1
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ Are you talking about the holonomy group of the total space of $OM$ with its usual (Sasaki-Moek/O'Neill) Riemannian metric? $\endgroup$ Nov 14, 2012 at 18:37
  • $\begingroup$ Yes, although the Riemannian metric chosen on $OM$ doesn't play a role in this question. $\endgroup$ 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. $\endgroup$ Nov 15, 2012 at 15:33
  • $\begingroup$ Sorry, I apparently misunderstood your question. Bryant's answer is exactly what I was looking for. $\endgroup$ Nov 15, 2012 at 18:42

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thanks! Are there examples in each class? Any references? $\endgroup$ 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? $\endgroup$ 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. $\endgroup$ Nov 15, 2012 at 13:09
  • $\begingroup$ Once again, Thanks! This is great! $\endgroup$ Nov 15, 2012 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.