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I would like to see different ways of calculating the cohomology ring of $\mathbb{R}P^n$. I know there are several ways, for example, using Poincare Duality, Gysin sequence, etc... Sketch your favorite proof!

PD. Is there a direct way (maybe a geometric argument) of calculating this chomology ring without using any big theorems?

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up vote 3 down vote accepted

In the complex case, which is easier:

Using the fact that there is a cellular decomposition of $P^n\mathbb C$ using only one even dimensional cell per dimension, we know their classes freely generate homology. Looking at the decomposition, we see that each of those classes is in fact the fundamental class of a projective subspace. It is thus easy to compute by hand (even by picture!) intersections in homology, and then one can invoke duality to identify that with the cup product in cohomology. The resulting computation is almost completely geometric. You can find it (for $n=3$ only, iirc) in Greenberg's book on algebraic topology.

The real case is similar, of course.

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