Some Riemannian manifolds are expressed as a product manifold. Recently, I have read two articles about space-times. In both articles, the authors prove that a Riemannian manifold $\bar{M}^n$ is expressed as a product of the form $I\times M^{n-1}$. Both authors use similar techniques, namely integrable distribution, in this decomposition. Really, I do not understand this technique. But it is enough to know a characterization of Riemannian manifolds which we can express it as a product manifold $M_1\times M_2$.

Q1 Does this characterization exist?(if yes, a reference is required)

Q2 What conditions and proof hints could one think of to characterize these manifolds?

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    $\begingroup$ The de Rham theorem is the key theorem you probably want. It states that if a manifold is simply connected and the holonomy preserves a connect-sum decomposition of a tangent space, then your manifold is a direct product. $\endgroup$ – Ryan Budney Mar 14 '15 at 23:11
  • $\begingroup$ @RyanBudney Thank you. $\endgroup$ – Semsem Mar 15 '15 at 4:09

The most general result ensuring the decomposition of a manifold is the following (see R.A. Blumenthal and J.J. Hebda. Ehresmann connection for foliations, Indiana Math. J. 33 (1984), 597-612):

Theorem: Let M be a simply connected manifold furnished with a foliation. If it admits an integrable Ehresmann connection, then M is diffeomorphic to the product of two leaves and the foliations are identified with the canonical foliations of the product.

Observe that no metric is assumed in the above theorem.

The classical result is the De Rham decomposition theorem: Let M be a complete and simply connected manifold Riemannian manifold. If M is reducible, then it split as a direct product manifold $M_1\times M_2$.

A wide generalization were obtained by N. Koike (Totally umbilic foliations and decomposition theorems, Saitama Math. J. 8 (1990), 1-18)

Theorem: Let M be a simply connected semi-Riemannian manifold and $(F_1, F_2)$ two complementary, orthogonal and umbilic foliations. If the leaves of $F_1$ are complete and $dim F_1 \geq 3$, then M is isometric to a doubly twisted product of two leaves.

An also see R. Ponge and H. Reckziegel, Twisted product in pseudo-Riemannian geometry, Geom. Dedicata 49 (1993), 15-25, were it is proven the following.

Theorem: Let $M$ be a simply connected semi-Riemannian manifold with $(F_1, F_2)$ two orthogonal and complementary foliations. Suppose that $F_1$ is geodesic and with complete leaves. 1. If $F_2$ is umbilic then M is isometric to a twisted product. 2. If $F_2$ is spheric then M is isometric to a warped product. 3. If $F_2$ is geodesic then M is isometric to a direct product.

As you can see, completeness and simply connectedness are always assumed, although they are not necessary conditions. In the paper M. Gutierrez and B.Olea, Semi-Riemann manifold with a doubly warped structure, Rev. Mat. Iberoam. 28 (2012), no. 1, 1–24, some decomposition results are given without assuming the simply connectedness hypothesis.

If you are interested in decomposition with a one-dimensional factor, you can see the following papers:

Y. Tashiro, Complete Riemannian manifolds and some vector fields, Trans. Amer. Math. Soc. 117 (1965) 251–275.

M. Kanai, On a differential equation characterizing a Riemannian structure of a manifold, Tokyo J. Math. 6 (1983), 143-151.

E. Garc ́ıa-R ́ıo and D. N. Kupeli, Singularity versus splitting theorems for stably causal spacetimes, Ann. Global Anal. Geom. 14 (1996), 301- 312.

T. Sakai, On Riemannian manifolds admitting a function whose gradient is of constant norm, Kodai Math. J. 19 (1996), 39-51.

M. Gutierrez, B. Olea, Global decomposition of a Lorentzian manifold as a generalized Robertson–Walker space, Differential Geom. Appl. 27 (2009) 146–156.

  • $\begingroup$ Thank you for your interesting answer. $\endgroup$ – Semsem Mar 15 '15 at 4:07

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