My question is:
Can we judge a manifold that can admit a (p,q) metric?
I only know the case that the existen of a lorentz metric is equivalent to Euler Character is zero
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1$\begingroup$ @Siqi: To be slightly more precise, you should say that a closed manifold admits a Lorentz metric if and only if its Euler characteristic vanishes. Any noncompact manifold admits a Lorentz metric (and perhaps this is the type of manifold more interesting for a model of the universe). For existence of $(p,q)$-metrics on closed manifolds, as pointed out below, the situation is more complicated. Perhaps the simplest case is that of spheres: quite a bit is known, mostly due to Steenrod (see e.g., arxiv.org/pdf/1008.4986.pdf, Thm 1.127, p.40, for details and references). $\endgroup$– Renato G. BettiolCommented Sep 3, 2012 at 14:38
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2 Answers
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The criterion for existence of a $(p,q)$ metric is (assuming $p+q=dim X$) that the tangent bundle splits as a direct sum of two subbundles of dimensions $p$ and $q$.
EDIT : I doubt there is an easy algebraic topology criterion in general, as characteristic classes [EDIT after Lennart Meier's comment: other than Euler class, which cannot help if $p,q>1$] are stable invariants of vector bundles. At least, one has necessary conditions, such as : the total Stiefel-Whitney and Ponttryagin (Chern of complexified bundle) class of $TX$ is the product of two (inhomogeneous) classes of degrees at most $p$ and $q$. Perhaps real $K$-theory (and its operations) gives better necessary conditions (although still not sufficient in general). But it's not easy to compute.
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$\begingroup$ The Euler class isn't stable; in fact, it just detects whether a trivial line bundle splits off. $\endgroup$ Commented Jul 29, 2012 at 11:32
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$\begingroup$ You're right, I phrased it badly. What I meant is that for $p$ and $q$ at least 2, Euler class can't help, and the rest is stable. $\endgroup$– BS.Commented Jul 30, 2012 at 16:02
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I believe that you're a bit mistaken about the final claim. The correct statement should be that:
There exists a "time-orientable lorentz" (i.e. a lorentzian metric with a nowhere vanishing timelike vector field) metric if and only if there exists a nowhere vanishing vector field (which happens if and only if the euler characteristic is zero).
The proof of this is easy:
(the only if direction is trivial) Suppose there is a nowhere vanishing vector field $X$. Pick any Riemannian metric $g$ and let $\omega$ be the dual $1$-form with $X$ with respect to $g$. Then, defining $$ \tilde g = - 2\omega\cdot\omega + g, $$ this is a "time orientable lorentz metric" as desired.
The same proof shows that
There is a $(p,q)$-metric with $p$ linearly independent non-vanishing vector fields $X_i$ with $\tilde g(X_i,X_i) < 0$ if and only if there are $p$ linearly independent nonvanishing vector fields on the manifold.
This is something which can be detected by Euler classes, I think.
I agree that this second conclusion is a bit unsatisfying, because it is natural to restrict to time orientable Lorentz metrics for physical reasons, but here it is not clear that it is a natural restriction.
answered Jul 27, 2012 at 16:26
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$\begingroup$ Can these statements be generalized to the existence of a splitting of the tangent bundle into a direct sum of rank $p$ and $q$ sub-bundles? $\endgroup$ Commented Jul 27, 2012 at 16:41
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1$\begingroup$ I think that its fine, just take the metric and subtract off twice the two tensor which is the metric on the $p$-subbundle and zero otherwise. $\endgroup$ Commented Jul 27, 2012 at 17:40
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$\begingroup$ Yes, exactly. I was going to suggest that you add this to your answer but it appears that BS has already posted the same thing. $\endgroup$ Commented Jul 27, 2012 at 18:00