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- Newsgroups: sci.physics
- Path: sparky!uunet!europa.asd.contel.com!darwin.sura.net!spool.mu.edu!agate!linus!linus.mitre.org!faron!wdh
- From: wdh@faron.mitre.org (Dale Hall)
- Subject: Re: Can space-time intersect itself?
- Message-ID: <1992Dec14.211427.9559@linus.mitre.org>
- Followup-To: sci.physics
- Summary: be careful with embedding theorems
- Keywords: embedding, isometry
- Sender: Dale Hall
- Nntp-Posting-Host: faron.mitre.org
- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- References: <1992Dec2.232639.13222@galois.mit.edu> <398@moene.indiv.nluug.nl>
- Distribution: net
- Date: Mon, 14 Dec 1992 21:14:27 GMT
- Lines: 48
-
- In article <398@moene.indiv.nluug.nl> toon@moene.indiv.nluug.nl (Toon
- Moene) writes:
- >In article <1992Dec2.232639.13222@galois.mit.edu> jbaez@riesz.mit.edu
- >(John C. Baez) writes:
- >> In article <1992Dec2.014924.8076@netcom.com> raicu@netcom.com (Eugen
- >Raicu) writes:
- >> >In principle, can there be any mass distribution that would result in a
- >> >space-time which intersects itself?
- >>
- >> Not as far as I can tell, if I understand the question. The point is
- >> that spacetime is not embedded in some higher-dimensional space so it
- >> can't "bump into itself". This is precisely why it's nice to have an
- >> intrinsic definition of manifolds, by the way!
- >
- >Now that's interesting - I'm sure I remember from my Differential Geometry
- >course that ANY manifold can be embedded in Rn. A five minutes worth of
- >flapping through the pages of my DG book* tells me that I have to modify
- >this to: ANY manifold that has a COUNTABLE base (so-called seperable
- >manifold).
-
- This is of course true, but for embedding results you require
- a certain number of "extra dimensions".
-
- For plain-old smooth embeddings a manifold of dimension n can
- be embedded in R^(2n+1). (Actually if n>2 it's R^(2n) by
- something called the Whitney trick. But then, there are only
- so many 2-dimensional manifolds [besides the 2-sphere, it's
- just connected sums of tori & projective planes], and they all
- embed into R^4, so R^(2n) works for every compact n-manifold.)
-
- However, if you're married to the notion of a particular
- Riemannian metric, then the best you can do is R^(n^2+10n+3),
- according to Gromov ("Partial Differential Relations").
- Notice the quadratic in the dimension: it tells you that to
- have all the luxury of home, it'll cost you plenty. Who
- woulda thunka needing R^42 for a dumbole M^3? If the version
- for Lorentz metrics were true, you would be worrying about
- R^59 to hold spacetime. Damb.
-
- The fact that manifolds can be embedded in Euclidean space
- appears (to me) to be over-used and under-understood by people
- in general. While embedding can be useful and fun, it does
- carry a certain price, in this example, in terms of additional
- coordinates that may appear. Whether this additional load of
- scientific baggage constitutes a good investment is not an
- obvious question.
-
-
-