# Cohomology of real projective space

This article describes the value (and the process used to compute it) of some homotopy invariant(s) for a topological space or family of topological spaces. The invariant is cohomology group and the topological space/family is real projective space
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## Cohomology groups in piecewise form

### Odd-dimensional projective space with coefficients in integers

$H^p(\mathbb{P}^n(\R); \mathbb{Z}) = \left\lbrace \begin{array}{rl} \Z, & \qquad p=0,n\\ \Z/2\Z, &\qquad p \ \operatorname{even}, 0 < p < n\\ 0, & \qquad \operatorname{otherwise}\end{array}\right.$

### Even-dimensional projective space with coefficients in integers

$H^p(\mathbb{P}^n(\R); \mathbb{Z}) = \left\lbrace \begin{array}{rl} \Z, & \qquad p=0\\ \Z/2\Z, & \qquad p \ \operatorname{even}, 0 < p \le n\\ 0, & \qquad \operatorname{otherwise}\end{array}\right.$

### Odd-dimensional projective space with coefficients in an abelian group

For an abelian group $M$, the cohomology is given by:

$H^p(\mathbb{P}^n(\R); M) = \left\lbrace \begin{array}{rl} M, & \qquad p=0,n\\ M/2M, &\qquad p \ \operatorname{even}, 0 < p < n\\ T, & \qquad p \ \operatorname{odd}, 0 < p < n \\ 0, & \qquad \operatorname{otherwise}\end{array}\right.$

Here, $T$ denotes the 2-torsion subgroup, i.e., the subgroup comprising elements of order dividing 2.

### Even-dimensional projective space with coefficients in an abelian group

For an abelian group $M$, the cohomology is given by:

$H^p(\mathbb{P}^n(\R); M) = \left\lbrace \begin{array}{rl} M, & \qquad p=0\\ M/2M, & \qquad p \ \operatorname{even}, 0 < p \le n\\ T, & \qquad p \ \operatorname{odd}, 0 < p < n \\ 0, & \qquad \operatorname{otherwise}\end{array}\right.$

Here, $T$ denotes the 2-torsion subgroup, i.e., the subgroup comprising elements of order dividing 2.

### Coefficients in a module over a 2-divisible ring

If we consider the cohomology with coefficients in a module $M$ over a ring $R$ where 2 is invertible, then we have:

$H^p(\mathbb{P}^n(\R);M) := \left\lbrace\begin{array}{rl} M, & p = 0 \\ M, & p = n, n \ \operatorname{odd}\\ 0, & p = n, n \ \operatorname{even}\\ 0, & p \ne 0,n \\\end{array}\right.$

In particular, these results are valid over the field of rational numbers or over any field of characteristic zero, or indeed any characteristic other than 2.

### Coefficients in characteristic two

Suppose $M$ is an elementary abelian 2-group, i.e., a group in which the double of every element is zero. Then, $2M = 0$ (so $M/2M \cong M$) and $T = M$, and we get:

$H^p(\mathbb{P}^n(\R);M) = \left\lbrace\begin{array}{rl}M, & \qquad 0 \le p \le n \\ 0, & \qquad \operatorname{otherwise}\\\end{array}\right.$

This in particular applies to the case that $M$ is the group $\mathbb{Z}/2\mathbb{Z}$, i.e., when we are taking coefficients in the field of two elements.

## Cohomology groups in tabular form

### Coefficients in integers

We illustrate how the cohomology groups work for small values of $n$. Note that for $p > n$, all cohomology groups $H^p$ are zero, so we omit those cells for visual ease.

$n$ Real projective space $\R\mathbb{P}^n$ Orientable? $H^0$ $H^1$ $H^2$ $H^3$ $H^4$ $H^5$
1 circle Yes $\mathbb{Z}$ $\mathbb{Z}$
2 real projective plane No $\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$
3 RP^3 Yes $\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ $\mathbb{Z}$
4 RP^4 No $\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$
5 RP^5 Yes $\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ $\mathbb{Z}$

### Coefficients in an abelian group

We let the abelian group be $M$. Denote by $T$ the 2-torsion of $M$ and by $2M$ the submodule comprising doubles of elements.

$n$ Real projective space $\R\mathbb{P}^n$ Orientable? $H^0$ $H^1$ $H^2$ $H^3$ $H^4$ $H^5$
1 circle Yes $M$ $M$
2 real projective plane No $M$ $T$ $M/2M$
3 RP^3 Yes $M$ $T$ $M/2M$ $M$
4 RP^4 No $M$ $T$ $M/2M$ $T$ $M/2M$
5 RP^5 Yes $M$ $T$ $M/2M$ $T$ $M/2M$ $M$

### Coefficients in a module over a 2-divisible ring

Suppose $M$ has the structure of a module over a unital ring $R$ where 2 is invertible. Then, in particular, we know that $2M = M$ and $T = 0$. Thus, both $M/2M$ and $T$ are equal to $0$. We get:

$n$ Real projective space $\R\mathbb{P}^n$ Orientable? $H^0$ $H^1$ $H^2$ $H^3$ $H^4$ $H^5$
1 circle Yes $M$ $M$
2 real projective plane No $M$ 0 0
3 RP^3 Yes $M$ 0 0 $M$
4 RP^4 No $M$ 0 0 0 0
5 RP^5 Yes $M$ 0 0 0 0 $M$

### Coefficients in characteristic two

Suppose $M$ is an elementary abelian 2-group, i.e., a group in which the double of every element is zero. Then, $2M = 0$ (so $M/2M \cong M$) and $T = M$, and we get:

$n$ Real projective space $\R\mathbb{P}^n$ Orientable? $H^0$ $H^1$ $H^2$ $H^3$ $H^4$ $H^5$
1 circle Yes $M$ $M$
2 real projective plane No $M$ $M$ $M$
3 RP^3 Yes $M$ $M$ $M$ $M$
4 RP^4 No $M$ $M$ $M$ $M$ $M$
5 RP^5 Yes $M$ $M$ $M$ $M$ $M$ $M$

## Cohomology ring structure

### Over the integers for even $n$

The cohomology ring $H^*(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$ is the ring $\mathbb{Z}[x]/\langle 2x, x^{(n/2)+1} \rangle$, where $x$ is the unique non-identity element in $H^2(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$. $x^r$ in turn is the unique non-identity element in $H^{2r}(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$ for $1 \le r \le n/2$. The coefficients ring (i.e., the constant terms) is $H^0$.

Note that that is almost the same as the ring $\mathbb{Z}/2\mathbb{Z}[x]/\langle x^{(n/2)+1} \rangle$, with the only difference being that for the constant terms, we are allowed to use the ring $\mathbb{Z}$ rather than the quotient ring $\mathbb{Z}/2\mathbb{Z}$.

### Over the integers for odd $n$

The cohomology ring $H^*(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$ is the ring $\mathbb{Z}[x,y]/\langle 2x, x^{(n+1)/2}, xy, y^2\rangle$ where $x$ is the unique non-identity element in $H^2(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$ and $y$ is a generator of $H^n(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$. $x^r$ in turn is the unique non-identity element in $H^{2r}(\mathbb{R}\mathbb{P}^n;\mathbb{Z})$ for $1 \le r \le (n-1)/2$. The coefficients ring (i.e., the constant terms) is $H^0$.

Note that that is almost the same as the ring $\mathbb{Z}/2\mathbb{Z}[x,y]/\langle x^{(n/2)+1},y^2,xy \rangle$, with the only difference being that for the constant terms, we are allowed to use the ring $\mathbb{Z}$ rather than the quotient ring $\mathbb{Z}/2\mathbb{Z}$.

## Reality checks

General assertion Verification in this case See also ...
For any compact connected $n$-dimensional manifold, the top cohomology group $H^n$ is $\mathbb{Z}$ if the space is orientable and is (?) (finite group?) otherwise. $n$ odd: In this case, the space is obtained by taking the quotient of the orientable manifold $S^n$ by the antipodal action, which is orientation-preserving (one way of seeing it is that is given by a scalar matrix of $-1$s in dimension $n + 1$, so has determinant 1). The quotient is thus also orientable. Indeed, for $n$ odd, the top cohomology is $\mathbb{Z}$.
$n$ even: In this case, the space is obtained by taking the quotient of the orientable manifold $S^n$ by the antipodal action, which is orientation-reversing (one way of seeing it is that is given by a scalar matrix of $-1$s in dimension $n + 1$, so has determinant -1). The quotient is thus non-orientable. Indeed, for $n$ even, the top cohomology is $\mathbb{Z}/2\mathbb{Z}$.
?
For a compact connected orientable manifold of dimension $n$, the Poincare duality theorem says that the homology group of dimension $p$ is isomorphic to the cohomology group of dimension $n - p$. Case $n$ odd: As noted above, the manifold is orientable. The top and bottom homology and cohomology groups are isomorphic to $\mathbb{Z}$. The even-dimensional cohomology groups and odd-dimensional homology groups are both isomorphic to $\mathbb{Z}/2\mathbb{Z}$. The odd-dimensional cohomology groups and even-dimensional homology groups are both zero groups. homology of real projective space

## Proof using homology groups

### Case of odd dimension

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### Case of even dimension

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## Proof using cochain complex constructed from CW structure

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