11
$\begingroup$

I need a reference for a complete proof of the below theorem:

Let $RP^n$ be $n$-dimensional real projective space. Then $RP^n$ is a compact, connected manifold. (Consider the standard topology over $R$)

I'd appreciate it If you refer me to a reference with a non-complicated proof. Thanks in advance.

$\endgroup$

3 Answers 3

13
$\begingroup$

William Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry Rev 2nd ed. At the beginning of ch3. You can find a detailed proof that $\mathbb RP^n$ admits a differentiable manifold structure over the quotient topology induced by the natural projection $\pi:\mathbb R^{n+1}\to\mathbb RP^n$.

To see it is compact and connected, this is really a topological matter. Restrict the domain of $\pi$ to $S^n$. According to the definition of quotient topology, $\pi:\mathbb R^{n+1}\to\mathbb RP^n$ is a surjective continuous map. So $\pi|_{S^n}:S^n\to\mathbb RP^n$ with $S^n$ equipped with subspace topology is also a surjective continuous map. Surjectivity of $\pi|_{S^n}$ comes from the fact that $\forall [x_0,x_1,\dots,x_n]\in\mathbb RP^n$ has in its preimage a point $(x_0,\dots,x_n)/(\sum_{i=0}^nx_i^2)^{1/2}\in S^n$. Therefore. it preserves compactness and connectedness. So $\mathbb RP^n$ is compact and connected since $S^n$ is.

Appendix:

  • $S^n$ is compact, since it is closed and bounded in $\mathbb R^{n+1}$ by Heine-Borel theorem.
  • $S^n$ is path connected: any two points can be connected by an arc on a great circle. So it is also connected. (path connectedness implies connectedness).
  • All the topological facts can be picked up in James Munkres's Topology 2nd ed. Or you can find a quick review in Abraham, Marsden, Ratiu, Manifolds, Tensor Analysis and Applications (2007 draft recommended).
$\endgroup$
2
  • $\begingroup$ In the spirit of thoroughness, you might want to note that $\pi|_{S^n}$ is surjective when showing the compactness of $\mathbb{RP}^n$. $\endgroup$
    – Kyle
    Commented Aug 28, 2014 at 14:56
  • $\begingroup$ @squirrel Noted, thx. $\endgroup$
    – Troy Woo
    Commented Aug 28, 2014 at 15:10
9
$\begingroup$

$RP^n$ can be obtained as the quotient of $S^n$ by the symmetry $x\cong -x$. The spheres $S^n$ are compact and connected, hence the quotient is as well.

$\endgroup$
4
  • $\begingroup$ Should we need to verify that $\mathbb{RP}^n$ is Hausdorff too? $\endgroup$
    – Ho-Oh
    Commented Nov 6, 2022 at 2:36
  • $\begingroup$ The quotient inherits the compactness because the relation is and open relation, is not true in general $\endgroup$ Commented Feb 15 at 20:18
  • $\begingroup$ I dont understand your comment @ManuelBonanno. The image of a compact space under a continuous map is compact. This shows that a quotient space of a compact space is compact. Nothing else is needed $\endgroup$
    – Thomas Rot
    Commented Feb 16 at 7:42
  • $\begingroup$ You're right, i'm just confusing $\endgroup$ Commented Feb 16 at 8:59
0
$\begingroup$

Just look at a surjective map $S^n \to RP^n$. (which is locally a homeomorphism; for the manifold part)

Try to figure out, why all of your concerns follow by the existence of such a map.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .