Re: torus?

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

David Bowman,

Thanks for the long post, it helps to clear up some issues I had with what
you were writing earlier about the 3-torus, which I know little about.

I think the problem I had was in not taking into account the difference
between topology and local geometry; Which is to say I had been interpreting
T^2 = S1XS1 as the surface of the donut.  And it apparently may be or it may
not be as you pointed out.

for the interest of others, I'm quoting a conversation from another list
which I found today while trying to research the question; it says what D
Bowman says, but in different language.



> Jacques Distler writes:
> In article <38017e7e.42110866@news.gte.net>, dmorgan@suenet.qu.net wrote:

> ...the standard metric on the torus has
> everywhere-vanishing Riemann curvature. That is what I (and general
> relativists) mean by "flat".

> > Certainly it has both intrinsic and extrinsic curvature,

> No. The curvature you are talking about is an artifact of the embedding
> in R^3.  The intrinsic (Riemann) curvature vanishes.

I can see how this sort of disagreement arises, since words
like "standard" and "intrinsic" can have various
connotations. Let me take a crack at confusing the issue.

The torus, as a manifold, has no geometry, that is, no
curvature defined. It is just a topological space with a
differentiable structure. You can put infinitely many
metrics on the torus to give it some "shape". One metric,
Distler's "standard metric" is flat (although there are
infinitely many of those too!). Another pretty standard
metric, apparently the one described by dmorgan, is
obtained by embedding the torus in a flat R^3: the surface
of a doughnut.  This metric is most definitely NOT flat -
and here I DO mean the intrinsic Riemann curvature. (So,
geodesics on the doughnut DO focus.) There is negative and
positive (intrinsic, Riemann) curvature on the doughnut.
Of course, the scalar curvature averages out to zero, in
accord with Gauss-Bonnet.   The "standard metric" cannot be
induced by embedding in flat R^3, but it can be induced by
embedding in flat R^4.  Of course, in each of these
examples we have the same torus - as a manifold - but a
different torus geometrically. One sometimes says that the
two tori are diffeomorphic, but not isometric.

In the midst of a debate about which manifolds are
diffeomorphic, it is always worth mentioning that there are
infinitely many manifolds that are topologically R^4 but
still fail to be diffeomorphic. Seems to me that physics on
these different R^4's should be different...but how?

Charles Torre