What I'm asking for is how you visualize this.

Definition. Define an equivalence relation on $S^n$ by $x \sim \pm x$ for all $x \in S^n$. Then the quotient space $S^n/ \sim$ is called the (real) projective $n$-space and is denoted $P^n$.

Lecturer gave a construction of $P^2$ and showed it was like gluing a Möbius band to the edges of a disk.

However, he was unable to say what $P^3$ would be constructed in general and then he said that there were big problems in (real) projective spaces and he said it is hard to visualize.

Intuitively speaking, I feel like $P^3$ would be a Möbius bottle and you are gluing that onto a solid ball. Well gluing that onto the surface of a solid ball.

Anyway, intuitive way to visualize this space in higher dimensions?

Please can you describe it in terms of topological objects. In a way I don't care much about realizing the gluing. Well, ideally would want to know what it looks like say it looks like a Möbius strip with a disk. Even if the gluing itself you need to construct something that intersects itself three times at a point.

  • $\begingroup$ Do you know how the projective plane works in projective or algebraic geometry? $\endgroup$ – Henning Makholm Oct 28 '11 at 23:59
  • $\begingroup$ No. I haven't came across projective planes. I also don't know any algebraic geometry. Well, I hoping someone gives what projective n-space breaks down into in terms of topological objects i.e. what you need to glue together to get $P^n$ $\endgroup$ – simplicity Oct 29 '11 at 0:02

Projective $2$-space, as your lecturer points out, can be thought of as a disk with a twisted band attached, to which another disk is attached along the boundary. This generalizes to higher dimensions using the language of handle theory. In general, in $n$ dimensions, a $k$-handle is homeomorphic to $D^k\times D^{n-k}$ (which is itself homeomorphic to $D^n$, where $D^\ell$ denotes a closed $\ell$-dimensional disk. The attaching region of a $k$-handle is $(\partial D^k)\times D^{n-k}= S^{k-1}\times D^{n-k}$. You can build up $n$-manifolds using handles by first taking a union of $0$-handles, then attaching $1$-handles to the boundary of the $0$-handles along the attaching regions of the $1$-handles. In general, one you have all the $\leq k$-handles, you attach $(k+1)$-handles to the boundary of what you have so far, along the attaching regions of the $(k+1)$-handles. Now in $2$ dimensions, a $0$-handle is a disk, a $1$-handle is a rectangular strip whose attaching region is two opposite ends of the strip, and a $2$-handle is a disk with its entire boundary as an attaching region. Then the projective plane can be written as a single $0$-handle, to which a single $1$-handle attaches (with a twist), to which a single $2$-handle attaches. In fact, every surface has a handle decomposition. Indeed any smooth manifold has such a decomposition.

Now, for projective $n$-space $\mathbb{RP}^n$, you can show that there is a handle decomposition with a single $k$-handle for every $0\leq k\leq n$. One way to see this is to construct a handle decomposition for the sphere with two of every index of handle, such that the quotient by $x\sim -x$ identifies the handles in pairs.

Specifically for $\mathbb{RP}^3$, first you can think of this a quotient of $D^3$ by identifying antipodal points on $\partial D^3=S^2$. That's not a bad way to visualize it, but to get to the handle decomposition, put a little $0$-handle (ball) in the middle of $D^3$. Then extend a $1$-handle (solid tube) out to $\partial D^3$ along the $x$-axis, which then comes out the other side along the negative $x$-axis to attach again to the $0$-handle. Now attach a $2$-handle (penny) to fill in a neighborhood of the $xy$-plane. (You have to convince yourself that this really is homeomorphic to $D^2\times I$.) So now fill in the rest with a $3$-handle (3-ball) which joins the top part of $D^3$ with the bottom part using the boundary identifications. I guess you would say that the union of $0$ and $1$-handle is a solid Möbius bottle, but you also need to glue on a $2$-handle before you complete it by adding a ball.

  • $\begingroup$ That's a very good answer. Thanks for that. I'm having a hard time seeing how you would prove that projective n-space has a only one decomposition into a k-handle. But, yeah thanks for that. $\endgroup$ – simplicity Oct 29 '11 at 0:19
  • $\begingroup$ The proof works the same. Find a handle decomposition of $S^n$ which respects the $\mathbb Z_2$-action with two $k$-handles of every index $k$. Then only one $k$-handle will survive to the quotient. $\endgroup$ – Cheerful Parsnip Oct 29 '11 at 0:21

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