How should I visualise RP^n?

Question

So I did some algebraic topology at university, including homotopy theory and basic simplicial homology, as well as some differential geometry; and now I'm coming back to the subject for fun via Hatcher's textbook. A problem I had in the past and still have now is how to understand projective space RP^n - I just can't visualise it or think about it in any concrete way. Any ideas?

edit: Essentially RP^n is always the example I don't understand. So when for example Hatcher says that S^n is of a CW complex with two cells e^0 and e^n, I can picture what's going on because I know what spheres look like and I can imagine the attachment in some concrete-ish way. But when he says "we see that RP^n is obtained from RP^{n-1} by attaching an n-cell [...] it follows by induction on n that RP^n has a cell complex structure e^0 U e^1 U ... e^n" - my brain just gives up.

Answer 1

You can "visualize" the cell structure on $\mathbb{R}P^n$ rather explicitly as follows. The set of tuples $(x_0, ... x_n) \in \mathbb{R}^{n+1}$, not all equal to zero, under the equivalence relation where we identify two tuples that differ by multiplication by a nonzero real number, can be broken up into pieces depending on which of the $x_i$ are equal to zero.

• If $x_0 \neq 0$, the corresponding points can be written $(1, x_1, ... x_n)$, and they form a subspace isomorphic to $\mathbb{R}^n$.

• If $x_0 = 0$ and $x_1 \neq 0$, the corresponding points can be written $(0, 1, x_2, ... x_n)$, and they form a subspace isomorphic to $\mathbb{R}^{n-1}$.

And so forth. One way to say this is that the tuples where $x_0 \neq 0$ form an affine slice or affine cover of $\mathbb{R}P^n$ and the tuples where $x_0 = 0$ constitute the "points at infinity," which themselves form a copy of $\mathbb{R}P^{n-1}$.

Answer 2

The set of lines passing through the origin in $R^n$. From this you can see that it's equivalent to the unit sphere modulo reflections. Or the upper half of the unit sphere with the antipodal points on the boundary identified with each other.

Also, you can parameterize a dense open set using the hyperplane $x^n = 1$; the only lines you miss are the ones parallel to this hyperplane, i.e., the lines through the origin lying inside $x^n = 0$. The set of missing lines is therefore itself an $RP^{n-2}$. So $RP^{n-1}$ can be viewed as the union of $R^{n-1}$ with $RP^{n-2}$ (which is usually called the "hyperplane at infinity").

Obviously, you can do this with each co-ordinate, giving you $n$ dense open sets, each with a natural map to $R^{n-1}$ (called "affine co-ordinates"), that cover $RP^{n-1}$.

I assume that these descriptions don't do it for you. Can you say more about what you want?

Answer 3

You might enjoy Problem 1B from Milnor and Stasheff's book Characteristic Classes. The problem produces an explicit (if rather high-dimensional) embedding of $RP^n$ into Euclidean space. Specifically, it turns out that $RP^n$ can be viewed as the space of $(n+1)\times(n+1)$ real symmetric projection matrices with trace 1. This may not be so helpful for visualization, but it is quite concrete.

This embedding comes from the map sending a vector $x = (x_1, \ldots, x_{n+1})\in \mathbb{R}^{n+1}$ to the matrix with $(i,j)^{\textrm{th}}$ entry $(x_i x_j)/|x|^2$.

Answer 4

I find it easiest to get $\mathbb RP^n$ from $S^n$ with two cells in every dimension. That is to say, you start with two points for $S^0$, attach two one-cells oriented opposite ways for $S^1$, attach one 2-cell on top and one on the bottom for $S^2$, and so on until you have $S^n$. Then the antipodal map actually switches the two cells in each dimension, so it defines a free $\mathbb Z/2$ action on $S^n$. If you quotient out by that action, then you get a cell structure on $\mathbb RP^n$ with one cell in each dimension, and you have a universal double cover ($n$>$1$) for it.

I like this construction for a couple of reasons. First, it displays the (co)boundary maps in cellular (co)homology: the two $k$-cells share the same boundary (up to sign $(-1)^k$) which is the two $k-1$-cells. Because it's a covering space you can sort of see how after quotienting you get one $k$-cell whose boundary folds twice onto the $k-1$-cell. So that also shows you why the maps alternate between the 0 map and the multiplication by 2, or are all the 0 map in $\mathbb Z/2$ coefficients.

Also, this generalizes to a construction of $S^\infty$ and thus $\mathbb RP^\infty$, with $\mathbb RP^n$ inside it just by taking the $n$-skeleton. It looks like you could modify Qiaochu's definition to account for the infinite case by taking a $k$-cell to be the points that vanish after the first $k+1$ dimensions and are scaled to 1 in the $k+1$st.

Answer 5

A Point in $RP^n$ corresponds to a pair of antipodal points on $S^n$ - so just practice visualizing two antipodal points on a sphere every time you say Point. Such an approach is clearly equivalent to other definitions, but I find it good for "seeing." The standard cell structure is then defined by a Point being in the interior of the $i$-cell if and only if the first $n-i$ coordinates of the corresponding pair of points vanish.

You can also visualize Lens spaces in this way: a single point in a Lens space is seen as a collection of some $k$ points on the unit sphere in some ${\mathbb C}^n$, which are all related by multiplication by some $k$th rood of unity.

Answer 6

For $\mathbb{R}P^2$, I like Boy's surface, which is a particularly symmetric immersion of the projective plane into $\mathbb{R}^3$.

Also, see this Java-based model.

You can build a piecewise-linear version of one of these out of paper. If you cut out a disk-shaped window (to see inside to the triple point), what you have is a model of the Mobius band for which the boundary circle is really a round circle!

Of course, this doesn't really help with all of the rest of them $(n \ge 3)$!