Compact n-manifolds as Compactifications of R^n

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  • #1
Hi, I am trying to show that RP^n is the
compactification of R^n. I have some , but not all I

I have also heard the claim that every compact n-manifold is
a compactification of R^n, but I cannot find a good general
argument . I can see, e.g., for n=2, we can construct a manifold
by using an n-gon, and identifying some edges. Then the interior
seems clearly to be homeo. to R^n.

In some other cases, e.g., S^n, this seems easy to show
using the stereo projection, and some work with the topology
defined on the (Alexandroff) 1-pt. compactification ( use
as open sets in S^n, all open sets in C --or in R^n --together
with complements of compact subsets K of C.) Then R^n is open
in S^n, and its closure is necessarily S^n itself, i.e., R^n is
embedded densely in the compact space S^n ( S^n is
compact by construction.)

For the sake of practicing my pointset topology and some
basic diff. geometry, I tried to show that RP^n is also a
compactification of R^n, before trying a more general argument:

i) I know RP^n is compact, as the quotient of the
compact space S^n by a continuous (quotient) map.

ii) I think I can show R^n is dense in RP^n, since
the identification of S^n is made only in the
boundary of S^n, and the interior of S^n is homeo.
to R^n, i.e., missing some boundary points. The
interior of the disk (homeo. to R^n ) is dense in the
disk. Is the image
(under the quotient map) also dense in RP^n?

Do I need something else?

Let me be more specific:
The quotient map
q:S^n-->S^n/~ is continuous, by construction. Then
the continuous image of S^n is compact in the quotient
topology. Right?

Maybe more rigorously, we should start by giving
S^n the subspace topology of R^(n+1). Then, as a closed
subset/subspace of a compact space, it is
itself compact.

Now: How you have R^n in RP^n (i.e. which embedding) and
why it is
dense in the quotient topology (without hand waving)?

Let's see: if we consider RP^n as the set of
points in S^n/~ ( x~y iff x=ty ; t non-zero ),
then R^n would be everything except for the equator
D^n_ /\D_n+ , i.e., in a "standard" coordinate system
X1,..,Xn, R^n={x=(x1,...,xn);||x||=1 and xn>0 }

Then R^n is a saturated open subset of S^n , under ~
(i.e., the map q(x)=x/~ )so that (since quotient maps
send saturated open sets to open sets) the image
p(R^n) is open in S^n/~

But Cl(R^n)={x=(x1,..,xn);||x||=1, xn>=0 }
is a saturated closed set re q(x)=x/~. So
q(Cl(R^n)) is closed in RP^n.

Now we need to show that Cl(q(R^n))=RP^n

But for any point z in q(Cl(R^n)), where
xn=0 for z, we have:

q^-1( q(Cl(R^n))-{z} ) is not closed in
(S^n, subspace) (though I don't have a good
argument to this effect)

Then q((Cl(R^n)) =RP^n is the closure of
(the embedded image of ) R^n in RP^n, and
so R^n is dense in RP^n.

Thanks For any Comments.

Answers and Replies

  • #2
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This seems quite good. However, compactifications are usually required to be Hausdorff. Non-Hausdorff compactifications are quite useless in my opinion. So it would be good if you also showed that RP^n is Hausdorff. But I guess you already showed this by showing that RP^n is a manifold?
  • #3
Thanks, Micromass, for going through that messy post:

To show Hausdorff, we could take, given any x,y in RP^n , q-saturated

'hoods (neighborhoods) U_x, U_y in S^n (quotient maps send saturated open

sets to open sets ), of arc-length less than Pi each, and such that the total length

of the union is less than Pi , so that there are no antipodal pairs in U_x\/U_y.

Then , for q(x)=x/`~ , the natural projection, O_x=q(U_x) and O_y= q(U_y) are open

in RP^n. And I claim they are disjoint too: if there was a [z] in the overlap

then either:

i) there is a z in S^1 common to both U_x, U_y. Not possible by construction

ii) There is a z in U_x , and its antipode -z in U_y . Also not possible by construction.

Still, I am looking for a general argument for why a compact n-manifold --and therefore

Hausdorff-- is a compactification of R^n
  • #4
Science Advisor
Gold Member
The covering map of the sphere onto projective space is continuous and onto. Since the sphere is compact, this implies that projective space is also compact.

A fundamental domain of the covering map is any closed hemisphere. A closed hemisphere is homeomorphic to an n-ball.

I think that your problem is more accurately stated by saying that a compact manifold is always obtained from a close n-ball via identifications on its boundary n-1 sphere. I got a little confused when you said that is was a compactification of Euclidean space.

This theorem, if it is true, I would think could be proved by following the exponential mapping at some point out along geodesics until you obtain a maximal n-ball i.e. an open n ball in the tangent space that is mapped diffeomorphically into the manifold and which can not be enlarged and still be a diffeomorphism. For instance for the sphere you get a ball around the origin in the tangent space whose boundary is identified to a single point.
Last edited:
  • #5
Lavinia said, in part:

"I think that your problem is more accurately stated by saying that a compact manifold is always obtained from a close n-ball via identifications on its boundary n-1 sphere. I got a little confused when you said that is was a compactification of Euclidean space."

By a compactification K of Euclidean space R^n, I mean that X is a compact manifold,
and that R^n is densely-embedded in K, i.e., that there is an embedding e: R^n-->K
with Cl(e(R^n))=K, with Cl the closure operator.

And an alternative, more topological argument, may be possible if we just consider
compact triangulatable ( I don't know if this is an actual word) manifolds, and then we
can check a finite number of triangles/simplices.

But, yes, if we could show that every compact n-manifold can be obtained by
identification, that would do it, but I did not know how to generalize the process
of identifying the edges of fundamental surfaces, nor how to prove that every
compact manifold K could be obtained that way.

And your idea of the exponential map sounds good, thanks, I will try it out.

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