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Let $M$ be a compact Riemannian manifold and $TM$ be its tangent bundle. Given a initial point-vector $(x,v) \in TM$ and a curve $\alpha:[0,1] \to M$ starting at $x$ we can parallel transport $(x,v)$ along $\alpha$ to obtain a point vector $(y,w)$.

The most natural question is which vectors can be connected in this way. Obviously the norm of the two vectors must be the same.

One situation is that there is a smaller dimensional subbundle of $TM$ which is preserved by parallel transport. In this case de Rhams theorem says that $M = K\times L$ for some Riemannian manifolds $K$ and $L$, so this case is uninteresting.

In the case where there is no invariant subbundle of $TM$, Berger's theorem states that either $M$ is a locally symmetric space with rank 2 or more or one can take any vector to any other of the same norm. In the second case we say holonomy is transitive on $M$ (or more precisely on the unit tangent bundle $SM$).

I'd like to consider the orthogonal frame bundle $OM$. This is a connected manifold if $M$ is non-orientable and has two components if $M$ is orientable.

My question is the following: If holonomy is transitive on the unit tangent bundle $SM$ is it necesarilly transitive on each component of the orthogonal frame bundle $OM$?

Thanks for the help!

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  • $\begingroup$ Are you talking about the holonomy group of the total space of $OM$ with its usual (Sasaki-Moek/O'Neill) Riemannian metric? $\endgroup$
    – Paul Reynolds
    Commented Nov 14, 2012 at 18:37
  • $\begingroup$ Yes, although the Riemannian metric chosen on $OM$ doesn't play a role in this question. $\endgroup$
    – Pablo Lessa
    Commented Nov 15, 2012 at 10:30
  • $\begingroup$ Given that you've accepted Robert Bryant's answer below, I think the answer to my question should actually be no. Unless I've misunderstood completely, your question is answered by the notion of holonomy subbundle. $\endgroup$
    – Paul Reynolds
    Commented Nov 15, 2012 at 15:33
  • $\begingroup$ Sorry, I apparently misunderstood your question. Bryant's answer is exactly what I was looking for. $\endgroup$
    – Pablo Lessa
    Commented Nov 15, 2012 at 18:42

1 Answer 1

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The answer is 'no' in general, when the dimension of the manifold is $n>2$. The holonomy group $H_x\subset\mathrm{O}(T_xM)$ could act transitively on the unit sphere in $T_xM$ and its identity component be conjugate to any of the following subgroups

  1. $\mathrm{SO}(n)$

  2. $\mathrm{U}({\tfrac12}n)$, ($n$ even)

  3. $\mathrm{SU}({\tfrac12}n)$ ($n$ even)

  4. $\mathrm{Sp}({\tfrac14}n)$, ($n$ divisible by $4$)

  5. $\mathrm{Sp}({\tfrac14}n)\mathrm{Sp}(1)$ ($n$ divisible by $4$)

  6. $\mathrm{G_2}$ ($n=7$)

  7. $\mathrm{Spin(7)}$ ($n=8$)

  8. $\mathrm{Spin(9)}$ ($n=16$, but this only happens for symmetric spaces).

In none of these cases except the first does the holonomy act transitively on the full orthonormal frame bundle.

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  • $\begingroup$ Thanks! Are there examples in each class? Any references? $\endgroup$
    – Pablo Lessa
    Commented Nov 15, 2012 at 10:32
  • $\begingroup$ Also: examples 2 and 3 are complex matrix groups, What is the action on $\mathbb{R}^n$ under consideration? $\endgroup$
    – Pablo Lessa
    Commented Nov 15, 2012 at 10:40
  • $\begingroup$ @Pablo: Yes, it is now known that there are many examples of each kind, except in Case 8, where the only examples are locally equivalent to either the Cayley plane $\mathbb{OP}^2$ and its noncompact dual. One reference (though there are, of course, very many) is my survey article Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), Sémin. Congr., vol. 1 (1996), pp. 93–165, Soc. Math. France, Paris. The action in all cases is the lowest dimensional faithful representation of that group. $\endgroup$
    – Robert Bryant
    Commented Nov 15, 2012 at 13:09
  • $\begingroup$ Once again, Thanks! This is great! $\endgroup$
    – Pablo Lessa
    Commented Nov 15, 2012 at 18:31

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