From: abergman@Princeton.EDU (Aaron J. Bergman)
Subject: Re: Who's Afraid of Lorentzian Metrics?
Date: Mon, 01 Jan 2001 14:28:55 GMT
Newsgroups: sci.physics.research,sci.math.research
Summary: Which 4-manifolds have Lorentzian metrics?

In article <92lcnn$nf2$1@nnrp1.deja.com>, squark@my-deja.com wrote:
>What 4-manifolds have Lorentzian* metrics? Intuitively, it seems as
>though not all of them - can you put a one on a sphere, for instance?
>Maybe I'm wrong, though...

You can put a Lorentzian metric on a manifold iff it has a
nondegeneerate direction field, ie at every point there is set of
vectors {v,-v}.

Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
==============================================================================

From: galatius+usenet@imf.au.dk (S¯ren Galatius Smith)
Subject: Re: Who's Afraid of Lorentzian Metrics?
Date: Mon, 01 Jan 2001 14:28:55 GMT
Newsgroups: sci.physics.research,sci.math.research

squark@my-deja.com writes:

> What 4-manifolds have Lorentzian* metrics? Intuitively, it seems as
> though not all of them - can you put a one on a sphere, for instance?
> Maybe I'm wrong, though...

You cannot have one on S^4.

In general, if E is a smooth n+m-dimensional vectorfield on the smooth
manifold M, you can have a symmentric nondegenerate bilinear form of
signature (n,m) on E iff E splits as a direct sum E=E1 + E2 of n-
resp. m-dimensional bundles over M (given such a form w, one may
choose any positively definite inner product <-,-> on E and write
w(x,y) = <x,Ay> for a symmetric (wrt <-,->) section A of
Hom(E,E). Then E splits as the sum of positive eigenspaces for A plus
the sum of negative eigenspaces of A.)

If n=1, we ask if the bundle has a linebundle as direct summand. If in
addition M is simply-connected this will be the same as having a
non-zero section. And the tangent vectorfield on S^4 does not have
such a thing (hairy ball theorem).

Søren

--
 Søren Galatius Smith       http://www.imf.au.dk/~galatius/
==============================================================================

From: Steve Carlip <carlip@dirac.ucdavis.edu>
Subject: Re: Who's Afraid of Lorentzian Metrics?
Date: Mon, 1 Jan 2001 00:54:12 +0000 (UTC)
Newsgroups: sci.physics.research

squark@my-deja.com wrote:

> What 4-manifolds have Lorentzian* metrics? 

Any paracompact 4-manifold admits a Riemannian metric.  
It will admit a Lorentzian metric iff it admits a direction
field, that is, a continuous choice of an unordered pair (n,-n) 
of nowhere vanishing vectors.  The proof is easy.  If g is a
Lorentzian metric, choose an arbitrary auxiliary Riemannian
metric h, and consider the eigenvalue equation
  g_{ab}n^b = \lambda h_{ab} n^b
This will have one negative eigenvalue, and the corresponding
eigenvector can be normalized by   g_{ab}n^an^b = -1 to give
a direction field.  Conversely, if the manifold admits a line
element field (-n,n), we can choose an auxiliary Riemannian
metric h, normalized to h_{ab}n^an^b = 1; then
  g_{ab} = h_{ab} - 2(h_{ac}n^c)(h_{bd}n^d)
is a Lorentzian metric.

A time orientation on M amounts to a continuous choice of sign
of n.  So M will admit a time-orientable Lorentzian metric iff
it admits a nowhere vanishing vector field.

So, when does a manifold admit a nowhere vanishing vector field?  
If M is noncompact, always: a typical vector field on M will have 
zeros, but these can be ``pushed off to infinity'' by a suitable sequence
of diffeomorphisms.  If M is closed, then by the Poincare-Hopf 
index theorem, it admits a  nowhere vanishing vector field iff its
Euler number is zero.  For manifolds with boundary, the results
are more complicated (if, for example, you demand that the
boundaries by spacelike), and I don't remember them off hand;
you might look at R. Sorkin, Int. J. Theor. Phys. 25 (1986) 877,
which I think has a discussion of this kind of issue.

Steve Carlip