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It is possible that on a sphere $S^n$ there is a natural Riemannian metric in $R^(n+1)$. But it is not always possible for pseudo Riemann metric since the sum of two symmetric matrix which are not positive definite but may have rank different from the two matrix. So I wonder what is the sufficient and necessary condition for the dimension of a sphere which can be endowed with a Lorentz metric.

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Manifolds have Lorentz metrics if and only if they have a section of the corresponding projective unit tangent bundle. If you want details, please post to the math.stackexchange website listed in the FAQ as this is more of an upper-level undergraduate mathematics issue. –  Ryan Budney Nov 4 '10 at 18:52
    
If you don't want to post your question on math.stackexchange, take a look in the nearest manifolds textbook at the proof that abstract manifolds admit Riemann metrics (i.e. avoid proofs that use embeddings in euclidean space). This is a paracompactness/bump function/partition of unity argument. Now consider how you would perturb that proof for Lorentz metrics. –  Ryan Budney Nov 4 '10 at 21:16
    
@Ryan: should one understand that with a partition of unity argument you can show the existence of Lorenz metrics on every (paracompact) manifold? I have serious doubts about that... –  Andrei Moroianu Nov 26 '10 at 18:00
    
@Andrei: it appears you did not read what I said in my first comment "Manifolds have Lorentz metrics if and only if they have a section of the corresponding projective unit tangent bundle". –  Ryan Budney Nov 26 '10 at 18:43
    
Andrei, I think Ryan is suggesting that when you try to adapt the proof for a Riemannian metric to a Lorentzian one, you run into anecessary condition for the Lorentzian metric to exist. Then you can check that the necessary condition is also sufficient. –  Deane Yang Nov 26 '10 at 18:44
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2 Answers 2

up vote 16 down vote accepted

A compact simply connected manifold carriez a Lorenz metric iff its Euler characteristic vanishes.

Proof: If $\chi(M)=0$, $M$ carries a nowhere vanishing vector field $X$. Pick up a Riemannian metric $g$ on $M$ (using a partition of unity argument) and denote by $\eta$ the 1-form dual to $X$: $\eta(Y):=g(X,Y)$ for all $Y\in TM$. Then
$$g-2\frac{\eta\otimes\eta}{g(X,X)}$$ is a Lorenz metric.

Conversely, if $M$ has a Lorenz metric $h$ of signature $(n-1,1)$, pick again a Riemannian metric $g$ and consider the symmetric endomorphism $A$ of $TM$ defined by $h(.,.)=g(A.,.)$. The eigenspaces of $A$ corresponding to the unique negative eigenvalue define a line sub-bundle of $TM$ which is trivial if $M$ is simply connected, so $\chi(M)=0$.

Therefore, the answer to your question is: $S^n$ carries a Lorenz metric iff $n$ is odd.

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Does this theorem have a name? Thanks! –  user34669 Apr 22 at 12:59
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I would just like to point out that a slightly more general version of the statement in Moroianu's answer holds:

A compact orientable manifold $M$ admits a Lorentzian metric if and only if $\chi(M)=0$.

Clearly, simply-connected manifolds are orientable, hence the above statement is more general. But, e.g., tori clearly admit product Lorentzian metrics.

For the proof of the above statement, recall $M$ admits a semi-Riemannian metric of index $k$ if and only if the tangent bundle $TM$ admits a sub-bundle, or distribution, of rank $k$. Thus, $M$ admits a Lorentzian metric if and only if it admits a line bundle (which is also equivalent to $M$ admitting a vector field that never vanishes). The obstruction to the existence of a line bundle on $M$ is given by the Euler class of $M$ (see e.g. Davis and Kirk "Lecture notes in Algebraic Topology"). By the Gauss-Bonnet-Chern Theorem, the Euler characteristic is the Euler class evaluated on the fundamental class of $M$, so that, for compact orientable manifolds, the obstruction is precisely $\chi(M)$.

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