## De Rham decomposition theorem, generalisations and good references

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De Rham decomposition theorem states that every simply-connected Riemannian manifold $M$ that admits complementary sub-bundles $T'(M)$ and $T''(M)$ of its tangent bundle parallel with respect to the levi-Chevita connection is isometric to the direct product of two Riemanninan manifolds $M'\times M''$.

Question 1. In the first place I would like to have a good reference for a clear "modern" and complete proof of this theorem, if it exists (more recent than Kobayshi-Nomizu pp. 187-193) (Note, that Besse 10.44 claimed that no simple proof exists yet).

Eddited.

Question 2. Secondly it seems to me that there should be some statement much more general than de Rham theorem. Namely, suppose we have a metric space $X$ that is locally decomposable as an isometric product of two in such a way that this decomposition is "coherent" in a appropriate sense, i.e. forms something like a presheaf. When will we be able to say that $X=Y\times Z$? (I am interested only in the cases when this will work, not when this will fail). As corollary of such a general statement one should be able to deduce de Rham theorem for example, for Finsler of polyhedral manifold, ect.

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intlpress.com/JDG/archive/1972/7-1&2-161.pdf – Steve Huntsman Jun 13 2010 at 16:44
ams.org/journals/proc/1998-126-10/… – Steve Huntsman Jun 13 2010 at 16:46
emis.ams.org/journals/GM/vol5/reck.ps – Steve Huntsman Jun 13 2010 at 16:47
springerlink.com/content/k41141205w063k47 – Steve Huntsman Jun 13 2010 at 16:50

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Steve, it appears to me that you're just doing a search on Riemannian decomposition on mathscinet, google, or somewhere and listing some of the more promising results. I think your response would be more helpful, if you could actually reveal how you're finding these papers and saying more about what you know about them. Otherwise, you're not providing any more help to Dmitri than what he could easily do himself. – Deane Yang Jun 13 2010 at 18:40
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## 4 Answers

For the first question, I can provide some literature:de Rham's original proof, Wu Hongxi's Ph.D thesis On the de Rham decomposition theorem. I don't know which reference would be earliest one. http://dspace.mit.edu/handle/1721.1/11601

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I suppose we have to ask the Riemannian manifold to be complete. Otherwise $\mathbb R^3 - \lbrace 0 \rbrace$ would be a counterexample.

I do not have an answer to question 2, but you might be interested in variations of De Rham decomposition Theorem to the realm of compact complex manifolds. There, it is natural to ask, as Beauville did, if a holomorphic decomposition of the holomorphic tangent bundle implies that the universal covering is isomorphic to a product(no metric assumption).

Without further assumptions there is no hope since Hopf surfaces provide examples with decomposable tangent bundle but with universal covering isomorphic to $\mathbb C^2 - \lbrace 0 \rbrace$.

If one assume that the manifold is projective or Kahler then there are some positive results, the first of which can be found in Beaviulle's paper linked to above. In the projective case you can also look at this, this, and this paper. In the Kahler case you can look at here.

The general problem seems to be wide open, in the above results either one assumes that one of the factors is one-dimensional or imposes strong conditions on the ambient variety itself (dimension $\le 3$ or uniruledness).

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what does it means that a sub-bundle of the tangent bundle is parallel with respect to the levi-Civita connection? I found that the for each couple of vector fields $X,Y$ of the subbundle the covariant derivative is in the subbundle too. Is it correct?

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This means, that if $X'$ ($X''$) is in the sub-bundle $TM'$ ($TM''$) then its covariant derivative along any other field is also in this sub bundle.   – Dmitri Apr 26 2011 at 15:09

what does it means that a sub-bundle of the tangent bundle is parallel with respect to the levi-Civita connection? I found that the for each couple of vector fields $X,Y$ of the subbundle the covariant derivative is in the subbundle too. Is it correct?

It is stupid, it is a question! Sorry

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