>--- Howard Iseri <hiseri@...> wrote:
> > > Hi Minh,
> > >
> > > The Euclidean plane can be thought of as a
> > > Riemannian manifold, so not all
> > > Riemannian manifolds are Smarandache geometries.
> > > That's true.
> >
> >Therefore a Smarandache manifold is a
> generalization
> >of the Riemannian manifold. Nice.
> >How can we prove that on a Riemannian manifold a
> axiom
> >is smarandachely denied?
>
>
> My s-manifolds are not quite generalizations of
> Riemannian manifolds.

I was referring to the general definition: a manifold that supports a
smarandache geometry is a smarandache manifold.

1) Can we say that an orbifold that support a smarandache geometry is
a smarandache orbifold?

2) I don't know how to relate/compare the smoothness (riemann
manifold) with smarandachely denying?

> They
> are generalizations of flat Riemannian 2-manifolds.
> S-denying an axiom
> would be similar to what I did in my book, but
> Riemannian manifolds are
> much harder to work with.
>
>
> > >From internet I found the definition of an
> orbifold:
> >"An orbifold is Hausdorff space X which is locally
> >homeomorfic at each point x from X to R^n /Gx,
> where
> >Gx is some finite group, perhaps different for x
> >different. The set of x in X such that Gx
> different
> >from {1} (in other words, points no neighborhood of
> >which is homeomorphic to R^n) is called the
> singular
> >locus of orbifold X."
> >
> >The definition looks contradictory: in the first
> part
> >it is said that X is locally homeomorfic at each
> >point, and in te second part it is said that there
> are
> >points that are not homeomorfic... ?
>
> They mean homeomorphic everywhere except at some
> special points.
>
>
> >
> > > Riemannian manifolds must be smooth,
> >
> >Can this make any connection with the smarandachely
> >denied axioms?
> >
> > >and there is no
> > > such restriction on a
> > > Smarandache geometry, so there are a lot of
> > > S-geometries that are not
> > > Riemannian manifolds. Almost all of the
> s-manifolds
> > > in my book, for
> > > example, are not Riemannian manifolds.
> >
> >I understand it.
> >
> > >
> > > Howard.
> > >
> > > At 08:58 AM 2/18/03 -0800, you wrote:
> > >
> > > >--- Howard Iseri <hiseri@...>
> wrote:
> > > > > Hi Minh,
> > > > >
> > > > > At 08:53 PM 2/15/03 -0800, you wrote:
> > > > > >sorry for coming back that late...
> > > > > >
> > > > > >--- Howard Iseri <hiseri@...>
> > > wrote:
> > > > > > > Basically what I know is this. A
> Riemannian
> > > > > manifold
> > > > > > > is a manifold that has
> > > > > > > the same local geometry as an elliptic,
> > > > > Euclidean,
> > > > > > > or hyperbolic geometry
> > > > > > > (in some particular dimension).
> > > > > >
> > > > > >does it mean that a riemann manifold should
> > > have at
> > > > > >least one of them (either elliptic, or
> > > euclidean,
> > > > > or
> > > > > >hiperbolic geometry)?
> > > > >
> > > > >
> > > > > A Riemannian manifold has a smoothly varying
> > > > > curvature. If a particular
> > > > > point has negative curvature, then the
> region
> > > around
> > > > > it will behave
> > > > > similarly to a hyperbolic space. In the
> > > s-manifolds
> > > > > in my book, the
> > > > > curvature is zero everywhere (Euclidean),
> except
> > > at
> > > > > the cone points where
> > > > > the curvature is infinite, but in a positive
> or
> > > > > negative way.
> > > >
> > > >I understand that a Riemannian manifold can
> have
> > > local
> > > >geometry which are all only elliptic? Or all
> only
> > > >hyperbolic? Or all only euclidean?
> > > >Therefore, not all Riemannian manifolds support
> a
> > > >Smarandache geometry. Am I wrong?
> > > >What about the opposite: all smarandache
> > > geonmetries
> > > >are riemannian manifolds?
> > > >
> > > >In a small dictionary of math it is said that a
> > > >riemannian manifold is a manifold with a metric
> > > >tensor.
> > > >
> > > > > > >An orbifold
> > > > > > > generalizes this by allowing
> > > > > > > local geometries that are like one of
> these
> > > mod
> > > > > some
> > > > > > > symmetry group.
> > > > > >
> > > > > >i don't understand it well: how is a say
> local
> > > > > >elliptic geometry modulo a symmetry group?
> > > > >
> > > > >
> > > > > For example, in the plane, you can have the
> > > group
> > > > > generated by a 120 degree
> > > > > rotation about the origin. If you define two
> > > points
> > > > > to be equivalent if you
> > > > > can get from one to the other by a 120
> degree
> > > > > rotation about the origin,
> > > > > then this is roughly what is meant by modulo
> > > this
> > > > > group. The effect of this
> > > > > particular example is that the origin now
> only
> > > has
> > > > > 120 degrees around it,
> > > > > and it has a geometry essentially the same
> as
> > > that
> > > > > of a cone with 120
> > > > > degrees around the vertex. In my book, the
> cone
> > > > > points always had 300
> > > > > degrees or 420 degrees around them. In an
> > > orbifold
> > > > > the number of degrees
> > > > > would have to evenly divide into 360 (180,
> 120,
> > > 90,
> > > > > 72, etc.)
> > > > >
> > > > >
> > > > > >i guess / i feel the smarandache geometries
> can
> > > be
> > > > > put
> > > > > >in orbifolds somehow...
> > > > >
> > > > >
> > > > > There is more variation in the curvature of
> an
> > > > > orbifold than a riemannian
> > > > > manifold or an s-manifold, so yes, most of
> these
> > > > > will be smarandache
> > > > > geometries. The few of Hilbert's axioms that
> > > were
> > > > > not S-denied in my book,
> > > > > however, will problably not be S-denied in
> an
> > > > > orbifold either. I'm pretty
> > > > > sure about that.
> > > >
> > > >What can be found if in your book models for a
> > > >smarandache geometry one replace the stright
> lines