What Is the Cup-Product Ring of Projective Space?

The phrase “projective space” most often means real projective space, denoted $\mathbb{RP}^n$, but it can also mean complex projective space, $\mathbb{CP}^n$. Their cohomology rings have different generators and coefficient conventions. The main formula below is for real projective space with coefficients in $\mathbb{F}_2$, the field with two elements.

The cup-product ring of real projective space

For $n\geq 1$, the ring is

$$H^*(\mathbb{RP}^n;\mathbb{F}_2) \cong \mathbb{F}_2[x]/(x^{n+1}), \qquad |x|=1.$$

Here $|x|=1$ means that $x$ lies in cohomological degree one. The formula says that $1,x,x^2,\ldots,x^n$ are nonzero, while $x^{n+1}=0$. Since the ring is graded, $x^k$ lies in degree $k$.

This is more information than the cohomology groups alone provide. The groups tell you that there is one copy of $\mathbb{F}_2$ in each degree from zero through $n$. The cup product tells you that the degree-one class generates all those groups by taking powers.

Why the powers are nonzero

A geometric way to understand the generator is to take a projective hyperplane $\mathbb{RP}^{n-1}\subset\mathbb{RP}^n$. Its cohomology class is $x$. Intersecting $k$ hyperplanes in general position gives a projective subspace of codimension $k$, representing the dual of $x^k$. In particular, up to $n$ hyperplanes can meet in a point, counted modulo two, so $x^n\neq 0$. There is no nonzero cohomology in degree $n+1$, so $x^{n+1}=0$.

The use of $\mathbb{F}_2$ matters: counting intersections modulo two works without choosing orientations, including when real projective space is nonorientable.

Worked example: $\mathbb{RP}^4$

Set $n=4$ in the formula:

$$H^*(\mathbb{RP}^4;\mathbb{F}_2) \cong \mathbb{F}_2[x]/(x^5).$$

The nonzero classes and their degrees are

Degree	Class
0	$1$
1	$x$
2	$x^2$
3	$x^3$
4	$x^4$

For example, multiplying the degree-one class by itself three times gives $x^3$, a nonzero class in degree three. Multiplying once more gives $x^4$, still nonzero; multiplying again gives $x^5=0$.

What changes with integer coefficients?

The mod-2 formula should not be reused unchanged for integer coefficients. For $\mathbb{RP}^n$, the positive-degree integral groups are different: there is a $\mathbb{Z}/2$ group in each positive even degree at most $n$, and, when $n$ is odd, an additional $\mathbb{Z}$ in top degree $n$.

More explicitly, for $n\geq 2$, let $a\in H^2(\mathbb{RP}^n;\mathbb{Z})$ be a generator of the degree-two group. Its powers $a^j$ generate the degree-$2j$ groups as long as $2j\leq n$; these groups have order two. If $n$ is odd, there is also a top-degree generator $b\in H^n(\mathbb{RP}^n;\mathbb{Z})\cong\mathbb{Z}$. Products whose degrees exceed $n$ vanish. This describes the groups and products without confusing the integral degree-two generator with the mod-2 degree-one generator.

For comparison, complex projective space has the particularly simple integral ring

$$H^*(\mathbb{CP}^n;\mathbb{Z})\cong\mathbb{Z}[u]/(u^{n+1}), \qquad |u|=2.$$

The degree-two generator is dual to a complex hyperplane $\mathbb{CP}^{n-1}$. Intersections of $k$ hyperplanes give the nonzero powers $u^k$ through $k=n$.

Common mistakes

• Leaving out the coefficient system: The ring for $\mathbb{RP}^n$ over $\mathbb{F}_2$ is not the same as its ring over $\mathbb{Z}$.
• Using the wrong generator degree: The standard real-projective-space generator over $\mathbb{F}_2$ has degree one. The standard generator for complex projective space has degree two.
• Assuming groups determine products: Knowing that cohomology is nonzero in degrees zero through $n$ does not by itself show that the classes are powers of one generator. The cup-product calculation supplies that fact.
• Treating real and complex projective spaces as interchangeable: Their dimensions, coefficient behavior, and generator degrees differ.

If a text says only “projective space,” check its notation and coefficients. The notation $\mathbb{RP}^n$ or $\mathbb{CP}^n$ and the coefficient group settle which ring formula applies.