Cohomology of real projective space: Difference between revisions

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{{homotopy invariant computation|
invariant = homology group|
invariant = cohomology group|
space = real projective space}}
space = real projective space}}


==Statement==
==Cohomology groups in piecewise form==


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


<math>H_p(\mathbb{P}^n(\R)) = \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.</math>
<math>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.</math>


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


<math>H_p(\mathbb{P}^n(\R)) = \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.</math>
<math>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.</math>
 
===Odd-dimensional projective space with coefficients in an abelian group===
 
For an abelian group <math>M</math>, the cohomology is given by:
 
<math>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.</math>
 
Here, <math>T</math> 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 <math>M</math>, the cohomology is given by:
 
<math>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.</math>
 
Here, <math>T</math> 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 <math>M</math> over a ring <math>R</math> where 2 is invertible, then we have:
 
<math>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.</math>
 
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 <math>M</math> is an elementary abelian 2-group, i.e., a group in which the double of every element is zero. Then, <math>2M = 0</math> (so <math>M/2M \cong M</math>) and <math>T = M</math>,  and we get:
 
<math>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.</math>
 
This in particular applies to the case that <math>M</math> is the group <math>\mathbb{Z}/2\mathbb{Z}</math>, 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 <math>n</math>. Note that for <math>p > n</math>, all cohomology groups <math>H^p</math> are zero, so we omit those cells for visual ease.
 
{| class="sortable" border="1"
! <math>n</math> !! Real projective space <math>\R\mathbb{P}^n</math> !! Orientable? !! <math>H^0</math> !! <math>H^1</math> !! <math>H^2</math> !! <math>H^3</math> !! <math>H^4</math> !! <math>H^5</math>
|-
| 1 || [[circle]] || Yes || <math>\mathbb{Z}</math> || <math>\mathbb{Z}</math>
|-
| 2 || [[real projective plane]] || No || <math>\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math>
|-
| 3 || [[RP^3]] || Yes || <math>\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math>  || <math>\mathbb{Z}</math>
|-
| 4 || [[RP^4]] || No || <math>\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math>
|-
| 5 || [[RP^5]] || Yes || <math>\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math> || 0 || <math>\mathbb{Z}/2\mathbb{Z}</math> || <math>\mathbb{Z}</math>
|}
 
===Coefficients in an abelian group===
 
We let the abelian group be <math>M</math>. Denote by <math>T</math> the 2-torsion of <math>M</math> and by <math>2M</math> the submodule comprising doubles of elements.
 
{| class="sortable" border="1"
! <math>n</math> !! Real projective space <math>\R\mathbb{P}^n</math> !! Orientable? !! <math>H^0</math> !! <math>H^1</math> !! <math>H^2</math> !! <math>H^3</math> !! <math>H^4</math> !! <math>H^5</math>
|-
| 1 || [[circle]] || Yes || <math>M</math> || <math>M</math>
|-
| 2 || [[real projective plane]] || No || <math>M</math> || <math>T</math> || <math>M/2M</math>
|-
| 3 || [[RP^3]] || Yes || <math>M</math> || <math>T</math> || <math>M/2M</math> || <math>M</math>
|-
| 4 || [[RP^4]] || No || <math>M</math> || <math>T</math> || <math>M/2M</math> || <math>T</math> || <math>M/2M</math>
|-
| 5 || [[RP^5]] || Yes || <math>M</math> || <math>T</math> || <math>M/2M</math> || <math>T</math> || <math>M/2M</math> || <math>M</math>
|}
 
===Coefficients in a module over a 2-divisible ring===
 
Suppose <math>M</math> has the structure of a module over a unital ring <math>R</math> where 2 is invertible. Then, in particular, we know that <math>2M = M</math> and <math>T = 0</math>. Thus, both <math>M/2M</math> and <math>T</math> are equal to <math>0</math>. We get:
 
{| class="sortable" border="1"
! <math>n</math> !! Real projective space <math>\R\mathbb{P}^n</math> !! Orientable? !! <math>H^0</math> !! <math>H^1</math> !! <math>H^2</math> !! <math>H^3</math> !! <math>H^4</math> !! <math>H^5</math>
|-
| 1 || [[circle]] || Yes || <math>M</math> || <math>M</math>
|-
| 2 || [[real projective plane]] || No || <math>M</math> || 0 || 0
|-
| 3 || [[RP^3]] || Yes || <math>M</math> || 0 || 0 || <math>M</math>
|-
| 4 || [[RP^4]] || No || <math>M</math> || 0 || 0 || 0 || 0
|-
| 5 || [[RP^5]] || Yes || <math>M</math> || 0 || 0 || 0 || 0 || <math>M</math>
|}
 
===Coefficients in characteristic two===
 
Suppose <math>M</math> is an elementary abelian 2-group, i.e., a group in which the double of every element is zero. Then, <math>2M = 0</math> (so <math>M/2M \cong M</math>) and <math>T = M</math>,  and we get:
 
{| class="sortable" border="1"
! <math>n</math> !! Real projective space <math>\R\mathbb{P}^n</math> !! Orientable? !! <math>H^0</math> !! <math>H^1</math> !! <math>H^2</math> !! <math>H^3</math> !! <math>H^4</math> !! <math>H^5</math>
|-
| 1 || [[circle]] || Yes || <math>M</math> || <math>M</math>
|-
| 2 || [[real projective plane]] || No || <math>M</math> || <math>M</math> || <math>M</math>
|-
| 3 || [[RP^3]] || Yes || <math>M</math> || <math>M</math> || <math>M</math> || <math>M</math>
|-
| 4 || [[RP^4]] || No || <math>M</math> || <math>M</math> || <math>M</math> || <math>M</math> || <math>M</math>
|-
| 5 || [[RP^5]] || Yes || <math>M</math> || <math>M</math> || <math>M</math> || <math>M</math>|| <math>M</math>|| <math>M</math>
|}
 
==Cohomology ring structure==
 
===Over the integers for even <math>n</math>===
 
The cohomology ring <math>H^*(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math> is the ring <math>\mathbb{Z}[x]/\langle 2x, x^{(n/2)+1} \rangle</math>, where <math>x</math> is the unique non-identity element in <math>H^2(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math>. <math>x^r</math> in turn is the unique non-identity element in <math>H^{2r}(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math> for <math>1 \le r \le n/2</math>. The coefficients ring (i.e., the constant terms) is <math>H^0</math>.
 
Note that that is ''almost'' the same as the ring <math>\mathbb{Z}/2\mathbb{Z}[x]/\langle x^{(n/2)+1} \rangle</math>, with the only difference being that for the constant terms, we are allowed to use the ring <math>\mathbb{Z}</math> rather than the quotient ring <math>\mathbb{Z}/2\mathbb{Z}</math>.
 
===Over the integers for odd <math>n</math>===
 
The cohomology ring <math>H^*(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math> is the ring <math>\mathbb{Z}[x,y]/\langle 2x, x^{(n+1)/2}, xy, y^2\rangle</math> where <math>x</math> is the unique non-identity element in <math>H^2(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math> and <math>y</math> is a generator of <math>H^n(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math>. <math>x^r</math> in turn is the unique non-identity element in <math>H^{2r}(\mathbb{R}\mathbb{P}^n;\mathbb{Z})</math> for <math>1 \le r \le (n-1)/2</math>. The coefficients ring (i.e., the constant terms) is <math>H^0</math>.
 
Note that that is ''almost'' the same as the ring <math>\mathbb{Z}/2\mathbb{Z}[x,y]/\langle x^{(n/2)+1},y^2,xy \rangle</math>, with the only difference being that for the constant terms, we are allowed to use the ring <math>\mathbb{Z}</math> rather than the quotient ring <math>\mathbb{Z}/2\mathbb{Z}</math>.
 
==Reality checks==
 
{| class="sortable" border="1"
! General assertion !! Verification in this case !! See also ...
|-
| For any compact connected <math>n</math>-dimensional manifold, the top cohomology group <math>H^n</math> is <math>\mathbb{Z}</math> if the space is orientable and is (?) (finite group?) otherwise. || <math>n</math> odd: In this case, the space is obtained by taking the quotient of the orientable manifold <math>S^n</math> by the antipodal action, which is orientation-preserving (one way of seeing it is that is given by a scalar matrix of <math>-1</math>s in dimension <math>n + 1</math>, so has determinant 1). The quotient is thus also orientable. Indeed, for <math>n</math> odd, the top cohomology is <math>\mathbb{Z}</math>.<br><math>n</math> even: In this case, the space is obtained by taking the quotient of the orientable manifold <math>S^n</math> by the antipodal action, which is orientation-reversing (one way of seeing it is that is given by a scalar matrix of <math>-1</math>s in dimension <math>n + 1</math>, so has determinant -1). The quotient is thus non-orientable. Indeed, for <math>n</math> even, the top cohomology is <math>\mathbb{Z}/2\mathbb{Z}</math>. || ?
|-
| For a compact connected orientable manifold of dimension <math>n</math>, the [[Poincare duality theorem]] says that the homology group of dimension <math>p</math> is isomorphic to the cohomology group of dimension <math>n - p</math>. || Case <math>n</math> odd: As noted above, the manifold is orientable. The top and bottom homology and cohomology groups are isomorphic to <math>\mathbb{Z}</math>. The even-dimensional cohomology groups and odd-dimensional homology groups are both isomorphic to <math>\mathbb{Z}/2\mathbb{Z}</math>. The odd-dimensional cohomology groups and even-dimensional homology groups are both zero groups. || [[homology of real projective space]]
|}
 
==Facts used==
 
# [[uses::Homology of real projective space]]
# [[uses::Dual universal coefficients theorem]]
# [[uses::CW structure of real projective space]]
 
==Proof using homology groups==
 
===Case of odd dimension===
 
{{fillin}}
 
===Case of even dimension===
 
{{fillin}}
==Proof using cochain complex constructed from CW structure==
 
{{fillin}}

Latest revision as of 21:07, 30 September 2011

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
Get more specific information about real projective space | Get more computations of cohomology group

Cohomology groups in piecewise form

Odd-dimensional projective space with coefficients in integers

Hp(Pn(R);Z)={Z,p=0,nZ/2Z,peven,0<p<n0,otherwise

Even-dimensional projective space with coefficients in integers

Hp(Pn(R);Z)={Z,p=0Z/2Z,peven,0<pn0,otherwise

Odd-dimensional projective space with coefficients in an abelian group

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

Hp(Pn(R);M)={M,p=0,nM/2M,peven,0<p<nT,podd,0<p<n0,otherwise

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:

Hp(Pn(R);M)={M,p=0M/2M,peven,0<pnT,podd,0<p<n0,otherwise

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:

Hp(Pn(R);M):={M,p=0M,p=n,nodd0,p=n,neven0,p0,n

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/2MM) and T=M, and we get:

Hp(Pn(R);M)={M,0pn0,otherwise

This in particular applies to the case that M is the group Z/2Z, 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 Hp are zero, so we omit those cells for visual ease.

n Real projective space RPn Orientable? H0 H1 H2 H3 H4 H5
1 circle Yes Z Z
2 real projective plane No Z 0 Z/2Z
3 RP^3 Yes Z 0 Z/2Z Z
4 RP^4 No Z 0 Z/2Z 0 Z/2Z
5 RP^5 Yes Z 0 Z/2Z 0 Z/2Z 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 RPn Orientable? H0 H1 H2 H3 H4 H5
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 RPn Orientable? H0 H1 H2 H3 H4 H5
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/2MM) and T=M, and we get:

n Real projective space RPn Orientable? H0 H1 H2 H3 H4 H5
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*(RPn;Z) is the ring Z[x]/2x,x(n/2)+1, where x is the unique non-identity element in H2(RPn;Z). xr in turn is the unique non-identity element in H2r(RPn;Z) for 1rn/2. The coefficients ring (i.e., the constant terms) is H0.

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

Over the integers for odd n

The cohomology ring H*(RPn;Z) is the ring Z[x,y]/2x,x(n+1)/2,xy,y2 where x is the unique non-identity element in H2(RPn;Z) and y is a generator of Hn(RPn;Z). xr in turn is the unique non-identity element in H2r(RPn;Z) for 1r(n1)/2. The coefficients ring (i.e., the constant terms) is H0.

Note that that is almost the same as the ring Z/2Z[x,y]/x(n/2)+1,y2,xy, with the only difference being that for the constant terms, we are allowed to use the ring Z rather than the quotient ring Z/2Z.

Reality checks

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

Facts used

  1. Homology of real projective space
  2. Dual universal coefficients theorem
  3. CW structure of real projective space

Proof using homology groups

Case of odd dimension

Fill this in later

Case of even dimension

Fill this in later

Proof using cochain complex constructed from CW structure

Fill this in later