundle, its total space equals $$ P= ({\mathbb R}\times SU(2))/\pi_1(S^1), $$ where $\pi_1(S^1)$ acts on the first factor as the group of covering transformations ${\mathbb R}\to S^1$ and on the second factor via left multiplication, $L_\gamma x= \rho(\gamma)x$. (This is all quite standard.) The bundle $P\to S^1$ has a natural flat connection (the projection of the trivial flat connection on ${\mathbb R}\times SU(2)$) whose holonomy representation is $\rho$ and the holonomy group is $\rho(\pi_1(S^1))$. The latter is clearly not closed in $SU(2)$ (its closure is a copy of $U(1)$ in $SU(2)$). I am sure both Donaldson and Kronheimer are aware of such examples.

There is a more interesting side question about Riemannian holonomy: Is it true that the holonomy group of a Riemannian manifold is always closed? This turns out to be also false, but an example is nontrivial, constructed by Burkhard Wilking (I can find the reference if somebody is interested).

Answer 2

The holonomy group of a flat connection is clearly not closed (as you say, it has countable image, and that is almost never finite). The group that actually matters for Donaldson is the stabilizer of $A$ inside of the gauge group of $E$, which is isomorphic to the centralizer of the holonomy group $H_A$ inside $SU(2)$. On the other hand, $C(H_A)$ is a closed subgroup of $SU(2)$ (defined by the equation $gh = hg$ for all $h \in H_A$); in fact $C(H_A) = C(\overline{H_A})$. So all that actually matters is $\overline{H_A}$, which is a closed subgroup of $SU(2)$.

When talking about instantons I usually take "reducible" to mean your third definition, that $\Gamma_A$ is a larger subgroup than $\Bbb Z/2$. This is pretty restrictive; the only groups that arise as centralizers in $SU(2)$ are the center $\Bbb Z/2$, the maximal abelian subgroup $U(1)$, and $SU(2)$ itself. In particular for a flat connection $A$, this is only true if $A$ has holonomy contained inside $U(1)$.

In practice, you're not just thinking about flat connections. For non-flat connections, the Lie algebra of the holonomy group is generated by the curvature of the connection at a point. I guess it's not unreasonable that the Lie algebra could be 1-dimensional but that contribution from the fundamental group could lead to the holonomy group again being noncompact (and hence non-closed).

In any case, I wouldn't be surprised if there was some assumption of simple-connectedness somewhere.

Answer 3

I think the authors of [KN] are certainly aware of the fact that holonomy group need not be closed. For example, on p.53 they talk about reduced subbundle $P^\prime(M^\prime, G^\prime)$ of a principal bundle $P(M, G)$ where $G^\prime$ is a Lie subgroup of $G$, and then they add "Note that we do not require in general that $G^\prime$ is a closed subgroup of $G$. This generality is needed in the theory of connections."

Their definition of a Lie subgroup on p.40 is somewhat unclear to me. Tracing father back to the definition of a submanifold (p.9) note that "a submanifold may or may not be a closed subset". It seems that in [KN] a submanifold is the image of an injective immersion, e.g., the image of the map $\mathbb Z\to S^1$ given by $n\to e^{in}$ is a submanifold.

So I suspect that all the constructions in [KN] work for this (now non-standard) definition of Lie subgroup. I have not checked this of course.

Another piece of evidence is the paper On the degree of differentiability of curves used in the definition of the holonomy group by Nomuzu and Ozeki where they write "a Lie group can admit a Lie subgroup of lower dimension which is everywhere dense."

Without a doubt Nomizu was aware of holonomy groups which are not closed. He generalized the main result of the paper On the holonomy groups of linear connections by Hano and Ozeki where such example appears. The generalization appears in the same journal.