# Numbers of charts needed to make an atlas of real projective plane?

I think it must be two? clearly if you divide the sphere into stereographic projection from upper and lower part of the sphere i.e. say (1,0,0) and (0,1,0) to make an atlas of sphere.

Why can't you just naively take the same chart and claim it is an atlas of real projective plane?

However, he said you get in a big problem.

But, my lecturer says you can't do that and it must be three charts are needed.

Well, in general why shouldn't one chart cover real projective plane? the point is that opposite points get identified. So you can imagine being an ant and just navigating the space by taking everything you are in z is negative just treating it like z is positive, hence being in the chart of the upper hemisphere?

Is the minimum number of charts a topological property?

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Hint: A chart is never a compact space, while the real projective plane (and the spehere) is. –  Thomas Andrews Dec 9 '11 at 14:45
@ThomasAndrews I see your point. Changed it. Really wanted to say that. –  simplicity Dec 9 '11 at 14:47
So, a chart of a space $X$ is an open $U\subset X$ and a homeomorphism of $U$ and $\mathbb R^n$ for some $n$. If $U=X$, then $X$ must be homeomorphic with $\mathbb R^n$. –  Thomas Andrews Dec 9 '11 at 14:55
Think of $\mathbb{RP}^2$ as a $\mathbb{D}^2$ modulo the antipodal map on the boundary. If you could cover $\mathbb{RP}^2$ with two charts, what would they look like on this model? –  Neal Dec 9 '11 at 15:05
Well, what happens is simply that they fail to be charts anymore. You can create all sorts of spaces that no longer are "chart-able" when you identify points. For example, take $\mathbb C$ and identify $-1$ and $1$. Then there is no chart around $\pm 1$. –  Thomas Andrews Dec 9 '11 at 15:39

Obviously three charts suffice, since the standard atlas consists of three charts with domains $x_i\neq 0$, homeomeorphic to $\mathbb R^2$.
Namely, do you want your charts to just give homeomorphisms with open subsets of $\mathbb R^2$ or do you insist that they give homeomorphisms with (balls in) $\mathbb R^2$?
However for $\mathbb P^2(\mathbb R)$ you must use at least 3 charts in either interpretation.
More generally, M.Hopkins has computed the minimal number of charts in the first interpretation for all projective spaces $\mathbb P^n(\mathbb R)$ (with a possible little uncertainty if $n= 31$ or$47$) here