# Alternative (easier) Proof of Ambrose Singer Holonomy theorem

Let $$P(M,G)$$ be a principal bundle. Giving a connection on $$P(M,G)$$ means two equivalent things. One as an assignment of subspace of $$T_pP$$ for each $$p\in P$$ and another as a $$\mathfrak{g}$$ valued $$1$$ form.

• Given connection as a $$\mathfrak{g}$$ valued $$1$$ form $$\omega$$ on $$P(M,G)$$ we have associated a $$2$$ form (differential of $$\omega$$ and modifying it little bit) which we called the curvature form for the connection, denoted by $$\Omega$$.

• Given connection as a distribution $$p\mapsto H_pP\subseteq T_pP$$, we know what it means to say a curve is horizontal curve. We write $$v\sim u$$ to mean that $$v$$ and $$u$$ are related by horizontal curve. Given $$u\in P$$ we have defined what is called a Holonomy bundle $$P(u)=\{v\in P:v\sim u\}$$ based at $$u$$ and what is called a Holonomy group $$\Phi(u)=\{a\in G:u\sim ua\}$$ based at $$u$$.

Given $$p\in P$$ we have $$\Omega(p)(X(p),Y(p))\in \mathfrak{g}$$ for all $$X(p),Y(p)\in T_pP$$.

As $$\Phi(u)$$ is a Lie subgroup of $$G$$, we have, Lie algebra of $$\Phi(u)$$ to be subalgebra of $$\mathfrak{g}$$.

So, from curvature form, we get a collection that is a subset of $$\mathfrak{g}$$ and from holonomy group/holonomy bundle we get a collection that is a subset of $$\mathfrak{g}$$ and considering suitable restrictions, it is natural to expect these two sets to be same. Ambrose-Singer Holonomy theorem what is the precise relation between these. The set $$\Omega(p)(X(p),Y(p))$$ where $$p\in P(u)$$ and $$X(p),Y(p)\in H_pP$$ generated the Lie algebra of $$\Phi(u)$$.

More precisely, we have the following.

Let $$P(M,G)$$ be a Principal bundle where $$M$$ is connected and paracompact. Let $$\Gamma$$ be a connection in $$P$$, $$\Omega$$ the curvature form, $$\Phi(u)$$ the holonomy group with reference point $$u\in P$$ and $$P(u)$$ the holonomy bundle through $$u$$ of $$\Gamma$$. Then the Lie algebra of holonomy group $$\Phi(u)$$ is equal to the subspace of $$\mathfrak{g}$$, Lie algebra of $$G$$, generated by all elements of the form $$\Omega_v(X,Y)$$ where $$v\in P(u)$$ and $$X$$ and $$Y$$ are arbitrary horizontal vectors at $$v$$.

This is very natural thing one can expect but the proof is not that simple (I am not saying I could have guessed the statement, I am saying that if some one show me this statement and ask me to guess if that is true or not I would have said that it is mostly true). The proof given in Kobayashi involves some thing called involutive distribution. I want to know if there is any alternative proof for this result.

We have to prove $$\mathfrak{g}'$$ which is Lie algebra of $$\Phi(u)$$ is spanned by elements of the form $$\Omega(p)(X(p),Y(p))$$ where $$p\in P(u)$$.

Let $$v\in \mathfrak{g}'$$ i.e., there exists a smooth curve $$\alpha:[-1,1]\rightarrow \Phi(u)$$ such that $$\alpha(0)=e$$ and $$\alpha'(0)=v$$. Then, I want to prove $$v=\sum a_i \Omega(p_i)(X(p_i),Y(p_i))$$. So, I have to look for some elements $$p_i\in P(u)$$ and then some horizontal vectors $$X(p_i),Y(p_i)$$. It is not clear what choice of $$p_i$$ should one make. I am not very sure if this approach works or not. Any suggestion on making this approach work are welcome. Any other alternative proof is also welcome. An exposition of proof given in Kobayashi and Nomizu is also welcome.

• By "involutive distribution" they just mean "vector subbundle of the tangent bundle whose smooth sections are closed under Lie bracket", as in the Frobenius theorem. – Ben McKay Sep 24 '18 at 17:54
• There is a nice proof in section 4.8, theorem 6, of Koszul, TIFR Lectures on Fiber Bundles and Differential Geometry. I doubt if it is clearer or simpler than the other proofs; they all seem to be essentially the same. – Ben McKay Sep 24 '18 at 18:01
• I only know definition and $1$ or $2$ things about involutive distribution. A distribution on a manifold can be seen as a subbundle of tangent bundle. Given any two sections of tangent bundle (vector fields on $M$) $X,Y:M\rightarrow TM$ we can talk about their Lie bracket $[X,Y]$. I call a distribution involutive if Lie bracket $[X,Y]$ has its image in $E$ if both $X$ and $Y$ have their image in $E$.. I will check that notes :) Thank you @BenMcKay – Praphulla Koushik Sep 24 '18 at 18:05
• I do not completely understand when you say "To correct the statement. you need to parallel translate curvature from other points"... Can you please elaborate.. @BenMcKay – Praphulla Koushik Sep 24 '18 at 18:07
• A complete correct and concise statement in modern notation is in Joyce, Riemannian Holonomy Groups and Calibrated Geometry, Theorem 2.4.3, p. 31. He doesn't give a proof, but instead refers for proof to the original paper of Ambrose and Singer and to the incorrect statement you give above from Kobayashi and Nomizu. – Ben McKay Sep 24 '18 at 18:13

I always find it easier to work with the vector bundle induced by a linear representation of the structure group. I believe this theorem is a consequence of the following loop formula (a terse proof can be found in Holonomy Equals Curvature):

If \begin{align*} M &= \text{smooth manifold}\\ E &= \text{rank n vector bundle over M}\\ \nabla &= \text{connection on E}\\ R &= \text{curvature tensor of \nabla}\\ \Omega &=\text{curvature 2-form of \nabla}\\ \gamma: [0,1]\rightarrow M &=\text{ null-homotopic loop in }M\\ p &= \gamma(0) = \gamma(1)\\ \Gamma: [0,1]\times[0,1]\rightarrow M &=\text{ smooth homotopy from p to \gamma}\\ P_\gamma: E_p\rightarrow E_p &=\text{ parallel translation around \gamma}, \end{align*} then $$\int_{[0,1]\times[0,1]} \pi_{(s,t)}^{-1}\Omega\hat\pi_{(s,t)} = P_\gamma - 1,$$ where $$\pi_{(s,t)}: E_{p} \rightarrow E_{\Gamma(s,t)}$$ is parallel translation along the curve $$\Gamma(\cdot,t)$$ from $$p$$ to $$\Gamma(s,t)$$ and $$\hat\pi_{(s,t)}: E_p \rightarrow E_{\Gamma(s,t)}$$ is parallel translation first along $$\Gamma(1,\cdot)$$ from $$p$$ to $$\gamma(t) = \Gamma(1,t)$$ and then along $$\Gamma(\cdot,t)$$ from $$\Gamma(1,t)$$ to $$\Gamma(s,t)$$.

• I find this to be exceptionally clean. – Mike Miller Sep 25 '18 at 22:50
• Well, to be honest, that's mostly because I've hidden the mess inside the notation. – Deane Yang Sep 25 '18 at 23:12
• A sign of good notation! – Mike Miller Sep 25 '18 at 23:14
• @MartinGisser, thanks for your comments. Unfortunately, I don't know what the integrals you cite are. I use only a double integral over a rectangle. The key observation is that you can still pull everything back from the manifold, even if the map is not an immersion. Although others do similar calculations, they don't justify them as carefully as I like. I would add that I learned this approach from writings of people such as Karcher, Heintze, Eschenburg, and Jost. They all have also written elegant expositions of how to use geodesics and Jacobi fields effectively. – Deane Yang Oct 26 '19 at 17:39
• On "prettifying Ambrose-Singer" methinks it should indeed be boiled down to an introductory "higher elementary calculus" course accessible to the gifted high school student. Like, extending the appendix to Einsteins little book link.springer.com/book/10.1007/978-3-540-87777-6 ... Now I need to take a day or two of distance from all this beauty :-) – Martin Gisser Oct 26 '19 at 19:10

I think of this result in terms of Sussmann's orbit theorem. I wrote a paper about his theorem (https://arxiv.org/pdf/math/0508121.pdf), in which I show that a map of manifolds $$P \to M$$ which takes some family of vector fields on $$P$$ to a family on $$M$$ (in a suitable sense) takes orbits to orbits, so that each orbit in $$P$$ becomes a fiber bundle over the associated orbit in $$M$$. The proof is fairly easy to read, I hope. This is the fundamental idea behind Ambrose-Singer. As families of vector fields, you take all vector fields on $$M$$, and take the associated horizontal vector fields they lift to on the bundle $$P$$. We can add Lie brackets of vector fields to any family of vector fields without altering orbits. Applied to horizontal lifts of vector fields, these brackets compute out curvature and its covariant derivatives. So applying curvature and its covariant derivatives to vector fields in our family doesn't change the orbits. Use the flow of any one vector field in our family on $$P$$ to move around the others, without changing orbits. So the gives us parallel transport of the curvature and its covariant derivatives. I guess that a complete proof would not be difficult along those lines. It is not much different to Kobayashi and Nomizu.

• Thanks for the answer... This is not completely clear but I hope I can understand after spending some more time.. Thanks.. – Praphulla Koushik Sep 25 '18 at 4:35

There is another approach in JP. Magnot; Structure Groups and Holonomy in Infinite Dimensions, Bull. Sci. Math. 128 (2004) 513–529 based on the very nice book from Lichnerowicz Théorie globale des connexions et des groupes d’holonomie, Cremonese, Roma, 1956, where this theorem is extended to an infinite-dimensional setting.

The best overview on holonomy is perhaps in Besse's Einstein Manifolds. Regarding "Ambrose-Singer-Nijenhuis" they (pl) have a remark 10.59 about Nijenhuis' "more intuitive" approach with some "double integral". The best and easiest pure ODE writeup of this (no Green theorem for a hybrid double product-sum integral, no principal bundle wizardry (culminating with Peter Michor :-) ) is perhaps Werner Ballmann's p.26 here: http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/conncurv1999.pdf

(At some places in differential geometry one has to ask: Do we need Frobenius here? More often than not, the answer is no. With Ambrose-Singer things can culminate in a double Frobenius (e.g. Koszul's Tata lectures). But here: One Frobenius is indeed differential geometry (and not just a cop-out or historical ballast). It is in Freundenthal's simplification of the Kuranishi-Yamabe theorem in the differentiable case (Kobayashi-Nomizu Vol.I App. 4). This is one necessary fulcrum of the proof of Ambrose-Singer.)

• I agree. Ballman's exposition, including the proof of the Ambrose-Singer-Nijenhuis theorem, is simple and elegant. Thanks. – Deane Yang Oct 26 '19 at 17:31
• Thanks for agreeing. I found Ballmann's writeup quite a relief after half a dozen other expositions. -- You are the older master here :-) Regarding superfluous Frobeniuses methinks it gets even worse with superfluous Jacobi fields... – Martin Gisser Oct 26 '19 at 18:27