Re: Simple queries on intersection of hypersurfaces

In article <20090714.025525@xxxxxxxx>,
Rob Johnson <rob@xxxxxxxxxxxxxx> wrote:
In article <20090714013540.A16805@xxxxxxxxxxxxxxx>,
William Elliot <marsh@xxxxxxxxxxxxxxxx> wrote:
On Mon, 13 Jul 2009, stargene wrote:

Our universe is sometimes conjectured to be a four-dimensional hyper-
surface. Imagining that it may be a four-dimensional surface of a
hyper- sphere, I am trying to get a rough picture of the locus of the
intersection of such a universe with another four-dimensional universe
of the same kind, ie: which is also the surface of a hypersphere.

Would the zone of intersection be a 4-1 dimensional object by definition
(having three dimensions), and yet would it also be a surface?

A n-sphere with radius r, centered at the origin is the set
{ (x1,.. x_n) | x1^2 +..+ (x_n)^2 = r^2 }

When n = 2 or 3, you'll see that a 2-sphere is a circle and
a 3-sphere is a sphere.

A non-tangent intersection of two spheres is a circle.
A non-tangent intersection of two 4-spheres is a sphere.

In general, a non-tangent intersection of two (n+1)-spheres
is a n-sphere.

In general, the "usual" intersection of two subspaces results in a
subspace whose codimension is the sum of the codimensions of the
intersecting subspaces.

For example, in R^n, the intersection of two n-1 hyperplanes (both
codimension 1) is an n-2 hyperplane (codimension 2). In tangential
cases, this fails; for example, parallel hyperplanes do not intersect
and the intersection of two coincident hyperplanes is either
hyperplane.

This same rule holds in a restricted sense for submanifolds.

For some examples in R^3, the usual intersection of two spheres
(codimension 1) is a circle (codimension 1). The intersection of a

A circle has codimension 2, which is good since 1+1 = 2.

plane (codimension 1) and a circle (codimension 2) is two points
(codimension 3).

Rob Johnson <rob@xxxxxxxxxxxxxx>
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