# Homology 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 homology group and the topological space/family is real projective space
Get more specific information about real projective space | Get more computations of homology group

## Statement

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

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

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

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

Thus the key difference between even and odd dimensional projective spaces is that the top homology vanishes in even-dimensional projective spaces. This is related to the fact that even-dimensional projective space is non-orientable, while odd-dimensional projective space is orientable.

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

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

where $T$ is the 2-torsion submodule of $M$, i.e. the submodule comprising elements whose double is zero.

### Even-dimensional projective space with coefficients in an abelian group or module $M$

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

where $T$ is the 2-torsion submodule of $M$, i.e. the submodule comprising elements whose double is zero.

### Coefficients in a 2-divisible ring

If we consider the homology 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.

## Homology groups with integer coefficients in tabular form

We illustrate how the homology groups work for small values of $n$. Note that for $p > n$, all homology 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 $\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z}$ 0
3 RP^3 Yes $\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}$
4 RP^4 No $\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ 0
5 RP^5 Yes $\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}/2\mathbb{Z}$ 0 $\mathbb{Z}$

## Reality checks

For any path-connected space, the first homology group $H_1$ with coefficients in $\mathbb{Z}$ is the abelianization of the fundamental group. (This is part of the Hurewicz theorem). If the fundamental group is abelian, it is isomorphic to the first homology group. $n = 1$: The space is homeomorphic to the circle $S^1$, and the fundamental group is isomorphic to $\mathbb{Z}$. This is also the first homology group.
$n > 1$: The fundamental group is $\mathbb{Z}/2\mathbb{Z}$, because its double cover $S^n$ is simply connected (see n-sphere is simply connected for n greater than 1). The first homology group $H_1$ is also $\mathbb{Z}/2\mathbb{Z}$.
homotopy of real projective space
For any compact connected $n$-dimensional manifold, the top homology group $H_n$ is $\mathbb{Z}$ if the space is orientable and is $0$ 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 homology 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 homology is $0$.
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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. cohomology of real projective space

## Related invariants

These are all invariants that can be computed in terms of the homology groups.

Invariant General description Description of value for real projective space
Betti numbers The $k^{th}$ Betti number $b_k$ is the rank of the torsion-free part of the $k^{th}$ homology group. $b_0 = 1$. $b_n = 1$ if $n$ is odd and $b_n = 0$ if $n$ is even. All other $b_k$s are zero.
Poincare polynomial Generating polynomial for Betti numbers $1 + x^n$ if $n$ is odd. $1$ if $n$ is even.
Euler characteristic $\sum_{k=0}^\infty (-1)^k b_k$ $0$ if $n$ is odd. $1$ if $n$ is even. Note that the Euler characteristic is half the Euler characteristic of the sphere $S^n$, which is its double cover.

## Facts used

1. CW structure of real projective space

## Proof

### Explication of chain complex

The proof follows from fact (1). By fact (1), we note that the homology of real projective space $\mathbb{P}^n(\R)$ is the same as the homology of the following chain complex, obtained as its cellular chain complex:

• For $n$ even:

$0 \to 0 \to \dots \to 0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z} \to \dots \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z}$

where the largest nonzero chain group is the $n^{th}$ chain group.

• For $n$ odd:

$0 \to 0 \to \dots \to 0 \to \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \dots \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z}$

Note that the multiplication maps alternate between multiplication by two and multiplication by zero. In particular, in both cases, the map is multiplication by two if going down from an even to an odd index and multiplication by zero if going down from an odd to an even index. The key difference between the odd and even case is whether we start with a multiplication by two map or a multiplication by zero map.

### Homology computation over integers

Case for $n$ Case for $p$ Fragment of relevance in chain complex ($(p+1)^{th}$ to $p^{th}$ to $(p-1)^{th}$) Cycle group (kernel from $p^{th}$ to $(p-1)^{th}$ Boundary group (image from $(p+1)^{th}$ group to $p^{th}$ group Homology group = cycle group/boundary group
both odd and even 0 $\mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z} \stackrel{0}{\to} 0$ $\mathbb{Z}$ 0 $\mathbb{Z}$
both odd and even odd, strictly less than $n$ $\mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z}$ $\mathbb{Z}$ $2\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z}$
both odd and even even, positive, strictly less than $n$ $\mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z}$ 0 0 0
even equal to $n$ $0 \stackrel{0}{\to} \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z}$ 0 0 0
odd equal to $n$ $0 \stackrel{0}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z}$ $\mathbb{Z}$ 0 $\mathbb{Z}$
both odd and even greater than $n$ $0 \stackrel{0}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} 0$ (the last term becomes $\mathbb{Z}$ when $p = n + 1$) 0 0 0

### Homology computation over an abelian group or module $M$

The chain complex remains the same, but each $\mathbb{Z}$ is replaced by $M$.

Denote by $T$ the 2-torsion submodule of $M$ and by $M/2M$ the quotient of $M$ by the submodule $2M$ comprising the doubles of elements.

Case for $n$ Case for $p$ Fragment of relevance in chain complex ($(p+1)^{th}$ to $p^{th}$ to $(p-1)^{th}$) Cycle group (kernel from $p^{th}$ to $(p-1)^{th}$ Boundary group (image from $(p+1)^{th}$ group to $p^{th}$ group Homology group = cycle group/boundary group
both odd and even 0 $M \stackrel{\cdot 0}{\to} M \stackrel{0}{\to} 0$ $M$ 0 $M$
both odd and even odd, strictly less than $n$ $M \stackrel{\cdot 2}{\to} M \stackrel{\cdot 0}{\to} \mathbb{Z}$ $M$ $2M$ $M/2M$
both odd and even even, positive, strictly less than $n$ $M \stackrel{\cdot 0}{\to} M \stackrel{\cdot 2}{\to} M$ $T$ 0 $T$
even equal to $n$ $0 \stackrel{0}{\to} M \stackrel{\cdot 2}{\to} M$ $T$ 0 $T$
odd equal to $n$ $0 \stackrel{0}{\to} M \stackrel{\cdot 0}{\to} M$ $M$ 0 $M$
both odd and even greater than $n$ $0 \stackrel{0}{\to} M \stackrel{\cdot 0}{\to} 0$ (though the last term becomes $M$ when $p = n + 1$) 0 0 0